JEE Main Mathematics — Calculus previous year questions with solutions.
If $y=y(x)$ is the solution of the differential equation, $x\frac{dy}{dx}+2y={x}^{2}$ satisfying $y(1)=1,$ then $y(\frac{1}{2})$ is equal to
The area of the region $A={(x, y): 0\leq y\leq x|x|+1\mathrm{and}-1\leq x\leq 1}$ in sq. units, is
If the function $f(x)={\begin{matrix}a|\pi -x|+1, x\leq 5 \\ b|x-\pi |+3, x>5\end{matrix}$ is continuous at $x=5,$ then the value of $a-b$ is:
$\int \frac{sin\frac{5x}{2}}{sin\frac{x}{2}}dx$, is equal to
If $m$ is the minimum value of $k$ for which the function $f(x)=x\sqrt{kx-{x}^{2}}$ is increasing in the interval $[0,3]$ and $M$ is the maximum value of $f$ in $[0,3]$ when $k=m,$ then the ordered pair $(m, M)$ is equal to:
The value of the integral $\int_{-2}^{2} \frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$ (where $[x]$ denotes the greatest integer less than or equal to x) is
The area (in sq. units) of the region $A={(x,y):{x}^{2} \leq y\leq x+2}$ is
The area (in sq. units) of the region bounded by the curve $x^{2}=4 y$ and the straight line $x=4 y-2$ is :
Let $f(x)={e}^{x}-x$ and $g(x)={x}^{2}-x, \forall x \epsilon R$ . Then the set of all $x \epsilon R$ , where the function $h(x)=(fog)(x)$ is increasing, is:
Let $f(x)$ be a differentiable function such that ${f}^{'}(x)=7-\frac{3}{4}\frac{f(x)}{x}, (x>0)$ and $f(1)\neq 4.$ Then $\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}x f(\frac{1}{x})$
$\int se{c}^{2}x\cdot {\mathrm{cot}}^{\frac{4}{3}}xdx$ is equal to
Let $f:R\rightarrow R$ be a function defined as $f(x)={\begin{matrix} 5, if x\leq 1 \\ a+bx, if 1<x<3 \\ b+5x, if 3\leq x<5 \\ 30, if x\geq 5\end{matrix}$ Then $f$ is:
If $f:R\rightarrow R$ is a differentiable function and $f(2)=6,$ then $\underset{x\rightarrow 2}{lim}{\int }_{6}^{f(x)}\frac{2tdt}{(x-2)}$ is:
The area (in sq. units) of the region $A={(x,y)\in R\times R|0\leq x\leq 3, 0\leq y\leq 4,y\leq {x}^{2}+3x}$ is
Let $f:[-1,3]\rightarrow R$ be defined as $f(x)={\begin{matrix}|x|+[x], \\ x+|x|, \\ x+[x],\end{matrix}\begin{matrix}-1\leq x<1 \\ 1\leq x<2 \\ 2\leq x\leq 3,\end{matrix}$ Where $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is discontinuous at:
If ${e}^{y}+xy=e,$ the ordered pair $(\frac{dy}{dx},\frac{{d}^{2}y}{d{x}^{2}})$ at $x=0$ is equal to
If ${S}_{1}$ and ${S}_{2}$ are respectively the sets of local minimum and local maximum points of the function, $f(x)=9{x}^{4}+12{x}^{3}-36{x}^{2}+25,x\in R,$ then
$\underset{x\rightarrow {1}^{-}}{\mathrm{lim}}\frac{\sqrt{\pi }-\sqrt{2{\mathrm{sin}}^{-1}x}}{\sqrt{1-x}}$ is equal to
Let $f:(-1, 1)\rightarrow R$ be a function defined by $f(x)=max{-|x|, -\sqrt{1-{x}^{2}}}.$ If $K$ be the set of all points at which $f$ is not differentiable, then $K$ has exactly
If $\underset{x\rightarrow 1}{lim}\frac{{x}^{4}-1}{x-1}=\underset{x\rightarrow k}{lim}\frac{{x}^{3}-{k}^{3}}{{x}^{2}-{k}^{2}}$ , then $k$ is
If $\underset{x\rightarrow 1}{\mathrm{lim}}\frac{{x}^{2}-ax+b}{x-1}=5,$ then $a+b$ is equal to:
$\underset{x\rightarrow \frac{\pi }{4}}{lim}\frac{co{t}^{3}x-tanx}{cos(x+\frac{\pi }{4})}$ is
If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx}=(tanx-y){sec}^{2}x$ , $x\in (-\frac{\pi }{2}, \frac{\pi }{2})$ , such that $y(0)=0$, then $y(-\frac{\pi }{4})$ is equal to:
Let $f(x)={\begin{matrix}max(|x|,{x}^{2}), & |x|\leq 2 \\ 8-2|x|, & 2<|x|\leq 4\end{matrix}.$ Let $S$ be the set of points in the interval $(-4,4)$ at which $f$ is not differentiable. Then $S$