JEE Main 2013 — Calculus

A spherical balloon is being inflated at the rate of $35 \mathrm{cc} / \mathrm{min}$. The rate of increase in the surface area (in $\mathrm{cm}^2 / \mathrm{min}$.) of the balloon when its diameter is $14 \mathrm{~cm}$, is :

At present, a firm is manufacturing $2 00 0$ items. It is estimated that the rate of change of production $P$ w.r.t. additional number of workers $x$ is given by $\frac{dP}{dx}=100-12\sqrt{x}$. If the firm employs $2 5$ more workers, then the new level of production of items is

Consider the differential equation : $$ \frac{d y}{d x}=\frac{y^3}{2\left(x y^2-x^2\right)} $$ Statement-1: The substitution $z=y^2$ transforms the above equation into a first order homogenous differential equation. Statement-2: The solution of this differential equation is $y^2 e^{-y^2} / x=C$.

Consider the function : $f(x)=[x]+|1-x|,-1 \leq x \leq 3$ where $[x]$ is the greatest integer function. Statement 1: $f$ is not continuous at $x=0,1,2$ and 3 Statement 2:f(x)= $=\begin{array}{cc}-x, & -1 \leq x < 0 \\ 1-x, & 0 \leq x < 1 \\ 1+x, & 1 \leq x < 2 \\ 2+x, & 2 \leq x \leq 3\end{array}$

For $a>0, t \in\left(0, \frac{\pi}{2}\right)$, let $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$, Then, $1+\left(\frac{d y}{d x}\right)^2$ equals :

For $0 \leq x \leq \frac{\pi}{2}$, the value of $$ \int_0^{\sin ^2 x} \sin ^{-1}(\sqrt{t}) d t+\int_0^{\cos ^2 x} \cos ^{-1}(\sqrt{t}) d t \text { equals : } $$

If $f(x)=\sin (\sin x)$ and $f^{\prime \prime}(x)+\tan x f^{\prime}(x)+g(x)$ $=0$, then $g(x)$ is :

If the integral $$ \int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x=A \cos 8 x+k $$ where $k$ is an arbitrary constant, then $\mathrm{A}$ is equal to:

If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8 \mathrm{~cm}^2 / \mathrm{s}$, then the rate of change of its volume is:

If $y=\mathrm{sec}({\mathrm{tan}}^{-1}x)$, then $\frac{dy}{dx}$ at $x=1$ is equal to

If $\int \frac{d x}{x+x^7}=p(x)$ then, $\int \frac{x^6}{x+x^7} d x$ is equal to:

If $\int \frac{x^2-x+1}{x^2+1} e^{\cot ^{-1} x} d x=A(x) e^{\cot ^{-1} x}+C$, then $A(x)$ is equal to :

If $\int f(x)dx=\psi (x)$, then $\int {x}^{5}f({x}^{3})dx$, is equal to

Let $f(1)=-2$ and $f^{\prime}(x) \geq 4.2$ for $1 \leq x \leq 6$. The possible value of $f(6)$ lies in the interval :

Let $f(x)=-1+|x-2|$, and $g(x)=1-|x|$; then the set of all points where $f_{o g}$ is discontinuous is :

Let $f$ be a composite function of $x$ defined by $f(u)=\frac{1}{u^2+u-2}, u(x)=\frac{1}{x-1}$. Then the number of points $x$ where $f$ is discontinuous is :

Let $f:[-2,3] \rightarrow[0, \infty)$ be a continuous function such that $f(1-x)=f(x)$ for all $x \in[-2,3]$. If $\mathrm{R}_1$ is the numerical value of the area of the region bounded by $y=f(x), x=-2, x=3$ and the axis of $x$ and $R_2=\int_{-2}^3 x f(x) d x$, then :

Let $f(x)=\frac{x^2-x}{x^2+2 x} x \neq 0,-2$. Then $\frac{d}{d x}\left[f^{-1}(x)\right]$ (wherever it is defined) is equal to:

Statement - I : The value of the integral $\int _{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\mathrm{tan}x}}$ is equal to $\frac{\pi }{6}.$ Statement - II : $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx.$

Statement-1: The equation $x \log x=2-x$ is satisfied by at least one value of $x$ lying between 1 and 2. Statement-2: The function $f(x)=x \log x$ is an increasing function in $[1,2]$ and $g(x)=2-x$ is a decreasing function in $[1,2]$ and the graphs represented by these functions intersect at a point in $[1,2]$

Statement-1: The function $x^2\left(e^x+e^{-x}\right)$ is increasing for all $x>0$. Statement-2: The functions $x^2 e^x$ and $x^2 e^{-x}$ are increasing for all $x>0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b)$.

The area bounded by the curve $y=\ln (x)$ and the lines $y=0, y=\ln (3)$ and $x=0$ is equal to:

The area (in square units) bounded by the curves $y=\sqrt{x},2y-x+3=0$, $X$-axis and lying in the first quadrant is

The area of the region (in sq. units), in the first quadrant bounded by the parabola $y=9 x^2$ and the lines $x=0, y=1$ and $y=4$, is :

The area under the curve $y=|\cos x-\sin x|$, $0 \leq x \leq \frac{\pi}{2}$, and above $x$-axis is :

The cost of running a bus from $A$ to $B$, is $₹\left(a v+\frac{b}{v}\right)$, where $v \mathrm{~km} / \mathrm{h}$ is the average speed of the bus. When the bus travels at $30 \mathrm{~km} / \mathrm{h}$, the cost comes out to be $₹ 75$ while at $40 \mathrm{~km} / \mathrm{h}$, it is $₹ 65$. Then the most economical speed (in $\mathrm{km} / \mathrm{h}$ ) of the bus is :

The equation of the curve passing through the origin and satisfying the differential equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2$ is

The integral $\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}$ equals :

The integral $\int_{7 \pi / 4}^{7 \pi / 3} \sqrt{\tan ^2 x} d x$ is equal to :

The maximum area of a right angled triangle with hypotenuse $h$ is :

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x$ is :

The value of $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{(1-\mathrm{cos}2x)(3+\mathrm{cos}x)}{x\mathrm{tan}4x}$ is equal to

The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\frac{\pi}{4}\right]$ is :