NTA JEE Main 16th March 2021 Shift 2

Let $$f : S \to S$$ where $$S = (0, \infty)$$ be a twice differentiable function such that $$f(x+1) = xf(x)$$. If $$g : S \to R$$ be defined as $$g(x) = \log_e f(x)$$, then the value of $$|g''(5) - g''(1)|$$ is equal to:

Let $$f$$ be a real valued function, defined on $$R - \{-1, 1\}$$ and given by $$f(x) = 3\log_e\left|\frac{x-1}{x+1}\right| - \frac{2}{x-1}$$. Then in which of the following intervals, function $$f(x)$$ is increasing?

Consider the integral $$I = \int_0^{10} \frac{[x]e^{[x]}}{e^{x-1}}dx$$ where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. Then the value of $$I$$ is equal to:

Let $$P(x) = x^2 + bx + c$$ be a quadratic polynomial with real coefficients such that $$\int_0^1 P(x)dx = 1$$ and $$P(x)$$ leaves remainder 5 when it is divided by $$(x-2)$$. Then the value of $$9(b+c)$$ is equal to:

If $$y = y(x)$$ is the solution of the differential equation $$\frac{dy}{dx} + (\tan x)y = \sin x$$, $$0 \leq x \leq \frac{\pi}{3}$$, with $$y(0) = 0$$, then $$y\left(\frac{\pi}{4}\right)$$ is equal to:

Let $$C_1$$ be the curve obtained by the solution of differential equation $$2xy\frac{dy}{dx} = y^2 - x^2$$, $$x > 0$$. Let the curve $$C_2$$ be the solution of $$\frac{2xy}{x^2-y^2} = \frac{dy}{dx}$$. If both the curves pass through $$(1, 1)$$, then the area (in sq. units) enclosed by the curves $$C_1$$ and $$C_2$$ is equal to:

Let $$\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$$ and $$\vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$$. If $$\vec{r} \times \vec{a} = \vec{b} \times \vec{r}$$, $$\vec{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$$ and $$\vec{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$$, $$\alpha \in R$$, then the value of $$\alpha + |\vec{r}|^2$$ is equal to:

If $$(x, y, z)$$ be an arbitrary point lying on a plane $$P$$ which passes through the point $$(42, 0, 0)$$, $$(0, 42, 0)$$ and $$(0, 0, 42)$$, then the value of expression $$3 + \frac{x-11}{(y-19)^2(z-12)^2} + \frac{y-19}{(x-11)^2(z-12)^2} + \frac{z-12}{(x-11)^2(y-19)^2} - \frac{x+y+z}{14(x-11)(y-19)(z-12)}$$ is

If the foot of the perpendicular from point $$(4, 3, 8)$$ on the line $$L_1: \frac{x-a}{l} = \frac{y-3}{3} = \frac{z-b}{4}$$, $$l \neq 0$$ is $$(3, 5, 7)$$, then the shortest distance between the line $$L_1$$ and line $$L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$ is equal to:

Let $$A$$ denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event $$A$$ is equal to: