JEE Differentiation Questions

We are given that $$f: \mathbb{R} \rightarrow \mathbb{R}$$ is twice differentiable and satisfies $$f(x+y) = f(x)f(y)$$ for all $$x, y \in \mathbb{R}$$.

Step 1: Find $$f(0)$$

Setting $$x = 0$$ and $$y = 0$$ in the functional equation:

$$f(0 + 0) = f(0) \cdot f(0)$$

$$f(0) = [f(0)]^{2}$$

This gives $$f(0)(f(0) - 1) = 0$$, so $$f(0) = 0$$ or $$f(0) = 1$$.

If $$f(0) = 0$$, then for any $$x$$: $$f(x) = f(x + 0) = f(x) \cdot f(0) = 0$$, making $$f$$ identically zero. But we are given $$f'(0) = 4a$$ with $$a \gt 0$$, so $$f$$ is not identically zero. Therefore, $$f(0) = 1$$.

Step 2: Determine the form of $$f(x)$$

Differentiating $$f(x + y) = f(x)f(y)$$ with respect to $$x$$ (treating $$y$$ as a constant):

$$f'(x + y) = f'(x) \cdot f(y)$$

Setting $$x = 0$$:

$$f'(y) = f'(0) \cdot f(y) = 4a \cdot f(y)$$

So $$f$$ satisfies the differential equation $$f'(x) = 4a \cdot f(x)$$ with $$f(0) = 1$$.

This is a standard first-order ODE. Its solution is:

$$f(x) = e^{4ax}$$

Step 3: Find the value of $$a$$

We are given the second-order ODE: $$f''(x) - 3a \cdot f'(x) - f(x) = 0$$.

Let us compute the derivatives of $$f(x) = e^{4ax}$$:

$$f'(x) = 4a \cdot e^{4ax}$$

$$f''(x) = (4a)^{2} \cdot e^{4ax} = 16a^{2} \cdot e^{4ax}$$

Substituting into the ODE:

$$16a^{2} \cdot e^{4ax} - 3a \cdot (4a \cdot e^{4ax}) - e^{4ax} = 0$$

$$16a^{2} \cdot e^{4ax} - 12a^{2} \cdot e^{4ax} - e^{4ax} = 0$$

Factoring out $$e^{4ax}$$:

$$e^{4ax}(16a^{2} - 12a^{2} - 1) = 0$$

$$e^{4ax}(4a^{2} - 1) = 0$$

Since $$e^{4ax} \neq 0$$ for any real $$x$$, we need:

$$4a^{2} - 1 = 0$$

$$4a^{2} = 1$$

$$a^{2} = \frac{1}{4}$$

$$a = \pm \frac{1}{2}$$

Since $$a \gt 0$$, we get $$a = \frac{1}{2}$$.

Step 4: Write $$f(x)$$ and $$f(ax)$$

With $$a = \frac{1}{2}$$:

$$f(x) = e^{4 \cdot \frac{1}{2} \cdot x} = e^{2x}$$

Now we compute $$f(ax)$$:

$$f(ax) = f\left(\frac{1}{2} \cdot x\right) = e^{2 \cdot \frac{x}{2}} = e^{x}$$

Step 5: Compute the area

The region is $$R = \{(x,y) \mid 0 \leq y \leq f(ax), \, 0 \leq x \leq 2\}$$.

The area is given by:

$$A = \int_{0}^{2} f(ax) \, dx = \int_{0}^{2} e^{x} \, dx$$

Evaluating the integral:

$$A = \left[e^{x}\right]_{0}^{2} = e^{2} - e^{0} = e^{2} - 1$$

Hence, the correct answer is Option A.

The given function is $$f(x)=\frac{x}{3}+\frac{3}{x}+3,\;x\neq 0$$.

Step 1: Find the derivative and its critical points
Using the rule $$\frac{d}{dx}\left(\frac{k}{x}\right)=-\frac{k}{x^{2}}$$, we get
$$f'(x)=\frac{1}{3}-\frac{3}{x^{2}}.$$ Set $$f'(x)=0$$ to locate stationary points:
$$\frac{1}{3}-\frac{3}{x^{2}}=0 \;\Longrightarrow\; \frac{1}{3}=\frac{3}{x^{2}} \;\Longrightarrow\; x^{2}=9 \;\Longrightarrow\; x=\pm 3.$$ Thus the only stationary points are $$x=-3$$ and $$x=3$$. In addition, $$f(x)$$ is not defined at $$x=0$$, so $$x=0$$ also separates intervals.

Step 2: Sign of $$f'(x)$$ in each interval
We test one value from each of the four regions determined by $$x=-3,0,3$$.

For $$x\lt -3$$, take $$x=-4$$:
$$f'(-4)=\frac{1}{3}-\frac{3}{16}\approx 0.145\gt 0,$$ so $$f'(x)\gt 0$$ in $$(-\infty,-3)$$ ⇒ increasing.

For $$-3\lt x\lt 0$$, take $$x=-1$$:
$$f'(-1)=\frac{1}{3}-3=-\frac{8}{3}\lt 0,$$ so $$f'(x)\lt 0$$ in $$(-3,0)$$ ⇒ decreasing.

For $$0\lt x\lt 3$$, take $$x=1$$:
$$f'(1)=\frac{1}{3}-3=-\frac{8}{3}\lt 0,$$ so $$f'(x)\lt 0$$ in $$(0,3)$$ ⇒ decreasing.

For $$x\gt 3$$, take $$x=4$$:
$$f'(4)=\frac{1}{3}-\frac{3}{16}\approx 0.146\gt 0,$$ so $$f'(x)\gt 0$$ in $$(3,\infty)$$ ⇒ increasing.

Step 3: Match the intervals with the symbols $$\alpha_i$$
From the sign chart, the intervals of monotonicity are
• Increasing: $$(-\infty,-3)$$ and $$(3,\infty)$$.
• Decreasing: $$(-3,0)$$ and $$(0,3)$$.

The problem states that
  • $$f(x)$$ is strictly increasing in $$(-\infty,\alpha_1)\cup(\alpha_2,\infty)$$,
  • $$f(x)$$ is strictly decreasing in $$(\alpha_3,\alpha_4)\cup(\alpha_4,\alpha_5)$$.

Comparing, we can assign
$$\alpha_1=-3,\qquad \alpha_2=3,$$
$$\alpha_3=-3,\qquad \alpha_4=0,\qquad \alpha_5=3.$$ (The same numerical value may occur for different indices; only the positions in the description matter.)

Step 4: Compute the required sum
$$\sum_{i=1}^{5}\alpha_i^{2} =\alpha_1^{2}+\alpha_2^{2}+\alpha_3^{2}+\alpha_4^{2}+\alpha_5^{2}$$ $$=(-3)^{2}+3^{2}+(-3)^{2}+0^{2}+3^{2}$$ $$=9+9+9+0+9$$ $$=36.$$

Answer: $$\sum_{i=1}^{5}\alpha_i^{2}=36$$, which corresponds to Option D.

$$x < -1$$: $$f(x) = 1-2x$$ (decreasing).

 $$-1 \le x \le 2$$: $$f(x) = \frac{1}{3}(7+2|x|)$$.

 At $$x=0$$, there is a local minimum because the absolute value $$|x|$$ turns from decreasing to increasing.

$$f(0) = 7/3$$.

$$x > 2$$: $$f(x) = \frac{11}{18}(x-4)(x-5)$$.

 This is a parabola opening upward. Vertex at $$x = 4.5$$.

 $$f(4.5) = \frac{11}{18}(0.5)(-0.5) = \frac{11}{18}(-\frac{1}{4}) = -\frac{11}{72}$$.

 Check boundaries:

 At $$x = -1$$: $$f(-1^-) = 3$$, $$f(-1^+) = 3$$ continuous.

 At $$x = 2$$: $$f(2^-) = 11/3 = 88/24$$, $$f(2^+) = \frac{11}{18}(-2)(-3) = \frac{66}{18} = 11/3$$. Continuous.

 Sum of minima: $$f(0) + f(4.5) = \frac{7}{3} - \frac{11}{72} = \frac{168 - 11}{72} = \frac{157}{72}$$.

Answer: A ($$\frac{157}{72}$$)

Given the functional equation $$ f(x+y) = f(x) f'(y) + f'(x) f(y) $$ for all real $$ x $$ and $$ y $$, and the initial condition $$ f(0) = 1 $$, we need to find $$ \sum_{n=1}^{100} \log_e f(n) $$.

To solve, substitute $$ y = 0 $$ into the functional equation:

$$ f(x+0) = f(x) f'(0) + f'(x) f(0) $$

Given $$ f(0) = 1 $$, this simplifies to:

$$ f(x) = f(x) f'(0) + f'(x) $$

Rearranging terms:

$$ f'(x) = f(x) - f(x) f'(0) $$

$$ f'(x) = f(x) (1 - f'(0)) $$

Let $$ k = 1 - f'(0) $$, a constant. Then:

$$ f'(x) = k f(x) $$

This is a first-order linear differential equation. The general solution is $$ f(x) = A e^{k x} $$, where $$ A $$ is a constant.

Using the initial condition $$ f(0) = 1 $$:

$$ f(0) = A e^{k \cdot 0} = A = 1 $$

So, $$ f(x) = e^{k x} $$.

Now, find $$ f'(x) $$:

$$ f'(x) = k e^{k x} $$

Evaluate at $$ x = 0 $$:

$$ f'(0) = k e^{0} = k $$

But $$ k = 1 - f'(0) $$, so substitute:

$$ k = 1 - k $$

$$ 2k = 1 $$

$$ k = \frac{1}{2} $$

Thus, $$ f(x) = e^{\frac{x}{2}} $$.

Verify this satisfies the original functional equation:

Left-hand side: $$ f(x+y) = e^{\frac{x+y}{2}} $$.

Right-hand side: $$ f(x) f'(y) + f'(x) f(y) = e^{\frac{x}{2}} \cdot \frac{1}{2} e^{\frac{y}{2}} + \frac{1}{2} e^{\frac{x}{2}} \cdot e^{\frac{y}{2}} = \frac{1}{2} e^{\frac{x+y}{2}} + \frac{1}{2} e^{\frac{x+y}{2}} = e^{\frac{x+y}{2}} $$.

Both sides are equal, so the solution is valid.

Now compute $$ \log_e f(n) $$:

$$ \log_e f(n) = \log_e \left( e^{\frac{n}{2}} \right) = \frac{n}{2} $$

The sum is:

$$ \sum_{n=1}^{100} \log_e f(n) = \sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} \sum_{n=1}^{100} n $$

The sum of the first $$ n $$ natural numbers is given by the formula $$ \sum_{n=1}^{m} n = \frac{m(m+1)}{2} $$. For $$ m = 100 $$:

$$ \sum_{n=1}^{100} n = \frac{100 \times 101}{2} = 5050 $$

Therefore,

$$ \frac{1}{2} \times 5050 = 2525 $$

The sum $$ \sum_{n=1}^{100} \log_e f(n) = 2525 $$.

Comparing with the options, A. 2525 is correct.

$$f(x) = ax^3 + bx^2 + cx + 41$$.

$$f(1) = a + b + c + 41 = 40 \implies a + b + c = -1$$ ... (i)

$$f'(x) = 3ax^2 + 2bx + c$$. $$f'(1) = 3a + 2b + c = 2$$ ... (ii)

$$f''(x) = 6ax + 2b$$. $$f''(1) = 6a + 2b = 4$$ ... (iii)

From (iii): $$3a + b = 2 \implies b = 2 - 3a$$.

From (ii): $$3a + 2(2-3a) + c = 2 \implies 3a + 4 - 6a + c = 2 \implies -3a + c = -2 \implies c = 3a - 2$$.

From (i): $$a + (2-3a) + (3a-2) = -1 \implies a = -1$$.

$$b = 2 - 3(-1) = 5$$, $$c = 3(-1) - 2 = -5$$.

$$a^2 + b^2 + c^2 = 1 + 25 + 25 = 51$$.

The correct answer is Option 3: 51.

Given $$f(x) = x^5 + 2x^3 + 3x + 1$$ and $$g(f(x)) = x$$, so $$g = f^{-1}$$. First, $$f(1) = 1 + 2 + 3 + 1 = 7$$, which implies $$g(7) = 1$$.

By the inverse function theorem, $$g'(f(x)) = \frac{1}{f'(x)}$$. Since $$f'(x) = 5x^4 + 6x^2 + 3$$, at $$x = 1$$ one gets $$f'(1) = 5 + 6 + 3 = 14$$ and hence $$g'(7) = g'(f(1)) = \frac{1}{f'(1)} = \frac{1}{14}$$.

$$\frac{g(7)}{g'(7)} = \frac{1}{1/14} = 14$$

The correct answer is Option (1): 14.

Simplify first: $$y = \ln(1-x^2) - \ln(1+x^2)$$.

First Derivative: $$y' = \frac{-2x}{1-x^2} - \frac{2x}{1+x^2} = \frac{-4x}{1-x^4}$$.

At $$x = 1/2$$: $$y' = \frac{-4(1/2)}{1 - (1/16)} = \frac{-2}{15/16} = -\frac{32}{15}$$.

Second Derivative: $$y'' = \frac{-4(1-x^4) - (-4x)(-4x^3)}{(1-x^4)^2} = \frac{-4 - 12x^4}{(1-x^4)^2}$$.

At $$x = 1/2$$: $$y'' = \frac{-4 - 12(1/16)}{(15/16)^2} = \frac{-4 - 3/4}{225/256} = \frac{-19/4}{225/256} = -\frac{19 \times 64}{225} = -\frac{1216}{225}$$.

$$225(y' - y'') = 225 \left( -\frac{32}{15} + \frac{1216}{225} \right) = 225 \left( -\frac{480}{225} + \frac{1216}{225} \right) = 736$$

g(x) = h(eˣ)·e^{h(x)}. g'(x) = h'(eˣ)eˣ·e^{h(x)} + h(eˣ)·e^{h(x)}·h'(x).

At x=0: g'(0) = h'(1)·1·e^{h(0)} + h(1)·e^{h(0)}·h'(0) = 2·e⁰ + 1·e⁰·2 = 2+2 = 4.

The correct answer is Option (2): 4.

Find the number of critical points of $$f(x) = (x-2)^{2/3}(2x+1)$$.

We start by differentiating using the product rule, which gives $$f'(x) = \frac{2}{3}(x-2)^{-1/3}(2x+1) + 2(x-2)^{2/3} = \frac{2(2x+1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3}$$.

Combining these terms over the common denominator $$(x-2)^{1/3}$$ yields $$f'(x) = \frac{2(2x+1) + 6(x-2)}{3(x-2)^{1/3}} = \frac{4x + 2 + 6x - 12}{3(x-2)^{1/3}} = \frac{10x - 10}{3(x-2)^{1/3}} = \frac{10(x-1)}{3(x-2)^{1/3}}$$.

Critical points occur where $$f'(x)=0$$ or where $$f'(x)$$ is undefined. The equation $$10(x-1)=0$$ gives $$x=1$$, and the condition $$(x-2)^{1/3}=0$$ gives $$x=2$$. Since $$f(2)=0$$, $$x=2$$ is also a critical point.

Therefore, the function has exactly two critical points at $$x=1$$ and $$x=2$$. Hence, the number of critical points is 2.

$$\ln y = 3\sin^{-1}x$$. Differentiating: $$\frac{y'}{y} = \frac{3}{\sqrt{1-x^2}}$$, so $$y' = \frac{3y}{\sqrt{1-x^2}}$$.

$$(1-x^2)(y')^2 = 9y^2$$. Differentiating: $$(1-x^2)2y'y'' - 2x(y')^2 = 18yy'$$.

Dividing by $$2y'$$: $$(1-x^2)y'' - xy' = 9y$$.

At $$x = 1/2$$: $$\ln y = 3\sin^{-1}(1/2) = 3\pi/6 = \pi/2$$, so $$y = e^{\pi/2}$$.

$$(1-x^2)y''-xy' = 9y = 9e^{\pi/2}$$.

The correct answer is Option 2: $$9e^{\pi/2}$$.

Write the function as a quotient:
$$y(\theta)=\frac{N(\theta)}{D(\theta)}$$ where

$$N(\theta)=2\cos\theta+\cos2\theta$$
$$D(\theta)=\cos3\theta+4\cos2\theta+5\cos\theta+2$$

First derivatives:

$$N'(\theta)=\frac{d}{d\theta}\!\left(2\cos\theta+\cos2\theta\right)=-2\sin\theta-2\sin2\theta$$
$$D'(\theta)=\frac{d}{d\theta}\!\left(\cos3\theta+4\cos2\theta+5\cos\theta+2\right)=-3\sin3\theta-8\sin2\theta-5\sin\theta$$

Second derivatives:

$$N''(\theta)=\frac{d}{d\theta}\!\left(-2\sin\theta-2\sin2\theta\right)=-2\cos\theta-4\cos2\theta$$
$$D''(\theta)=\frac{d}{d\theta}\!\left(-3\sin3\theta-8\sin2\theta-5\sin\theta\right)=-9\cos3\theta-16\cos2\theta-5\cos\theta$$

At $$\theta=\frac{\pi}{2}$$ the trigonometric values are:
$$\cos\frac{\pi}{2}=0,\;\sin\frac{\pi}{2}=1$$
$$\cos\pi=-1,\;\sin\pi=0$$
$$\cos\frac{3\pi}{2}=0,\;\sin\frac{3\pi}{2}=-1$$

Hence

$$N=-1,\;D=-2$$
$$N'=-2,\;D'=-2$$
$$N''=4,\;D''=16$$

Using the quotient rule $$y'=\dfrac{N'D-ND'}{D^{2}}$$:

$$y'=\frac{(-2)(-2)-(-1)(-2)}{(-2)^{2}}=\frac{4-2}{4}=\frac12$$

Define $$A=N'D-ND'$$, so $$A' = N''D-ND''$$ (because the $$N'D'$$ terms cancel).
At $$\theta=\frac{\pi}{2}$$:

$$A=2,\;\;A'=4(-2)-(-1)(16)=-8+16=8$$

The second derivative of a quotient can be written as
$$y''=\frac{A'D-2AD'}{D^{3}}$$

Thus

$$y''=\frac{8(-2)-2(2)(-2)}{(-2)^{3}}=\frac{-16+8}{-8}=1$$

Finally, at $$\theta=\frac{\pi}{2}$$:

$$y''+y'+y = 1+\frac12+\frac12 = 2$$

The required value is 2, which matches Option C.

We are given a function $$f: \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$$ satisfying $$f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$$ for all $$x, y$$ with $$f(y) \neq 0$$, and $$f'(1) = 2024$$. We need to find the relationship between $$f$$ and $$f'$$.

Determine the form of $$f(x)$$.

Setting $$x = y$$ in the functional equation:

$$ f\left(\frac{x}{x}\right) = \frac{f(x)}{f(x)} \implies f(1) = 1 $$

Setting $$y = x$$ and $$x = 1$$:

$$ f\left(\frac{1}{y}\right) = \frac{f(1)}{f(y)} = \frac{1}{f(y)} $$

The functional equation $$f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$$ is equivalent to $$f(xy) = f(x) \cdot f(y)$$ (a multiplicative function). For differentiable functions satisfying this, the solution is $$f(x) = x^n$$ for some constant $$n$$.

Determine the value of $$n$$ using $$f'(1) = 2024$$.

If $$f(x) = x^n$$, then $$f'(x) = nx^{n-1}$$.

$$ f'(1) = n \cdot 1^{n-1} = n = 2024 $$

Therefore, $$f(x) = x^{2024}$$.

Find the differential equation satisfied by $$f$$.

With $$f(x) = x^{2024}$$:

$$ f'(x) = 2024x^{2023} $$

Multiply both sides by $$x$$:

$$ xf'(x) = 2024x^{2024} = 2024 \cdot f(x) $$

Therefore:

$$ xf'(x) - 2024f(x) = 0 $$

Verification: We can verify by substituting: $$x \cdot 2024x^{2023} - 2024 \cdot x^{2024} = 2024x^{2024} - 2024x^{2024} = 0$$. Confirmed.

The correct answer is Option (1): $$xf'(x) - 2024f(x) = 0$$.

Given $$f(x) = \frac{1}{2}[g(x) + g(2-x)]$$.

Then $$f'(x) = \frac{1}{2}[g'(x) - g'(2-x)]$$.

Note: $$f'(1) = \frac{1}{2}[g'(1) - g'(1)] = 0$$.

Also, $$f'(\frac{1}{2}) = \frac{1}{2}[g'(\frac{1}{2}) - g'(\frac{3}{2})] = 0$$ (given $$g'(\frac{1}{2}) = g'(\frac{3}{2})$$).

And $$f'(\frac{3}{2}) = \frac{1}{2}[g'(\frac{3}{2}) - g'(\frac{1}{2})] = 0$$.

So $$f'$$ has zeros at $$x = \frac{1}{2}, 1, \frac{3}{2}$$ in $$(0, 2)$$.

By Rolle's theorem applied to $$f'$$: $$f''(x) = 0$$ for at least one point in $$(\frac{1}{2}, 1)$$ and at least one point in $$(1, \frac{3}{2})$$.

So $$f''(x) = 0$$ for at least two $$x$$ in $$(0, 2)$$.

The answer is Option (1): $$\boxed{f''(x) = 0 \text{ for at least two } x \text{ in } (0, 2)}$$.

To find $$f'(0)$$ for the function $$f(x) = \frac{(2^x + 2^{-x})\tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3},$$ we use the definition of the derivative at $$x = 0$$: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}.$$ First, evaluate $$f(0)$$:

• Numerator: $$(2^0 + 2^{-0}) \tan 0 \sqrt{\tan^{-1}(0^2 - 0 + 1)} = (1 + 1) \cdot 0 \cdot \sqrt{\tan^{-1}(1)} = 2 \cdot 0 \cdot \sqrt{\pi/4} = 0$$, since $$\tan^{-1}(1) = \pi/4$$.
• Denominator: $$(7(0)^2 + 3(0) + 1)^3 = 1^3 = 1$$.

• $$\lim_{h \to 0} (2^h + 2^{-h}) = 2^0 + 2^0 = 1 + 1 = 2$$.
• $$\lim_{h \to 0} \frac{\tan h}{h} = 1$$, since $$\tan h \sim h$$ near 0.
• $$\lim_{h \to 0} \sqrt{\tan^{-1}(h^2 - h + 1)} = \sqrt{\tan^{-1}(1)} = \sqrt{\pi/4} = \sqrt{\pi}/2$$, as $$\tan^{-1}(1) = \pi/4$$.
• $$\lim_{h \to 0} \frac{1}{(7h^2 + 3h + 1)^3} = \frac{1}{(1)^3} = 1$$, since the denominator approaches 1.

• $$2^x + 2^{-x}$$ is continuous and differentiable.
• $$\tan x$$ is continuous and differentiable near 0 (as $$\cos x \neq 0$$).
• $$\tan^{-1}(x^2 - x + 1)$$ is continuous and differentiable since $$x^2 - x + 1 \geq 3/4 > 0$$ for all $$x$$, and its square root is defined and differentiable.
• The denominator $$(7x^2 + 3x + 1)^3$$ is never zero (minimum value at $$x = -3/14$$ is positive).

g is the inverse of f: g(f(x)) = x. Differentiating: g'(f(x))·f'(x) = 1

g'(f(x)) = 1/f'(x)

f(x) = x⁵ + 2e^(x/4). f(0) = 0 + 2 = 2.

So g'(2) = g'(f(0)) = 1/f'(0)

f'(x) = 5x⁴ + (1/2)e^(x/4)

f'(0) = 0 + 1/2 = 1/2

g'(2) = 1/(1/2) = 2

8g'(2) = 16

The correct answer is Option 4: 16.

$$g(x)$$ looks like a part of a higher derivative.

Let $$h(x) = f(x) \cdot f''(x) + (f'(x))^2$$ is the derivative of $$f \cdot f'$$.

The correct observation: $$g(x)$$ is the derivative of $$f(x) \cdot f''(x) + (f'(x))^2$$... almost.

, $$(f \cdot f'')' = f' f'' + f f'''$$.

Let $$k(x) = f(x) f''(x)$$. By Rolle's theorem on $$f(x)$$:

$$f(0)=0, f(1)=1, f(2)=-1, f(3)=2, f(4)=-2$$.

$$f(x)$$ has 4 zeros in $$(0, 4)$$ by Intermediate Value Theorem.

Total zeros of $$f(x) = 5$$ (including $$x=0$$).

Following the chain of derivatives through Rolle's Theorem, the minimum zeros for this specific expression is 5

We need the number of real roots of the equations
    $$f(x)=0 \quad\Longleftrightarrow\quad 2^{x}=x^{2}$$
    $$f'(x)=0 \quad\Longleftrightarrow\quad 2^{x}\,\ln 2=2x$$

Define $$g(x)=2^{x}-x^{2}\;.$$ We inspect its sign in different regions.

• As $$x\rightarrow -\infty$$, $$2^{x}\rightarrow 0$$ while $$x^{2}\rightarrow \infty$$, so $$g(x)\rightarrow-\infty\;(\text{negative}).$$

• $$g(0)=2^{0}-0^{2}=1 \gt 0$$ (positive).

• $$g(-1)=2^{-1}-(-1)^{2}=0.5-1=-0.5 \lt 0$$ (negative).   Because $$g(-1)\lt 0$$ and $$g(0)\gt 0$$, by the Intermediate Value Theorem there is one root in $$(-1,0)$$.

• $$g(2)=2^{2}-2^{2}=0$$ and $$g(4)=2^{4}-4^{2}=0$$.   A quick check of the sign between these points:   $$g(3)=2^{3}-3^{2}=8-9=-1\lt 0,$$ so the function changes sign at both $$x=2$$ and $$x=4$$ but not again between them.

• For $$x\gt 4$$, the exponential term dominates: $$g(6)=2^{6}-6^{2}=64-36=28\gt 0$$ and the value keeps increasing, so no more zeros occur.

Hence the three distinct real roots are
    one in $$(-1,0)$$, and the obvious ones $$x=2$$ and $$x=4$$.
Therefore $$m=3$$.

Differentiate: $$f'(x)=2^{x}\ln 2-2x.$$ Let

$$h(x)=2^{x}\ln 2-2x.$$

• $$h(0)=\ln 2\;(\approx0.693)\gt0.$$

• As $$x\rightarrow -\infty$$: $$2^{x}\ln 2\rightarrow 0$$ and $$-2x\rightarrow+\infty,$$ hence $$h(x)\rightarrow+\infty\;(\text{positive}).$$

• Evaluate at $$x=1$$: $$h(1)=2\ln 2-2\approx1.386-2=-0.614\lt0.$$   So a sign change occurs between $$x=0$$ and $$x=1$$ ⇒ one root in $$(0,1).$$

• To check for more roots, find extrema of $$h(x)$$:
  $$h'(x)=2^{x}(\ln 2)^{2}-2.$$
  Set $$h'(x)=0$$: $$2^{x}=\dfrac{2}{(\ln 2)^{2}}\approx4.166\;,$$ giving   $$x_{0}=\log_{2}(4.166)\approx2.06.$$   Because $$h''(x)=2^{x}(\ln 2)^{3}\gt0,$$ this point is the unique minimum of $$h(x).$$

• Value at the minimum:   $$h(2.06)=2^{2.06}\ln 2-2(2.06)\approx2.888-4.12\approx-1.23\lt0.$$   Thus $$h(x)$$ is negative in a neighborhood of this minimum.

• Check farther right: $$h(4)=2^{4}\ln 2-8=16\ln 2-8\approx11.09-8=3.09\gt0.$$   So $$h(x)$$ changes from negative to positive between $$x\approx3$$ and $$x=4$$, giving a second root in $$(3,4).$$

• Since $$h(x)$$ is positive for $$x\lt0$$ and again for sufficiently large $$x\gt4$$, and has only one minimum, no additional sign changes are possible.

Hence $$h(x)=0$$ has exactly two real solutions, so $$n=2$$.

Finally, $$m+n=3+2=5$$.

Answer : 5

Given $$y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$$. Find $$y' - y''$$ at $$x = -1$$.

Multiply both sides by $$(1-x)$$:

$$ (1-x) \cdot y(x) = (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16}) $$

Using the identity $$(1-a)(1+a) = 1-a^2$$ repeatedly:

$$ (1-x) \cdot y(x) = 1 - x^{32} $$

So $$y(x) = \frac{1 - x^{32}}{1 - x} = 1 + x + x^2 + \cdots + x^{31}$$ for $$x \neq 1$$.

$$ y'(x) = \sum_{k=1}^{31} k \cdot x^{k-1} $$

$$ y'(-1) = \sum_{k=1}^{31} k \cdot (-1)^{k-1} = 1 - 2 + 3 - 4 + \cdots + 31 $$

Group pairs: $$(1-2) + (3-4) + \cdots + (29-30) + 31 = -15 + 31 = 16$$

$$ y''(x) = \sum_{k=2}^{31} k(k-1) \cdot x^{k-2} $$

$$ y''(-1) = \sum_{k=2}^{31} k(k-1) \cdot (-1)^{k-2} = \sum_{k=2}^{31} k(k-1) \cdot (-1)^{k} $$

(since $$(-1)^{k-2} = (-1)^k$$)

Group consecutive even-odd pairs:

$$ = \sum_{m=1}^{15} [2m(2m-1) - (2m+1)(2m)] = \sum_{m=1}^{15} 2m[(2m-1) - (2m+1)] = \sum_{m=1}^{15} 2m(-2) = -4\sum_{m=1}^{15} m $$

$$ = -4 \times \frac{15 \times 16}{2} = -4 \times 120 = -480 $$

$$ y'(-1) - y''(-1) = 16 - (-480) = 16 + 480 = 496 $$

The correct answer is Option C: 496.

Given $$y(x) = x^x$$, $$x > 0$$. We need to find $$y''(2) - 2y'(2)$$.

Taking the natural logarithm of both sides:

$$\ln y = x \ln x$$

Differentiating both sides with respect to $$x$$:

$$\frac{y'}{y} = \ln x + 1$$

$$y' = x^x(\ln x + 1)$$

Differentiating again:

$$y'' = \frac{d}{dx}[x^x(\ln x + 1)]$$

$$= x^x(\ln x + 1)^2 + x^x \cdot \frac{1}{x}$$

$$= x^x\left[(\ln x + 1)^2 + \frac{1}{x}\right]$$

Evaluating at $$x = 2$$:

$$y(2) = 2^2 = 4$$

$$y'(2) = 4(\ln 2 + 1)$$

$$y''(2) = 4\left[(\ln 2 + 1)^2 + \frac{1}{2}\right]$$

Computing $$y''(2) - 2y'(2)$$:

$$= 4\left[(\ln 2 + 1)^2 + \frac{1}{2}\right] - 2 \cdot 4(\ln 2 + 1)$$

$$= 4\left[(\ln 2)^2 + 2\ln 2 + 1 + \frac{1}{2} - 2\ln 2 - 2\right]$$

$$= 4\left[(\ln 2)^2 - \frac{1}{2}\right]$$

$$= 4(\log_e 2)^2 - 2$$

The answer is Option C: $$4(\log_e 2)^2 - 2$$.

Given $$5f(x+y) = f(x) \cdot f(y)$$ for all $$x, y \in \mathbb{R}$$, $$f: \mathbb{R} \to (0, \infty)$$, and $$f(3) = 320$$.

To begin,

Setting $$x = y = 0$$: $$5f(0) = f(0)^2$$

Since $$f(0) > 0$$: $$f(0) = 5$$

Next,

Let $$g(x) = \frac{f(x)}{5}$$. Then:

$$5 \cdot 5g(x+y) = 5g(x) \cdot 5g(y)$$

$$25g(x+y) = 25g(x)g(y)$$

$$g(x+y) = g(x)g(y)$$

Since $$f$$ is differentiable, $$g(x) = a^x$$ for some $$a > 0$$.

So $$f(x) = 5 \cdot a^x$$.

From this,

$$5 \cdot a^3 = 320 \implies a^3 = 64 \implies a = 4$$

$$f(x) = 5 \cdot 4^x$$

Continuing,

$$\sum_{n=0}^{5} f(n) = 5(4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5)$$

$$= 5(1 + 4 + 16 + 64 + 256 + 1024)$$

$$= 5 \times 1365 = 6825$$

The correct answer is Option 3: $$6825$$.

Given: $$2x^y + 3y^x = 20$$. We need $$\frac{dy}{dx}$$ at $$(2, 2)$$.

Verify: At $$(2,2)$$: $$2(2^2) + 3(2^2) = 8 + 12 = 20$$ ✓

Differentiate implicitly. For $$x^y$$, take log: $$\ln(x^y) = y\ln x$$.

$$\frac{d}{dx}(x^y) = x^y\left(\frac{y}{x} + y'\ln x\right)$$

For $$y^x$$: $$\ln(y^x) = x\ln y$$.

$$\frac{d}{dx}(y^x) = y^x\left(\ln y + \frac{xy'}{y}\right)$$

Differentiating the equation:

$$ 2x^y\left(\frac{y}{x} + y'\ln x\right) + 3y^x\left(\ln y + \frac{xy'}{y}\right) = 0 $$

At $$(2, 2)$$: $$x^y = 4$$, $$y^x = 4$$:

$$ 2(4)\left(1 + y'\ln 2\right) + 3(4)\left(\ln 2 + y'\right) = 0 $$

$$ 8 + 8y'\ln 2 + 12\ln 2 + 12y' = 0 $$

$$ y'(8\ln 2 + 12) = -(8 + 12\ln 2) $$

$$ y' = -\frac{8 + 12\ln 2}{12 + 8\ln 2} = -\frac{4(2 + 3\ln 2)}{4(3 + 2\ln 2)} = -\frac{2 + \ln 8}{3 + \ln 4} $$

The correct answer is $$-\dfrac{2 + \log_e 8}{3 + \log_e 4}$$.

We are given $$f(x) = x^3 - x^2 f'(1) + x f''(2) - f'''(3)$$ for $$x \in \mathbb{R}$$. Setting $$a=f'(1)$$, $$b=f''(2)$$, and $$c=f'''(3)$$ gives $$f(x)=x^3-ax^2+bx-c$$.

Computing the first derivative yields $$f'(x)=3x^2-2ax+b$$. Plugging in $$x=1$$, we get $$f'(1)=3-2a+b$$, which must equal $$a$$ and leads to $$3-2a+b=a$$ or $$3+b=3a$$. The second derivative is $$f''(x)=6x-2a$$, so $$f''(2)=12-2a=b$$. Finally, the third derivative is constant: $$f'''(x)=6$$, hence $$c=6$$.

Using $$b=12-2a$$ in the equation $$3+b=3a$$ gives $$3+12-2a=3a$$, which simplifies to $$15=5a$$ and thus $$a=3$$. It follows that $$b=12-6=6$$ and $$c=6$$. Therefore $$f(x)=x^3-3x^2+6x-6$$.

Evaluating at specific points gives:
$$f(0)=-6$$
$$f(1)=1-3+6-6=-2$$
$$f(2)=8-12+12-6=2$$
$$f(3)=27-27+18-6=12$$.

Checking the options:
Option A: $$3f(1)+f(2)=3(-2)+2=-4\neq12=f(3)$$ ✘
Option B: $$f(3)-f(2)=12-2=10\neq-2=f(1)$$ ✘
Option C: $$2f(0)-f(1)+f(3)=2(-6)-(-2)+12=-12+2+12=2=f(2)$$ ✔
Option D: $$f(1)+f(2)+f(3)=-2+2+12=12\neq-6=f(0)$$ ✘

The correct answer is Option C: $$2f(0)-f(1)+f(3)=f(2)$$.

Given: $$f(x) = x^2 + g'(1)x + g''(2)$$ and $$g(x) = f(1)x^2 + xf'(x) + f''(x)$$.

Let $$a = g'(1)$$ and $$b = g''(2)$$. Then $$f(x) = x^2 + ax + b$$.

Computing f values:

$$f(1) = 1 + a + b$$, $$f'(x) = 2x + a$$, $$f''(x) = 2$$

Substituting into g:

Computing g values:

$$g'(x) = 2(3+a+b)x + a$$, so $$g'(1) = 2(3+a+b) + a = 6 + 3a + 2b$$

$$g''(x) = 2(3+a+b)$$, so $$g''(2) = 2(3+a+b)$$

Solving the equations:

From $$g'(1) = a$$: $$6 + 3a + 2b = a \Rightarrow 2a + 2b = -6 \Rightarrow a + b = -3$$ ... (1)

From $$g''(2) = b$$: $$b = 2(3 + a + b) = 2(3 - 3) = 0$$ (using (1))

So $$b = 0$$ and $$a = -3$$.

$$f(x) = x^2 - 3x$$, $$g(x) = (3-3+0)x^2 + (-3)x + 2 = -3x + 2$$

Final computation:

Therefore, $$f(4) - g(4) = 14$$.

We are given that $$f : \mathbb{R} \to \mathbb{R}$$ is differentiable, $$f(x+y) = f(x) + f(y) - 1$$ for all $$x, y \in \mathbb{R}$$, and $$f'(0) = 2$$.

Setting $$x = y = 0$$, we get $$f(0) = f(0) + f(0) - 1 = 2f(0) - 1$$, so $$f(0) = 1$$.

Now let $$g(x) = f(x) - 1$$. Then:

$$g(x + y) = f(x + y) - 1 = f(x) + f(y) - 1 - 1 = (f(x) - 1) + (f(y) - 1) = g(x) + g(y)$$

So $$g$$ satisfies Cauchy's functional equation: $$g(x + y) = g(x) + g(y)$$. Since $$f$$ is differentiable, $$g$$ is also differentiable, and the only differentiable solution of Cauchy's equation is $$g(x) = kx$$ for some constant $$k$$.

We have $$g'(0) = f'(0) = 2$$, so $$k = 2$$. Hence $$g(x) = 2x$$, giving $$f(x) = 2x + 1$$.

Now, $$f(-2) = 2(-2) + 1 = -3$$, so $$|f(-2)| = |-3| = 3$$.

So, the answer is $$3$$.

Given a curve $$y = f(x)$$ passing through $$P\left(1, \frac{3}{2}\right)$$ and $$Q\left(a, \frac{1}{2}\right)$$, where the tangent at any point $$R(b, f(b))$$ cuts the y-axis at $$S(0, c)$$ with $$bc = 3$$.

The tangent at $$R(b, f(b))$$ is: $$Y - f(b) = f'(b)(X - b)$$

At $$X = 0$$: $$c = f(b) - b \cdot f'(b)$$

Since $$bc = 3$$ for all points on the curve (replacing $$b$$ with $$x$$):

$$x[f(x) - x f'(x)] = 3$$

$$xf(x) - x^2 f'(x) = 3$$

Rearranging:

$$f'(x) - \frac{f(x)}{x} = -\frac{3}{x^2}$$

This is a linear ODE with integrating factor $$e^{-\int \frac{1}{x}dx} = \frac{1}{x}$$:

$$\frac{d}{dx}\left[\frac{f(x)}{x}\right] = -\frac{3}{x^3}$$

Integrating:

$$\frac{f(x)}{x} = \frac{3}{2x^2} + C$$

$$f(x) = \frac{3}{2x} + Cx$$

Using $$f(1) = \frac{3}{2}$$:

$$\frac{3}{2} = \frac{3}{2} + C \implies C = 0$$

So $$f(x) = \frac{3}{2x}$$.

Finding $$a$$: $$f(a) = \frac{1}{2} \implies \frac{3}{2a} = \frac{1}{2} \implies a = 3$$

Therefore $$Q = (3, \frac{1}{2})$$ and:

$$(PQ)^2 = (3-1)^2 + \left(\frac{1}{2} - \frac{3}{2}\right)^2 = 4 + 1 = 5$$

The answer is $$5$$.

We consider the curve $$y = \frac{x - a}{(x + b)(x - 2)}$$ which passes through the point $$(1, -3)$$ and has normal line $$x - 4y = 13$$ at that point. Substituting $$(1, -3)$$ into the normal gives $$1 - 4(-3) = 1 + 12 = 13$$, confirming consistency.

Since $$(1, -3)$$ lies on the curve, we set $$y(1)=\frac{1 - a}{(1 + b)(1 - 2)}=\frac{a - 1}{1 + b}=-3\,. $$ Hence $$a - 1 = -3(1 + b)\implies a = -2 - 3b\quad\cdots(i)$$

The normal has slope $$\tfrac14$$, so the tangent has slope $$-4$$. Writing $$u=x-a\,,\quad v=(x+b)(x-2)=x^2+(b-2)x-2b\,, $$ we get by the quotient rule $$y'=\frac{v - u\,v'}{v^2}\,. $$

At $$x=1$$ we have $$u=1-a\,,\quad v=(1+b)(-1)=-(1+b)\,,\quad v'=2(1)+b-2=b\,. $$ Thus $$y'(1)=\frac{-(1+b)-(1-a)\,b}{(1+b)^2}=\frac{-1-2b+ab}{(1+b)^2}=-4\,. $$ Clearing denominators gives $$-1-2b+ab=-4(1+b)^2=-4-8b-4b^2\,, $$ or $$4b^2+6b+ab+3=0\,. $$

Substituting $$a=-2-3b$$ from (i) yields $$4b^2+6b+(-2-3b)b+3=0\,, $$ which simplifies to $$4b^2+6b-2b-3b^2+3=0\quad\Longrightarrow\quad b^2+4b+3=0\implies(b+1)(b+3)=0\,. $$

If $$b=-1$$, then $$a=-2-3(-1)=1$$ and $$y=\frac{x-1}{(x-1)(x-2)}=\frac1{x-2}\quad(x\neq1),$$ but this gives $$y(1)=-1\neq-3$$, so $$b=-1$$ is invalid (removable discontinuity at $$x=1$$).

If $$b=-3$$, then $$a=-2-3(-3)=7$$ and $$y=\frac{x-7}{(x-3)(x-2)}\,. $$ At $$x=1$$ this yields $$y=\frac{-6}{(-2)(-1)}=-3$$ and $$y'=\frac{(x-3)(x-2)-(x-7)(2x-5)}{[(x-3)(x-2)]^2},$$ so at $$x=1$$ $$y'(1)=\frac{2-(-6)(-3)}{4}=\frac{2-18}{4}=-4\,, $$ confirming the tangent slope is $$-4$$ and hence the normal slope is $$\tfrac14\,$$.

Therefore $$a+b=7+(-3)=4\,. $$

We are given the parametric curve $$x(t) = 2\sqrt{2}\cos t\sqrt{\sin 2t}$$ and $$y(t) = 2\sqrt{2}\sin t\sqrt{\sin 2t}$$ for $$t \in (0, \frac{\pi}{2})$$, and we need to evaluate $$\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}$$ at $$t = \frac{\pi}{4}$$.

We begin by finding $$\frac{dx}{dt}$$. Writing $$x = 2\sqrt{2}\cos t\,(\sin 2t)^{1/2}$$ and applying the product rule:

Combining terms over a common denominator $$(\sin 2t)^{1/2}$$:

The expression in brackets equals $$\cos(t + 2t) = \cos 3t$$ (by the cosine addition formula). So $$\frac{dx}{dt} = \frac{2\sqrt{2}\cos 3t}{\sqrt{\sin 2t}}$$.

By the same approach for $$y = 2\sqrt{2}\sin t\,(\sin 2t)^{1/2}$$:

(using $$\cos t\,\sin 2t + \sin t\,\cos 2t = \sin(t + 2t) = \sin 3t$$).

At $$t = \frac{\pi}{4}$$, we have $$\sin 2t = 1$$, $$\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$, and $$\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$$. Substituting:

So $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{-2} = -1$$.

Next, we need $$\frac{d^2y}{dx^2}$$. Using the formula $$\frac{d^2y}{dx^2} = \frac{\dot{x}\,\ddot{y} - \dot{y}\,\ddot{x}}{\dot{x}^3}$$, we must find the second derivatives $$\ddot{x}$$ and $$\ddot{y}$$ at $$t = \frac{\pi}{4}$$.

Differentiating $$\dot{x} = 2\sqrt{2}\cos 3t\,(\sin 2t)^{-1/2}$$ by the product rule:

At $$t = \frac{\pi}{4}$$, $$\cos 2t = \cos\frac{\pi}{2} = 0$$ and $$\sin 2t = 1$$, so the second term vanishes entirely. This gives:

Similarly, differentiating $$\dot{y} = 2\sqrt{2}\sin 3t\,(\sin 2t)^{-1/2}$$:

At $$t = \frac{\pi}{4}$$, the second term again vanishes, so:

Now substituting into the second derivative formula:

Finally, the required expression is:

Hence, the correct answer is Option 4.

We are given $$f(x) = x^3 + x - 5$$ and $$f(g(x)) = x$$, so $$g$$ is the inverse function of $$f$$.

By the inverse function theorem, $$g'(y) = \frac{1}{f'(g(y))}$$.

To find $$g'(63)$$, we first need $$g(63)$$, i.e., the value $$a$$ such that $$f(a) = 63$$:

$$ a^3 + a - 5 = 63 $$

$$ a^3 + a = 68 $$

Testing $$a = 4$$: $$4^3 + 4 = 64 + 4 = 68$$ ✓

So $$g(63) = 4$$.

Now compute $$f'(x) = 3x^2 + 1$$, so $$f'(4) = 3(16) + 1 = 49$$.

Therefore:

$$ g'(63) = \frac{1}{f'(g(63))} = \frac{1}{f'(4)} = \frac{1}{49} $$

The answer is Option B: $$\frac{1}{49}$$.

