Calculus Questions

Let $F(x)=f(x)+f(1/x)$, where $f(x)={\int }_1^x\frac{{\log }_{e}t}{1+t}dt$. Then $F(e)$ equals:

The area enclosed between the curves $y^2=x$ and $y=|x|$ is:

$\int \frac{dx}{\cos x+\sqrt{3}\sin x}$ equals:

The solution for $x$ of the equation ${\int }_{\sqrt{2}}^x\frac{dt}{t\sqrt{t^2-1}}=\frac{\pi }{12}$ is:

The value of the integral ${\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$ is:

${\int }_0^{\pi }xf(\sin x)dx$ is equal to:

${\int }_{-\pi /2}^{\pi /2}\left[[x+\pi ]+\cos (x+3\pi )\right]dx$ is equal to (where $[\cdot ]$ denotes the greatest integer function):

The function $f(x)=\frac{x}{2}+\frac{2}{x}$ has a local minimum at:

Angle between the tangents to the curve $y=x^2-5x+6$ at the points $(2,0)$ and $(3,0)$ is:

The set of points where $f(x)=\frac{x}{1+|x|}$ is differentiable is:

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $x$. The maximum area enclosed by the park is:

The value of ${\int }_1^a[x]f^{\prime }(x)dx,a>1$, where $[x]$ denotes the greatest integer not exceeding $x$, is:

The differential equation whose solution is $A{x}^2+B{y}^2=1$, where $A$ and $B$ are arbitrary constants, is of:

If $x^m{y}^n=(x+y{)}^{m+n}$, then $\frac{dy}{dx}$ is:

A particle located at $x=0$ at time $t=0$, starts moving along the positive $x$ -direction with a velocity $v$ that varies as $v=\alpha \sqrt{x}$. The displacement of the particle varies with time as:

A value of $c$ for which the conclusion of Mean Value Theorem holds for the function $f(x)={\log }_{e}x$ on the interval $[1,3]$ is:

1: The set $\{x:f(x)=f^{-1}(x)\}=\{0,-1\}$ where $f(x)=(x+1{)}^2-1,x\ge -1$.
2: $f$ is a bijection.

${\int }_0^{\pi }[\cot x]dx$, where $[\cdot ]$ denotes the greatest integer function, is equal to:

For real $x$, let $f(x)=x^3+5x+1$, then:

The differential equation which represents the family of curves $y=c_1{e}^{c_{2}x}$, where $c_1$ and $c_2$ are arbitrary constants is:

Let $y$ be an implicit function of $x$ defined by $x^{2x}-2x^x\cot y-1=0$. Then $y^{\prime }(1)$ equals:

The area of the region bounded by the parabola $(y-2{)}^2=x-1$, the tangent to the parabola at the point $(2,3)$ and the x-axis is:

Given $P(x)=x^4+a{x}^3+b{x}^2+cx+d$ such that $x=0$ is the only real root of $P^{\prime }(x)=0$. If $P(-1)<P(1)$, then in the interval $[-1,1]$:

The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is:

The domain of the function $f(x)=\frac{1}{\sqrt{|x|-x}}$ is:

The limit ${\lim }_{n\to \infty }\frac{1}{n}\left[{\sec }^2\frac{1}{n^2}+{\sec }^2\frac{4}{n^2}+\cdots +{\sec }^{2}1\right]$ equals:

Consider the function $f(x)=|x-2|+|x-5|,x\in \mathbb{R}$.Statement 1: $f^{\prime }(4)=0$ Statement 2: $f$ is continuous in $[2,5]$, differentiable in $(2,5)$ and $f(2)=f(5)$.

If $g(x)={\int }_0^x{\cos }^{4}tdt$, then $g(x+\pi )$ equals:

If $f:\mathbb{R}\to \mathbb{R}$ is a function defined by $f(x)=[x]\cos \left(\frac{2x-1}{2}\pi \right)$, where $[x]$ denotes the greatest integer function, then $f$ is:

Let $a,b\in \mathbb{R}$ be such that the function $f$ given by $f(x)=\ln |x|+b{x}^2+ax,x\ne 0$ has extreme values at $x=-1$ and $x=2$.Statement 1: $f$ has local maximum at $x=-1$ and at $x=2$.Statement 2: $a=\frac{1}{2}$ and $b=-\frac{1}{4}$.

