JEE Main 2018 — Calculus

$\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}$ equals.

$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{(27+x)}^{\frac{1}{3}}-3}{9-{(27+x)}^{\frac{2}{3}}}$ equals

For each $t\in R,$ let $[t]$ be the greatest integer less than or equal to $t$. Then$\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}x([\frac{1}{x}]+[\frac{2}{x}]+\ldots +[\frac{15}{x}])$

If a right circular cone, having maximum volume, is inscribed in a sphere of radius $3cm$, then the curved surface area (in ${\mathrm{cm}}^{2}$) of this cone is :

If a right circularcone having maximum volume, is inscribed in a sphere of radius $3 \mathrm{~cm}$, then the curved surface area (in $\mathrm{cm}^2$ ) of this cone is

If $I_1=\int_0^1 e^{-x} \cos ^2 x d x ; I_2=\int_0^1 e^{-x^2} \cos ^2 x d x$ and $I_3=\int_0^1 e^{-x^3} d x$; then

If $x=\sqrt{{2}^{{\mathrm{cosec}}^{-1}t}}$ and $y=\sqrt{{2}^{{\mathrm{sec}}^{-1}t}},(|t|\geq 1),$ then $\frac{dy}{dx}$ is equal to

If $\int \frac{\mathrm{tan}x}{1+\mathrm{tan}x+{\mathrm{tan}}^{2}x}dx=x-\frac{K}{\sqrt{A}}{\mathrm{tan}}^{-1}(\frac{K\mathrm{tan}x+1}{\sqrt{A}})+C$, ($C$ is a constant of integration), then the ordered pair $(K,A)$ is equal to

If $f(x)$ is a quadratic expression such that $f(1)+f$ (2) $=0$, and $-1$ is a root of $f(x)=0$, then the other root of $f(x)=0$ is

If the area of the region bounded by the curves, $y={x}^{2},y=\frac{1}{x}$ and the lines $y=0$ and $x=t(t>1)$ is $1\mathrm{sq}.\mathrm{unit}$, then $t$ is equal to

If the function $f$ defined as $f(x)=\frac{1}{x}-\frac{k-1}{{e}^{2x}-1},x\neq 0$ is continuous at $x=0$, then ordered pair $(k,f(0))$ is equal to

If $f(x)={\int }_{0}^{x}t(\mathrm{sin}x-\mathrm{sin}t)dt$, then

If $f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right)$, then $f^{\prime}\left(-\frac{1}{2}\right)$ equals.

If $f(\frac{x-4}{x+2})=2x+1,(x\in R-{1,-2})$, then$\int f(x)dx$ is equal to

If $f\left(\frac{x-4}{x+2}\right)=2 x+1,(x \in R=\{1,-2\})$, then $\int f$ ( $x$ ) $d x$ is equal to (where $C$ is a constant of integration)

If $x^2+y^2+\sin y=4$, then the value of $\frac{d^2 y}{d x^2}$ at the point $(-2,0)$ is

If ${x}^{2}+{y}^{2}+\mathrm{sin}y=4$, then the value of $\frac{{d}^{2}y}{d{x}^{2}}$ at the point $(-2,0)$ is :

Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum values of the function, $f(x)=2{x}^{3}-9{x}^{2}+12x+5$ in the interval $[0,3]$ . Then $M-m$ is equal to

Let $g(x)=\mathrm{cos}{x}^{2}, f(x)=\sqrt{x},$ and $\alpha ,\beta (\alpha <\beta )$ be the roots of the quadratic equation $18{x}^{2}-9\pi x+{\pi }^{2}=0$. Then the area (in sq. units) bounded by the curve $y=(gof)(x)$ and the lines $x=\alpha ,x=\beta$ and $y=0,$ is

Let $f(x)={x}^{2}+\frac{1}{{x}^{2}}$ and $g(x)=x-\frac{1}{x}, x\in R-{-1, 0, 1}.$ If $h(x)=\frac{f(x)}{g(x)}$ , then the local minimum value of $h(x)$ is:

Let $f(x)$ be a polynomial of degree 4 having extreme values at $x=1$ and $x=2$. If $\lim _{x \rightarrow 0}\left(\frac{f(x)}{x^2}+1\right)=3$ then $f(-1)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+2 y=f(x)$, where $$ f(x)=\left\{\begin{array}{lc} 1, & x \in[0,1] \\ 0, & \text { otherwise } \end{array}\right. $$ If $y(0)=0$, then $y\left(\frac{3}{2}\right)$ is

Let $y=y(x)$ be the solution of the differential equation $\mathrm{sin}⁡x\frac{dy}{dx}+y\mathrm{cos}x⁡=4x, x\in (0, \pi ).$ If $y(\frac{\pi }{2})=0,$ then $y(\frac{\pi }{6})$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+2y=f(x)$, where $f(x)= {\begin{matrix}1, & x\in [0, 1] \\ 0, & otherwise\end{matrix}$. If $y(0)=0$, then $y (\frac{3}{2})$ is

Let $S=\left\{(\lambda, \mu) \in R \times R: f(t)=\left(|\lambda| e^t-\mu\right) \cdot \sin (2|t|)\right.$, $t \in R$, is a differentiable function $\}$. Then $S$ is a subest of?

Let $S={(\lambda ,\mu )\in R\times R :f(t)=(|\lambda |{e}^{|t|}-\mu )\mathrm{sin}⁡(2|t|),t\in R$ is a differential function}. Then, $S$ is a subset of :

Let $f(x)=\left\{\begin{array}{cc}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\ k, & x=2\end{array}\right.$ The value of $k$ for which $f$ is continuous at $x=2$ is

Let $S={t\in R:f(x)=|x-\pi | \cdot ({e}^{|x|}-1)\mathrm{sin}|x| \text{is not differentiable at} t}.$ Then, the set $S$ is equal to:

The area (in sq. units) of the region ${x\in R:x\geq 0,y\geq 0,y\geq x-2$ and $y\leq \sqrt{x}}$ is

The area (in sq. units) of the region $\{x \in R: x \geq 0, y \geq 0, y \geq x-2$ and $y \leq \sqrt{x}\}$, is

The curve satisfying the differential equation, $\left(x^2-y^2\right) d x+2 x y d y=0$ and passing through the point $(1,1)$ is

The integral $\int \frac{{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}{{({\mathrm{sin}}^{5}x+{\mathrm{cos}}^{3}x{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{3}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{5}x)}^{2}}dx$, is equal to (where $C$ is the constant of integration).

The value of integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x$ is

The value of the integral $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x \text { is } $$

The value of the integral $\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\mathrm{sin}}^{4}⁡x(1+\mathrm{ln}⁡(\frac{2+\mathrm{sin}⁡x}{2-\mathrm{sin}⁡x}))dx$ is

The values of $\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}⁡x}{1+{2}^{x}}dx$ is

$$ \int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+C $$ (where $C$ is a constant of integration), then the ordered pair $(A, B)$ is equal to