Calculus Questions (384)

If $f\left(\frac{xy}{2}\right) = \frac{f(x).f(y)}{2}, \forall x, y \in R, f(1) = f'(1) = 2$. Then, $\frac{f(3)}{f'(3)}$ is_____.

The third derivative of a function $f(x)$ vanishes for all $x$. If $f(0) = 1, f'(1) = 2$ and $f''(1) = -1$, then find $f''(x)$ at $x = 3$.

Let $f(x)$ be a differentiable function, $f(1) = 0, f'(1) = 2$ then the value of $\lim_{x \to 1}\frac{\int_1^x \sin(t(f(t)))dt}{(x-1)^2}$ is_____.

Let $K > 0$ and $\lambda = \lim_{x \to 0} \frac{K\left(1-4\sqrt{K^2-x^2}\right)}{x^2\sqrt{K^2-x^2}}$ is finite then the value of $\lambda K$ is_____.

If the independent variable $x$ is changed to $y$, then the expression $x\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 - \frac{dy}{dx} = 0$ is transformed to $x\frac{d^2x}{dy^2} + \left(\frac{dx}{dy}\right)^2 = k\frac{dx}{dy}$ then $k$ equals.

Let $g(x) = \int \frac{1 + 2\cos x}{(\cos x + 2)^2} dx$ and $g(0) = 0$, then the value of $8g(\pi/2)$ is

$\text{Limit}_{x \to \infty} \frac{\cot^{-1}(\sqrt{x+1}-\sqrt{x})}{\sec^{-1}\left(\frac{2x+1}{x-1}\right)}$ is equal to:

If $\int\left[\left(\frac{x^{-6}-64}{4+2x^{-1}+x^{-2}}\right)\left(\frac{x^2}{4-4x^{-1}+x^{-2}}\right) - \frac{4x^2(2x+1)}{(1-2x)}\right] dx$ is equal to $f(x)$ where $f(1) = 2$ then $f(3) = $

To evaluate $\int \frac{dx}{(x-1)\sqrt{-x^2 + 3x - 2}}$, one of the most suitable substitution could be:

The value of $\lambda$ for which $2\left(\text{Lim}_{x \to 0} f(x^3-x^2)\right) = \lambda \left(\text{Lim}_{x \to 0} f(2x^4-x^3)\right)$ is:

In the interval $(a,b)$ there exists at least one point $c$, for any two differentiable function $f$ and $g$ such that $\begin{vmatrix} f(a) & f(b) \\ \phi(a) & \phi(b) \end{vmatrix} - \lambda^2(b-a)\begin{vmatrix} f(a) & f'(c) \\ \phi(a) & \phi'(c) \end{vmatrix}$, then sum of absolute value of $\lambda$ is_____.

The area bounded by the $x$-axis, part of the curve $y=1+x^{-2}$ and the ordinates $x=1$, $x=2$ is divided into equal parts by the ordinate at $x=a$ such that $a=\dfrac{3+\sqrt{p}}{8}$. The value of $p$ is

If $f(x) = \begin{cases} \frac{\left(\frac{\pi}{2} - \sin^{-1}\left|1-\{x\}^2\right|\right) \sin^{-1}(1-\{x\})}{\sqrt{2}\left(\{x\} - \{x\}^3\right)} & x > 0 \\ k & x = 0 \\ \frac{A\sin^{-1}(1-\{x\})\cos^{-1}(1-\{x\})}{\sqrt{2}\{x\}(1-\{x\})} & x < 0 \end{cases}$ is continuous at $x = 0$, then the value of $A$ is______. (where $\{.\}$ denotes fractional part of $x$).

$\int \frac{(x + \sqrt{1 + x^2})^{15}}{\sqrt{1 + x^2}} dx$ is equal to:

If $9 + f''(x) + f'(x) = x^2 + f^2(x)$, where $f(x)$ is twice differentiable function such that $f''(x) \neq 0 \forall x \in R$ and let $P$ be the point of maxima of $f(x)$ then find the number of tangents which can be drawn from $P$ to the circle $x^2 + y^2 = 9$.