The population $p(t)$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{dp(t)}{dt}=0.5p(t)-450$. If $p(0)=850$, then the time at which the population becomes zero is:

If the integral $\int \frac{5\tan x}{\tan x-2}dx=x+a\ln |\sin x-2\cos x|+k$, then $a$ is equal to:

For $x\in (0,5\pi /2)$, define $f(x)={\int }_0^{x}t\sin tdt$. Then $f$ has:

The area of the region enclosed by the curves $y=x$, $x=e$, $y=1/x$ and the positive $x$ -axis is:

The values of $p$ and $q$ for which the function
$$f(x)=\{\begin{array}{ll}\frac{\sin (p+1)x+\sin x}{x},& x<0\\ q,& x=0\\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3/2}},& x>0\end{array}$$
is continuous for all $x$ in $\mathbb{R}$, are:

Let $I$ be the purchase value of an equipment and $V(t)$ be the value after it has been used for $t$ years. The value $V(t)$ depreciates at a rate given by differential equation $\frac{dV(t)}{dt}=-k(T-t)$, where $k>0$ is a constant and $T$ is the total life in years of the equipment. Then the scrap value $V(T)$ of the equipment is:

A particle of mass $m$ is at rest at the origin at time $t=0$. It is subjected to a force $F(t)=F_0{e}^{-bt}$ in the $x$ direction. Its speed $v(t)$ is depicted by which of the following curves?

The value of ${\int }_0^{1}8\frac{\ln (1+x)}{1+x^2}dx$ is:

If $\frac{dy}{dx}=y+3>0$ and $y(0)=2$, then $y(\ln 2)$ is equal to:

$\frac{d^{2}x}{d{y}^2}$ equals:

The value of ${\lim }_{x\to 2}\frac{\sqrt{1-\cos \{2(x-2)\}}}{x-2}$:

An object, moving with a speed of $6.25\text{m/s}$, is decelerated at a rate given by:
$$\frac{dv}{dt}=-2.5\sqrt{v}$$
where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be:

Let $f:(-1,1)\to \mathbb{R}$ be a differentiable function with $f(0)=-1$ and $f^{\prime }(0)=1$. Let $g(x)=[f(2f(x)+2){]}^2$. Then $g^{\prime }(0)=$

Let $f:\mathbb{R}\to \mathbb{R}$ be a positive increasing function with ${\lim }_{x\to \infty }\frac{f(3x)}{f(x)}=1$. Then ${\lim }_{x\to \infty }\frac{f(2x)}{f(x)}=$

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function defined by $f(x)=\frac{1}{e^x+2e^{-x}}$.Statement-1: $f(c)=\frac{1}{3}$, for some $c\in \mathbb{R}$.Statement-2: $0<f(x)\le \frac{1}{2\sqrt{2}}$, for all $x\in \mathbb{R}$.

Let $f:\mathbb{R}\to \mathbb{R}$ be defined by $f(x)=\{\begin{array}{ll}k-2x,& \text{if }x\le -1\\ 2x+3,& \text{if }x>-1\end{array}$. If $f$ has a local minimum at $x=-1$, then a possible value of $k$ is:

The equation of the tangent to the curve $y=x+\frac{4}{x^2}$, that is parallel to the $x$ -axis, is:

1: $g\circ f$ is differentiable at $x=0$ and its derivative is continuous at that point, where $f(x)=x|x|$ and $g(x)=\sin x$.
2: $g\circ f$ is twice differentiable at $x=0$.

The area enclosed between the curve $y={\log }_e(x+e)$ and the coordinate axes is:

If the equation $a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x=0,a_1\ne 0,n\ge 2$, has a positive root $x=\alpha$, then the equation $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\cdots +a_1=0$ has a positive root, which is:

The value of ${\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$ is:

Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y=f(x)$, x-axis and the ordinates $x=\pi /4$ and $x=\beta >\pi /4$ is $\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta$. Then $f(\pi /2)$ is:

Let $f:\mathbb{R}\to \mathbb{R}$ be a differentiable function having $f(2)=6,f^{\prime }(2)=\frac{1}{48}$. Then ${\lim }_{x\to 2}\frac{{\int }_6^{f(x)}4t^{3}dt}{x-2}$ equals:

$\int {\left\{\frac{(\log x-1)}{1+(\log x{)}^2}\right\}}^{2}dx$ is equal to:

A spherical iron ball $10\text{ cm}$ in radius is coated with a layer of ice of uniform thickness that melts at a rate of $50{\text{ cm}}^3\text{/min}$. When the thickness of ice is $5\text{ cm}$, then the rate at which the thickness of ice decreases, is:

If $x\frac{dy}{dx}=y(\log y-\log x+1)$, then the solution of the equation is:

