JEE Main 2016 — Calculus

The value of the integral $\int _{4}^{10}\frac{[{x}^{2}]}{[{x}^{2}-28x+196]+[{x}^{2}]}dx,$ where $[x]$ denotes the greatest integer less than or equal to $x$, is

For $x\in R,f(x)=|\mathrm{log}2-\mathrm{sin}x|$ and $g(x)=f(f(x))$, then

A wire of length $2$ units is cut into two parts which are bent respectively to form a square of side $=x$ units and a circle of radius $=r$ units. If the sum of the areas of the square and the circle so formed is minimum, then

If $f(x)$ is a differentiable function in the interval $(0, \infty )$ such that $f(1)=1$ and $\underset{t\rightarrow x}{\mathrm{lim}}⁡\frac{{t}^{2} f(x)-{x}^{2}f(t)}{t-x}=1,$for each $x>0,$ then $f(\frac{3}{2})$ is equal to

$\underset{x\rightarrow 0}{\mathrm{lim}}⁡\frac{{(1-\mathrm{cos}⁡2x)}^{2}}{2x\mathrm{tan}⁡x-x\mathrm{tan}⁡2x}$is

$\underset{n\rightarrow \infty }{\mathrm{lim}}{(\frac{(n+1) (n+2)\ldots .3n}{{n}^{2n}})}^{\frac{1}{n}}$ is equal to

If the function $f(x)= {\begin{matrix}-x, & x<1 \\ a+{\mathrm{cos}}^{-1}⁡(x+b), & 1 \leq x \leq 2\end{matrix}$ is differentiable at $x=1$, then $\frac{a}{b}$ is equal to

Let $a,b\in R,(a\neq 0).$ If the function $f$, defined as $f(x)={\begin{matrix}\frac{2{x}^{2}}{a},0\leq x<1 \\ a,1\leq x<\sqrt{2} \\ \frac{2{b}^{2}-4b}{{x}^{3}},\sqrt{2}\leq x<8\end{matrix}$,is continuous in the interval $[0, \infty )$, then an ordered pair $(a,b)$ can be

The minimum distance of a point on the curve $y={x}^{2}-4$ from the origin is

Let $f(x)={\mathrm{sin}}^{4}⁡x+{\mathrm{cos}}^{4}⁡x.$ Then, $f$ is an increasing function in the interval:

If $\underset{x\rightarrow \infty }{\mathrm{lim}}⁡{(1+\frac{a}{x}-\frac{4}{{x}^{2}})}^{2x}={e}^{3}$, then $a$ is equal to

The integral $\int \frac{dx}{(1+\sqrt{x})\sqrt{x-{x}^{2}}}$ is equal to

The integral $\int \frac{2{x}^{12}+5{x}^{9}}{{({x}^{5}+{x}^{3}+1)}^{3}}dx$, is equal to

If $\int \frac{dx}{{\mathrm{cos}}^{3}x \sqrt{2\mathrm{sin}2x}}={(\mathrm{tan}x)}^{A}+C{(\mathrm{tan}x)}^{B}+k$, where $k$ is a constant of integration, then $A+B+C$ equals

For $x\in R,x\neq 0,$ if $y(x)$ is a differentiable function such that $x\int _{1}^{x}y(t)dt=(x+1) \int _{1}^{x}ty(t)dt,$ then $y(x)$ equals (where $C$ is a constant)

The area (in $\mathrm{sq}.\mathrm{units}$) of the region ${(x, y):{y}^{2}\geq 2x$ and ${x}^{2}+{y}^{2}\leq 4x, x \geq 0, y \geq 0}$ is

The area (in sq. units) of the region described by $A={(x,y)|y\geq {x}^{2}-5x+4,x+y\geq 1,y\leq 0}$ is

The solution of the differential equation $\frac{dy}{dx}+\frac{y}{2}\mathrm{sec}⁡x=\frac{\mathrm{tan}⁡x}{2y}$, where $0\leq x<\frac{\pi }{2}$ and $y(0)=1$, is given by

Let $P=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}{(1+{\mathrm{tan}}^{2}\sqrt{x} )}^{\frac{1}{2x}}$, then $\mathrm{log}P$ is equal to

If $2\int _{0}^{1}{\mathrm{tan}}^{-1}xdx=\int _{0}^{1}{\mathrm{cot}}^{-1}(1-x+{x}^{2})dx$, then $\int _{0}^{1}{\mathrm{tan}}^{-1}(1-x+{x}^{2})dx$ is equal to

If a curve $y=f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation, $y (1+xy)dx=x dy,$ then $f(-\frac{1}{2})$ is equal to