If $\int(\sin(2020x))(\sin^{2018} x) dx$ is equal to $\frac{(\sin(a)).(\sin x)^b}{c} + k$ (where $k$ is integration constant) then $\frac{a+b+c}{3} = $

If $f(x) = \int_0^x [f(t)]^{-1} dt$ and $\int_0^1 [f(x)]^{-1} dx = \sqrt{2}$, then:

If the primitive of the function $f(x) = \frac{x^{2009}}{(1 + x^2)^{1006}}$ w.r.t $x$ is equal to $\frac{1}{n}\left(\frac{x^2}{1 + x^2}\right)^m + C$ then find the value of $(m + n)$ (where $m,n \in \mathbb{N}$)

Let $f: [-1,1] \to \left[-\frac{\pi}{4}, \tan 1 + \tan^{-1}\right]$ defined by $f(x) = \tan x + \tan^{-1} x$ and the derivative of $f^{-1}(x)$ at $x = 0$ is $'k'$ then the value of $\frac{4}{k}$ is_____.

The number of point where $|f(x)| + |x-2| - 1$ is non-differentiable in $x \in (0, 3\pi)$, where $f(x) = \prod_{k=1}^{n}\frac{\left(1+2\cos\left(\frac{2x}{3^k}\right)\right)}{3}$ is_____.

Let $f : R \to R$ is a function satisfying $f(10-x) = f(x)$ and $f(2-x) = f(2+x), \forall x \in R$. If $f(0) = 101$. Then, the minimum possible number of values of $x$ satisfying $f(x) = 101, x \in [0, 25]$ is_____.

Let $f_1(x)$ and $f_2(x)$ be twice differentiable function. Where $F(x) = f_1(x) + f_2(x)$ and $G(x) = f_1(x) - f_2(x)$, $\forall x \in \mathbb{R}$. $f_1(0) = 2$ and $f_2(0) = 1$. If $f_1'(x) = f_2(x)$ and $f_2'(x) = f_1(x)$, $\forall x \in \mathbb{R}$, then the number of solutions of the equation $(F(x))^2 = \frac{9x^4}{G(x)}$ is______.

Let $f(n) = \left[\sqrt{n} + \frac{1}{2}\right]$, where $[.]$ denotes greatest integer function, $\forall n \in \mathbb{N}$. Then $\sum_{n=1}^{\infty} \frac{2^{f(n)} + 2^{-f(n)}}{2^n}$ is equal to______.

If $\displaystyle\int e^{\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)}\cdot\frac{x^4+x^2-1}{x^2}\,dx = f(x)+c$, then $\bigl(f(\sqrt{2})\bigr)^4$ is

Let $P(x_0, y_0)$ be a point on the curve $C : (x^2 - 11)(y + 1) + 4 = 0$ where $x_0, y_0 \in N$. If area of the triangle formed by the normal drawn to the curve 'C' at $P$ and the co-ordinate axes is $\left(\frac{a}{b}\right), a,b \in N$ then the least value of $(a - 6b)$.

$\int\dfrac{x(x\tan^{-1}x+(\ln x)(\ln(\ln x)))+\tan^{-1}x}{(x^3+x)\ln x}=f(x)+c$ where $f(e)=0$. Then $\left[\lim_{x\to 1^+}\dfrac{f(x)}{\tan\left(\frac{\pi x}{2}\right)}+\frac{11}{10}\right]=$

$\displaystyle\int\dfrac{\cos 2x-1}{\cos 2x+1}dx$

Area bounded by $y=\lfloor e^x\rfloor$, $y=\lfloor e^{-x}\rfloor$, and $|x|=1$ is

If $\displaystyle\int e^{\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)}\cdot\frac{x^4+x^2-1}{x^2}\,dx = f(x)+c$, then $\bigl(f(\sqrt{2})\bigr)^4$ is

Area enclosed by the curve $y=\dfrac{1}{1+x^2}$ and the $x$-axis from $x=-\infty$ to $x=+\infty$ is

$$\int \frac{(ax^2 - b)dx}{\sqrt{c^2x^2 - (ax^2 + b)^2}} =$$

Consider curves $C_1: y^2-x=0$; $C_2: y-x^2=0$; $0\leq x\leq\frac{\sqrt{3}}{2}$ and $C_3: y=f(x)$; $f(x)<0$ $\forall x\in\left(0,\frac{\sqrt{3}}{2}\right)$. From any point $P$ on $C_2$, lines parallel to coordinate axes intersect $C_1$ at $Q$ and $C_3$ at $R$. If area of region $OPRO$ is twice the area of region $OPQO$ (O = origin), then $\left|32f\!\left(\frac{1}{2}\right)\right|$ is

