JEE Continuity & Differentiability PYQs with Video Solutions

To find the value of the limit $$L$$, we first determine the derivative function $$f'(x)$$ using the first principles of differentiation and the given functional equation:

1. Find the general expression for the derivative $$f'(x)$$:
By the definition of the derivative, we have:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

Using the given functional equation to substitute for $$f(x + h)$$, we get:
$$f'(x) = \lim_{h \to 0} \frac{f(x)e^{h} + f(h)e^{x} + 4xh - f(x)}{h}$$

Rearrange and group the terms to isolate the limits:
$$f'(x) = \lim_{h \to 0} \frac{f(x)(e^{h} - 1) + f(h)e^{x} + 4xh}{h}$$
$$f'(x) = f(x) \cdot \lim_{h \to 0} \left(\frac{e^{h} - 1}{h}\right) + e^{x} \cdot \lim_{h \to 0} \left(\frac{f(h) - f(0)}{h}\right) + 4x$$

Since $$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$$ and $$\lim_{h \to 0} \frac{f(h) - f(0)}{h} = f'(0) = 2$$, substituting these standard limits yields the first-order differential equation:
$$f'(x) = f(x) + 2e^{x} + 4x$$

2. Find the second derivative function $$f''(x)$$:
Differentiate both sides of our newly found expression with respect to $$x$$:
$$f''(x) = f'(x) + 2e^{x} + 4$$

3. Evaluate the given limit expression $$L$$:
We are asked to evaluate the limit:
$$L = \lim_{x \to 1} \frac{f'(x) - f'(1)}{x^2 - 1}$$

We can factor the denominator as $$(x - 1)(x + 1)$$ and rewrite the expression to reveal a standard derivative structure:
$$L = \lim_{x \to 1} \left[ \frac{f'(x) - f'(1)}{x - 1} \cdot \frac{1}{x + 1} \right]$$

By the definition of the second derivative, the first part of the limit is exactly $$f''(1)$$:
$$L = f''(1) \cdot \frac{1}{1 + 1} = \frac{f''(1)}{2}$$

4. Compute the value of $$f''(1)$$:
Using the differential equations established in the earlier steps, we substitute $$x = 1$$:
$$f''(1) = f'(1) + 2e^{1} + 4$$

Now substitute the expression for $$f'(1)$$ into the equation above:
$$f'(1) = f(1) + 2e^{1} + 4(1)$$
$$f''(1) = (f(1) + 2e + 4) + 2e + 4 = f(1) + 4e + 8$$

To find $$f(1)$$, we solve the linear differential equation $$f'(x) - f(x) = 2e^x + 4x$$. The integrating factor is $$e^{-x}$$.
Multiplying both sides by the integrating factor gives:
$$\frac{d}{dx}[f(x)e^{-x}] = 2 + 4xe^{-x}$$

Integrating both sides from 0 to 1:
$$\left[f(x)e^{-x}\right]_0^1 = \int_0^1 (2 + 4xe^{-x}) dx$$
$$f(1)e^{-1} - f(0)e^0 = [2x]_0^1 + 4 \cdot \left([-xe^{-x}]_0^1 + \int_0^1 e^{-x} dx\right)$$
$$f(1)e^{-1} = 2 + 4 \cdot \left(-e^{-1} + [-e^{-x}]_0^1\right)$$
$$f(1)e^{-1} = 2 + 4 \cdot (-e^{-1} - e^{-1} + 1)$$
$$f(1)e^{-1} = 2 - 8e^{-1} + 4 = 6 - 8e^{-1}$$

Multiplying the entire equation by $$e$$ gives:
$$f(1) = 6e - 8$$

Now, substitute $$f(1)$$ back into the expression for $$f''(1)$$:
$$f''(1) = (6e - 8) + 4e + 8 = 10e$$

$$ L = 5e $$

$$ \frac{L}{e} = 5$$

We evaluate the left and right behaviors around the origin.
For $$x > 0$$, $$\text{sgn}(x) = 1 \implies f(x) = \frac{\ln(1 + \sin^2 x)}{x}$$.
Using Taylor expansions as $$x \to 0^+$$: $$\ln(1 + \sin^2 x) \approx \sin^2 x \approx x^2$$.
$$RHL = \lim_{x \to 0^+} \frac{x^2}{x} = 0$$.