The parabolas $y^2=4x$ and $x^2=4y$ divide the square region bounded by the lines $x=4,y=4$ and the coordinate axes into three parts. If $S_1,S_2,S_3$ are respectively the areas of these parts numbered from top to bottom, then $S_1:S_2:S_3$ is:

The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to:

If $I_1={\int }_0^1{2}^{x^2}dx,I_2={\int }_0^1{2}^{x^3}dx,I_3={\int }_1^2{2}^{x^2}dx$ and $I_4={\int }_1^2{2}^{x^3}dx$, then:

If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)|\le (x-y{)}^2$ for $x,y\in \mathbb{R}$ and $f(0)=0$, then $f(1)$ equals:

Let $f$ be differentiable for all $x$. If $f(1)=-2$ and $f^{\prime }(x)\ge 2$ for $x\in [1,6]$, then:

Suppose $f(x)$ is differentiable at $x=1$ and ${\lim }_{h\to 0}\frac{f(1+h)}{h}=5$, then $f^{\prime }(1)$ equals:

Let $\alpha$ and $\beta$ be the distinct roots of $a{x}^2+bx+c=0$, then ${\lim }_{x\to \alpha }\frac{1-\cos (a{x}^2+bx+c)}{(x-\alpha {)}^2}$ is equal to:

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?

The normal to the curve $x=a(\cos \theta +\theta \sin \theta )$, $y=a(\sin \theta -\theta \cos \theta )$ at any point $\theta$ is such that:

If $f(x)=x^n$, then the value of $f(1)-\frac{f^{\prime }(1)}{1!}+\frac{f^{\prime \prime }(1)}{2!}-\frac{f^{\prime \prime \prime }(1)}{3!}+\cdots +(-1{)}^n\frac{f^n(1)}{n!}$ is:

Let $f(x)$ be a function satisfying $f^{\prime }(x)=f(x)$ with $f(0)=1$ and $g(x)$ be a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral ${\int }_0^{1}f(x)g(x)dx$, is:

The area of the region bounded by the curves $y=|x-1|$ and $y=3-|x|$ is:

Let $\frac{d}{dx}F(x)=\frac{e^{\sin x}}{x}$, $x>0$. If ${\int }_1^4\frac{3e^{\sin x^3}}{x}dx=F(k)-F(1)$, then one of the possible values of $k$, is:

${\lim }_{n\to \infty }\left[\frac{1^3+2^3+3^3+\cdots +n^3}{n^4}\right]-{\lim }_{n\to \infty }\left[\frac{1^4+2^4+3^4+\cdots +n^4}{n^5}\right]$ is:

The value of the integral $I={\int }_0^{1}x(1-x{)}^{n}dx$ is:

The value of ${\lim }_{x\to 0}\frac{{\int }_0^{x^2}{\sec }^{2}tdt}{x\sin x}$ is:

If $f(a+b-x)=f(x)$, then ${\int }_a^{b}xf(x)dx$ is equal to:

If $f(y)=e^y$, $g(y)=y$ for $y>0$ and $F(t)={\int }_0^{t}f(t-y)g(y)dy$, then:

If the function $f(x)=2x^3-9a{x}^2+12a^{2}x+1$, where $a>0$, attains its maximum and minimum at $p$ and $q$ respectively such that $p^2=q$, then $a$ equals:

If $f(x)=\{\begin{array}{ll}x{e}^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)},& x\ne 0\\ 0,& x=0\end{array}$ then $f(x)$ is:

Let $f(a)=g(a)=k$ and their $n$ th derivatives $f^n(a),g^n(a)$ exist and are not equal for some $n$. Further if ${\lim }_{x\to a}\frac{f(a)g(x)-f(a)g(a)-g(a)f(x)+g(a)f(a)}{f(x)-g(x)}=4$, then the value of $k$ is:

If ${\lim }_{x\to 0}\frac{\log (3+x)-\log (3-x)}{x}=k$, the value of $k$ is:

${\lim }_{x\to \pi }\frac{\tan [1-\sin (x/2)]}{[\pi -2x{]}^2}$ is:

If $x=e^{y+e^{y+...to\cdot \infty }},$ $x>0$ then $\frac{dy}{dx}$ is

A function $y=f(x)$ has a second order derivative $f^{\prime \prime }(x)=6(x-1).$ If its graph passes through the point $(2,1)$ and at that point the tangent to the graph is $y=3x-5,$ then the function is

Let $f(x)=\frac{1-\tan x}{4x-\pi },x\ne \frac{\pi }{4},x\in [0,\frac{\pi }{2}].$ If $f(x)$ is continuous in $[0,\frac{\pi }{2}],$ then $f(\frac{\pi }{4})$ is