Value of $\displaystyle\int_0^1 x^6(x^3-1)^{2022}\,dx$ is

If $\int \frac{dx}{\sqrt[3]{(x - 1)^3(x + 2)^2}} = k\sqrt[3]{\frac{x - 1}{x + 2}} + c$, then $3k$ is equal to

The set of all real values of $x$ where $f(x)=\sin|x|-|x|+2(x-\pi)\cos|x|$ is not differentiable is

If $I=\int(x^2+1)\left((x+1)e^x\right)^2dx=A(f(x))^2+C$, where $C$ is constant and $f(-1)=\dfrac{2}{e}$, then $2A+f(0)$ is

Let $f(x) = x^3 - \frac{3x^2}{2} + x + \frac{1}{4}$. Then the value of $\left[\int_{1/4}^{3/4} f(f(x))dx\right]^{-1}$ is _____.

Which must be true for: I) For $x\in\mathbb{R}$, let $\tan^{-1}x\in(-\pi/2,\pi/2)$. Then minimum value of $f(x)=\int_0^{x\tan^{-1}x}\frac{e^{t-\cos t}}{1+t^{2023}}dt$ is zero. II) If $f(x)$ is periodic with period $T$, then $\int_{a+nT}^{b+nT}f(x)dx=\int_a^b f(x)dx$ for $n\in\mathbb{Z}$. III) If $\psi(x)\le\phi(x)$ for $a\le x\le b$, then $\int_a^b|\psi(x)|dx\le\int_a^b|\phi(x)|dx$.

The intercepts on the x-axis made by tangents to the curve $y=\displaystyle\int_0^x|t|\,dt$, $x\in\mathbb{R}$, which are parallel to $y=2x$, are equal to

Let $f(x) = x + \sin x$. Suppose $g$ denotes the inverse function of $f$. If the value of $g'\left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right)$ is $l$ then $2l = $

Let $e$ be the eccentricity of a hyperbola and $f(e)$ be the eccentricity of its conjugate hyperbola. Then $\int(f(e)+f(f(e)))de=g(e)$ and $g(\sqrt{2})=2$, then $g(e)=$

Let $I_1=\int_0^x e^{tx}\cdot e^{-t^2}\,dt$ and $I_2=\int_0^x e^{-t^2/4}\,dt$ where $x>0$. Then the value of $\dfrac{I_1}{I_2}$ is

$\displaystyle\int_1^{81}\dfrac{dx}{\sqrt{x}+\sqrt[4]{x}}$

Consider: I) If $f(x)$ is bounded for $x\in[a,b]$, then area bounded by $y=f(x)$, $x$-axis, $x=a$ and $x=b$ is $\int_a^b f(x)dx$. II) If $f(x)$ is bounded and differentiable on $[a,b]$, then the area between $y=f(x)$ and $y=f^{-1}(x)$ is equal to double the value of $\int_a^b|f(x)-x|dx$.

If $x\displaystyle\int_0^x f(t)\,dt = (x+1)\int_0^x tf(t)\,dt$ for $x>0$, and $f(1)=\dfrac{1}{e}$, then $f(-1)$ is

Value of $\displaystyle\int_0^1 \frac{\sin x}{x}\,dx$ lies in the interval

$\int_{-5}^{5}\dfrac{x^{10}}{x^{10}+(5-x)^{10}}\cdot\dfrac{dx}{x^2+16}$

Let $y=f(x)$ be a differentiable function satisfying $\int_2^x f(t)\,dt+2=\dfrac{x^2}{2}+\int_x^2 t^2 f(t)\,dt$. Then $\int_{-\pi/4}^{\pi/4}\dfrac{f(x)+x^9-x^3+x+1}{\cos^2 x}\,dx=$

$\displaystyle\int_0^{10}[x^2]dx$ (where $[.]$ is GIF) equals

$\displaystyle\int\sqrt{\dfrac{x}{1-x^3}}dx=$