For $$x < 0$$, $$\text{sgn}(x) = -1 \implies f(x) = \frac{\ln(1 - \sin^2 x)}{x} = \frac{\ln(\cos^2 x)}{x} = \frac{2 \ln(\cos x)}{x}$$.
Using expansions near zero: $$\ln(\cos x) \approx \ln(1 - \frac{x^2}{2}) \approx -\frac{x^2}{2}$$.
$$LHL = \lim_{x \to 0^-} \frac{2(-\frac{x^2}{2})}{x} = \lim_{x \to 0^-} (-x) = 0$$.
Since $$LHL = RHL = f(0) = 0$$, the function is continuous at $$x = 0$$.

Now evaluate differentiability using first principles:
$$RHD = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\ln(1 + \sin^2 h)}{h^2} = 1$$.
$$LHD = \lim_{h \to 0^+} \frac{f(-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{\frac{2\ln(\cos(-h))}{-h} - 0}{-h} = \lim_{h \to 0^+} \frac{2\ln(\cos h)}{h^2}$$.
Using our expansion: $$\lim_{h \to 0^+} \frac{2(-\frac{h^2}{2})}{h^2} = -1$$.
Since $$RHD = 1$$ and $$LHD = -1$$, the derivatives do not match. The function is continuous but not differentiable.

We analyze the differentiability of a sum of absolute value functions.

The individual components are $$g_1(x) = |x - 1|$$, $$g_2(x) = |x - 2|$$, and $$g_3(x) = |x - 3|$$.
A simple absolute value function of the form $$|x - c|$$ is continuous everywhere, but fails to be differentiable precisely at its sharp turning corner point, which occurs where the expression inside the modulus vanishes ($$x = c$$). At this corner, the left-hand derivative and right-hand derivative do not match.

For our composite sum function:
- $$|x - 1|$$ is non-differentiable at $$x = 1$$
- $$|x - 2|$$ is non-differentiable at $$x = 2$$
- $$|x - 3|$$ is non-differentiable at $$x = 3$$

Since the remaining added modulus components are smooth and differentiable linear curves around these individual critical roots, the non-differentiable behavior at each corner point is preserved in the combined sum function.

Therefore, the function $$f(x)$$ fails to be differentiable at exactly three discrete locations:
$$S = \{1, 2, 3\}$$

The total number of elements (cardinality) of the set $$S$$ is 3.

We analyze the points of non-differentiability for the upper boundary envelope of two real-valued functions: a reflected parabola and a shifted absolute value cone.

Let $$g(x) = |x^2 - 4x + 3| = |(x-1)(x-3)|$$ and $$h(x) = c - |x - 2|$$. Both graphs are perfectly symmetric about the vertical line $$x = 2$$. We simplify the algebra by shifting our coordinate frame origin to this line of symmetry using the substitution $$u = x - 2$$:
- $$g(u + 2) = |(u+1)(u-1)| = |u^2 - 1|$$
- $$h(u + 2) = c - |u|$$

Both component functions are even in terms of $$u$$, which means the composite maximum envelope $$f(u+2) = \max\{|u^2 - 1|, c - |u|\}$$ is also symmetric about $$u = 0$$ ($$x = 2$$). Points of non-differentiability on this profile can arise from:
1. Inherent corner roots where either independent curve is itself non-differentiable.
2. Transition corner crossings where the two functions intersect, forcing the upper envelope to switch profiles.

Let us analyze the behavior for the right-hand half ($$u \ge 0$$):
- Component 1: $$g(u) = |u^2 - 1|$$ contains an inherent corner at $$u = 1$$.
- Component 2: $$h(u) = c - u$$ contains an inherent peak corner at $$u = 0$$.

For the active domain region $$u > 0$$, the curve $$g(u)$$ behaves as $$1 - u^2$$ on the sub-interval $$0 \le u \le 1$$ and flips to $$u^2 - 1$$ for $$u > 1$$. The line $$h(u) = c - u$$ is a downward sloping line with a constant derivative slope of $$-1$$.

To minimize the total number of sharp non-differentiable transitions, the linear profile of $$h(u)$$ must cleanly mask the inner corner of $$g(u)$$ at $$u = 1$$. This is achieved when the line $$h(u) = c - u$$ is exactly tangent to the inner inverted segment $$g(u) = 1 - u^2$$ on the open interval $$0 < u < 1$$.

