Calculus Questions (384)

If $P = \int_0^\infty \frac{x^2}{1+x^4} dx$; $Q = \int_0^\infty \frac{xdx}{1+x^4}$ and $R = \int_0^\infty \frac{dx}{1+x^4}$, then:

Let $u = \int_0^{\pi/4} \left(\frac{\cos x}{\sin x + \cos x}\right)^2 dx$ and $v = \int_0^{\pi/4} \left(\frac{\sin x + \cos x}{\cos x}\right)^2 dx$, then:

Let $g(x) = x^c e^{2x}$ & let $f(x) = \int_0^x e^{2t}(3t^2+1)^{1/2} dt$. For a certain value of 'c', the limit of $\frac{f'(x)}{g'(x)}$ as $x \to \infty$ is finite and non-zero, then:

If $\int_0^2 \frac{\ln(1+2x)}{1+x^2} dx = (\tan^{-1}a)(\ln\sqrt{b})$ where $a,b \in \mathbb{N}$, then:

Which of the following is/are true?

Let $f$ be continuous and differentiable in $(x_1, x_2)$. If $f(x)f'(x) \geq x\sqrt{1-[f(x)]^4}$ and $\lim_{x\to x_1}(f(x))^2=1$, $\lim_{x\to x_2}(f(x))^2=\frac{1}{2}$. Then minimum value of $\left[x_1^2 - x_2^2\right]$ is ........... (where $[\cdot]$ denotes GIF)

Let $f$ be a function defined on $(-\pi/2, \pi/2)$ as follows: $f(x) = \begin{cases} \frac{2^{[1/n]} - [x] - \frac{[x]}{[n2-1]}}{x\tan x} & x \neq 0 \\ k & x = 0 \end{cases}$. The value of $k$ so that $f$ is continuous at $x = 0$ is:

The function $f(x) = [x] + \sqrt{\{x\}}$, where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part function respectively, is discontinuous at

Define $f: [0, \pi] \to \mathbb{R}$ by $f(x) = \begin{cases} \tan^2 x + \sqrt{2\sin^2 x + 3\sin x + 4 - \sqrt{\sin^2 x + 6\sin x + 2}} & x \neq \pi/2 \\ k & x = \pi/2 \end{cases}$ is continuous at $x = \pi/2$, then $k$ is equal to:

Let $f: R \to R$ be defined by $f(x) = \begin{cases} x + 2x^2 \sin \frac{1}{x} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$ then

Let $f(x) = \lim_{n \to \infty} \frac{2x^{2n} \sin + x}{1 + x^{2n}}$ then which of the following alternative(s) is/are correct?

Assume that $\lim_{\theta \to 1} f(0)$ exists and $\frac{\theta^2 + 0 - 2}{\theta + 3} \leq \frac{f(0)}{\theta^2} \leq \frac{\theta^2 + 20 - 1}{\theta + 3}$ holds for certain interval containing the point $\theta = -1$ then $\lim_{\theta \to 1} f(0)$ and $\lim_{\theta \to 1} \frac{f(0)}{\theta^2}$ is :

Let $f(x) = \begin{cases} \frac{\tan^2 [x]}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{[x]} \cot [x] & \text{for } x < 0 \end{cases}$ where $[x]$ is the step up function and $\{x\}$ is the fractional part function of $x$, then :

$\lim_{x \to c} f(x)$ does not exist when:

The function, $f(x) = [x] - [[x]]$, where $[ ]$ denotes greatest integer function:

The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

$f$ is a continuous function in $[a, b]$; $g$ is a continuous function in $[b, c]$. A function $h(x)$ is defined as: $h(x) = f(x)$ for $x \in [a, b]$ $= g(x)$ for $x \in [b, c]$ if $f(b) = g(b)$, then

In which of the following cases the given equations has atleast one root in the indicated interval?

If $f(x) = \begin{cases} \frac{x \cdot \ln(\cos x)}{\ln(1+x^2)} & x \neq 0 \\ 0 & x = 0 \end{cases}$ then:

Which of the following limits vanish?

Let $f(x) = |x - 1|([x] - [-x])$, then which of the following statement(s) is/are correct. (where $[.]$ denotes greatest integer function.)

