JEE Main Mathematics — Calculus previous year questions with solutions.
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
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
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:
$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{(27+x)}^{\frac{1}{3}}-3}{9-{(27+x)}^{\frac{2}{3}}}$ equals
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?
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
If $f(x)={\int }_{0}^{x}t(\mathrm{sin}x-\mathrm{sin}t)dt$, then
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 $y=y(x)$ 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
The area (in sq. units) of the region ${(x, y):x\geq 0, x+y\leq 3, {x}^{2}\leq 4y and y\leq 1+\sqrt{x}}$ is
$\underset{x\rightarrow 3}{\mathrm{lim}}\frac{\sqrt{3x}-3}{\sqrt{2x-4}- \sqrt{2}}$ is equal to
The function $f$ defined by $f(x)={x}^{3}-3{x}^{2}+5x+7$ is:
If $y={[x+\sqrt{{x}^{2}-1}]}^{15}+{[x-\sqrt{{x}^{2}-1}]}^{15}$, then $({x}^{2}-1)\frac{{d}^{2}y}{d{x}^{2}}+x\frac{dy}{dx}$ is equal to
The value of $k$ which the function $f(x)= {\begin{matrix}{(\frac{4}{5})}^{\frac{\mathrm{tan}4x}{\mathrm{tan}5x}}, & 0<x<\frac{\pi }{2} \\ k+\frac{2}{5}, & x=\frac{\pi }{2}\end{matrix}$ is continuous at $x=\frac{\pi }{2},$ is
The integral $\int _{\frac{\pi }{12}}^{\frac{\pi }{4}}\frac{8\mathrm{cos}2x}{{(\mathrm{tan}x+\mathrm{cot}x)}^{3}}dx$ equals
If $\int _{1}^{2}\frac{dx}{{({x}^{2}-2x+4)}^{\frac{3}{2}}}=\frac{k}{k+5}$, then $k$ is equal to
If for $x\in (0,\frac{1}{4}),$ the derivative of ${\mathrm{tan}}^{-1}(\frac{6x\sqrt{x}}{1-9{x}^{3}})$ is $\sqrt{x} \cdot g(x)$ , then $g(x)$ equals:
The integral $\int \sqrt{1+2\mathrm{cot}x(\mathrm{cosec}x+\mathrm{cot}x)}dx, (0<x<\frac{\pi }{2})$is equal to
The integral $\int _{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{1+\mathrm{cos}x}$ is equal to
The curve satisfying the differential equation, $ydx-(x+3{y}^{2})dy=0$ and passing through the point $(1,1)$ also passes through the point
If $(2+\mathrm{sin}x)\frac{dy}{dx}+(y+1)\mathrm{cos}x=0$ and $y(0)=1$, then $y(\frac{\pi }{2})$ is equal to
Let, ${I}_{n}=\int {\mathrm{tan}}^{n}xdx(n>1)$ . If ${I}_{4}+{I}_{6}=a{\mathrm{tan}}^{5}x+b{x}^{5}+c$, then the ordered pair $(a,b)$, is equal to
$\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}\frac{\mathrm{cot}x-\mathrm{cos}x}{{(\pi -2x)}^{3}}$ equals
If $2x={y}^{\frac{1}{5}}+{y}^{-\frac{1}{5}}$ and $({x}^{2}-1)\frac{{d}^{2}y}{d{x}^{2}}+\lambda x\frac{dy}{dx}+ky=0$, then $\lambda +k$ is equal to