Step 1: Match the derivative slopes to satisfy tangency
$$\frac{d}{du}(1 - u^2) = \frac{d}{du}(c - u)$$
$$-2u = -1 \implies u = \frac{1}{2}$$

This provides our exact point of contact at $$u = \frac{1}{2}$$, which validly lies inside the interval $$(0, 1)$$.

Step 2: Match the function values at the point of contact to solve for $$c$$
$$1 - \left(\frac{1}{2}\right)^2 = c - \left(\frac{1}{2}\right)$$
$$1 - \frac{1}{4} = c - \frac{1}{2}$$
$$\frac{3}{4} = c - \frac{1}{2} \implies c = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$$

Step 3: Verify the total elements when $$c = \frac{5}{4}$$
- At $$u = 0$$, $$h(0) = \frac{5}{4} > g(0) = 1$$. The maximum choice retains the sharp peak at $$u = 0$$ ($$x = 2$$).
- At $$u = 1$$, $$h(1) = \frac{5}{4} - 1 = \frac{1}{4} > g(1) = 0$$. Because the absolute value line sits strictly above the parabola vertex boundary here, it seamlessly overrides and eliminates the inherent non-differentiable corner of $$g(u)$$ at $$u = 1$$.
- For $$u > 1$$, the rising parabola curve $$g(u) = u^2 - 1$$ crosses the falling line $$h(u) = \frac{5}{4} - u$$ at exactly one coordinate point ($$u_0 \approx 1.081$$), creating a single outer transition corner.

By left-to-right even symmetry, the total points where the function is not differentiable reduce to exactly three positions: $$u \in \{0, u_0, -u_0\}$$. Any deviation from this precise value of $$c$$ fails to mask the inner corner or introduces additional crossing points, increasing the size of set $$S$$.

Thus, the minimizing constant is:
$$\alpha = \frac{5}{4}$$

Evaluating the required expression value:
$$4\alpha = 4 \times \frac{5}{4} = 5$$

For

$$x\ne0$$

given function is

$$f(x)=x^p\sin\left(\frac1x\right)$$

Differentiate using product rule.

We know,

$$\frac d{dx}\left(\sin\frac1x\right)=\cos\left(\frac1x\right)\left(-\frac1{x^2}\right)$$

Therefore,

$$f'(x)=\frac d{dx}\left(x^p\right)\sin\left(\frac1x\right)+x^p\frac d{dx}{\left(\sin\frac1x\right)}$$

$$=px^{p-1}\sin\left(\frac1x\right)-x^{p-2}\cos\left(\frac1x\right)$$

Now differentiate again.

Differentiate first term :

$$\frac d{dx}\left[px^{p-1}\sin\left(\frac1x\right)\right]$$

Using product rule,

$$=p(p-1)x^{p-2}\sin\left(\frac1x\right)+px^{p-1}\cos\left(\frac1x\right)\left(-\frac1{x^2}\right)$$

$$=p(p-1)x^{p-2}\sin\left(\frac1x\right)-px^{p-3}\cos\left(\frac1x\right)$$

Now differentiate second term :

$$\frac d{dx}\left[-x^{p-2}\cos\left(\frac1x\right)\right]$$

Again using product rule,

$$=-(p-2)x^{p-3}\cos\left(\frac1x\right)-x^{p-2}\left[-\sin\left(\frac1x\right)\left(-\frac1{x^2}\right)\right]$$

$$=-(p-2)x^{p-3}\cos\left(\frac1x\right)-x^{p-4}\sin\left(\frac1x\right)$$

Now combine both parts.

$$f''(x)=p(p-1)x^{p-2}\sin\left(\frac1x\right)-px^{p-3}\cos\left(\frac1x\right)$$

$$-(p-2)x^{p-3}\cos\left(\frac1x\right)-x^{p-4}\sin\left(\frac1x\right)$$

Group similar terms :

$$f''(x)=\left[p(p-1)x^{p-2}-x^{p-4}\right]\sin\left(\frac1x\right)$$

$$-(2p-2)x^{p-3}\cos\left(\frac1x\right)$$

Now check existence of $$f''(0)$$.