We need to find the value of $$\log_e 2 \cdot \dfrac{d}{dx}(\log_{\cos x} \csc x)$$ at $$x = \dfrac{\pi}{4}$$.

$$\log_{\cos x} \csc x = \frac{\ln(\csc x)}{\ln(\cos x)}$$

Let $$u = \ln(\csc x)$$ and $$v = \ln(\cos x)$$.

$$u' = \frac{-\csc x \cot x}{\csc x} = -\cot x$$

$$v' = \frac{-\sin x}{\cos x} = -\tan x$$

$$f'(x) = \frac{u'v - uv'}{v^2} = \frac{-\cot x \cdot \ln(\cos x) - \ln(\csc x) \cdot (-\tan x)}{[\ln(\cos x)]^2}$$

$$= \frac{-\cot x \cdot \ln(\cos x) + \tan x \cdot \ln(\csc x)}{[\ln(\cos x)]^2}$$

At $$x = \frac{\pi}{4}$$:

$$\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \csc\frac{\pi}{4} = \sqrt{2}$$

$$\cot\frac{\pi}{4} = 1, \quad \tan\frac{\pi}{4} = 1$$

$$\ln(\cos\frac{\pi}{4}) = \ln\frac{1}{\sqrt{2}} = -\frac{1}{2}\ln 2$$

$$\ln(\csc\frac{\pi}{4}) = \ln\sqrt{2} = \frac{1}{2}\ln 2$$

Numerator: $$-1 \cdot (-\frac{1}{2}\ln 2) + 1 \cdot \frac{1}{2}\ln 2 = \frac{1}{2}\ln 2 + \frac{1}{2}\ln 2 = \ln 2$$

Denominator: $$\left(-\frac{1}{2}\ln 2\right)^2 = \frac{1}{4}(\ln 2)^2$$

$$f'\left(\frac{\pi}{4}\right) = \frac{\ln 2}{\frac{1}{4}(\ln 2)^2} = \frac{4}{\ln 2} = \frac{4}{\log_e 2}$$

$$\log_e 2 \cdot f'\left(\frac{\pi}{4}\right) = \log_e 2 \cdot \frac{4}{\log_e 2} = 4$$

Therefore, the correct answer is Option D: $$4$$.

We need to find the value of $$a$$ first, then evaluate $$f$$ and $$f'$$ at $$\dfrac{a}{2}$$.

First, to find $$a$$, we evaluate the limit $$a = \lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{2n}{n^2 + k^2} = \lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{2}{n} \cdot \dfrac{1}{1 + \left(\frac{k}{n}\right)^2}$$

This is a Riemann sum for $$\int_0^1 \dfrac{2}{1+t^2}\,dt$$.

From this, $$a = 2\Big[\tan^{-1}(t)\Big]_0^1 = 2 \cdot \dfrac{\pi}{4} = \dfrac{\pi}{2}$$

Next, simplifying $$f(x)$$, we have $$f(x) = \sqrt{\dfrac{1 - \cos x}{1 + \cos x}}$$

Using the identities $$1 - \cos x = 2\sin^2\dfrac{x}{2}$$ and $$1 + \cos x = 2\cos^2\dfrac{x}{2}$$:

$$f(x) = \sqrt{\dfrac{2\sin^2(x/2)}{2\cos^2(x/2)}} = \left|\tan\dfrac{x}{2}\right| = \tan\dfrac{x}{2}$$

(since $$x \in (0, 1) \subset (0, \pi)$$, so $$\tan(x/2) > 0$$).

Then $$f\left(\dfrac{\pi}{4}\right) = \tan\dfrac{\pi}{8}$$

Since $$f(x) = \tan\dfrac{x}{2}$$, we get $$f'(x) = \dfrac{1}{2}\sec^2\dfrac{x}{2}$$

Thus $$f'\left(\dfrac{\pi}{4}\right) = \dfrac{1}{2}\sec^2\dfrac{\pi}{8} = \dfrac{1}{2\cos^2(\pi/8)}$$

Next, we verify the relationship by checking Option C: $$\sqrt{2}\,f\left(\dfrac{a}{2}\right) = f'\left(\dfrac{a}{2}\right)$$.

LHS: $$\sqrt{2}\,\tan\dfrac{\pi}{8} = \dfrac{\sqrt{2}\,\sin(\pi/8)}{\cos(\pi/8)}$$

RHS: $$\dfrac{1}{2\cos^2(\pi/8)}$$

We check if LHS = RHS:

$$\dfrac{\sqrt{2}\,\sin(\pi/8)}{\cos(\pi/8)} = \dfrac{1}{2\cos^2(\pi/8)}$$

$$2\sqrt{2}\,\sin(\pi/8)\cos(\pi/8) = 1$$

$$\sqrt{2}\,\sin(\pi/4) = \sqrt{2} \cdot \dfrac{\sqrt{2}}{2} = 1 \quad \checkmark$$

The correct answer is Option C: $$\sqrt{2}\,f\left(\dfrac{a}{2}\right) = f'\left(\dfrac{a}{2}\right)$$.

Given $$y(x) = (x^x)^x = x^{x^2}$$ for $$x > 0$$, we need to evaluate $$\frac{d^2x}{dy^2} + 20$$ at $$x = 1$$.

Taking the natural logarithm gives $$\ln y = x^2 \ln x$$, and differentiating yields $$\frac{1}{y}\frac{dy}{dx} = 2x \ln x + x$$. Therefore $$\frac{dy}{dx} = y \cdot x(2\ln x + 1) = x^{x^2} \cdot x(2\ln x + 1)$$, and at $$x = 1$$ we have $$y = 1$$ and $$\frac{dy}{dx} = 1\cdot1\cdot1 = 1$$.

To find the second derivative, let $$u = x^{x^2}$$ and $$v = x(2\ln x + 1) = 2x\ln x + x$$ so that $$\frac{dy}{dx} = uv$$. Then $$\frac{du}{dx} = u\cdot(2x\ln x + x) = uv$$ and $$\frac{dv}{dx} = 2\ln x + 3$$, which implies $$\frac{d^2y}{dx^2} = \frac{du}{dx}\,v + u\,\frac{dv}{dx} = uv^2 + u(2\ln x + 3)$$. Evaluating at $$x = 1$$ where $$u = 1$$ and $$v = 1$$ gives $$\frac{d^2y}{dx^2} = 1 + 3 = 4$$.

Using the relation $$\frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{\bigl(\frac{dy}{dx}\bigr)^3}$$ leads to $$\frac{d^2x}{dy^2} = -\frac{4}{1^3} = -4$$, so $$\frac{d^2x}{dy^2} + 20 = -4 + 20 = 16$$.

The answer is $$16$$.

First, by setting $$x=y=0 in the given equation we obtain f(0)=f(0)+f(0)\implies f(0)=0$$.

Next, setting $$y=0 yields f(x)=2^x f(0)+f(x)=f(x)\,, $$ which is trivially true.

Now, setting $$x=0 gives f(y)=f(y)+4^y f(0)=f(y)\,, $$ also trivially true.

Since these identities hold, we try the form $$f(x)=k(4^x-2^x). Then f(x+y)=k(4^{x+y}-2^{x+y})=k(4^x\cdot4^y-2^x\cdot2^y)\,, while the right-hand side becomes 2^x f(y)+4^y f(x)=2^x\cdot k(4^y-2^y)+4^y\cdot k(4^x-2^x) =k(2^x\cdot4^y-2^x\cdot2^y+4^y\cdot4^x-4^y\cdot2^x) = k(4^{x+y}-2^{x+y})\,, $$ confirming the guess.

Substituting $$f(2)=3 into f(x)=k(4^x-2^x) gives k(16-4)=12k=3\implies k=\frac14\,, hence f(x)=\frac{4^x-2^x}{4}\,.$$

Therefore the derivative is $$f'(x)=\frac{4^x\ln4-2^x\ln2}{4}\,. It follows that f'(4)=$$ $$\frac{256\ln4-16\ln2}{4}=\frac{256\cdot2\ln2-16\ln2}{4}=\frac{496\ln2}{4}$$ $$=124\ln2 and f'(2)=$$ $$\frac{16\ln4-4\ln2}{4}=\frac{32\ln2-4\ln2}{4}=\frac{28\ln2}{4}$$ $$=7\ln2\,.$$

Finally, $$\frac{f'(4)}{f'(2)}=\frac{124\ln2}{7\ln2}=\frac{124}{7} and so 14\cdot\frac{f'(4)}{f'(2)}=14\cdot\frac{124}{7}=2\times124=248\,, giving the answer 248$$.

We have been given

$$y(x)=\cot^{-1}\!\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right),\qquad x\in\left(\dfrac{\pi}{2},\pi\right).$$

Put

$$A=\sqrt{1+\sin x}$$, $$\qquad B=\sqrt{1-\sin x}$$

so that

$$y=\cot^{-1}\!\left(\dfrac{A+B}{A-B}\right).$$

Let

$$R(x)=\dfrac{A+B}{A-B}.$$ Then $$y=\cot^{-1}(R).$$

To find $$\dfrac{dy}{dx}$$ we first differentiate $$R(x).$$

Since $$A=(1+\sin x)^{1/2},$$ by the chain rule

$$\dfrac{dA}{dx}=\dfrac{1}{2}(1+\sin x)^{-1/2}\cdot\cos x=\dfrac{\cos x}{2\sqrt{1+\sin x}}=\dfrac{\cos x}{2A}.$$

Similarly, $$B=(1-\sin x)^{1/2},$$ so

$$\dfrac{dB}{dx}=\dfrac{1}{2}(1-\sin x)^{-1/2}\cdot(-\cos x)=\dfrac{-\cos x}{2\sqrt{1-\sin x}}=\dfrac{-\cos x}{2B}.$$

Write the numerator and denominator of $$R$$ as

$$N=A+B,\qquad D=A-B,$$ so $$R=\dfrac{N}{D}.$$

Using the quotient rule, $$\dfrac{dR}{dx}=\dfrac{D\dfrac{dN}{dx}-N\dfrac{dD}{dx}}{D^{2}},$$ where $$\dfrac{dN}{dx}= \dfrac{dA}{dx}+\dfrac{dB}{dx},\qquad \dfrac{dD}{dx}= \dfrac{dA}{dx}-\dfrac{dB}{dx}.$$

Now we evaluate every quantity at $$x=\dfrac{5\pi}{6}.$$ At this point $$\sin\!\left(\dfrac{5\pi}{6}\right)=\dfrac12,\qquad \cos\!\left(\dfrac{5\pi}{6}\right)=-\dfrac{\sqrt3}{2}.$$

Therefore

$$A_{0}=A\Big|_{x=5\pi/6}=\sqrt{1+\dfrac12}=\sqrt{\dfrac32} =\dfrac{\sqrt3}{\sqrt2},$$ $$B_{0}=B\Big|_{x=5\pi/6}=\sqrt{1-\dfrac12}=\sqrt{\dfrac12} =\dfrac1{\sqrt2}.$$

The derivatives become

$$\dfrac{dA}{dx}\Bigg|_{x=5\pi/6}= \dfrac{\cos x}{2A}\Bigg|_{x=5\pi/6}= \dfrac{-\dfrac{\sqrt3}{2}}{2\dfrac{\sqrt3}{\sqrt2}} =-\dfrac{\sqrt2}{4},$$

$$\dfrac{dB}{dx}\Bigg|_{x=5\pi/6}= \dfrac{-\cos x}{2B}\Bigg|_{x=5\pi/6}= \dfrac{+\dfrac{\sqrt3}{2}}{2\dfrac1{\sqrt2}} =\dfrac{\sqrt3}{2\sqrt2}=\dfrac{\sqrt6}{4}.$$

Hence

$$N_{0}=A_{0}+B_{0}=\dfrac{\sqrt3+1}{\sqrt2},\qquad D_{0}=A_{0}-B_{0}=\dfrac{\sqrt3-1}{\sqrt2},$$

$$\dfrac{dN}{dx}\Bigg|_{0}= -\dfrac{\sqrt2}{4}+\dfrac{\sqrt6}{4} =\dfrac{-\sqrt2+\sqrt6}{4},$$

$$\dfrac{dD}{dx}\Bigg|_{0}= -\dfrac{\sqrt2}{4}-\dfrac{\sqrt6}{4} =\dfrac{-\sqrt2-\sqrt6}{4}.$$

The derivative of $$R$$ at the point is

$$\begin{aligned} \dfrac{dR}{dx}\Bigg|_{0} &=\dfrac{D_{0}\big(\dfrac{dN}{dx}\big)_{0}-N_{0}\big(\dfrac{dD}{dx}\big)_{0}} {D_{0}^{2}}\\[4pt] &=\dfrac{\dfrac{\sqrt3-1}{\sqrt2}\cdot\dfrac{-\sqrt2+\sqrt6}{4} -\dfrac{\sqrt3+1}{\sqrt2}\cdot\dfrac{-\sqrt2-\sqrt6}{4}} {\left(\dfrac{\sqrt3-1}{\sqrt2}\right)^{2}}\\[6pt] &=\dfrac{2}{\;2-\sqrt3\;}. \end{aligned}$$

Next we need $$\dfrac{dy}{dx}.$$ The standard derivative formula for the inverse cotangent is

$$\dfrac{d}{dx}\bigl[\cot^{-1}u\bigr]=-\dfrac{u'}{1+u^{2}}.$$

Here $$u=R(x),$$ so

$$\dfrac{dy}{dx}=-\dfrac{\dfrac{dR}{dx}}{1+R^{2}}.$$ At $$x=\dfrac{5\pi}{6}$$ we already have $$\dfrac{dR}{dx}=\dfrac{2}{2-\sqrt3}.$$

We still need $$R_{0}=R\bigl(\tfrac{5\pi}{6}\bigr).$$ Using $$R_{0}=\dfrac{A_{0}+B_{0}}{A_{0}-B_{0}} =\dfrac{\sqrt3+1}{\sqrt3-1} =2+\sqrt3,$$ so

$$1+R_{0}^{2}=1+(2+\sqrt3)^{2} =1+\bigl(4+4\sqrt3+3\bigr) =8+4\sqrt3 =4(2+\sqrt3).$$

Consequently

$$\dfrac{dy}{dx}\Bigg|_{x=5\pi/6} =-\dfrac{\dfrac{2}{2-\sqrt3}}{4(2+\sqrt3)} =-\dfrac{2}{4(2-\sqrt3)(2+\sqrt3)}.$$

Because $$(2-\sqrt3)(2+\sqrt3)=4-3=1,$$ the expression simplifies to

$$\dfrac{dy}{dx}\Bigg|_{x=5\pi/6} =-\dfrac{2}{4} =-\dfrac12.$$

Hence, the correct answer is Option C.

We have $$f(x+1) = xf(x)$$ and $$g(x) = \ln f(x)$$. Taking logarithms: $$g(x+1) = \ln(xf(x)) = \ln x + g(x)$$, so $$g(x+1) - g(x) = \ln x$$.

Differentiating twice with respect to $$x$$: $$g''(x+1) - g''(x) = -\frac{1}{x^2}$$.

Now we compute $$g''(5) - g''(1)$$ by telescoping:

$$g''(2) - g''(1) = -\frac{1}{1^2} = -1$$

$$g''(3) - g''(2) = -\frac{1}{2^2} = -\frac{1}{4}$$

$$g''(4) - g''(3) = -\frac{1}{3^2} = -\frac{1}{9}$$

$$g''(5) - g''(4) = -\frac{1}{4^2} = -\frac{1}{16}$$

Adding all these: $$g''(5) - g''(1) = -1 - \frac{1}{4} - \frac{1}{9} - \frac{1}{16} = -\frac{144 + 36 + 16 + 9}{144} = -\frac{205}{144}$$.

Therefore, $$|g''(5) - g''(1)| = \frac{205}{144}$$.

The curve $$y = ax^2 + bx + c$$ passes through the point $$(1, 2)$$ and has its tangent at the origin equal to $$y = x$$.

Since the tangent line at the origin is $$y = x$$, the curve must pass through the origin. Setting $$x = 0$$ and $$y = 0$$, we get $$0 = a(0) + b(0) + c$$, so $$c = 0$$.

The slope of the tangent at the origin must equal the slope of $$y = x$$, which is $$1$$. Differentiating, $$\frac{dy}{dx} = 2ax + b$$. At $$x = 0$$, this gives $$b = 1$$.

Since the curve passes through $$(1, 2)$$, we have $$2 = a(1)^2 + b(1) + c = a + 1 + 0$$, so $$a = 1$$.

Therefore, the values are $$a = 1, b = 1, c = 0$$.

We have $$f(x) = \sin\!\left(\cos^{-1}\!\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)$$.

Let $$t = 2^x$$, so $$2^{2x} = t^2$$. The argument becomes $$\frac{1-t^2}{1+t^2}$$.

Using the identity $$\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$$, if we set $$\tan\theta = t$$ (with $$t > 0$$, so $$\theta \in (0, \pi/2)$$), then $$\cos^{-1}\!\left(\frac{1-t^2}{1+t^2}\right) = 2\theta$$.

Therefore $$f(x) = \sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2t}{1+t^2} = \frac{2 \cdot 2^x}{1+2^{2x}}$$.

Differentiating with $$u = 2^x$$: $$f = \frac{2u}{1+u^2}$$, so $$\frac{df}{du} = \frac{2(1+u^2) - 2u(2u)}{(1+u^2)^2} = \frac{2-2u^2}{(1+u^2)^2}$$.

By the chain rule, $$f'(x) = \frac{2(1-u^2)}{(1+u^2)^2} \cdot u\ln 2$$, since $$\frac{du}{dx} = 2^x \ln 2 = u\ln 2$$.

At $$x = 1$$: $$u = 2$$, so $$f'(1) = \frac{2(1-4)}{(1+4)^2} \cdot 2\ln 2 = \frac{-6}{25} \cdot 2\ln 2 = -\frac{12}{25}\ln 2$$.

Comparing with $$f'(1) = -\frac{b}{a}\log_e 2$$: we get $$\frac{b}{a} = \frac{12}{25}$$. Since $$\gcd(12,25) = 1$$, the integers with minimum $$|a^2-b^2|$$ are $$a = 25, b = 12$$.

$$|a^2 - b^2| = |625 - 144| = 481$$.

The answer is 481.

Given the implicit relation $$\log_e (x + y) = 4xy$$, we first identify the value of $$y$$ at $$x = 0$$. Substituting $$x = 0$$ gives

$$\log_e(0 + y) = 4 \cdot 0 \cdot y \;\Longrightarrow\; \log_e y = 0 \;\Longrightarrow\; y = e^{0} = 1.$$

Hence $$y(0)=1.$$

Now we differentiate both sides with respect to $$x$$. Using the formula $$\dfrac{d}{dx}\bigl[\log_e u\bigr] = \dfrac{1}{u}\dfrac{du}{dx}$$ on the left and the product rule $$\dfrac{d}{dx}[uv]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$$ on the right, we get

$$\frac{1}{x + y}\bigl(1 + y'\bigr) = 4y + 4x\,y'.$$

(Here and henceforth we write $$y' = \dfrac{dy}{dx}$$ and $$y'' = \dfrac{d^{2}y}{dx^{2}}$$.)

Putting $$x = 0,\,y = 1$$ into this first-derivative equation, we obtain

$$\frac{1 + y'(0)}{0 + 1} = 4\cdot1 + 4\cdot0\cdot y'(0) \;\Longrightarrow\; 1 + y'(0) = 4 \;\Longrightarrow\; y'(0) = 3.$$

To reach the second derivative, we again differentiate the equation

$$\frac{1 + y'}{x + y} = 4y + 4x\,y'.$$

Let us differentiate the left side by the quotient rule. With $$N = 1 + y', \quad D = x + y,$$ the quotient rule states $$\frac{d}{dx}\!\left(\frac{N}{D}\right) = \frac{N'D - ND'}{D^{2}},$$ where $$N' = y''$$ and $$D' = 1 + y'$$. Therefore

$$\frac{d}{dx}\!\left(\frac{1 + y'}{x + y}\right) = \frac{y''(x + y) - (1 + y')^{2}}{(x + y)^{2}}.$$

The right side $$4y + 4x\,y'$$ differentiates to

$$4y' + 4y' + 4x\,y'' = 8y' + 4x\,y''.$$ (Here we used the product rule on $$4x\,y'$$.)

Equating the two derivatives gives

$$\frac{y''(x + y) - (1 + y')^{2}}{(x + y)^{2}} = 8y' + 4x\,y''.$$ Multiplying across by $$(x + y)^{2}$$ simplifies this to

$$y''(x + y) - \bigl(1 + y'\bigr)^{2} = \bigl(8y' + 4x\,y''\bigr)(x + y)^{2}.$$

We wish to evaluate at $$x = 0$$. Substituting $$x = 0,\;y = 1,\;y' = 3$$ into the simpler fractional form is easiest:

$$\frac{y''(0 + 1) - (1 + 3)^{2}}{(0 + 1)^{2}} = 8\cdot3 + 4\cdot0\cdot y'',$$ which reduces to

$$y'' - 16 = 24.$$

Solving for $$y''$$ yields

$$y'' = 40.$$

Therefore, the second derivative of $$y$$ with respect to $$x$$ at $$x = 0$$ is $$40$$.

So, the answer is $$40$$.

We start from the relation $$y^{1/4}+y^{-1/4}=2x.$$

Put $$t=y^{1/4}\;(\Rightarrow y=t^{4}).$$ The given relation becomes

$$t+\frac1t=2x\qquad\Longrightarrow\qquad x=\frac{t+\dfrac1t}{2}.$$

First we differentiate this relation to connect the derivatives of $$t$$ and $$x$$. Using the chain rule,

$$\frac{d}{dx}\left(t+\frac1t\right)=2\;\Longrightarrow\;\left(1-\frac1{t^{2}}\right)\frac{dt}{dx}=2,$$

so

$$\frac{dt}{dx}=\frac{2t^{2}}{t^{2}-1}.$$

Because $$y=t^{4},$$ we have by the chain rule

$$\frac{dy}{dx}=4t^{3}\frac{dt}{dx}=4t^{3}\left(\frac{2t^{2}}{t^{2}-1}\right)=\frac{8t^{5}}{t^{2}-1}.$$

Again differentiating, and writing $$y'=\frac{dy}{dx},\;y''=\frac{d^{2}y}{dx^{2}},$$ we find

$$y'=\frac{8t^{5}}{t^{2}-1}\;\Longrightarrow\;\frac{dy'}{dt}= \frac{d}{dt}\left(\frac{8t^{5}}{t^{2}-1}\right) =\frac{40t^{4}(t^{2}-1)-8t^{5}\cdot2t}{(t^{2}-1)^{2}} =\frac{24t^{6}-40t^{4}}{(t^{2}-1)^{2}} =\frac{8(3t^{6}-5t^{4})}{(t^{2}-1)^{2}}.$$

Hence

$$y''=\frac{dy'}{dx}=\frac{dy'}{dt}\frac{dt}{dx} =\frac{8(3t^{6}-5t^{4})}{(t^{2}-1)^{2}}\cdot\frac{2t^{2}}{t^{2}-1} =\frac{16t^{6}(3t^{2}-5)}{(t^{2}-1)^{3}}.$$

Next we need $$x^{2}-1$$ in terms of $$t$$. From $$x=\dfrac{t+\dfrac1t}{2},$$

$$x^{2}=\frac{(t+\frac1t)^{2}}{4} =\frac{t^{2}+2+\dfrac1{t^{2}}}{4},$$

so

$$x^{2}-1=\frac{t^{2}+\dfrac1{t^{2}}-2}{4} =\frac{(t-\frac1t)^{2}}{4} =\frac{(t^{2}-1)^{2}}{4t^{2}}.$$

We now substitute $$y,\;y',\;y''$$ and $$x^{2}-1$$ into the differential equation

$$(x^{2}-1)y''+\alpha xy'+\beta y=0.$$

The first term is

$$(x^{2}-1)y'' =\frac{(t^{2}-1)^{2}}{4t^{2}}\cdot\frac{16t^{6}(3t^{2}-5)}{(t^{2}-1)^{3}} =\frac{4t^{4}(3t^{2}-5)}{t^{2}-1}.$$

The second term uses $$xy'=\frac{t+\dfrac1t}{2}\cdot\frac{8t^{5}}{t^{2}-1} =\frac{4(t^{6}+t^{4})}{t^{2}-1},$$ so

$$\alpha xy'=\alpha\cdot\frac{4(t^{6}+t^{4})}{t^{2}-1}.$$

The third term is simply $$\beta y=\beta t^{4}.$$

Multiplying the whole equation by $$t^{2}-1$$ to clear denominators gives

$$4t^{4}(3t^{2}-5)+4\alpha(t^{6}+t^{4})+\beta t^{4}(t^{2}-1)=0.$$

Collecting like powers of $$t$$:

$$\bigl[12+4\alpha+\beta\bigr]t^{6}+\bigl[-20+4\alpha-\beta\bigr]t^{4}=0.$$

This must hold for all (non-zero) $$t$$, so the coefficients of both $$t^{6}$$ and $$t^{4}$$ must vanish:

$$\begin{cases} 12+4\alpha+\beta=0,\\ -20+4\alpha-\beta=0. \end{cases}$$

Adding the two equations gives $$8\alpha-8=0\;\Longrightarrow\;\alpha=1.$$

Substituting $$\alpha=1$$ into the first equation yields $$\beta=-12-4(1)=-16.$$

Therefore

$$|\alpha-\beta|=\bigl|1-(-16)\bigr|=17.$$

Hence, the correct answer is Option 17.

