JEE Main Mathematics — Calculus previous year questions with solutions.
The value of the integral $\int\limits_{-1}^{1} \left(\dfrac{x^3 + |x| + 1}{x^2 + 2|x| + 1}\right) dx$ is equal to :
Let the solution curve of the differential equation $x d y-y d x=\sqrt{x^{2}+y^{2}} d x, x>0$, $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to
Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in \mathbf{R}$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to :
Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^{2}\right) \mathrm{d} y+\left(y-\tan ^{-1} x\right) d x=0, y(0)=1$. Then the value of $y(1)$ is :
If the area of the region $\left\{(x, y): 1-2 x \leqslant y \leqslant 4-x^{2}, x \geqslant 0, y \geqslant 0\right\}$ is $\frac{\alpha}{\beta}, \alpha, \beta \in \mathbf{N}, \operatorname{gcd}(\alpha, \beta)=1$, then the value of $(\alpha+\beta)$ is :
Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in \mathbf{R}$. Consider the following two statements : (I) $f(x)$ is discontinous at $x=1$. (II) $f(x)$ is continous at $x=-1$. Then,
The integral $\int_{0}^{1}\cot^{-1}(1+x+x^2)dx$ is equal to:
Let the line $x=-1$ divide the area of the region $\left\{(x, y): 1+x^{2} \leq y \leq 3-x\right\}$ in the ratio $m: n, \operatorname{gcd}(m, n)=1$. Then $m+n$ is equal to
The area of the region bounded by the curves $x+3y^2=0$ and $x+4y^2=1$ is equal to:
Let $f$ be a real polynomial of degree $n$ such that $f(x) = f'(x) f''(x)$, for all $x \in \mathbb{R}$. If $f(0) = 0$, then $36\left(f'(2) + f''(2) + \int_0^2 f(x)\,dx\right)$ is equal to:
Consider the following three statements for the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=\left|\log _{\mathrm{e}} x\right|-|x-1|$ : (I) $f$ is differentiable at all $x>0$. (II) $f$ is increasing in $(0,1)$. (III) $f$ is decreasing in $(1, \infty)$. Then.
Let $f(x)=x^{2025}-x^{2000}, x \in[0,1]$ and the minimum value of the function $f(x)$ in the interval $[0,1]$ be $(80)^{80}(n)^{-81}$. Then $n$ is equal to
If the area of the region $\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq a+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}\gt0\right\}$ is $\frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}$, then the value of $a$ is :
Let \(y=y(x)\) be the solution of the differential equation \(\cos x\left(\log _{\mathrm{e}}(\cos x)\right)^2 \mathrm{dy}+\left(\sin x-3 y \sin x \log _{\mathrm{e}}(\cos x)\right) \mathrm{d} x=0, x \in\left(0, \frac{\pi}{2}\right)\). If \(y\left(\frac{\pi}{4}\right)=\frac{-1}{\log _{\mathrm{e}} 2}\), then \(y\left(\frac{\pi}{6}\right)\) is equal to :
Let f be a differentiable function on $\mathbf{R}$ such that $\mathrm{f}(2) = 1$, $f^{\prime}(2)=4$. Let $\lim _{x \rightarrow 0}(f(2+x))^{3 / x}=e^\alpha$. Then the number of times the curve $y=4 x^3-4 x^2-4(\alpha-7) x-\alpha$ meets x -axis is :-
If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5, x \geq 1$, then at $x=2$, the value of cosy is:
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a function defined by $f(x)=\|x+2|-2| x\|$. If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$, then $m+n$ is
Let $f(x)=\int x^3 \sqrt{3-x^2} d x$. If $5 f(\sqrt{2})=-4$, then $f(1)$ is equal to
Let \(f:(0, \infty) \rightarrow \mathbf{R}\) be a twice differentiable function. If for some \(\mathrm{a} \neq 0, \int_0^1 f(\lambda x) \mathrm{d} \lambda=\mathrm{a} f(x), f(1)=1\) and \(f(16)=\frac{1}{8}\), then \(16-f^{\prime}\left(\frac{1}{16}\right)\) is equal to _______.
A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of $81 \mathrm{~cm}^3 / \mathrm{min}$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4 \pi} \mathrm{~cm} / \mathrm{min}$. The surface area (in $\mathrm{cm}^2$ ) of the chocolate ball (without the ice-cream layer) is :
Let the domain of the function $f(\mathrm{x})=\log _2 \log _4 \log _6\left(3+4 x-x^2\right)$ be $(\mathrm{a}, \mathrm{~b})$. If $\int_0^{\mathrm{b}-\mathrm{a}}\left[\mathrm{x}^2\right] \mathrm{dx}=\mathrm{p}-\sqrt{\mathrm{q}}-\sqrt{\mathrm{r}}, \mathrm{p}, \mathrm{q},$ $\mathrm{r} \in \mathbb{N}, \operatorname{gcd}(\mathrm{p}, \mathrm{q}, \mathrm{r})=1,$ where [$\cdot]$ is the greatest integer function, then $\mathrm{p}+\mathrm{q}+\mathrm{r}$ is equal to
The area (in sq. units) of the region $\left\{(x, y): 0 \leq \mathrm{y} \leq 2|x|+1,0 \leq \mathrm{y} \leq x^2+1,|x| \leq 3\right\}$ is
The value of $\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x$ is
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a polynomial function of degree four having extreme values at $x=4$ and $x=5$. If $\lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5$, then $f(2)$ is equal to :