JEE Main Mathematics — Calculus previous year questions with solutions.
Let $f$ and $g$ be continuous functions on $[0,a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$, then ${\int }_{0}^{a}f(x)g(x)dx$ is equal to
For each $t\in R,$ let $[t]$ be the greatest integer less than or equal to $t$. Then, $\underset{x\rightarrow {1}^{+}}{lim}\frac{(1-|x|+sin|1-x|)sin([1-x]\frac{\pi }{2})}{|1-x|[1-x]}$
For, ${x}^{2}\neq n\pi +1,n\in N$ (the set of natural numbers), the integral $\int x\sqrt{\frac{2\mathrm{sin}({x}^{2}-1)-\mathrm{sin}2({x}^{2}-1)}{2\mathrm{sin}({x}^{2}-1)+\mathrm{sin}2({x}^{2}-1)}}dx$, is equal to (where $c$ is a constant of integration).
Let $\mathrm{K}$ be the set of all real values of $x$ where the function $f(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$ is not differentiable. Then the set $K$ is equal to :
Let $x, y$ be positive real numbers and $m, n$ positive integers. The maximum value of the expression $\frac{x^{\mathrm{m}} y^{\mathrm{n}}}{\left(1+x^{2 \mathrm{~m}}\right)\left(1+y^{2 \mathrm{n}}\right)}$ is :
A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is ${tan}^{-1}(\frac{1}{2}).$ Water is poured into it at a constant rate of $5cubicm/min.$ Then the rate $($in $m/min),$ at which the level of water is rising at the instant when the depth of water in the tank is $10 m;$ is:
Let $f:R\rightarrow R$ be a differentiable function satisfying ${f}^{'}(3)+{f}^{'}(2)=0.$ Then $\underset{x\rightarrow 0}{lim}{(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)})}^{\frac{1}{x}}$ is equal to
Let $y=y(x)$ be the solution of the differential equation, $x\frac{dy}{dx}+y=x{\mathrm{log}}_{e}x, (x>1)$. If $2y(2)={\mathrm{log}}_{e}4-1$, then $y(e)$ is equal to
$\underset{x\rightarrow 0}{lim}\frac{x+2sinx}{\sqrt{{x}^{2}+2sinx+1} - \sqrt{{sin}^{2}x-x+1}}$ is
The integral $\int cos(\mathrm{lnx})dx$, is equal to
The integral ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{sec}^{\frac{2}{3}}x\cdot cose{c}^{\frac{4}{3}}xdx$ is equal to
The solution of the differential equation, $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^{2}$, when $y(1)=1,$ is:
If $\int \frac{dx}{{({x}^{2}-2x+10)}^{2}}=A({\mathrm{tan}}^{-1}(\frac{x-1}{3})+\frac{f(x)}{{x}^{2}-2x+10})+C$, then (where $C$ is a constant of integration)
The integral $\int \frac{3{x}^{13}+2{x}^{11}}{{(2{x}^{4}+3{x}^{2}+1)}^{4}}dx$, is equal to
If $\int {x}^{5}{e}^{-{x}^{2}}dx=g(x){e}^{-{x}^{2}}+c$, where $c$ is a constant of integration, then $g(-1)$ is equal to
If the function $f$ defined on $(\frac{\pi }{6},\frac{\pi }{3})$ by $f(x)={\begin{matrix}\frac{\sqrt{2}cosx-1}{cotx-1},x\neq \frac{\pi }{4} \\ k, x=\frac{\pi }{4}\end{matrix}$ is continuous, then $k$ is equal to
If $2y={({\mathrm{cot}}^{-1}(\frac{\sqrt{3}\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\sqrt{3}\mathrm{sin}x}))}^{2}\forall x\in (0,\frac{\pi }{2})$, then $\frac{dy}{dx}$is equal to
The maximum area (in sq. units) of a rectangle having its base on the $x-$ axis and its other two vertices on the parabola, $y=12-{x}^{2}$ such that the rectangle lies inside the parabola, is :
Let $f(x)=15–|x –10|;x\in R.$ Then the set of all values of $x$, at which the function $g(x)=f(f(x))$ is not differentiable, is:
Let $f:[0, 2]\rightarrow R$ be a twice differentiable function such that ${f}^{''}(x)>0,$ for all $x\in [0, 2].$ If $\phi (x)= f(x)+ f(2–x),$ then $\phi$ 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 :
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
The value of the integral $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x \text { is } $$
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