We have been given the function

$$y(\alpha)=\sqrt{\,2\left(\dfrac{\tan\alpha+\cot\alpha}{1+\tan^{2}\alpha}\right)+\dfrac{1}{\sin^{2}\alpha}}\,,\qquad\alpha\in\left(\dfrac{3\pi}{4},\pi\right).$$

Our objective is to find $$\dfrac{dy}{d\alpha}$$ at $$\alpha=\dfrac{5\pi}{6}.$$

First we simplify the expression inside the square root. We observe the identity

$$1+\tan^{2}\alpha=\sec^{2}\alpha.$$

Also,

$$\tan\alpha+\cot\alpha =\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha} =\frac{\sin^{2}\alpha+\cos^{2}\alpha}{\sin\alpha\cos\alpha} =\frac{1}{\sin\alpha\cos\alpha}.$$

Substituting these into the fraction we get

$$\dfrac{\tan\alpha+\cot\alpha}{1+\tan^{2}\alpha} =\dfrac{\dfrac{1}{\sin\alpha\cos\alpha}}{\dfrac{1}{\cos^{2}\alpha}} =\dfrac{1}{\sin\alpha\cos\alpha}\cdot\cos^{2}\alpha =\frac{\cos\alpha}{\sin\alpha} =\cot\alpha.$$

So the whole quantity under the square root becomes

$$2\cot\alpha+\frac{1}{\sin^{2}\alpha}.$$

Next we recall the Pythagorean identity

$$\csc^{2}\alpha=1+\cot^{2}\alpha.$$

Using this identity we write

$$2\cot\alpha+\frac{1}{\sin^{2}\alpha} =2\cot\alpha+\csc^{2}\alpha =2\cot\alpha+1+\cot^{2}\alpha =\cot^{2}\alpha+2\cot\alpha+1 =(\cot\alpha+1)^{2}.$$

Therefore

$$y(\alpha)=\sqrt{(\cot\alpha+1)^{2}} =|\cot\alpha+1|.$$

Now we determine the sign of $$\cot\alpha+1$$ in the given interval $$\alpha\in\left(\dfrac{3\pi}{4},\pi\right).$$ In this interval $$\sin\alpha\gt 0$$ and $$\cos\alpha\lt 0,$$ so $$\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\lt 0.$$ At the left‐end point $$\alpha=\dfrac{3\pi}{4},$$ $$\cot\alpha=-1,$$ making $$\cot\alpha+1=0.$$ As $$\alpha$$ increases beyond $$\dfrac{3\pi}{4},$$ $$\cot\alpha$$ becomes even more negative, hence $$\cot\alpha+1\lt 0$$ throughout the open interval. Thus

$$|\cot\alpha+1|=-(\cot\alpha+1).$$

So we can write, for all $$\alpha\in\left(\dfrac{3\pi}{4},\pi\right),$$

$$y(\alpha)=-(\cot\alpha+1)=-\cot\alpha-1.$$

Now we differentiate. The basic derivative we use is

$$\frac{d}{d\alpha}(\cot\alpha)=-\csc^{2}\alpha.$$

Therefore,

$$\frac{dy}{d\alpha} =-\frac{d}{d\alpha}(\cot\alpha)-\frac{d}{d\alpha}(1) =-(-\csc^{2}\alpha)-0 =\csc^{2}\alpha.$$

We must evaluate this at $$\alpha=\dfrac{5\pi}{6}.$$ We know

$$\sin\left(\dfrac{5\pi}{6}\right)=\frac{1}{2}\quad\Longrightarrow\quad \csc\left(\dfrac{5\pi}{6}\right)=2.$$

Hence

$$\csc^{2}\left(\dfrac{5\pi}{6}\right)=2^{2}=4.$$

Thus,

$$\left.\frac{dy}{d\alpha}\right|_{\alpha=\frac{5\pi}{6}}=4.$$

Hence, the correct answer is Option A.

We are given the parametric equations $$x = 2\sin\theta - \sin 2\theta$$ and $$y = 2\cos\theta - \cos 2\theta$$, and we need to find $$\frac{d^2y}{dx^2}$$ at $$\theta = \pi$$.

First, we find the first derivatives with respect to $$\theta$$:

$$\frac{dx}{d\theta} = 2\cos\theta - 2\cos 2\theta$$

$$\frac{dy}{d\theta} = -2\sin\theta + 2\sin 2\theta$$

Now, $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-\sin\theta + \sin 2\theta}{\cos\theta - \cos 2\theta}$$

Using sum-to-product formulas:

Numerator: $$\sin 2\theta - \sin\theta = 2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}$$

Denominator: $$\cos\theta - \cos 2\theta = 2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}$$

Therefore: $$\frac{dy}{dx} = \frac{2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}}{2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}} = \cot\frac{3\theta}{2}$$

(The $$\sin\frac{\theta}{2}$$ terms cancel; at $$\theta = \pi$$, $$\sin\frac{\pi}{2} = 1 \neq 0$$, so this is valid.)

For the second derivative, we use: $$\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}$$

$$\frac{d}{d\theta}\left(\cot\frac{3\theta}{2}\right) = -\csc^2\frac{3\theta}{2} \cdot \frac{3}{2} = -\frac{3}{2}\csc^2\frac{3\theta}{2}$$

At $$\theta = \pi$$:

$$\csc^2\frac{3\pi}{2} = \frac{1}{\sin^2\frac{3\pi}{2}} = \frac{1}{(-1)^2} = 1$$

So $$\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\bigg|_{\theta=\pi} = -\frac{3}{2}(1) = -\frac{3}{2}$$

Also, $$\frac{dx}{d\theta}\bigg|_{\theta=\pi} = 2\cos\pi - 2\cos 2\pi = 2(-1) - 2(1) = -4$$

Therefore: $$\frac{d^2y}{dx^2}\bigg|_{\theta=\pi} = \frac{-\frac{3}{2}}{-4} = \frac{3}{8}$$

The answer is Option A: $$\frac{3}{8}$$.

We have the implicit curve $$x^2 + 2xy - 3y^2 = 0$$ and the specific point $$(2,2)$$ on it. To obtain the slope of the tangent at this point, we first differentiate the given equation with respect to $$x$$, remembering that $$y = y(x)$$ is a function of $$x$$.

Differentiating term by term:

$$\dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(2xy) - \dfrac{d}{dx}(3y^2) = 0.$$

Using $$\dfrac{d}{dx}(x^2) = 2x$$, the product rule $$\dfrac{d}{dx}(2xy) = 2y + 2x\dfrac{dy}{dx}$$, and the chain rule $$\dfrac{d}{dx}(3y^2) = 6y\dfrac{dy}{dx}$$, we get

$$2x + (2y + 2x\dfrac{dy}{dx}) - 6y\dfrac{dy}{dx} = 0.$$

Collecting the $$\dfrac{dy}{dx}$$ terms together,

$$2x + 2y + (2x - 6y)\dfrac{dy}{dx} = 0.$$

Solving for the derivative $$\dfrac{dy}{dx}$$ (which is the slope of the tangent), we have

$$(2x - 6y)\dfrac{dy}{dx} = -\,(2x + 2y),$$

so

$$\dfrac{dy}{dx} = -\dfrac{2x + 2y}{2x - 6y}.$$

Now substitute the coordinates $$(2,2)$$:

$$\dfrac{dy}{dx}\Big|_{(2,2)} = -\dfrac{2\cdot 2 + 2\cdot 2}{2\cdot 2 - 6\cdot 2} = -\dfrac{4 + 4}{4 - 12} = -\dfrac{8}{-8} = 1.$$

Thus the slope of the tangent at $$(2,2)$$ is $$m_t = 1$$. The slope of the normal is the negative reciprocal, therefore

$$m_n = -\dfrac{1}{m_t} = -1.$$

The normal line passing through $$(2,2)$$ with slope $$-1$$ is obtained from the point-slope form $$y - y_1 = m(x - x_1)$$:

$$y - 2 = -1(x - 2).$$

Simplifying,

$$y - 2 = -x + 2 \quad\Longrightarrow\quad x + y - 4 = 0.$$

We must now find the perpendicular distance of the origin $$(0,0)$$ from this normal line. For a line written as $$Ax + By + C = 0$$, the distance from a point $$(x_0,y_0)$$ is given by the formula

$$\text{Distance} = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.$$

Here $$A = 1,\; B = 1,\; C = -4$$ and $$(x_0,y_0) = (0,0).$$ Substituting, we get

$$\text{Distance} = \dfrac{|1\cdot 0 + 1\cdot 0 - 4|}{\sqrt{1^2 + 1^2}} = \dfrac{|-4|}{\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 4 \times \dfrac{\sqrt{2}}{2} = 2\sqrt{2}.$$

Hence, the correct answer is Option D.

We are given a twice differentiable function $$f : \mathbb{R} \to \mathbb{R}$$ with $$f(0) = 0$$, $$f(1) = 0$$, and $$f'(0) = 0$$.

Step 1: Apply Rolle's Theorem to $$f$$ on $$[0, 1]$$.

Since $$f$$ is differentiable (hence continuous) on $$[0, 1]$$ and $$f(0) = f(1) = 0$$, by Rolle's Theorem there exists $$c_1 \in (0, 1)$$ such that:

$$f'(c_1) = 0$$

Step 2: Apply Rolle's Theorem to $$f'$$ on $$[0, c_1]$$.

Since $$f$$ is twice differentiable, $$f'$$ is differentiable on $$(0, c_1)$$ and continuous on $$[0, c_1]$$. We have:

$$f'(0) = 0 \quad \text{and} \quad f'(c_1) = 0$$

By Rolle's Theorem applied to $$f'$$, there exists $$c_2 \in (0, c_1) \subset (0, 1)$$ such that:

$$f''(c_2) = 0$$

Step 3: Verify with a counterexample that other options fail.

Consider $$f(x) = x^2(x - 1) = x^3 - x^2$$. One verifies:

$$f(0) = 0, \quad f(1) = 0, \quad f'(x) = 3x^2 - 2x \implies f'(0) = 0$$

The second derivative is $$f''(x) = 6x - 2$$, so:

$$f''(0) = -2 \neq 0 \quad (\text{Option C is false})$$

$$f''(x) = 0 \text{ only at } x = \tfrac{1}{3} \quad (\text{Option A is false, Option D is false})$$

Conclusion:

For every twice differentiable function satisfying the given conditions, $$f''(x) = 0$$ for some $$x \in (0, 1)$$. This is guaranteed by the double application of Rolle's Theorem.

The correct answer is Option B: $$f''(x) = 0$$, for some $$x \in (0, 1)$$.

We start from the implicit relation

$$\left(a+\sqrt{2b}\cos x\right)\left(a-\sqrt{2b}\cos y\right)=a^{2}-b^{2},\qquad a>b>0.$$

Here both $$x$$ and $$y$$ are variables linked through the above equation, so we regard $$x$$ as a function of $$y$$ and differentiate with respect to $$y$$.

The right-hand side is a constant, so its derivative is $$0$$. For the left-hand side we first state the product rule:

$$\dfrac{d}{dy}[UV]=U\dfrac{dV}{dy}+V\dfrac{dU}{dy},$$

where $$U=a+\sqrt{2b}\cos x$$ and $$V=a-\sqrt{2b}\cos y$$. Applying the rule, we obtain

$$\bigl(a-\sqrt{2b}\cos y\bigr)\dfrac{d}{dy}\!\bigl(a+\sqrt{2b}\cos x\bigr) \;+\;\bigl(a+\sqrt{2b}\cos x\bigr)\dfrac{d}{dy}\!\bigl(a-\sqrt{2b}\cos y\bigr)=0.$$

Now we differentiate each factor. Remember that $$a$$ and $$b$$ are constants, while $$x$$ depends on $$y$$:

$$\dfrac{d}{dy}\!\bigl(a+\sqrt{2b}\cos x\bigr)=\sqrt{2b}\,(-\sin x)\dfrac{dx}{dy},$$

because $$\dfrac{d}{dy}\cos x=-\sin x\dfrac{dx}{dy}.$$

$$\dfrac{d}{dy}\!\bigl(a-\sqrt{2b}\cos y\bigr)=-\sqrt{2b}\,(-\sin y)=\sqrt{2b}\sin y,$$ because here $$y$$ is the differentiation variable itself.

Substituting these derivatives back gives

$$\bigl(a-\sqrt{2b}\cos y\bigr)\,\sqrt{2b}(-\sin x)\dfrac{dx}{dy} \;+\;\bigl(a+\sqrt{2b}\cos x\bigr)\,\sqrt{2b}\sin y=0.$$

Each term contains a common factor $$\sqrt{2b}$$; dividing by it simplifies the equation:

$$-\bigl(a-\sqrt{2b}\cos y\bigr)\sin x\,\dfrac{dx}{dy} \;+\;\bigl(a+\sqrt{2b}\cos x\bigr)\sin y=0.$$

Rearranging, we isolate $$\dfrac{dx}{dy}$$:

$$\dfrac{dx}{dy}=\dfrac{\bigl(a+\sqrt{2b}\cos x\bigr)\sin y} {\bigl(a-\sqrt{2b}\cos y\bigr)\sin x}.$$

We must now evaluate this derivative at the point $$\bigl(x,y\bigr)=\left(\dfrac{\pi}{4},\dfrac{\pi}{4}\right).$$ First record the basic trigonometric values

$$\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2},\qquad \sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}.$$

Compute the auxiliary quantity

$$\sqrt{2b}\cos\dfrac{\pi}{4}=\sqrt{2b}\,\dfrac{\sqrt2}{2}=\sqrt{b}.$$

Hence at $$x=y=\dfrac{\pi}{4}$$ we have

$$a+\sqrt{2b}\cos x=a+\sqrt{b},\qquad a-\sqrt{2b}\cos y=a-\sqrt{b},$$

and also

$$\sin x=\sin y=\dfrac{\sqrt2}{2}.$$

Substituting all these values into the derivative formula gives

$$\left.\dfrac{dx}{dy}\right|_{\left(\frac{\pi}{4},\frac{\pi}{4}\right)} =\dfrac{(a+\sqrt{b})\left(\dfrac{\sqrt2}{2}\right)} {(a-\sqrt{b})\left(\dfrac{\sqrt2}{2}\right)} =\dfrac{a+\sqrt{b}}{a-\sqrt{b}}.$$

At first glance the answer still contains $$\sqrt{b}$$, yet none of the options do. We therefore use the original relation to find an extra link between $$a$$ and $$b$$ at this specific point. Putting $$x=y=\dfrac{\pi}{4}$$ into the given equation itself, we get

$$\bigl(a+\sqrt{b}\bigr)\bigl(a-\sqrt{b}\bigr)=a^{2}-b^{2}.$$

The left side expands to $$a^{2}-b,$$ so

$$a^{2}-b=a^{2}-b^{2}\quad\Longrightarrow\quad b=b^{2} \;\Longrightarrow\; b(1-b)=0.$$

Because $$b>0$$, the only admissible root is $$b=1.$$ With this value, the derivative simplifies to

$$\dfrac{dx}{dy}=\dfrac{a+1}{a-1} =\dfrac{a+b}{a-b},$$

since now $$b=1.$$

This matches option C. Hence, the correct answer is Option C.

We are told that $$f \circ g$$ is the identity function on $$\mathbb R$$. In symbols this means

$$f\bigl(g(x)\bigr)=x \quad\text{for every real }x.$$

To connect the derivatives of $$f$$ and $$g$$ we differentiate both sides of this equality with respect to $$x$$. First we recall the Chain Rule:

The Chain Rule states that if $$h(x)=u(v(x)),$$ then $$h'(x)=u'\bigl(v(x)\bigr)\,v'(x).$$

Applying the Chain Rule to $$h(x)=f\bigl(g(x)\bigr)$$ we obtain

$$\bigl(f \circ g\bigr)'(x)=f'\bigl(g(x)\bigr)\,g'(x).$$

Because $$f \circ g$$ is the identity function, its derivative is simply $$1$$ for every $$x$$, so we have

$$f'\bigl(g(x)\bigr)\,g'(x)=1.$$

Now we substitute the specific value $$x=a$$ that is mentioned in the question. This gives

$$f'\bigl(g(a)\bigr)\,g'(a)=1.$$

The problem states that $$g'(a)=5$$ and that $$g(a)=b$$, so we replace these quantities:

$$f'(b)\,\bigl(5\bigr)=1.$$

To isolate $$f'(b)$$ we divide both sides by $$5$$:

$$f'(b)=\frac{1}{5}.$$

Hence, the correct answer is Option A.

Let us denote

$$y(x)=\tan^{-1}\!\left(\dfrac{\sqrt{1+x^{2}}-1}{x}\right), \qquad z(x)=\tan^{-1}\!\left(\dfrac{2x\sqrt{1-x^{2}}}{1-2x^{2}}\right).$$

We have to find $$\dfrac{dy}{dz}$$ at $$x=\dfrac12.$$ Since both $$y$$ and $$z$$ are functions of $$x$$, the chain rule gives

$$\frac{dy}{dz}=\frac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}} =\frac{dy/dx}{dz/dx}.$$

Derivative of $$y$$

Write $$u=\dfrac{\sqrt{1+x^{2}}-1}{x},\quad\text{so}\quad y=\tan^{-1}u.$$

The derivative formula $$\dfrac{d}{dx}\tan^{-1}u=\dfrac{1}{1+u^{2}}\dfrac{du}{dx}$$ yields

$$\frac{dy}{dx}=\frac{1}{1+u^{2}}\frac{du}{dx}.$$

To compute $$\dfrac{du}{dx}$$, note that $$u=\frac{\sqrt{1+x^{2}}-1}{x} =(\sqrt{1+x^{2}}-1)\,x^{-1}.$$

Using the product rule, $$\frac{du}{dx}=\frac{d}{dx}\!\bigl(\sqrt{1+x^{2}}-1\bigr)\,x^{-1} +(\sqrt{1+x^{2}}-1)\,\frac{d}{dx}(x^{-1}).$$

Now $$\frac{d}{dx}\sqrt{1+x^{2}} =\frac{1}{2}(1+x^{2})^{-1/2}\cdot2x =\frac{x}{\sqrt{1+x^{2}}},$$ and $$\frac{d}{dx}(x^{-1})=-\frac1{x^{2}}.$$

Thus

$$\frac{du}{dx} =\frac{x}{\sqrt{1+x^{2}}}\cdot\frac1{x} -(\sqrt{1+x^{2}}-1)\frac1{x^{2}} =\frac1{\sqrt{1+x^{2}}} -\frac{\sqrt{1+x^{2}}-1}{x^{2}}.$$

Derivative of $$z$$

Put $$v=\dfrac{2x\sqrt{1-x^{2}}}{1-2x^{2}},\quad\text{so}\quad z=\tan^{-1}v.$$

Again, the same formula gives

$$\frac{dz}{dx}=\frac{1}{1+v^{2}}\frac{dv}{dx}.$$

Write $$N=2x\sqrt{1-x^{2}},\qquad D=1-2x^{2},\qquad v=\frac{N}{D}.$$

Differentiating $$N$$: $$\frac{dN}{dx}=2\sqrt{1-x^{2}} +2x\cdot\frac{-x}{\sqrt{1-x^{2}}} =2\sqrt{1-x^{2}} -\frac{2x^{2}}{\sqrt{1-x^{2}}} =\frac{2(1-2x^{2})}{\sqrt{1-x^{2}}}.$$

Differentiating $$D$$: $$\frac{dD}{dx}=-4x.$$

Using the quotient rule $$\frac{dv}{dx} =\frac{(dN/dx)\,D-N\,(dD/dx)}{D^{2}} =\frac{\dfrac{2(1-2x^{2})}{\sqrt{1-x^{2}}}(1-2x^{2}) +2x\sqrt{1-x^{2}}\,(4x)} {(1-2x^{2})^{2}}.$$

Combining the terms over the common denominator $$\sqrt{1-x^{2}}$$ gives

$$\frac{dv}{dx} =\frac{2(1-2x^{2})^{2}+8x^{2}(1-x^{2})} {\sqrt{1-x^{2}}\,\bigl(1-2x^{2}\bigr)^{2}}.$$

Evaluating every quantity at }$$x=\dfrac12$$

First compute the needed numbers:

$$x=\frac12,\quad x^{2}=\frac14,\quad 1+x^{2}=\frac54,\quad \sqrt{1+x^{2}}=\frac{\sqrt5}{2},$$

$$1-x^{2}=\frac34,\quad \sqrt{1-x^{2}}=\frac{\sqrt3}{2},\quad 1-2x^{2}=\frac12.$$

For the $$y$$-part: $$u=\frac{\sqrt5/2-1}{1/2}=\sqrt5-2,$$ $$u^{2}=(\sqrt5-2)^{2}=9-4\sqrt5,$$ $$1+u^{2}=10-4\sqrt5,$$ $$\frac{du}{dx} =\frac{2}{\sqrt5}-2\sqrt5+4.$$

For the $$z$$-part: $$N=\frac{\sqrt3}{2},\quad D=\frac12,\quad v=\frac{\sqrt3}{2}\big/\frac12=\sqrt3,$$ $$1+v^{2}=1+3=4.$$

The earlier simplified formula for $$\dfrac{dv}{dx}$$ now gives

$$\frac{dv}{dx} =\frac{2}{\dfrac{\sqrt3}{2}\cdot\left(\dfrac12\right)^{2}} =\frac{2}{\dfrac{\sqrt3}{8}} =\frac{16}{\sqrt3}.$$

Assembling }$$\dfrac{dy}{dz}$$

Because $$\frac{dy}{dx}=\frac{1}{1+u^{2}}\frac{du}{dx},\qquad \frac{dz}{dx}=\frac{1}{1+v^{2}}\frac{dv}{dx},$$ we have $$\frac{dy}{dz} =\frac{\dfrac{du}{dx}}{\dfrac{dv}{dx}} \cdot\frac{1+v^{2}}{1+u^{2}} =\frac{\dfrac{2}{\sqrt5}-2\sqrt5+4}{\dfrac{16}{\sqrt3}} \cdot\frac{4}{10-4\sqrt5} =\frac{\sqrt3}{16}\left(4+\frac{2}{\sqrt5}-2\sqrt5\right) \cdot\frac{4}{10-4\sqrt5}.$$

Simplifying step by step, first factor $$2$$ from the bracket:

$$4+\frac{2}{\sqrt5}-2\sqrt5 =2\!\left(2+\frac1{\sqrt5}-\sqrt5\right) =\frac{2(-2+\sqrt5)}{\sqrt5},$$

so that

$$\frac{dy}{dz} =\frac{\sqrt3}{16}\cdot\frac{2(-2+\sqrt5)}{\sqrt5} \cdot\frac{4}{10-4\sqrt5} =\frac{\sqrt3(-2+\sqrt5)}{2\sqrt5(5-2\sqrt5)}.$$

Notice the common factor $$5-2\sqrt5$$ in the denominator: $$5-2\sqrt5= \frac{10\sqrt5-20}{2\sqrt5}= \frac{10(\sqrt5-2)}{2\sqrt5},$$ so

$$\frac{dy}{dz} =\sqrt3\;\frac{\sqrt5-2}{2\sqrt5}\; \frac{2\sqrt5}{10(\sqrt5-2)} =\sqrt3\;\frac{1}{10} =\frac{\sqrt3}{10}.$$

Hence, the correct answer is Option D.

We have the implicit curve $$y=(1+x)^{2y}+\cos^{2}\!\bigl(\sin^{-1}x\bigr).$$

First we evaluate the point of contact by putting $$x=0.$$

The trigonometric term simplifies because $$\cos\!\bigl(\sin^{-1}x\bigr)=\sqrt{1-x^{2}},$$ so $$\cos^{2}\!\bigl(\sin^{-1}x\bigr)=1-x^{2}.$$

Therefore, at any $$x$$ we can write

$$y=(1+x)^{2y}+1-x^{2}.$$

Putting $$x=0$$ gives

$$y=(1+0)^{2y}+1-0^{2}=1^{2y}+1=1+1=2.$$

So the point on the curve is $$(0,2).$$

To find the slope of the tangent, we differentiate both sides with respect to $$x$$. We rewrite the power term by the exponential form so that a known formula can be applied:

$$(1+x)^{2y}=e^{2y\ln(1+x)}.$$

The derivative of $$e^{g(x)}$$ is $$e^{g(x)}\,g'(x).$$ Using this, we obtain

$$\frac{d}{dx}\bigl[(1+x)^{2y}\bigr]=(1+x)^{2y}\,\frac{d}{dx}\!\bigl[2y\ln(1+x)\bigr].$$

The derivative inside is found by the product rule:

$$\frac{d}{dx}\!\bigl[2y\ln(1+x)\bigr]=2\,\frac{dy}{dx}\,\ln(1+x)+\frac{2y}{1+x}.$$

The derivative of the right-hand side of the implicit equation is therefore

$$\frac{dy}{dx}=(1+x)^{2y}\Bigl[2\,\frac{dy}{dx}\,\ln(1+x)+\frac{2y}{1+x}\Bigr]-2x.$$

Now we substitute $$x=0$$ and $$y=2.$$ We note that $$\ln(1+0)=0,\quad(1+0)^{2y}=1,\quad\text{and}\quad-2x=-0.$$ Hence

$$\frac{dy}{dx}=1\Bigl[2\,\frac{dy}{dx}\,(0)+\frac{2\cdot2}{1}\Bigr]-0=4.$$

Thus the slope of the tangent at $$(0,2)$$ is $$m_t=4.$$ The slope of the normal is the negative reciprocal:

$$m_n=-\frac{1}{4}.$$

Using the point-slope form of a line, $$y-y_1=m(x-x_1),$$ with $$(x_1,y_1)=(0,2)$$ and $$m=m_n=-\frac{1}{4},$$ we get

$$y-2=-\frac{1}{4}(x-0).$$

Multiplying by $$4$$ to clear the fraction gives

$$4y-8=-x,$$

which rearranges to

$$x+4y=8.$$

Hence, the correct answer is Option C.

We are told that the differentiable function $$f(x)$$ satisfies the functional equation

$$f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y$$ $$\text{for every real }x,y.$$

The presence of the mixed terms $$xy^{2}+x^{2}y$$ reminds us of the identity for the cube of a sum. Indeed, the algebraic formula

$$(x+y)^{3}=x^{3}+y^{3}+3x^{2}y+3xy^{2}$$

implies, after dividing by 3, that

$$\frac{(x+y)^{3}-x^{3}-y^{3}}{3}=x^{2}y+xy^{2}.$$

This observation suggests subtracting one-third of $$x^{3}$$ from $$f(x)$$ so that the mixed terms cancel. We therefore introduce a new function

$$g(x)=f(x)-\frac{x^{3}}{3}.$$

Now we compute $$g(x+y)$$:

$$\begin{aligned} g(x+y)&=f(x+y)-\frac{(x+y)^{3}}{3}\\[4pt] &=\Bigl[f(x)+f(y)+xy^{2}+x^{2}y\Bigr]-\frac{x^{3}+y^{3}+3x^{2}y+3xy^{2}}{3}\\[4pt] &=f(x)-\frac{x^{3}}{3}+f(y)-\frac{y^{3}}{3} +\underbrace{\bigl[xy^{2}+x^{2}y-\bigl(x^{2}y+xy^{2}\bigr)\bigr]}_{=\,0}\\[4pt] &=g(x)+g(y). \end{aligned}$$

Thus $$g(x)$$ is additive: $$g(x+y)=g(x)+g(y).$$ Because $$f$$ (hence $$g$$) is differentiable, the classical result for differentiable additive functions tells us that $$g(x)$$ must be linear. Hence there exists a constant $$k$$ such that

$$g(x)=kx\quad\text{for all }x.$$

Returning to $$f(x)$$, we therefore have

$$f(x)=g(x)+\frac{x^{3}}{3}=kx+\frac{x^{3}}{3}.$$

The problem also gives us the limit condition

$$\lim_{x\to 0}\frac{f(x)}{x}=1.$$

We substitute our expression for $$f(x)$$ into this limit:

$$\begin{aligned} \frac{f(x)}{x}&=\frac{kx+\dfrac{x^{3}}{3}}{x} =k+\frac{x^{2}}{3}. \end{aligned}$$

Taking the limit as $$x\to 0$$ simply removes the $$x^{2}$$ term, so

$$\lim_{x\to 0}\frac{f(x)}{x}=k=1.$$

Therefore $$k=1$$, and the explicit formula for $$f(x)$$ becomes

$$f(x)=x+\frac{x^{3}}{3}.$$

Because $$f$$ is differentiable, we differentiate term by term, using the basic rule $$\dfrac{d}{dx}(x^{n})=nx^{n-1}$$:

$$f'(x)=\frac{d}{dx}\Bigl(x\Bigr)+\frac{d}{dx}\Bigl(\frac{x^{3}}{3}\Bigr) =1+\frac{3x^{2}}{3}=1+x^{2}.$$

Finally, we evaluate the derivative at $$x=3$$:

$$f'(3)=1+3^{2}=1+9=10.$$

So, the answer is $$10$$.

We are given the function $$f(x)=\log_e\frac{1-x}{1+x}$$ with the condition $$|x|<1$$, and we have to evaluate $$f\!\left(\dfrac{2x}{1+x^{2}}\right).$$

First we recall the definition of the function. For any admissible argument $$t$$ we have

$$f(t)=\log_e\frac{1-t}{1+t}.$$

Now we substitute $$t=\dfrac{2x}{1+x^{2}}$$ in this definition. So

$$f\!\left(\dfrac{2x}{1+x^{2}}\right)=\log_e\frac{1-\dfrac{2x}{1+x^{2}}}{1+\dfrac{2x}{1+x^{2}}}.$$

We simplify the fraction inside the logarithm by writing the two numerators over the common denominator $$1+x^{2}.$$ We have

$$1-\dfrac{2x}{1+x^{2}}=\dfrac{1+x^{2}-2x}{1+x^{2}},\qquad 1+\dfrac{2x}{1+x^{2}}=\dfrac{1+x^{2}+2x}{1+x^{2}}.$$

Hence the whole expression becomes

$$\log_e\frac{\dfrac{1+x^{2}-2x}{1+x^{2}}}{\dfrac{1+x^{2}+2x}{1+x^{2}}}.$$

Because both the numerator and the denominator have the same factor $$1+x^{2}$$ in the denominator, that factor cancels out:

$$\log_e\frac{1+x^{2}-2x}{1+x^{2}+2x}.$$

Next we recognize the two quadratic expressions as perfect squares:

$$1+x^{2}-2x=(x-1)^{2},\qquad 1+x^{2}+2x=(x+1)^{2}.$$

So the argument of the logarithm becomes

$$\frac{(x-1)^{2}}{(x+1)^{2}}.$$

Therefore

$$f\!\left(\dfrac{2x}{1+x^{2}}\right)=\log_e\frac{(x-1)^{2}}{(x+1)^{2}}.$$

We now use the well-known logarithmic rule $$\log_e a^{2}=2\log_e a$$ (that is, $$\log_e a^{k}=k\log_e a$$). Setting $$a=\dfrac{x-1}{x+1},$$ we get

$$\log_e\frac{(x-1)^{2}}{(x+1)^{2}}=\log_e\left[\left(\frac{x-1}{x+1}\right)^{2}\right]=2\log_e\frac{x-1}{x+1}.$$

Next, compare this with the original definition of $$f(x).$$ We have

$$f(x)=\log_e\frac{1-x}{1+x}=\log_e\left(\frac{-1(x-1)}{1+x}\right).$$

The factor $$-1$$ inside the logarithm only adds the constant $$\log_e(-1),$$ which is irrelevant when we are matching up to a multiplicative constant of $$f(x).$$ More straightforwardly, observe that multiplying numerator and denominator of $$\dfrac{1-x}{1+x}$$ by $$-1$$ gives

$$\frac{1-x}{1+x}=\frac{-(x-1)}{1+x}=\frac{x-1}{-(1+x)}.$$

Either way, the ratio $$\dfrac{x-1}{x+1}$$ differs from $$\dfrac{1-x}{1+x}$$ only by a sign in both numerator and denominator, so their logarithms differ only by an additive constant. That constant disappears when we multiply by 2. Consequently, we may write

$$2\log_e\frac{x-1}{x+1}=2\log_e\frac{1-x}{1+x}=2f(x).$$

Thus we have proved

$$f\!\left(\dfrac{2x}{1+x^{2}}\right)=2f(x).$$

Hence, the correct answer is Option 4.

$$2y = \left\{ \cot^{-1} \left( \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) \right\}^2$$

$$2y = \left\{ \cot^{-1} \left( \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} \right) \right\}^2$$

$$2y = \left\{ \cot^{-1} \left( \tan \left( \frac{\pi}{3} + x \right) \right) \right\}^2$$

$$2y = \left\{ \frac{\pi}{2} - \tan^{-1} \left( \tan \left( \frac{\pi}{3} + x \right) \right) \right\}^2 = \left( \frac{\pi}{2} - \left( \frac{\pi}{3} + x \right) \right)^2$$

$$2y = \left( x - \frac{\pi}{6} \right)^2$$

$$2y = x^2 - \frac{\pi}{3}x + \frac{\pi^2}{36}$$

$$y' = x - \frac{\pi}{6}$$

Hence The correct ans is Option C

We are given the implicit relation $$e^{y}+x\,y=e.$$

First we find the value of $$y$$ at the point where $$x=0.$$ Substituting $$x=0$$ gives

$$e^{y}+0\cdot y=e\;\;\Longrightarrow\;\;e^{y}=e\;\;\Longrightarrow\;\;y=1.$$

So, at $$x=0$$ we have the point $$(0,1).$$

Now we need the first derivative $$\dfrac{dy}{dx}.$$ We differentiate the original equation with respect to $$x$$. Using the chain rule for $$e^{y}$$ and the product rule for $$x\,y$$ we have

$$\frac{d}{dx}(e^{y})+\frac{d}{dx}(x\,y)=\frac{d}{dx}(e).$$

The derivative of $$e^{y}$$ is $$e^{y}\dfrac{dy}{dx}$$ (chain rule). The derivative of $$x\,y$$ is $$y+x\dfrac{dy}{dx}$$ (product rule). The derivative of the constant $$e$$ is $$0.$$ Hence

$$e^{y}\frac{dy}{dx}+y+x\frac{dy}{dx}=0.$$

Collecting the $$\dfrac{dy}{dx}$$ terms yields

$$(e^{y}+x)\frac{dy}{dx}+y=0.$$

Solving for $$\dfrac{dy}{dx}$$ gives

$$\frac{dy}{dx}=-\frac{y}{\,e^{y}+x\,}.$$

At $$x=0,\;y=1$$ we obtain

$$\left.\frac{dy}{dx}\right|_{(0,1)}=-\frac{1}{e^{1}+0}=-\frac{1}{e}.$$

Thus the first component of the ordered pair is $$-\dfrac1e.$$

Next we find the second derivative $$\dfrac{d^{2}y}{dx^{2}}.$$ We start from the differentiated equation

$$e^{y}\frac{dy}{dx}+y+x\frac{dy}{dx}=0$$

and differentiate it once more with respect to $$x$$. Each term is handled carefully:

• The derivative of $$e^{y}\dfrac{dy}{dx}$$ is $$e^{y}\left(\frac{dy}{dx}\right)^{2}+e^{y}\frac{d^{2}y}{dx^{2}}.$$ The first part comes from differentiating $$e^{y}$$ (chain rule), the second part from differentiating $$\dfrac{dy}{dx}.$$
• The derivative of $$y$$ is $$\dfrac{dy}{dx}.$$
• The derivative of $$x\dfrac{dy}{dx}$$ is $$\dfrac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}$$ (product rule).
• The derivative of the right‐hand side $$0$$ is $$0.$$

Adding these pieces we get

$$e^{y}\left(\frac{dy}{dx}\right)^{2}+e^{y}\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}+\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}=0.$$

That is

$$e^{y}\left(\frac{dy}{dx}\right)^{2}+e^{y}\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}=0.$$

Factor the second‐derivative terms:

$$(e^{y}+x)\frac{d^{2}y}{dx^{2}}+e^{y}\left(\frac{dy}{dx}\right)^{2}+2\frac{dy}{dx}=0.$$

Now solve for $$\dfrac{d^{2}y}{dx^{2}}:$$

$$\frac{d^{2}y}{dx^{2}}=-\frac{e^{y}\left(\dfrac{dy}{dx}\right)^{2}+2\dfrac{dy}{dx}}{e^{y}+x}.$$

We evaluate this at $$x=0,\;y=1.$$ We already know $$\dfrac{dy}{dx}=-\dfrac1e.$$ Hence

$$e^{y}=e,\qquad\left(\frac{dy}{dx}\right)^{2}=\left(-\frac1e\right)^{2}=\frac1{e^{2}}.$$

Compute the numerator:

$$e^{y}\left(\frac{dy}{dx}\right)^{2}+2\frac{dy}{dx}=e\left(\frac1{e^{2}}\right)+2\left(-\frac1e\right)=\frac1e-\frac2e=-\frac1e.$$

The denominator is

$$e^{y}+x=e+0=e.$$

Therefore

$$\left.\frac{d^{2}y}{dx^{2}}\right|_{(0,1)}=-\frac{-\dfrac1e}{e}=\frac1{e^{2}}.$$

Thus the ordered pair is

$$\left(\frac{dy}{dx},\frac{d^{2}y}{dx^{2}}\right)=\left(-\frac1e,\;\frac1{e^{2}}\right).$$

Hence, the correct answer is Option B.

We have a differentiable function that satisfies the first-order linear differential equation $$f'(x)=f(x)$$ for every real number $$x$$.

First, we recall the general solution of the differential equation $$y'=y$$. The standard result is:

$$\text{If } y'=y,\ \text{then } y=C e^{x},$$

where $$C$$ is a constant of integration. Applying this to our function, we obtain

$$f(x)=C e^{x}.$$

We now use the given initial value $$f(1)=2$$ to determine the constant $$C$$. Substituting $$x=1$$ and $$f(1)=2$$ in the expression for $$f(x)$$ we get

$$2 = C e^{1} \quad \Longrightarrow \quad C = \frac{2}{e}.$$

Hence the explicit form of the function is

$$f(x)=\frac{2}{e}\,e^{x}=2e^{x-1}.$$

Next, we define the composite function $$h(x)=f(f(x))$$. Our task is to compute $$h'(1)$$. To differentiate the composition, we employ the chain rule, which states:

$$\text{If } h(x)=f(g(x)),\ \text{then } h'(x)=f'(g(x))\cdot g'(x).$$

In our setting, $$g(x)=f(x)$$, so

$$h'(x)=f'(f(x))\cdot f'(x).$$

But we already know from the original differential equation that $$f'(x)=f(x)$$ for every $$x$$. Therefore we can rewrite the derivative as

$$h'(x)=f(f(x))\cdot f(x).$$

Now we evaluate this expression at $$x=1$$. We first compute individual values of $$f$$.

Since $$f(1)=2$$ (given), we have

$$f(1)=2.$$

Next we need $$f(2)$$. Using the explicit formula $$f(x)=2e^{x-1}$$, we substitute $$x=2$$:

$$f(2)=2e^{2-1}=2e.$$

The derivative of $$h$$ at $$x=1$$ is therefore

$$h'(1)=f\bigl(f(1)\bigr)\cdot f(1)=f(2)\cdot 2=(2e)\cdot 2=4e.$$

Hence, the correct answer is Option C.

We are given $$f(x) = \log_e(\sin x)$$ for $$0 < x < \pi$$ and $$g(x) = \sin^{-1}(e^{-x})$$ for $$x \geq 0$$.

We are told that $$\alpha$$ is a positive real number such that $$a = (f \circ g)'(\alpha)$$ and $$b = (f \circ g)(\alpha)$$.

Step 1: Find $$(f \circ g)(x)$$

$$(f \circ g)(x) = f(g(x)) = \log_e(\sin(\sin^{-1}(e^{-x})))$$

Since $$e^{-x} \in (0, 1]$$ for $$x \geq 0$$, we have $$\sin^{-1}(e^{-x}) \in (0, \pi/2]$$, so $$\sin(\sin^{-1}(e^{-x})) = e^{-x}$$.

Therefore:

$$(f \circ g)(x) = \log_e(e^{-x}) = -x$$

Step 2: Find $$(f \circ g)'(x)$$

$$(f \circ g)'(x) = \frac{d}{dx}(-x) = -1$$

Step 3: Compute $$a$$ and $$b$$

$$a = (f \circ g)'(\alpha) = -1$$

$$b = (f \circ g)(\alpha) = -\alpha$$

Step 4: Check each option

Substituting $$a = -1$$ and $$b = -\alpha$$ into Option D: $$a\alpha^2 - b\alpha - a$$

$$= (-1)(\alpha^2) - (-\alpha)(\alpha) - (-1)$$

$$= -\alpha^2 + \alpha^2 + 1$$

$$= 1$$

This matches Option D: $$a\alpha^2 - b\alpha - a = 1$$.

The correct answer is Option D.

We need to find the derivative of $$\tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right)$$ with respect to $$\frac{x}{2}$$, where $$x \in \left(0, \frac{\pi}{2}\right)$$.

Let $$y = \tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right)$$.

We simplify the argument inside the $$\tan^{-1}$$. Divide both numerator and denominator by $$\cos x$$:

$$\frac{\sin x - \cos x}{\sin x + \cos x} = \frac{\tan x - 1}{\tan x + 1}$$

We use the tangent subtraction formula. Recall that $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$.

Comparing with $$\frac{\tan x - 1}{1 + \tan x \cdot 1}$$, we recognise this as:

$$\frac{\tan x - \tan\frac{\pi}{4}}{1 + \tan x \cdot \tan\frac{\pi}{4}} = \tan\left(x - \frac{\pi}{4}\right)$$

Therefore:

$$y = \tan^{-1}\left(\tan\left(x - \frac{\pi}{4}\right)\right)$$

For $$x \in \left(0, \frac{\pi}{2}\right)$$, we have $$x - \frac{\pi}{4} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$$, which lies within $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$.

So $$\tan^{-1}(\tan(\theta)) = \theta$$ applies directly, and:

$$y = x - \frac{\pi}{4}$$

Now we find the derivative of $$y$$ with respect to $$\frac{x}{2}$$. Let $$u = \frac{x}{2}$$, so $$x = 2u$$.

Using the chain rule:

$$\frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du}$$

We have $$\frac{dy}{dx} = 1$$ (since $$y = x - \frac{\pi}{4}$$) and $$\frac{dx}{du} = 2$$ (since $$x = 2u$$).

Therefore:

$$\frac{dy}{d\left(\frac{x}{2}\right)} = 1 \times 2 = 2$$

The correct answer is Option A: 2.

We are given the functional relation for $$x \gt 1$$

$$ (2x)^{2y}=4\,e^{\,2x-2y}. $$

First take the natural logarithm (base $$e$$) of both sides. Using the law $$\ln a^{b}=b\ln a,$$ we obtain

$$ \ln\!\bigl((2x)^{2y}\bigr)=\ln\!\bigl(4e^{\,2x-2y}\bigr). $$

Thus

$$ 2y\,\ln(2x)=\ln 4 + 2x-2y. $$

We now differentiate implicitly with respect to $$x$$. Remember that $$y=y(x)$$ is a function of $$x$$, so whenever $$y$$ is differentiated we must multiply by $$\dfrac{dy}{dx}$$ (the chain rule).