If $y = f(x)$ defined parametrically by $x = 2t - |t - 1|$ and $y = 2t^2 + t|t|$, then:

$\lim f(x)$ does not exist when (where $[x]$ denotes the greatest integer less than or equal to $x$)

The function $f(x) = x^2 \left[x^2 - \frac{1}{x^2}\right], x \neq 0$ is ($[x]$ represents the greatest integer $\leq x$)

A function is defined as $f(x) = [\tan x] + \sqrt{\tan x - [\tan x]}$ $0 \leq x 2 \\ 5x - 7 & \text{if } x \leq 2 \end{cases}$ then:

If $F(x) = f(x)g(x)$ and $f'(x)g'(x) = c$, then (where $f$ and $g$ are thrice differentiable)

$f(x)$ is defined for $x \geq 0$ and has a continuous derivative. It satisfies $f(0) = 1, f'(0) = 0$ and $(1 + f(x))f''(x) = 1 + x$. The values $f(1)$ can't take is/are:

P and Q are two points on a circle of centre C and radius $a$, the angle PCQ being 20 then the radius of the circle inscribed in the triangle CPQ is maximum when

Let a function $f$ be defined as $f(x) = \begin{cases} \frac{|x-1|}{x^2+1} & \text{if } x > -1 \\ x^2 & \text{if } x \leq -1 \end{cases}$. Then the number of critical point(s) on the graph of this function is/are:

Consider $f(x) = \int \left(t + \frac{1}{t}\right) dt$ and $g(x) = f'(x)$ for $x \in \left[-3, -\frac{1}{2}\right]$. If P is a point on the curve $y = g(x)$ such that the tangent to this curve at P is parallel to a chord joining the points $\left(\frac{1}{2}, g\left(\frac{1}{2}\right)\right)$ and $(3, g(3))$ of the curve, then the coordinates of the point P

The integral $\int \frac{\sec^{3/2}\theta - \sec^{1/2}\theta}{2 + \tan^2\theta} \tan\theta d\theta$ is:

If $f(x) = \lim_{n \to \infty} \frac{x^n - x^{-n}}{x^n + x^{-n}}$, $0 < x < 1$, $n \in \mathbb{N}$ then $\int (\sin^{-1} x) f'(x) dx$ is equal to:

$$\int \frac{x^4 - 2}{x^2\sqrt{x^4 + x^2 + 2}} dx =$$

Anti-derivative of $$\frac{x - 1}{(x + 1)\sqrt{x^3 + x^2 + x}}$$ is:

The derivative of $x^4 + x^{-5}$ is $-\left(4x^{-5} + 5x^{-6}\right)$. So, $$\int \frac{5x^3 + 4x^5}{\left(x^5 + x + 1\right)^2} dx =$$

If $f'(x) = \frac{1}{-x + \sqrt{x^2 + 1}}$ and $f(0) = -\frac{1 + \sqrt{2}}{2}$, then the value of $f(5)$ will be :

Let $f(xy) = f(x) \cdot f(y)$, $\forall x > 0, y > 0$ and $f(1 + x) = 1 + x[1 + g(x)]$, where $\lim_{x \to 0} g(x) = 0$, then $$\int \frac{f(x)}{f'(x)} dx$$ is:

If $$\int \frac{dx}{x^2(x^n + 1)^{(n-1)/n}} = -[f(x)]^{1/n} + C$$, then $f(x)$ is

If $I_{m,n} = \int \cos^m x \sin nx dx$, then $7I_{4,3} - 4I_{3,2} =$$

$I = \int \frac{dx}{(\sin x - 2\cos x)(2\cos x + \sin x)}$ is equal to

If $\int \sqrt{\cos ecx + 1}\,dx = kfog(x) + c$, where $k$ is a real constant, then :

Let $f(x) = \frac{1}{4 - 3\cos^2 x + 5\sin^2 x}$ and its anti-derivative $F(x) = \frac{1}{3}\tan^{-1}(g(x)) + c$, then :

A function $f(x)$ continuous on $\mathbb{R}$ and periodic with $2\pi$ satisfies $f(x) + (\sin x) f(x + \pi) = \sin^2 x$ then,

$$\int \frac{dx}{\prod_{i=0}^{n}(x+r)} \text{ is equal to:}$$

$I_1 = \int f(x) dx$ and $I_2 = \int_0^1 f(x) dx$ where $f(x) = x^2 \ln\left(1-x^2\right)$, then:

$f(x) = \int e^{\tan^{-1}x}\left(1+x+x^2\right)d\left(\cot^{-1}x\right)$ is equal to:

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x+2y) = f(x)e^{2y} + f(2y)e^x + x^2\left(1-e^{2y}\right) + 4y^2\left(1-e^x\right) + 4xyy$ for all $x, y \in \mathbb{R}$ and $f'(0) = 1$, then:

If $\int\left[\frac{1}{1-x^8}\left\{\cos^{-1}\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right\}\right]dx$ has $p\tan^{-1}f(x)$ & $q\tan^{-1}g(x)$ terms and $x \in (-1, 1)$, then: (where $p$ and $q$ are constant)

If $A = \int e^{ax}\cos bx dx$ and $B = \int e^{ax}\sin bx dx$, then which of the following may be correct?

If $\int \sqrt{\frac{\cos x - \cos^3 x}{\left(1-\cos^3 x\right)}} dx = f(x) + c$, then $f(x)$ is equal to: