JEE Main Mathematics — Calculus previous year questions with solutions.
The value of $\int _{-\pi /2}^{\pi /2}\frac{dx}{[x]+[\mathrm{sin}x] + 4},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is
If $\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C,$ for a suitable chosen integer $\mathrm{m}$ and a function $\mathrm{A}(\mathrm{x})$, where $\mathrm{C}$ is a constant of integration, then $(\mathrm{A}(\mathrm{x}))^{\mathrm{m}}$ equals :
For $x>1$, if ${(2x)}^{2y}=4{e}^{2x-2y}$, then ${(1+{\mathrm{log}}_{e}2x)}^{2} \frac{dy}{dx}$ is equal to
Let $y=y(x)$ be the solution of the differential equation, ${({x}^{2}+1)}^{2} \frac{dy}{dx}+2x({x}^{2}+1)y=1$ such that $y(0)=0.$ If $\sqrt{a} y(1)=\frac{\pi }{32},$ then the value of $a$ is
The value of $\int _{0}^{2\pi }[\mathrm{sin}2x(1+\mathrm{cos}3x)]dx$ , where $[t]$ denotes the greatest integer function is
A spherical iron ball of radius $10cm$ is coated with a layer of ice of uniform thickness that melts at a rate of $50c{m}^{3}/min.$ When the thickness of the ice is $5cm,$ then the rate at which the thickness $($in $cm/min)$ of the ice decreases, is :
The region represented by $|x-y|\leq 2$ and $|x+y|\leq 2$ is bounded by a
$\lim _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^{2} x \cot ^{2}(2 x)}$ is equal to:
If $x=3 tant$ and $y=3sect,$ then the value of $\frac{{d}^{2}y}{d{x}^{2}}$ at $t=\frac{\pi }{4},$ is:
Let $S(\alpha )={(x,y):{y}^{2}\leq x, 0\leq x\leq \alpha }$ and $A(\alpha )$ is area of the region $S(\alpha ).$ If for a $\lambda ,0<\lambda <4, A(\lambda ):A(4)=2:5,$then $\lambda$ equals:
If $\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=f(x) \sqrt{2 x-1}+\mathrm{C},$ where $\mathrm{C}$ is a constant of integration, then $f(x)$ is equal to:
A curve amongst the family of curves represented by the differential equation, $({x}^{2}-{y}^{2}) dx+2xy dy=0$ which passes through $(1,1)$, is
If $\int _{0}^{\pi /3}\frac{\mathrm{tan}\theta }{\sqrt{2k \mathrm{sec}\theta }}d\theta =1-\frac{1}{\sqrt{2}}, (k>0)$ , then the value of $k$ is
If $f(x)=\frac{2-xcosx}{2+xcosx}$ and $g(x)={\mathrm{log}}_{e}x,$ then the value of the integral $\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}g(f(x))dx$ is
If the area enclosed between the curves $y=k{x}^{2}$ and $x=k{y}^{2},(k>0),$ is $1sq.unit.$ Then $k$ is
The integral $\int _{1}^{e}{{(\frac{x}{e})}^{2x}-{(\frac{e}{x})}^{x}}lo{g}_{e} x dx$ is equal to
If $f(x)=[x]-[\frac{x}{4}], x\in R,$ where $[x]$ denotes the greatest integer function, then:
$\underset{x\rightarrow 0}{lim} \frac{{sin}^{2}x}{\sqrt{2}-\sqrt{1+cosx}}$ equals
Let, $n\geq 2$ be a natural number and $0<\theta <\frac{\pi }{2}.$ Then $\int \frac{{(si{n}^{n}\theta -sin\theta )}^{\frac{1}{n}}cos\theta }{si{n}^{n+1}\theta }d\theta ,$ is equal to
The solution of the differential equation $x\frac{dy}{dx}+2y={x}^{2},(x\neq 0)$ with $y(1)=1$, is
Let $y=y(x)$ be the solution of the differential equation, $\frac{dy}{dx}+y\mathrm{tan}x=2x+{x}^{2}\mathrm{tan}x,x\in (-\frac{\pi }{2},\frac{\pi }{2}),$ such that $y(0)= 1.$ Then
If $f(x)=\int \frac{(5{x}^{8}+7{x}^{6})}{{({x}^{2}+1+2{x}^{7})}^{2}}dx$, $(x\geq 0),$ and $f(0)=0,$ then the value of $f(1)$ is
Let $f(x)=\int _{0}^{x}g(t)dt,$ where $g$ is a non-zero even function. If $f(x+5)=g(x),$ then $\int _{0}^{x}f(t)dt$ equals
Let $I={\int }_{a}^{b}({x}^{4}-2{x}^{2})dx.$ If $I$ is minimum then the ordered pair $(a, b)$ is