Derivative of the left side:

$$ \dfrac{d}{dx}\!\bigl[2y\,\ln(2x)\bigr]=2\frac{dy}{dx}\,\ln(2x)+2y\cdot\dfrac{d}{dx}\!\bigl[\ln(2x)\bigr]. $$

Since $$\dfrac{d}{dx}\bigl[\ln(2x)\bigr]=\dfrac{1}{2x}\cdot 2=\dfrac{1}{x},$$ the left derivative becomes

$$ 2\frac{dy}{dx}\,\ln(2x)+\dfrac{2y}{x}. $$

Derivative of the right side:

$$ \dfrac{d}{dx}\!\bigl[\ln 4 + 2x-2y\bigr]=0+2-2\frac{dy}{dx}. $$

Equating the two derivatives, we have

$$ 2\frac{dy}{dx}\,\ln(2x)+\dfrac{2y}{x}=2-2\frac{dy}{dx}. $$

Gathering the $$\dfrac{dy}{dx}$$ terms on one side:

$$ 2\frac{dy}{dx}\,\ln(2x)+2\frac{dy}{dx}=2-\dfrac{2y}{x}. $$

Factor out $$2\dfrac{dy}{dx}$$:

$$ 2\frac{dy}{dx}\,\bigl(\ln(2x)+1\bigr)=2-\dfrac{2y}{x}. $$

Divide by $$2\bigl(\ln(2x)+1\bigr)$$ to get

$$ \frac{dy}{dx}= \frac{1-\dfrac{y}{x}}{\ln(2x)+1}. $$

Because we ultimately need $$(1+\ln 2x)^2\dfrac{dy}{dx},$$ we next express $$y$$ explicitly in terms of $$x$$. Returning to the original logarithmic equation

$$ 2y\,\ln(2x)+2y=\ln 4+2x, $$

or equivalently

$$ 2y\bigl(\ln(2x)+1\bigr)=\ln 4+2x. $$

Hence

$$ y=\frac{\ln 4+2x}{2\bigl(\ln(2x)+1\bigr)}. $$

Now compute $$1-\dfrac{y}{x}$$:

$$ 1-\frac{y}{x}=1-\frac{\ln 4+2x}{2x\bigl(\ln(2x)+1\bigr)} =\frac{2x\bigl(\ln(2x)+1\bigr)-\bigl(\ln 4+2x\bigr)} {2x\bigl(\ln(2x)+1\bigr)} =\frac{2x\ln(2x)-\ln 4}{2x\bigl(\ln(2x)+1\bigr)}. $$

Therefore

$$ (1+\ln 2x)^2\frac{dy}{dx} =(1+\ln 2x)\Bigl(1-\frac{y}{x}\Bigr) =(1+\ln 2x)\, \frac{2x\ln(2x)-\ln 4}{2x\bigl(\ln(2x)+1\bigr)} =\frac{2x\ln(2x)-\ln 4}{2x}. $$

The factor $$1+\ln 2x$$ in numerator and denominator cancels neatly, leaving

$$ (1+\ln 2x)^2\frac{dy}{dx} =\frac{2x\ln(2x)-\ln 4}{2x} =\ln(2x)-\frac{\ln 2}{x}. $$

Since $$\ln 4=2\ln 2,$$ the final form may be written as

$$ \frac{x\,\ln(2x)-\ln 2}{x}, $$

which coincides exactly with Option B.

Hence, the correct answer is Option B.

We have to find the derivative at $$x = 1$$ of the function

$$g(x)=f\!\left(f\!\left(f(x)\right)\right)+\left(f(x)\right)^{2}.$$

The given data are

$$f(1)=1 \qquad\text{and}\qquad f'(1)=3.$$

First we tackle the derivative of the composite term $$f\!\left(f\!\left(f(x)\right)\right).$$

By the chain rule, which states that if $$h(x)=f\!\bigl(u(x)\bigr)$$ then $$h'(x)=f'\!\bigl(u(x)\bigr)\,u'(x),$$ we proceed step by step through every layer of the composition.

Let us denote successively

$$u(x)=f(x),$$

$$v(x)=f(u(x))=f\!\left(f(x)\right),$$

so that

$$f\!\left(f\!\left(f(x)\right)\right)=f(v(x)).$$

Applying the chain rule twice in succession we write

$$\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right) =f'\!\bigl(v(x)\bigr)\cdot v'(x).$$

Now we still need $$v'(x).$$ Again by the chain rule,

$$v'(x)=\frac{d}{dx}\,f\!\left(f(x)\right)=f'\!\bigl(u(x)\bigr)\cdot u'(x).$$

Finally $$u'(x)=f'(x)$$ directly from the definition of $$u(x).$$ Collecting all three factors, the complete derivative of the triple composition is

$$\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right) =f'\!\bigl(v(x)\bigr)\;f'\!\bigl(u(x)\bigr)\;f'(x).$$

In compact form, using the original names, this is

$$\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right) =f'\!\Bigl(f\!\bigl(f(x)\bigr)\Bigr)\;f'\!\bigl(f(x)\bigr)\;f'(x).$$

Now we evaluate this at $$x=1.$$

First compute the necessary inner values:

$$f(1)=1\quad\text{(given)}.$$

Therefore $$f\!\bigl(f(1)\bigr)=f(1)=1$$ again.

Now substitute these into each derivative factor one by one:

$$f'\!\Bigl(f\!\bigl(f(1)\bigr)\Bigr) =f'(1)=3,$$

$$f'\!\bigl(f(1)\bigr)=f'(1)=3,$$

and directly $$f'(1)=3.$$

Hence

$$\left.\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right)\right|_{x=1} =3 \times 3 \times 3 = 27.$$

Next, we differentiate the square term $$\bigl(f(x)\bigr)^{2}.$$ Using the rule for a square, namely $$\displaystyle\frac{d}{dx}\bigl(u^{2}\bigr)=2u\,u',$$ with $$u=f(x),$$ we get

$$\frac{d}{dx}\left(f(x)\right)^{2}=2\,f(x)\,f'(x).$$

Evaluating at $$x=1$$ gives

$$2\,f(1)\,f'(1)=2 \times 1 \times 3 = 6.$$

Now we add the two derivative parts to obtain the derivative of the whole function $$g(x)$$ at $$x=1$$:

$$g'(1)=27+6=33.$$

Hence, the correct answer is Option D.

We have the implicit relation

$$x\,\log_e\!\bigl(\log_e x\bigr)\;-\;x^{2}\;+\;y^{2}\;=\;4.$$

Here $$y$$ is a function of $$x$$ and we have to find $$\dfrac{dy}{dx}$$ at $$x=e$$, given that $$y>0$$.

First we differentiate both sides with respect to $$x$$. Whenever we differentiate a term containing $$y$$ we must multiply by $$\dfrac{dy}{dx}$$ (chain rule).

The derivative of the constant $$4$$ is $$0$$, so

$$\frac{d}{dx}\Bigl[x\log_e\!\bigl(\log_e x\bigr)\Bigr]\;-\;\frac{d}{dx}(x^{2})\;+\;\frac{d}{dx}(y^{2})\;=\;0.$$

Now we handle each derivative separately.

1. Derivative of $$x\log_e(\log_e x)$$. We use the product rule $$\dfrac{d}{dx}(uv)=u'\,v+u\,v'$$ with $$u=x,\;v=\log_e(\log_e x).$$ We have $$u'=1$$ and $$v'=\dfrac{1}{\log_e x}\cdot\dfrac{1}{x}=\dfrac{1}{x\log_e x}.$$ So

$$\frac{d}{dx}\Bigl[x\log_e(\log_e x)\Bigr]=1\cdot\log_e(\log_e x)+x\cdot\dfrac{1}{x\log_e x}=\log_e(\log_e x)+\frac{1}{\log_e x}.$$

2. Derivative of $$-x^{2}$$. Using the power rule $$\dfrac{d}{dx}(x^{n})=nx^{n-1}$$ we get

$$\frac{d}{dx}(-x^{2})=-2x.$$

3. Derivative of $$y^{2}$$. Using the chain rule $$\dfrac{d}{dx}(y^{2})=2y\dfrac{dy}{dx}.$$

Putting the three results together:

$$\Bigl[\log_e(\log_e x)+\frac{1}{\log_e x}\Bigr]\;-\;2x\;+\;2y\,\frac{dy}{dx}=0.$$

We now isolate $$\dfrac{dy}{dx}$$.

$$2y\,\frac{dy}{dx}=2x-\log_e(\log_e x)-\frac{1}{\log_e x},$$

$$\frac{dy}{dx}=\frac{2x-\log_e(\log_e x)-\dfrac{1}{\log_e x}}{2y}.$$

Next we substitute $$x=e$$.

Since $$\log_e e=1,$$ we obtain $$\log_e x=1$$ and therefore $$\log_e(\log_e x)=\log_e 1=0.$$ Substituting these values gives

$$\frac{dy}{dx}\Big|_{x=e}=\frac{2e-0-\dfrac{1}{1}}{2y}=\frac{2e-1}{2y}.$$

To complete the evaluation, we must find $$y$$ when $$x=e$$. We go back to the original equation and put $$x=e$$.

$$e\,\log_e(\log_e e)\;-\;e^{2}\;+\;y^{2}=4.$$

Again $$\log_e e=1$$ and $$\log_e 1=0$$, so the first term is $$0$$. Hence

$$-e^{2}+y^{2}=4,$$

$$y^{2}=4+e^{2}.$$

Because $$y>0$$, we take the positive square root:

$$y=\sqrt{\,4+e^{2}\,}.$$

Finally, substituting this value of $$y$$ into our derivative:

$$\frac{dy}{dx}\Big|_{x=e}=\frac{2e-1}{2\sqrt{\,4+e^{2}\,}}.$$

Therefore, $$\frac{dy}{dx}\Big|_{x=e}=\frac{(2e-1)}{2\sqrt{4+e^{2}}}.$$

Hence, the correct answer is Option B.

We are given that the function $$f:\mathbb R\to\mathbb R$$ satisfies, for every real $$x,$$

$$f(x)=x^{3}+x^{2}f'(1)+x\,f''(2)+f'''(3).$$

For convenience, let us denote the three unknown constants that appear on the right-hand side as

$$A=f'(1),\qquad B=f''(2),\qquad C=f'''(3).$$

With this notation the functional equation becomes

$$f(x)=x^{3}+Ax^{2}+Bx+C\qquad\forall x\in\mathbb R.$$

Thus $$f(x)$$ is itself the cubic polynomial

$$f(x)=x^{3}+Ax^{2}+Bx+C.$$

Now we differentiate this expression successively, always using the basic power-rule $$\dfrac{d}{dx}(x^{n})=nx^{n-1}.$$

First derivative:

$$f'(x)=3x^{2}+2Ax+B.$$

Second derivative:

$$f''(x)=6x+2A.$$

Third derivative:

$$f'''(x)=6.$$

Notice that $$f'''(x)=6$$ is a constant, independent of $$x.$$ Evaluating this at $$x=3$$ gives

$$C=f'''(3)=6.$$

Next, we evaluate $$f''(x)$$ at $$x=2$$ in order to relate $$B$$ and $$A$$($$\text{since }$$B=f''(2)).$$

$$f''(2)=6\cdot 2+2A=12+2A.$$

So we have

$$B=12+2A.$$

Finally, we evaluate $$f'(x)$$ at $$x=1$$ because $$A=f'(1).$$

$$f'(1)=3(1)^{2}+2A(1)+B=3+2A+B.$$

But by definition $$f'(1)=A,$$ hence

$$A=3+2A+B.$$

We now solve the two simultaneous equations

$$\begin{cases} B=12+2A,\\ A=3+2A+B. \end{cases}$$

Substituting $$B=12+2A$$ into the second equation gives

$$A=3+2A+(12+2A)=3+12+4A.$$

$$A=15+4A.$$

Bringing all terms to one side,

$$A-4A=15\;\Longrightarrow\;-3A=15\;\Longrightarrow\;A=-5.$$

Using $$A=-5$$ in $$B=12+2A$$ yields

$$B=12+2(-5)=12-10=2.$$

We already have $$C=6.$$

Therefore the explicit form of the function is

$$f(x)=x^{3}-5x^{2}+2x+6.$$

We are asked to find $$f(2).$$ Substituting $$x=2$$ in the polynomial,

$$\begin{aligned} f(2)&=(2)^{3}-5(2)^{2}+2(2)+6\\ &=8-5\cdot 4+4+6\\ &=8-20+4+6\\ &=18-20\\ &=-2. \end{aligned}$$

Hence, the correct answer is Option D.

We are given the parametric relations $$x = 3\tan t \quad \text{and} \quad y = 3\sec t.$$ Our goal is to find the second derivative $$\dfrac{d^{2}y}{dx^{2}}$$ at the particular value $$t = \dfrac{\pi}{4}.$$

For curves expressed parametrically, the first derivative with respect to $$x$$ is obtained through the chain rule as

$$\frac{dy}{dx} \;=\; \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}.$$

We therefore begin by differentiating $$x$$ and $$y$$ with respect to the parameter $$t$$.

Starting with $$x = 3\tan t$$, we use the standard derivative $$\dfrac{d}{dt}(\tan t) = \sec^{2}t$$. Hence

$$\frac{dx}{dt} \;=\; 3\sec^{2}t.$$

Next, for $$y = 3\sec t$$, we recall the derivative $$\dfrac{d}{dt}(\sec t) = \sec t\tan t$$. Thus

$$\frac{dy}{dt} \;=\; 3\sec t \tan t.$$

Substituting these two results into the first-derivative formula gives

$$\frac{dy}{dx} \;=\; \frac{3\sec t \tan t}{3\sec^{2}t} \;=\; \frac{\sec t \tan t}{\sec^{2}t} \;=\; \tan t \cdot \frac{\sec t}{\sec^{2}t} \;=\; \tan t \cdot \cos t \;=\; \sin t.$$

We now have the simple expression $$\dfrac{dy}{dx} = \sin t.$$

To obtain the second derivative, we first differentiate $$\dfrac{dy}{dx}$$ with respect to $$t$$:

$$\frac{d}{dt}\!\left(\frac{dy}{dx}\right) \;=\; \frac{d}{dt}(\sin t) \;=\; \cos t.$$

The standard formula for the second derivative in parametric form is

$$\frac{d^{2}y}{dx^{2}} \;=\; \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)} {\dfrac{dx}{dt}}.$$

Substituting the values we have already obtained, we write

$$\frac{d^{2}y}{dx^{2}} \;=\; \frac{\cos t}{3\sec^{2}t} \;=\; \cos t \cdot \frac{\cos^{2}t}{3} \;=\; \frac{\cos^{3}t}{3}.$$

We now evaluate this at the specified parameter value $$t=\dfrac{\pi}{4}$$. For $$t = \dfrac{\pi}{4}$$ we know $$\cos\!\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}$$. Therefore,

$$\cos^{3}\!\left(\frac{\pi}{4}\right) \;=\; \left(\frac{1}{\sqrt2}\right)^{3} \;=\; \frac{1}{2\sqrt2}.$$

Putting this into the expression for the second derivative, we obtain

$$\frac{d^{2}y}{dx^{2}}\Bigg|_{\,t=\frac{\pi}{4}} \;=\; \frac{1}{3} \cdot \frac{1}{2\sqrt2} \;=\; \frac{1}{6\sqrt2}.$$

Hence, the correct answer is Option B.

We have the function $$f(x)=\sin^{-1}\!\left(\dfrac{2\,3^{x}}{1+9^{x}}\right).$$ To differentiate it, we first note the standard derivative formula

$$\dfrac{d}{dx}\bigl[\sin^{-1}(u)\bigr]=\dfrac{u'}{\sqrt{1-u^{2}}},$$

where $$u=u(x).$$ So we set

$$u(x)=\dfrac{2\,3^{x}}{1+9^{x}}=\dfrac{2\,3^{x}}{1+3^{2x}}.$$

Now we compute $$u'$$ by the quotient rule, which states

$$\dfrac{d}{dx}\!\left(\dfrac{N}{D}\right)=\dfrac{N'D-ND'}{D^{2}},$$

where $$N=2\,3^{x}$$ and $$D=1+3^{2x}.$$

First we differentiate the numerator and the denominator separately.

For the numerator: $$N=2\,3^{x}\quad\Longrightarrow\quad N'=2\,3^{x}\ln 3.$$

For the denominator: $$D=1+3^{2x}\quad\Longrightarrow\quad D'=3^{2x}\cdot 2\ln 3=2\,3^{2x}\ln 3.$$

Substituting into the quotient rule gives

$$u'(x)=\dfrac{\bigl(2\,3^{x}\ln 3\bigr)(1+3^{2x})-\bigl(2\,3^{x}\bigr)\bigl(2\,3^{2x}\ln 3\bigr)}{(1+3^{2x})^{2}}.$$

We now evaluate every quantity at $$x=-\dfrac12.$$ We have $$3^{x}=3^{-1/2}=\dfrac1{\sqrt3},\qquad 3^{2x}=3^{-1}= \dfrac13.$$ Hence

$$N\Bigl(-\tfrac12\Bigr)=2\cdot\dfrac1{\sqrt3}=\dfrac2{\sqrt3},\quad N'\Bigl(-\tfrac12\Bigr)=2\cdot\dfrac1{\sqrt3}\ln 3=\dfrac{2\ln3}{\sqrt3},$$ $$D\Bigl(-\tfrac12\Bigr)=1+\dfrac13=\dfrac43,\quad D'\Bigl(-\tfrac12\Bigr)=2\cdot\dfrac13\ln 3=\dfrac{2\ln3}{3}.$$

So

$$u'\!\Bigl(-\tfrac12\Bigr)= \dfrac{\displaystyle\left(\dfrac{2\ln3}{\sqrt3}\right)\!\left(\dfrac43\right)-\left(\dfrac2{\sqrt3}\right)\!\left(\dfrac{2\ln3}{3}\right)} {\left(\dfrac43\right)^{2}} =\dfrac{\dfrac{8\ln3}{3\sqrt3}-\dfrac{4\ln3}{3\sqrt3}}{\dfrac{16}{9}} =\dfrac{\dfrac{4\ln3}{3\sqrt3}}{\dfrac{16}{9}} =\dfrac{4\ln3}{3\sqrt3}\cdot\dfrac{9}{16} =\dfrac{36\ln3}{48\sqrt3} =\dfrac{3\ln3}{4\sqrt3} =\dfrac{\sqrt3\,\ln3}{4}. $$

Next we need $$\sqrt{1-u^{2}}$$ at $$x=-\dfrac12.$$ First find $$u\bigl(-\tfrac12\bigr):$$

$$u\!\Bigl(-\tfrac12\Bigr)=\dfrac{2\cdot\dfrac1{\sqrt3}}{1+\dfrac13} =\dfrac{\dfrac2{\sqrt3}}{\dfrac43} =\dfrac2{\sqrt3}\cdot\dfrac34 =\dfrac6{4\sqrt3} =\dfrac3{2\sqrt3} =\dfrac{\sqrt3}{2}. $$

Therefore

$$1-u^{2} =1-\left(\dfrac{\sqrt3}{2}\right)^{2} =1-\dfrac34 =\dfrac14,$$ $$\sqrt{1-u^{2}}=\sqrt{\dfrac14}=\dfrac12.$$

Finally, using the derivative formula for $$\sin^{-1}u,$$ we obtain

$$f'\!\Bigl(-\tfrac12\Bigr) =\dfrac{u'\!\left(-\tfrac12\right)}{\sqrt{1-u^{2}}} =\dfrac{\dfrac{\sqrt3\,\ln3}{4}}{\dfrac12} =\dfrac{\sqrt3\,\ln3}{4}\times2 =\dfrac{\sqrt3\,\ln3}{2} =\sqrt3\;\ln\!\sqrt3.$$

The last equality follows because $$\ln\!\sqrt3=\dfrac12\ln 3,$$ so $$\sqrt3\,\ln\!\sqrt3=\sqrt3\cdot\dfrac12\ln3=\dfrac{\sqrt3\,\ln3}{2},$$ exactly the value we obtained.

Thus, $$f'\!\Bigl(-\dfrac12\Bigr)=\sqrt{3}\,\log_{e}\sqrt{3}.$$

Hence, the correct answer is Option A.

We have two functions of the same independent variable $$t$$, namely

$$x=\sqrt{2^{\cosec^{-1}t}}\quad\text{and}\quad y=\sqrt{2^{\sec^{-1}t}},\qquad |t|\ge 1.$$

Our aim is to determine $$\dfrac{dy}{dx}$$. Because both $$x$$ and $$y$$ are written in terms of $$t$$, we shall first find $$\dfrac{dx}{dt}$$ and $$\dfrac{dy}{dt}$$ separately and then divide them, using the chain-rule identity

$$\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}.$$

Let us begin with $$x$$. Rewrite the square root as a rational exponent:

$$x=\Bigl(2^{\cosec^{-1}t}\Bigr)^{1/2}=2^{\tfrac12\,\cosec^{-1}t}.$$

Taking natural logarithms on both sides gives

$$\ln x=\frac12\,\cosec^{-1}t\;\ln 2.$$

Differentiating this equation with respect to $$t$$ we obtain

$$\frac1x\,\frac{dx}{dt}=\frac{\ln 2}{2}\,\frac{d}{dt}\bigl(\cosec^{-1}t\bigr).$$

We now recall the standard derivative formula for the inverse cosecant:

$$\frac{d}{dt}\bigl(\cosec^{-1}t\bigr)=-\frac1{|t|\sqrt{t^{2}-1}}.$$

Substituting this value, we get

$$\frac1x\,\frac{dx}{dt}=\frac{\ln 2}{2}\left(-\frac1{|t|\sqrt{t^{2}-1}}\right).$$

Multiplying both sides by $$x$$ yields

$$\frac{dx}{dt}=x\;\frac{\ln 2}{2}\left(-\frac1{|t|\sqrt{t^{2}-1}}\right).$$

We carry out exactly the same procedure for $$y$$. First rewrite:

$$y=\Bigl(2^{\sec^{-1}t}\Bigr)^{1/2}=2^{\tfrac12\,\sec^{-1}t},$$

and hence

$$\ln y=\frac12\,\sec^{-1}t\;\ln 2.$$

Differentiating with respect to $$t$$ gives

$$\frac1y\,\frac{dy}{dt}=\frac{\ln 2}{2}\,\frac{d}{dt}\bigl(\sec^{-1}t\bigr).$$

The derivative formula for the inverse secant is

$$\frac{d}{dt}\bigl(\sec^{-1}t\bigr)=\frac1{|t|\sqrt{t^{2}-1}}.$$

Substituting this, we find

$$\frac1y\,\frac{dy}{dt}=\frac{\ln 2}{2}\left(\frac1{|t|\sqrt{t^{2}-1}}\right).$$

Therefore

$$\frac{dy}{dt}=y\;\frac{\ln 2}{2}\left(\frac1{|t|\sqrt{t^{2}-1}}\right).$$

Now we form the desired ratio:

$$\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{y\,\dfrac{\ln 2}{2}\left(\dfrac1{|t|\sqrt{t^{2}-1}}\right)}{x\,\dfrac{\ln 2}{2}\left(-\dfrac1{|t|\sqrt{t^{2}-1}}\right)}.$$

Notice that the factors $$\dfrac{\ln 2}{2}$$, $$|t|$$ and $$\sqrt{t^{2}-1}$$ appear in both numerator and denominator; they cancel out completely, leaving only the minus sign and the ratio of $$y$$ to $$x$$:

$$\frac{dy}{dx}=-\frac{y}{x}.$$

Hence, the correct answer is Option B.

We are given the implicit relation $$x^2 + y^2 + \sin y = 4$$ that connects the variables $$x$$ and $$y$$. Our objective is to find the second derivative $$\dfrac{d^2y}{dx^2}$$ at the specific point $$(-2,\,0)$$ lying on the curve.

First we differentiate the whole equation with respect to $$x$$. Using the rule $$\dfrac{d}{dx}(u+v+w)=\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}$$ together with the basic derivatives $$\dfrac{d}{dx}(x^2)=2x$$, $$\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}$$ (chain rule), and $$\dfrac{d}{dx}(\sin y)=\cos y\dfrac{dy}{dx}$$ (again chain rule), we obtain

$$2x \;+\; 2y\dfrac{dy}{dx} \;+\; \cos y\,\dfrac{dy}{dx} \;=\; 0.$$

Now we collect the terms that contain $$\dfrac{dy}{dx}$$:

$$2x \;+\;\left(2y+\cos y\right)\dfrac{dy}{dx}=0.$$

Rearranging to isolate the first derivative gives

$$\left(2y+\cos y\right)\dfrac{dy}{dx}=-2x,$$

so that

$$\dfrac{dy}{dx}=\dfrac{-2x}{\,2y+\cos y\,}.$$

Next, we evaluate this first derivative at the point $$(-2,\,0)$$. Substituting $$x=-2$$ and $$y=0$$, while noting that $$\cos0=1$$, we get

$$\dfrac{dy}{dx}\Big|_{(-2,0)}=\dfrac{-2(-2)}{\,2(0)+1\,}=\dfrac{4}{1}=4.$$

With the first derivative in hand, we proceed to the second derivative. Write the first derivative in a compact quotient form

$$\dfrac{dy}{dx}=\dfrac{N}{D},\quad\text{where }N=-2x\text{ and }D=2y+\cos y.$$

To differentiate a quotient we recall the quotient rule: if $$u=\dfrac{p}{q}$$, then $$\dfrac{du}{dx}=\dfrac{q\dfrac{dp}{dx}-p\dfrac{dq}{dx}}{q^2}.$$ Applying this rule to $$\dfrac{dy}{dx}=\dfrac{N}{D}$$ we have

$$\dfrac{d^2y}{dx^2}=\dfrac{D\dfrac{dN}{dx}-N\dfrac{dD}{dx}}{D^2}.$$

We now compute the required building blocks one by one.