First,

$$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}$$

$$=\lim_{h\to0}h^{p-1}\sin\left(\frac1h\right)$$

Since $$\sin\left(\frac1h\right)$$ is bounded,

$$f'(0)=0 \qquad \text{for } p>1$$

Now,

$$f''(0)=\lim_{h\to0}\frac{f'(h)-f'(0)}{h}$$

Since $$f'(0)=0$$,

$$f''(0)=\lim_{h\to0}\left[p\,h^{p-2}\sin\left(\frac1h\right)-h^{p-3}\cos\left(\frac1h\right)\right]$$

For this limit to exist, both terms must approach zero.

Therefore,

$$p-2>0,\qquad p-3>0$$

which gives

$$p>3$$

and hence

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

Now for continuity of $$f''(x)$$ at $$x=0$$

we need

$$\lim_{x\to0}f''(x)=f''(0)$$

Since $$\sin\left(\frac1x\right)$$ and $$\cos\left(\frac1x\right)$$ are bounded between $$-1$$ and $$1$$,

the powers of $$x$$ must force every term to go to zero.

The powers present are

$$p-2,\quad p-3,\quad p-4$$

The smallest power is

$$p-4$$

For the limit to exist and equal zero,

$$p-4>0$$

$$p>4$$

Therefore,

$$\lim_{x\to0}f''(x)=0=f''(0)$$

Hence,

$$f''(x)$$

is continuous at

$$x=0$$

when

$$\boxed{p>4}$$

To find the value of the given expression, we systematically apply the conditions of continuity at $$x = 1$$ and differentiability at $$x = 2$$.

Step 1: Analyze continuity at x = 1

\For $$f(x)$$ to be continuous at $$x = 1$$, the left-hand limit (LHL), right-hand limit (RHL), and functional value must be equal:
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$

Left-Hand Limit (LHL):

As $$x \to 1^-$$, we use the first branch. Notice that $$x^2 - 3x + 2 = (x-1)(x-2)$$. For $$x < 1$$, both factors are negative, making the product positive, so the absolute value drops directly:

$$\lim_{x \to 1^-} \left( \alpha + \beta (x^2 - 3x + 2) \right) = \alpha + \beta(1 - 3 + 2) = \alpha$$

Right-Hand Limit (RHL) and Functional Value f(1):

As $$x \to 1^+$$, we use the second branch:

$$\lim_{x \to 1^+} \frac{\sin(\pi x)}{x - 1}$$

Let $$x = 1 + h$$, where $$h \to 0^+$$. Substituting this variable gives:

$$\lim_{h \to 0^+} \frac{\sin(\pi(1 + h))}{h} = \lim_{h \to 0^+} \frac{\sin(\pi + \pi h)}{h} = \lim_{h \to 0^+} \frac{-\sin(\pi h)}{h}$$

Multiply the denominator by $$\pi$$ to evaluate the standard limit format:

$$\lim_{h \to 0^+} -\pi \left( \frac{\sin(\pi h)}{\pi h} \right) = -\pi(1) = -\pi$$

Equating the limits yields our first key parameter constraint value:

$$\alpha = -\pi$$

Step 2: Analyze continuity at x = 2

Differentiability at a point requires the function to be continuous at that point first. Therefore, $$f(x)$$ must be continuous at $$x = 2$$:

$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$$

Left-Hand Limit (LHL):

Using the second branch:

$$\lim_{x \to 2^-} \frac{\sin(\pi x)}{x - 1} = \frac{\sin(2\pi)}{2 - 1} = \frac{0}{1} = 0$$

Right-Hand Limit (RHL) and Functional Value f(2):

Using the third branch:

$$\lim_{x \to 2^+} \left( \gamma x^2 + \delta x + 1 \right) = \gamma(2)^2 + \delta(2) + 1 = 4\gamma + 2\delta + 1$$

Equating these two parameters yields our second system relationship:

$$4\gamma + 2\delta + 1 = 0 \implies 4\gamma + 2\delta = -1$$

Step 3: Solve for the parameters and target expression

From Step 2, we have the grouped variable value:

$$4\gamma + 2\delta = -1$$

From Step 1, we have: $$\alpha = -\pi$$

We evaluate the value of the target expression:

$$\text{Value} = 2\alpha + 4\gamma + 2\delta$$

$$\text{Value} = 2(-\pi) + (-1) = -2\pi - 1$$