1. The derivative of $$N=-2x$$ with respect to $$x$$ is

$$\dfrac{dN}{dx}=-2.$$

2. The derivative of $$D=2y+\cos y$$ with respect to $$x$$ requires the chain rule again:

$$\dfrac{dD}{dx}=2\dfrac{dy}{dx}-\sin y\,\dfrac{dy}{dx}=\left(2-\sin y\right)\dfrac{dy}{dx}.$$

We now evaluate every symbol at the point $$(-2,0)$$. We already have

$$x=-2,\quad y=0,\quad \dfrac{dy}{dx}=4,$$

and we also know $$\sin0=0$$ and $$\cos0=1$$. Hence,

$$N= -2(-2)=4,\qquad D=2(0)+1=1,$$

$$\dfrac{dN}{dx}=-2,\qquad \dfrac{dD}{dx}=\left(2-0\right)\cdot4 = 8.$$

Substituting everything into the quotient-rule expression gives

$$\dfrac{d^2y}{dx^2}\Big|_{(-2,0)}=\dfrac{(1)(-2)-\,(4)(8)}{1^2}=(-2)-32=-34.$$

Hence, the correct answer is Option A.

We have the two auxiliary functions $$f(x)=x^{2}+\dfrac{1}{x^{2}}$$ and $$g(x)=x-\dfrac{1}{x}$$ with the domain $$x\in\mathbb{R}\;-\;\{-1,0,1\}$$. The function whose local extrema we need is

$$h(x)=\dfrac{f(x)}{g(x)}=\dfrac{x^{2}+\dfrac{1}{x^{2}}}{\,x-\dfrac{1}{x}\,}.$$

First we simplify the numerator. We recall the algebraic identity

$$(a-b)^{2}=a^{2}-2ab+b^{2}.$$

Taking $$a=x$$ and $$b=\dfrac{1}{x}$$ gives

$$(x-\dfrac{1}{x})^{2}=x^{2}-2+\dfrac{1}{x^{2}}.$$

From this we immediately obtain

$$x^{2}+\dfrac{1}{x^{2}}=(x-\dfrac{1}{x})^{2}+2.$$

Substituting the expressions $$f(x)=g(x)^{2}+2$$ and $$g(x)=x-\dfrac{1}{x}$$ into $$h(x)$$ we arrive at

$$h(x)=\dfrac{g(x)^{2}+2}{g(x)}.$$

To avoid repeatedly writing the longer symbol, we now put $$t=g(x)=x-\dfrac{1}{x}.$$ The domain restriction $$x\neq -1,0,1$$ implies $$t\neq 0$$ (because $$x=1$$ or $$x=-1$$ would make $$t=0$$). Using this new variable the function simplifies beautifully to

$$h(x)=t+\dfrac{2}{t}\equiv\varphi(t).$$

So the original problem has been reduced to finding the local minimum of

$$\varphi(t)=t+\dfrac{2}{t},\qquad t\in\mathbb{R}\;-\;\{0\}.$$

To locate critical points we differentiate. The derivative of $$\dfrac{2}{t}$$ is $$-\dfrac{2}{t^{2}}$$, hence

$$\dfrac{d\varphi}{dt}=1-\dfrac{2}{t^{2}}.$$

Critical points occur where the derivative vanishes:

$$1-\dfrac{2}{t^{2}}=0\;\Longrightarrow\;t^{2}=2\;\Longrightarrow\;t=\sqrt{2}\;\text{or}\;t=-\sqrt{2}.$$

To decide the nature of these points we look at the second derivative. Differentiating once more, we obtain

$$\dfrac{d^{2}\varphi}{dt^{2}}=\dfrac{4}{t^{3}}.$$

Evaluating at the two critical values gives

$$\dfrac{d^{2}\varphi}{dt^{2}}\Bigg|_{t=\sqrt{2}}=\dfrac{4}{(\sqrt{2})^{3}}=\dfrac{4}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}\;>\;0,$$

$$\dfrac{d^{2}\varphi}{dt^{2}}\Bigg|_{t=-\sqrt{2}}=\dfrac{4}{(-\sqrt{2})^{3}}=-\dfrac{2}{\sqrt{2}}\;<\;0.$$

Because the second derivative is positive at $$t=\sqrt{2}$$, the point $$t=\sqrt{2}$$ gives a local minimum, while $$t=-\sqrt{2}$$ gives a local maximum.

We now compute the minimum value itself:

$$\varphi(\sqrt{2})=\sqrt{2}+\dfrac{2}{\sqrt{2}}=\sqrt{2}+\dfrac{2\sqrt{2}}{2}= \sqrt{2}+\sqrt{2}=2\sqrt{2}.$$

Finally, we verify that this minimum actually corresponds to some admissible $$x$$. Setting $$x-\dfrac{1}{x}=\sqrt{2}$$ and solving, we get

$$x^{2}-\sqrt{2}\,x-1=0\;\Longrightarrow\;x=\dfrac{\sqrt{2}\pm\sqrt{6}}{2},$$

neither of which equals $$-1,0,1$$, so the critical value is indeed attained within the domain.

Therefore the local minimum value of $$h(x)$$ is $$2\sqrt{2}$$.

Hence, the correct answer is Option A.

Let us denote the given function by

$$y \;=\; \tan^{-1}\!\!\left(\dfrac{6x\sqrt{x}}{1-9x^{3}}\right),\qquad 0<x<\dfrac14.$$

To find $$\dfrac{dy}{dx}$$ we recall the standard formula

$$\dfrac{d}{dx}\bigl(\tan^{-1}t\bigr)\;=\;\dfrac{t'}{1+t^{2}},$$

where $$t=t(x).$$ Here

$$t(x)=\dfrac{6x\sqrt{x}}{1-9x^{3}}=\dfrac{6x^{3/2}}{1-9x^{3}}.$$

We first differentiate $$t(x)$$. Writing $$t(x)=\dfrac{u}{v}$$ with

$$u=6x^{3/2},\qquad v=1-9x^{3},$$

we use the quotient rule $$\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}.$$

Compute $$u'$$:

$$u'=6\cdot\dfrac{3}{2}x^{1/2}=9x^{1/2}.$$
Compute $$v'$$:

$$v'=-27x^{2}.$$

Hence

$$ \begin{aligned} t'(x)&=\dfrac{u'v-uv'}{v^{2}}\\[4pt] &=\dfrac{9x^{1/2}(1-9x^{3})-6x^{3/2}(-27x^{2})}{(1-9x^{3})^{2}}\\[4pt] &=\dfrac{9x^{1/2}-81x^{7/2}+162x^{7/2}}{(1-9x^{3})^{2}}\\[4pt] &=\dfrac{9x^{1/2}+81x^{7/2}}{(1-9x^{3})^{2}}\\[4pt] &=\dfrac{9x^{1/2}\bigl(1+9x^{3}\bigr)}{(1-9x^{3})^{2}}. \end{aligned} $$

Now we need $$1+t^{2}$$. Since

$$t(x)=\dfrac{6x^{3/2}}{1-9x^{3}},\quad t^{2}(x)=\dfrac{36x^{3}}{(1-9x^{3})^{2}},$$

we get

$$ 1+t^{2}=1+\dfrac{36x^{3}}{(1-9x^{3})^{2}} =\dfrac{(1-9x^{3})^{2}+36x^{3}}{(1-9x^{3})^{2}}. $$

Observe that

$$ (1-9x^{3})^{2}+36x^{3}=1-18x^{3}+81x^{6}+36x^{3}=1+18x^{3}+81x^{6}=(1+9x^{3})^{2}. $$

Therefore

$$1+t^{2}=\dfrac{(1+9x^{3})^{2}}{(1-9x^{3})^{2}}.$$

Putting everything into the derivative formula, we have

$$ \begin{aligned} \dfrac{dy}{dx} &=\dfrac{t'}{1+t^{2}} =\dfrac{\displaystyle \dfrac{9x^{1/2}(1+9x^{3})}{(1-9x^{3})^{2}}} {\displaystyle \dfrac{(1+9x^{3})^{2}}{(1-9x^{3})^{2}}}\\[8pt] &=\dfrac{9x^{1/2}(1+9x^{3})}{(1-9x^{3})^{2}}\cdot \dfrac{(1-9x^{3})^{2}}{(1+9x^{3})^{2}}\\[6pt] &=\dfrac{9x^{1/2}}{1+9x^{3}}. \end{aligned} $$

Thus

$$\dfrac{dy}{dx}= \sqrt{x}\;\cdot\;\dfrac{9}{1+9x^{3}}.$$

Comparing with the given form $$\dfrac{dy}{dx}= \sqrt{x}\,g(x),$$ we finally obtain

$$g(x)=\dfrac{9}{1+9x^{3}}.$$

Hence, the correct answer is Option A.

First, we rewrite the given expression so that it becomes easier to differentiate. We notice that the form $$x \pm \sqrt{x^2-1}$$ suggests the substitution $$x=\cosh t$$ because

$$\cosh^2 t-\sinh^2 t = 1 \;\; \Rightarrow \;\; \sqrt{x^2-1}= \sqrt{\cosh^2 t-1}= \sinh t.$$

Hence

$$x+\sqrt{x^2-1}= \cosh t+\sinh t= e^{\,t}, \qquad x-\sqrt{x^2-1}= \cosh t-\sinh t= e^{-\,t}.$$

Substituting these in the definition of $$y$$, we obtain

$$y=\left[e^{\,t}\right]^{15}+\left[e^{-\,t}\right]^{15}=e^{15t}+e^{-15t}=2\cosh 15t.$$

We now need the first and second derivatives of $$y$$ with respect to $$x$$. Because $$x=\cosh t$$, we first find

$$\frac{dx}{dt}=\sinh t.$$

Using $$y=2\cosh 15t$$, we have

$$\frac{dy}{dt}=2\cdot15\sinh 15t=30\sinh 15t.$$

By the chain rule,

$$\frac{dy}{dx}= \frac{dy/dt}{dx/dt}= \frac{30\sinh 15t}{\sinh t}.$$

Next we differentiate $$\dfrac{dy}{dx}$$ with respect to $$x$$ to obtain $$\dfrac{d^2y}{dx^2}$$. First differentiate with respect to $$t$$:

We write $$\dfrac{dy}{dx}=30\bigl(\sinh 15t\bigr)\bigl(\sinh t\bigr)^{-1}$$ and use the quotient rule

$$\frac{d}{dt}\!\left[\frac{\sinh 15t}{\sinh t}\right]= \frac{(15\cosh 15t)(\sinh t)-(\sinh 15t)(\cosh t)}{(\sinh t)^2}.$$

Multiplying by the constant 30,

$$\frac{d}{dt}\!\left[\frac{dy}{dx}\right]= 30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t}{(\sinh t)^2}.$$

Now divide again by $$\dfrac{dx}{dt}=\sinh t$$ (chain rule) to get

$$\frac{d^2y}{dx^2}=30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t}{(\sinh t)^3}.$$

We are asked to evaluate $$(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}.$$ Because $$x=\cosh t$$ and $$x^2-1=\cosh^2 t-1=\sinh^2 t$$, we substitute:

$$\bigl(x^2-1\bigr)\frac{d^2y}{dx^2}= \sinh^2 t\; \frac{30\bigl(15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t\bigr)}{(\sinh t)^3} =30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t}{\sinh t}.$$

Similarly,

$$x\frac{dy}{dx}= \cosh t\;\frac{30\sinh 15t}{\sinh t}=30\;\frac{\cosh t\sinh 15t}{\sinh t}.$$

Adding the two results:

$$\begin{aligned} (x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx} &=30\;\frac{15\cosh 15t\,\sinh t-\sinh 15t\,\cosh t+\cosh t\sinh 15t}{\sinh t}\\[4pt] &=30\;\frac{15\cosh 15t\,\sinh t}{\sinh t}\\[4pt] &=30\cdot15\cosh 15t\\[4pt] &=450\cosh 15t. \end{aligned}$$

Recall that $$y=2\cosh 15t$$, so $$\cosh 15t=\dfrac{y}{2}$$. Substituting this back, we get

$$450\cosh 15t=450\left(\frac{y}{2}\right)=225y.$$

Hence, the correct answer is Option C.

Assume that the degree of the required polynomial is $$n$$. The left-hand expression $$f(3x)$$ clearly has degree $$n$$, while the right-hand expression $$f'(x)\cdot f''(x)$$ has degree $$(n-1)+(n-2)=2n-3$$. Equality for every real $$x$$ forces the two degrees to match, so

$$n = 2n-3 \;\;\Longrightarrow\;\; n=3.$$

Hence $$f$$ must be a cubic. Write it in its most general form

$$f(x)=ax^{3}+bx^{2}+cx+d.\quad -(1)$$

Because $$a\neq 0$$ (otherwise the degree would drop), calculate the first two derivatives

$$f'(x)=3ax^{2}+2bx+c,\quad -(2)$$

$$f''(x)=6ax+2b.\quad -(3)$$

Form the product demanded on the right-hand side of the functional equation:

$$\begin{aligned} f'(x)\,f''(x) &=(3ax^{2}+2bx+c)(6ax+2b) \\ &=3ax^{2}\cdot6ax + 3ax^{2}\cdot2b + 2bx\cdot6ax + 2bx\cdot2b + c\cdot6ax + c\cdot2b \\ &=18a^{2}x^{3} + 6abx^{2} + 12abx^{2} + 4b^{2}x + 6acx + 2bc \\ &=18a^{2}x^{3} + 18abx^{2} + (4b^{2}+6ac)x + 2bc. \quad -(4) \end{aligned}$$

Now expand the left-hand side $$f(3x)$$:

$$f(3x)=a(3x)^{3}+b(3x)^{2}+c(3x)+d =27ax^{3}+9bx^{2}+3cx+d.\quad -(5)$$

Since $$f(3x) = f'(x)\,f''(x)$$ for every $$x$$, equate the coefficients of corresponding powers of $$x$$ in (4) and (5).

Cubic term, $$x^{3}$$:

$$18a^{2}=27a \;\;\Longrightarrow\;\; 18a^{2}-27a=0 \;\;\Longrightarrow\;\; 9a(2a-3)=0 \;\;\Longrightarrow\;\; a=\dfrac32.\quad -(6)$$

Quadratic term, $$x^{2}$$:

$$18ab=9b \;\;\Longrightarrow\;\; 18ab-9b=0 \;\;\Longrightarrow\;\; 9b(2a-1)=0.$$

With $$a=\dfrac32$$ we have $$2a-1=2\neq0,$$ hence

$$b=0.\quad -(7)$$

Linear term, $$x^{1}$$:

$$4b^{2}+6ac=3c.$$

Insert $$b=0$$ to obtain $$6ac=3c,$$ that is

$$3c(2a-1)=0.$$

Again $$2a-1=2\neq0,$$ so

$$c=0.\quad -(8)$$

Constant term:

The constant on the right in (4) is $$2bc$$, which vanishes because $$b=0$$ or $$c=0$$ (in fact both). Hence

$$d=0.\quad -(9)$$

All coefficients are now fixed; the unique polynomial satisfying the given condition is therefore

$$f(x)=\dfrac32\,x^{3}.\quad -(10)$$

For the values asked in the options, compute

$$f(2)=\dfrac32\,(2)^{3}=\dfrac32\cdot8=12,\quad -(11)$$

$$f'(x)=\dfrac{9}{2}x^{2}\;\Longrightarrow\; f'(2)=\dfrac{9}{2}\,(2)^{2}=\dfrac{9}{2}\cdot4=18,\quad -(12)$$

$$f''(x)=9x\;\Longrightarrow\; f''(2)=9\cdot2=18.\quad -(13)$$

Check each option:

Option A: $$f(2)+f'(2)=12+18=30\neq28.$$

Option B: $$f''(2)-f'(2)=18-18=0,$$ which is true.

Option C: $$f(2)-f'(2)+f''(2)=12-18+18=12\neq10.$$

Option D: $$f''(2)-f(2)=18-12=6\neq4.$$

Only Option B is satisfied.

Hence, the correct answer is Option B.

We are told that the real-valued function $$f(x)$$ is differentiable for every $$x\gt 0$$, that it satisfies the initial condition $$f(1)=1$$ and, in addition, that for every $$x\gt 0$$ the following limit holds

$$\lim_{t\to x}\dfrac{t^{2}f(x)-x^{2}f(t)}{t-x}=1.$$

Because the limit is of the standard “difference quotient” type, we first rewrite the numerator so that a familiar derivative expression appears. We have

$$t^{2}f(x)-x^{2}f(t) =\bigl[t^{2}f(x)-x^{2}f(x)\bigr]-x^{2}\bigl[f(t)-f(x)\bigr] =(t^{2}-x^{2})f(x)-x^{2}\bigl[f(t)-f(x)\bigr].$$

Notice that $$t^{2}-x^{2}=(t-x)(t+x)$$, so dividing the whole numerator by $$t-x$$ gives

$$\dfrac{t^{2}f(x)-x^{2}f(t)}{t-x} =(t+x)f(x)-x^{2}\dfrac{f(t)-f(x)}{t-x}.$$

Now let $$t\to x$$. Using the continuity of addition and the definition of the derivative,

$$\lim_{t\to x}(t+x)=2x, \qquad \lim_{t\to x}\dfrac{f(t)-f(x)}{t-x}=f'(x).$$

Therefore the given limit statement becomes

$$2x\,f(x)-x^{2}\,f'(x)=1\quad\text{for every }x\gt 0.$$

We have obtained a first-order linear differential equation:

$$-x^{2}f'(x)+2x\,f(x)=1.$$ Multiplying by $$-1$$ to get the standard “$$f'$$ + ( something ) $$f$$” form, we find

$$x^{2}f'(x)-2x\,f(x)=-1,$$ or, after dividing by $$x^{2}$$ (which is always positive in the interval $$(0,\infty)$$),

$$f'(x)-\dfrac{2}{x}\,f(x)=-\dfrac{1}{x^{2}}.$$

This is a linear ordinary differential equation of the form $$f'(x)+P(x)f(x)=Q(x)$$ with

$$P(x)=-\dfrac{2}{x},\qquad Q(x)=-\dfrac{1}{x^{2}}.$$

First we find the integrating factor. By definition, the integrating factor $$\mu(x)$$ satisfies

$$\mu(x)=\exp\!\Bigl(\int P(x)\,dx\Bigr) =\exp\!\Bigl(\int -\dfrac{2}{x}\,dx\Bigr) =\exp(-2\ln x)=x^{-2}.$$

Multiplying the entire differential equation by this integrating factor gives

$$x^{-2}f'(x)-\dfrac{2}{x}\,x^{-2}f(x)=-x^{-4}.$$ But the left-hand side is exactly the derivative of $$x^{-2}f(x)$$, because

$$\dfrac{d}{dx}\bigl[x^{-2}f(x)\bigr] =x^{-2}f'(x)+(-2x^{-3})f(x) =x^{-2}f'(x)-\dfrac{2}{x}\,x^{-2}f(x).$$

Hence we may rewrite the equation compactly as

$$\dfrac{d}{dx}\bigl[x^{-2}f(x)\bigr]=-x^{-4}.$$

We now integrate both sides with respect to $$x$$:

$$\int\dfrac{d}{dx}\bigl[x^{-2}f(x)\bigr]\,dx =\int -x^{-4}\,dx.$$

The left integral simply returns the inside expression, while the right integral is computed using the power rule $$\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C$$:

$$x^{-2}f(x) =-\dfrac{x^{-3}}{-3}+C =\dfrac{1}{3}x^{-3}+C,$$ where $$C$$ is the constant of integration.

Finally we multiply by $$x^{2}$$ to solve for $$f(x)$$ explicitly:

$$f(x)=\dfrac{1}{3}x^{-1}+C\,x^{2}.$$

The initial condition $$f(1)=1$$ determines $$C$$. Substituting $$x=1$$ gives

$$1=f(1)=\dfrac{1}{3}(1)^{-1}+C(1)^{2} =\dfrac{1}{3}+C,$$ so

$$C=1-\dfrac{1}{3}=\dfrac{2}{3}.$$

Therefore the required function is

$$f(x)=\dfrac{2}{3}x^{2}+\dfrac{1}{3x}.$$

Now we must evaluate $$f\!\left(\dfrac{3}{2}\right)$$. Substituting $$x=\dfrac{3}{2}$$ yields

$$f\!\left(\dfrac{3}{2}\right) =\dfrac{2}{3}\left(\dfrac{3}{2}\right)^{2} +\dfrac{1}{3\left(\dfrac{3}{2}\right)} =\dfrac{2}{3}\cdot\dfrac{9}{4} +\dfrac{1}{\dfrac{9}{2}} =\dfrac{18}{12}+\dfrac{2}{9} =\dfrac{3}{2}+\dfrac{2}{9}.$$

To add the two fractions, convert both to the common denominator $$18$$:

$$\dfrac{3}{2}=\dfrac{27}{18},\qquad \dfrac{2}{9}=\dfrac{4}{18},$$

so

$$f\!\left(\dfrac{3}{2}\right) =\dfrac{27}{18}+\dfrac{4}{18} =\dfrac{31}{18}.$$

Hence, the correct answer is Option D.

We have the function

$$f(x)=\sin^{4}x+\cos^{4}x.$$

To find where $$f(x)$$ is increasing, we must examine the sign of its first derivative. The rule we will use is the power rule combined with the chain rule: if $$g(x)=\big(h(x)\big)^n,$$ then $$g'(x)=n\,\big(h(x)\big)^{\,n-1}\,h'(x).$$

First differentiate the two terms separately.

For the sine term:

$$\frac{d}{dx}\bigl(\sin^{4}x\bigr)=4\sin^{3}x\cdot\frac{d}{dx}(\sin x)=4\sin^{3}x\cos x.$$

For the cosine term, note that $$\frac{d}{dx}(\cos x)=-\sin x$$, so

$$\frac{d}{dx}\bigl(\cos^{4}x\bigr)=4\cos^{3}x\cdot\frac{d}{dx}(\cos x)=4\cos^{3}x(-\sin x)=-4\cos^{3}x\sin x.$$

Adding the two derivatives gives

$$f'(x)=4\sin^{3}x\cos x-4\cos^{3}x\sin x.$$

Both terms contain the common factor $$4\sin x\cos x,$$ so we factor it out:

$$f'(x)=4\sin x\cos x\bigl(\sin^{2}x-\cos^{2}x\bigr).$$

The derivative therefore breaks naturally into three factors:

$$\sin x,\qquad\cos x,\qquad\text{and}\qquad\sin^{2}x-\cos^{2}x.$$

To decide the sign of $$f'(x)$$ we examine each factor on the interval $$0<x<\pi$$ (all option intervals lie here).

1. In the first quadrant $$0<x<\frac{\pi}{2},$$ both $$\sin x$$ and $$\cos x$$ are positive, so their product $$\sin x\cos x$$ is positive.

2. In the second quadrant $$\frac{\pi}{2}<x<\pi,$$ we have $$\sin x>0$$ but $$\cos x<0,$$ hence $$\sin x\cos x<0.$$

3. The remaining factor $$\sin^{2}x-\cos^{2}x$$ changes sign precisely when it is zero. Setting it to zero,

$$\sin^{2}x-\cos^{2}x=0\quad\Longrightarrow\quad\sin^{2}x=\cos^{2}x\quad\Longrightarrow\quad\tan^{2}x=1,$$

which holds at $$x=\frac{\pi}{4}$$ inside the first quadrant (and at $$x=\frac{3\pi}{4}$$ in the second).

Choose test points to locate where $$\sin^{2}x-\cos^{2}x$$ is positive or negative:

• For $$x=\frac{\pi}{6}$$ (which is <$$\frac{\pi}{4}$$), $$\sin^{2}x<\cos^{2}x,$$ so $$\sin^{2}x-\cos^{2}x<0.$$
• For $$x=\frac{\pi}{3}$$ (which is >$$\frac{\pi}{4}$$ but <$$\frac{\pi}{2}$$), $$\sin^{2}x>\cos^{2}x,$$ so $$\sin^{2}x-\cos^{2}x>0.$$

We now combine the signs.

First quadrant, $$0<x<\frac{\pi}{2}$$

• For $$0<x<\frac{\pi}{4}:$$ $$\sin x\cos x>0$$ and $$\sin^{2}x-\cos^{2}x<0,$$ so $$f'(x)<0.$$ Thus $$f(x)$$ decreases here.

• For $$\frac{\pi}{4}<x<\frac{\pi}{2}:$$ $$\sin x\cos x>0$$ and $$\sin^{2}x-\cos^{2}x>0,$$ so $$f'(x)>0.$$ Thus $$f(x)$$ increases here.

Second quadrant, $$\frac{\pi}{2}<x<\pi$$

Here $$\sin x\cos x<0.$$ Meanwhile $$\sin^{2}x-\cos^{2}x$$ stays positive until $$x=\frac{3\pi}{4}$$ and becomes negative afterwards, but the product with the negative $$\sin x\cos x$$ ensures $$f'(x)$$ remains negative throughout the whole second quadrant. Consequently, $$f(x)$$ decreases for all $$\frac{\pi}{2}<x<\pi.$$

Putting everything together, the derivative is positive (so $$f(x)$$ is increasing) only on

$$\left(\frac{\pi}{4},\;\frac{\pi}{2}\right).$$

This matches Option C exactly.

Hence, the correct answer is Option C.

Rolle's theorem requires three conditions for a function $$ f(x) $$ on an interval $$[a, b]$$:

1. The function must be continuous on $$[a, b]$$.
2. The function must be differentiable on $$(a, b)$$.
3. The function values at the endpoints must be equal, i.e., $$ f(a) = f(b) $$.

Given the function $$ f(x) = 2x^3 + b x^2 + c x $$ on the interval $$[-1, 1]$$, and that Rolle's theorem holds at $$ x = \frac{1}{2} $$, we know:

• At $$ x = \frac{1}{2} $$, the derivative $$ f'(x) = 0 $$.
• The function values at the endpoints are equal: $$ f(-1) = f(1) $$.

First, compute $$ f(-1) $$ and $$ f(1) $$:

For $$ x = -1 $$:

$$ f(-1) = 2(-1)^3 + b(-1)^2 + c(-1) = 2(-1) + b(1) + c(-1) = -2 + b - c $$

For $$ x = 1 $$:

$$ f(1) = 2(1)^3 + b(1)^2 + c(1) = 2(1) + b(1) + c(1) = 2 + b + c $$

Set $$ f(-1) = f(1) $$:

$$ -2 + b - c = 2 + b + c $$

Solve for $$ c $$:

Subtract $$ b $$ from both sides:

$$ -2 - c = 2 + c $$

Add 2 to both sides:

$$ -c = 4 + c $$

Subtract $$ c $$ from both sides:

$$ -c - c = 4 \implies -2c = 4 $$

Divide both sides by -2:

$$ c = -2 $$

Now, find the derivative of $$ f(x) $$:

$$ f'(x) = \frac{d}{dx}(2x^3) + \frac{d}{dx}(b x^2) + \frac{d}{dx}(c x) = 6x^2 + 2b x + c $$

At $$ x = \frac{1}{2} $$, $$ f'\left( \frac{1}{2} \right) = 0 $$:

$$ f'\left( \frac{1}{2} \right) = 6 \left( \frac{1}{2} \right)^2 + 2b \left( \frac{1}{2} \right) + c = 0 $$

Compute $$ \left( \frac{1}{2} \right)^2 = \frac{1}{4} $$:

$$ 6 \times \frac{1}{4} + 2b \times \frac{1}{2} + c = 0 $$

Simplify:

$$ \frac{6}{4} + b + c = 0 \implies \frac{3}{2} + b + c = 0 $$

Substitute $$ c = -2 $$:

$$ \frac{3}{2} + b + (-2) = 0 $$

Simplify:

$$ \frac{3}{2} - 2 + b = 0 \implies \frac{3}{2} - \frac{4}{2} + b = 0 \implies -\frac{1}{2} + b = 0 $$

Add $$ \frac{1}{2} $$ to both sides:

$$ b = \frac{1}{2} $$

Now, compute $$ 2b + c $$:

$$ 2b + c = 2 \times \frac{1}{2} + (-2) = 1 - 2 = -1 $$

Hence, the correct answer is Option C.

We have two real-valued functions $$f$$ and $$g$$ which are differentiable on the open interval $$(0,1)$$ and continuous on the closed interval $$[0,1]$$. The given boundary values are

$$f(0)=2,\qquad f(1)=6, \qquad g(0)=0,\qquad g(1)=2.$$

Because both functions are continuous on $$[0,1]$$ and differentiable on $$(0,1)$$, we are allowed to apply the Cauchy Mean Value Theorem. First, let us recall the statement of that theorem.

Cauchy Mean Value Theorem. If functions $$p$$ and $$q$$ are continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, then there exists at least one point $$c\in(a,b)$$ such that

$$\frac{p'(c)}{q'(c)}=\frac{p(b)-p(a)}{q(b)-q(a)}.$$

We now take $$p(x)=f(x)$$ and $$q(x)=g(x)$$, with $$a=0$$ and $$b=1$$. All the hypotheses are satisfied, so there exists some $$c\in(0,1)$$ for which

$$\frac{f'(c)}{g'(c)}=\frac{f(1)-f(0)}{g(1)-g(0)}.$$

Next, we substitute the known endpoint values.

$$f(1)-f(0)=6-2=4,$$

$$g(1)-g(0)=2-0=2.$$

Hence the right-hand side of the Cauchy Mean Value Theorem becomes

$$\frac{f(1)-f(0)}{g(1)-g(0)}=\frac{4}{2}=2.$$

So we have obtained

$$\frac{f'(c)}{g'(c)}=2.$$

Multiplying both sides of this equation by $$g'(c)$$ (which is permissible because differentiability implies continuity and we may assume $$g'(c)\neq 0$$ at the chosen point), we get

$$f'(c)=2\,g'(c).$$

This matches exactly the relation given in option B.

Hence, the correct answer is Option B.

We are given that $$ y = e^{nx} $$. We need to find $$ \frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} $$.

First, we find $$ \frac{d^2y}{dx^2} $$. Since $$ y = e^{nx} $$, we start by finding the first derivative $$ \frac{dy}{dx} $$. The derivative of $$ e^{nx} $$ with respect to $$ x $$ is $$ n e^{nx} $$, so:

$$ \frac{dy}{dx} = n e^{nx} $$

But since $$ y = e^{nx} $$, we can write $$ \frac{dy}{dx} = n y $$. Now, to find the second derivative $$ \frac{d^2y}{dx^2} $$, we differentiate $$ \frac{dy}{dx} $$ again with respect to $$ x $$:

$$ \frac{d^2y}{dx^2} = \frac{d}{dx} (n y) $$

Using the chain rule, since $$ y $$ is a function of $$ x $$, we get:

$$ \frac{d}{dx} (n y) = n \frac{dy}{dx} = n \cdot (n y) = n^2 y $$

Alternatively, differentiating directly:

$$ \frac{d}{dx} (n e^{nx}) = n \cdot n e^{nx} = n^2 e^{nx} = n^2 y $$

So, $$ \frac{d^2y}{dx^2} = n^2 y $$.

Next, we need $$ \frac{d^2x}{dy^2} $$. This requires expressing $$ x $$ in terms of $$ y $$. Given $$ y = e^{nx} $$, we solve for $$ x $$:

Take the natural logarithm of both sides:

$$ \ln y = \ln(e^{nx}) $$

$$ \ln y = nx $$

Therefore,

$$ x = \frac{1}{n} \ln y $$

Now, find the first derivative $$ \frac{dx}{dy} $$:

$$ \frac{dx}{dy} = \frac{d}{dy} \left( \frac{1}{n} \ln y \right) = \frac{1}{n} \cdot \frac{1}{y} = \frac{1}{n y} $$

Now, find the second derivative $$ \frac{d^2x}{dy^2} $$ by differentiating $$ \frac{dx}{dy} $$ with respect to $$ y $$:

$$ \frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{1}{n y} \right) = \frac{1}{n} \cdot \frac{d}{dy} (y^{-1}) $$

The derivative of $$ y^{-1} $$ is $$ -y^{-2} $$, so:

$$ \frac{d^2x}{dy^2} = \frac{1}{n} \cdot (-y^{-2}) = -\frac{1}{n y^2} $$

Now, we multiply the two second derivatives:

$$ \frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = (n^2 y) \cdot \left( -\frac{1}{n y^2} \right) $$

Simplify the expression:

$$ = n^2 y \cdot \left( -\frac{1}{n y^2} \right) = -\frac{n^2}{n} \cdot \frac{y}{y^2} = -n \cdot \frac{1}{y} = -\frac{n}{y} $$

Since $$ y = e^{nx} $$, substitute back:

$$ -\frac{n}{y} = -\frac{n}{e^{nx}} = -n e^{-nx} $$

because $$ \frac{1}{e^{nx}} = e^{-nx} $$.

Therefore, $$ \frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = -n e^{-nx} $$.

Comparing with the options:

A. $$ ne^{-nx} $$

B. $$ -ne^{-nx} $$

C. $$ ne^{nx} $$

D. 1

Hence, the correct answer is Option B.

We are given the function $$ f(x) = \frac{x^2 - x}{x^2 + 2x} $$ for $$ x \neq 0, -2 $$. First, we simplify this function by factoring both the numerator and the denominator.

The numerator is $$ x^2 - x = x(x - 1) $$. The denominator is $$ x^2 + 2x = x(x + 2) $$. Substituting these factorizations, we get:

$$ f(x) = \frac{x(x - 1)}{x(x + 2)} $$

Since $$ x \neq 0 $$, we can cancel the $$ x $$ in the numerator and denominator:

$$ f(x) = \frac{x - 1}{x + 2} $$

Now, we need to find the derivative of the inverse function, $$ \frac{d}{dx} [f^{-1}(x)] $$. To do this, we first find the inverse function $$ f^{-1}(x) $$. Set $$ y = f(x) = \frac{x - 1}{x + 2} $$. We solve for $$ x $$ in terms of $$ y $$:

$$ y = \frac{x - 1}{x + 2} $$

Multiply both sides by $$ x + 2 $$:

$$ y(x + 2) = x - 1 $$

Expand the left side:

$$ yx + 2y = x - 1 $$

Bring all terms involving $$ x $$ to one side and constants to the other:

$$ yx - x = -1 - 2y $$

Factor out $$ x $$ on the left:

$$ x(y - 1) = -1 - 2y $$

Solve for $$ x $$:

$$ x = \frac{-1 - 2y}{y - 1} $$

Simplify by factoring out $$-1$$ in both numerator and denominator:

$$ x = \frac{-(1 + 2y)}{-(1 - y)} = \frac{1 + 2y}{1 - y} $$

Thus, the inverse function is $$ f^{-1}(x) = \frac{1 + 2x}{1 - x} $$.

Next, we differentiate $$ f^{-1}(x) $$ with respect to $$ x $$. Let $$ g(x) = f^{-1}(x) = \frac{1 + 2x}{1 - x} $$. We use the quotient rule for differentiation: if $$ g(x) = \frac{u}{v} $$, then $$ g'(x) = \frac{u'v - uv'}{v^2} $$, where $$ u = 1 + 2x $$ and $$ v = 1 - x $$.

Compute the derivatives: $$ u' = \frac{d}{dx}(1 + 2x) = 2 $$ and $$ v' = \frac{d}{dx}(1 - x) = -1 $$.

Apply the quotient rule:

$$ g'(x) = \frac{(2)(1 - x) - (1 + 2x)(-1)}{(1 - x)^2} $$

Simplify the numerator:

$$ 2(1 - x) - (1 + 2x)(-1) = 2 - 2x + (1 + 2x) $$

Because subtracting a negative is addition:

$$ = 2 - 2x + 1 + 2x = 3 $$

The $$ -2x $$ and $$ +2x $$ cancel out, leaving:

$$ g'(x) = \frac{3}{(1 - x)^2} $$

Therefore, $$ \frac{d}{dx} [f^{-1}(x)] = \frac{3}{(1 - x)^2} $$.

Comparing with the options:

A. $$ \frac{-1}{(1-x)^2} $$

B. $$ \frac{3}{(1-x)^2} $$

C. $$ \frac{1}{(1-x)^2} $$

D. $$ \frac{-3}{(1-x)^2} $$

Our result matches option B.

Hence, the correct answer is Option B.

Given $$ a > 0 $$ and $$ t \in \left(0, \frac{\pi}{2}\right) $$, we have $$ x = \sqrt{a^{\sin^{-1}t}} $$ and $$ y = \sqrt{a^{\cos^{-1}t}} $$. First, simplify the expressions for $$ x $$ and $$ y $$.

Rewrite $$ x $$ as: $$ x = \left( a^{\sin^{-1}t} \right)^{1/2} = a^{\frac{1}{2} \sin^{-1}t} $$ Similarly, rewrite $$ y $$ as: $$ y = \left( a^{\cos^{-1}t} \right)^{1/2} = a^{\frac{1}{2} \cos^{-1}t} $$

To find $$ \frac{dy}{dx} $$, use the chain rule since both $$ x $$ and $$ y $$ are functions of $$ t $$: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$

Compute $$ \frac{dx}{dt} $$ first. Set $$ u = \frac{1}{2} \sin^{-1}t $$, so $$ x = a^u $$. The derivative of $$ a^u $$ with respect to $$ u $$ is $$ a^u \ln a $$, and the derivative of $$ u $$ with respect to $$ t $$ is: $$ \frac{du}{dt} = \frac{1}{2} \cdot \frac{d}{dt} \left( \sin^{-1}t \right) = \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} $$ Thus: $$ \frac{dx}{dt} = \frac{dx}{du} \cdot \frac{du}{dt} = a^u \ln a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} $$ Since $$ a^u = a^{\frac{1}{2} \sin^{-1}t} = x $$, substitute: $$ \frac{dx}{dt} = x \ln a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} $$

Next, compute $$ \frac{dy}{dt} $$. Set $$ v = \frac{1}{2} \cos^{-1}t $$, so $$ y = a^v $$. The derivative of $$ a^v $$ with respect to $$ v $$ is $$ a^v \ln a $$, and the derivative of $$ v $$ with respect to $$ t $$ is: $$ \frac{dv}{dt} = \frac{1}{2} \cdot \frac{d}{dt} \left( \cos^{-1}t \right) = \frac{1}{2} \cdot \left( -\frac{1}{\sqrt{1-t^2}} \right) $$ Thus: $$ \frac{dy}{dt} = \frac{dy}{dv} \cdot \frac{dv}{dt} = a^v \ln a \cdot \left( -\frac{1}{2\sqrt{1-t^2}} \right) $$ Since $$ a^v = a^{\frac{1}{2} \cos^{-1}t} = y $$, substitute: $$ \frac{dy}{dt} = y \ln a \cdot \left( -\frac{1}{2\sqrt{1-t^2}} \right) $$

Now find $$ \frac{dy}{dx} $$: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{ y \ln a \cdot \left( -\frac{1}{2\sqrt{1-t^2}} \right) }{ x \ln a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} } $$ Simplify by canceling common factors. The $$ \ln a $$ terms cancel, and $$ \frac{1}{2} $$ and $$ \frac{1}{\sqrt{1-t^2}} $$ also cancel, leaving: $$ \frac{dy}{dx} = \frac{ y \cdot (-1) }{ x \cdot 1 } = -\frac{y}{x} $$

Now compute $$ \left( \frac{dy}{dx} \right)^2 $$: $$ \left( \frac{dy}{dx} \right)^2 = \left( -\frac{y}{x} \right)^2 = \frac{y^2}{x^2} $$

Then $$ 1 + \left( \frac{dy}{dx} \right)^2 $$ is: $$ 1 + \frac{y^2}{x^2} = \frac{x^2}{x^2} + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2} $$

Comparing with the options, this matches option D. Hence, the correct answer is Option D.

We are given the function $$ f(x) = \sin(\sin x) $$ and the equation $$ f''(x) + \tan x \cdot f'(x) + g(x) = 0 $$. We need to find $$ g(x) $$.

First, we find the first derivative of $$ f(x) $$. Using the chain rule, let $$ u = \sin x $$, so $$ f(x) = \sin u $$. Then:

$$ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \cos u \cdot \cos x $$

Substituting $$ u = \sin x $$ back in:

$$ f'(x) = \cos(\sin x) \cdot \cos x $$

Next, we find the second derivative $$ f''(x) $$. Since $$ f'(x) = \cos(\sin x) \cdot \cos x $$ is a product of two functions, we use the product rule:

$$ f''(x) = \frac{d}{dx} \left[ \cos(\sin x) \right] \cdot \cos x + \cos(\sin x) \cdot \frac{d}{dx} \left[ \cos x \right] $$

Compute each derivative separately. For $$ \frac{d}{dx} \left[ \cos(\sin x) \right] $$, let $$ v = \sin x $$, so:

$$ \frac{d}{dx} \cos v = -\sin v \cdot \frac{dv}{dx} = -\sin(\sin x) \cdot \cos x $$

And $$ \frac{d}{dx} \left[ \cos x \right] = -\sin x $$. Substituting these back:

$$ f''(x) = \left[ -\sin(\sin x) \cdot \cos x \right] \cdot \cos x + \cos(\sin x) \cdot \left[ -\sin x \right] $$

Simplify:

$$ f''(x) = -\sin(\sin x) \cdot \cos^2 x - \cos(\sin x) \cdot \sin x $$

Now, substitute $$ f''(x) $$ and $$ f'(x) $$ into the given equation:

$$ \left[ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x \right] + \tan x \cdot \left[ \cos(\sin x) \cos x \right] + g(x) = 0 $$

We know $$ \tan x = \frac{\sin x}{\cos x} $$, so simplify the term with $$ \tan x $$:

$$ \tan x \cdot f'(x) = \frac{\sin x}{\cos x} \cdot \cos(\sin x) \cos x = \sin x \cdot \cos(\sin x) $$

Substitute this back into the equation:

$$ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x + \sin x \cos(\sin x) + g(x) = 0 $$

Notice that $$ - \cos(\sin x) \sin x + \sin x \cos(\sin x) = 0 $$, so they cancel out:

$$ -\sin(\sin x) \cos^2 x + g(x) = 0 $$

Therefore, solving for $$ g(x) $$:

$$ g(x) = \sin(\sin x) \cos^2 x $$

We can write this as:

$$ g(x) = \cos^2 x \cdot \sin(\sin x) $$

Comparing with the options:

A. $$ \cos^2 x \cos(\sin x) $$

B. $$ \sin^2 x \cos(\cos x) $$

C. $$ \sin^2 x \sin(\cos x) $$

D. $$ \cos^2 x \sin(\sin x) $$

Our expression matches option D.

Hence, the correct answer is Option D.

We are given the function $$y=\sec\!\bigl(\tan^{-1}x\bigr)$$ and we wish to find its derivative with respect to $$x$$ and then evaluate that derivative at $$x=1$$.

First, let us introduce a new variable for clarity. We set $$\theta=\tan^{-1}x.$$ By definition of the inverse tangent, this means $$\tan\theta = x.$$

Now we re-express $$y$$ in terms of $$\theta$$:

$$y=\sec\theta.$$

To differentiate $$y$$ with respect to $$x$$ we must apply the Chain Rule. The Chain Rule states that if $$y=f(u)$$ and $$u=g(x)$$, then $$\frac{dy}{dx}= \frac{dy}{du}\cdot\frac{du}{dx}.$$

Here, $$f(u)=\sec u$$ and $$u=g(x)=\theta=\tan^{-1}x$$. We differentiate step by step:

• Derivative of the outer function.
We recall the standard derivative formula $$\frac{d}{du}\bigl[\sec u\bigr]=\sec u\,\tan u.$$

• Derivative of the inner function.
For the inverse tangent we use the formula $$\frac{d}{dx}\bigl[\tan^{-1}x\bigr]=\frac{1}{1+x^{2}}.$$

Putting these pieces together via the Chain Rule, we obtain

$$\frac{dy}{dx} = \sec\theta\,\tan\theta \;\cdot\; \frac{1}{1+x^{2}} = \frac{\sec\theta\,\tan\theta}{1+x^{2}}.$$

This derivative is expressed in terms of $$\theta$$, but we would like it entirely in terms of $$x$$. To do so, we convert $$\sec\theta$$ and $$\tan\theta$$ into algebraic expressions of $$x$$.

Because $$\tan\theta = x$$, we picture a right-angled triangle where the side opposite $$\theta$$ is $$x$$ and the side adjacent to $$\theta$$ is $$1$$. Using the Pythagorean theorem, the hypotenuse is

$$\sqrt{x^{2}+1}.$$

From this triangle we read off:

• $$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{x}{1}=x.$$

• $$\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{\sqrt{x^{2}+1}}{1} = \sqrt{x^{2}+1}.$$

Substituting these back into the derivative gives

$$\frac{dy}{dx} = \frac{\bigl(\sqrt{x^{2}+1}\bigr)\,(x)}{1+x^{2}} = \frac{x}{\sqrt{x^{2}+1}}.$$

Our next task is to evaluate this expression at $$x=1$$. We substitute $$x=1$$ into the derivative:

$$\left.\frac{dy}{dx}\right|_{x=1} = \frac{1}{\sqrt{1^{2}+1}} = \frac{1}{\sqrt{2}}.$$

Thus the numerical value of the derivative at $$x=1$$ is $$\dfrac{1}{\sqrt{2}}$$.

Examining the options, we see that $$\dfrac{1}{\sqrt{2}}$$ corresponds to Option C.

Hence, the correct answer is Option C.