Calculus PYQ — Page 15

Let $y=y(x)$ be the solution of the differential equation $\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0$ and $y\left(e^{-1}\right)=0$. Then, $y(e)$ is equal to

$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to

Let $S=(-1,\infty )$ and $f:S\rightarrow \mathbb{R}$ be defined as $f(x)=\int _{-1}^{x}{({e}^{t}-1)}^{11}{(2t-1)}^{5}{(t-2)}^{7}{(t-3)}^{12}{(2t-10)}^{61}dt$. Let $p=$ Sum of square of the values of $x$, where $f(x)$ attains local maxima on $S$. and $q=$Sum of the values of $x$, where $f(x)$ attains local minima on $S$. Then, the value of ${p}^{2}+2q$ is ________

The area (in sq. units) of the region described by $\left\{(x, y): y^2 \leq 2 x\right.$, and $\left.y \geq 4 x-1\right\}$ is

A function $y=f(x)$ satisfies $f(x)\mathrm{sin}2x+\mathrm{sin}x-(1+{\mathrm{cos}}^{2}x){f}^{'}(x)=0$ with condition $f(0)=0$. Then $f(\frac{\pi }{2})$ is equal to

The parabola $y^2=4 x$ divides the area of the circle $x^2+y^2=5$ in two parts. The area of the smaller part is equal to:

The integral $\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$ is equal to :

If the function $f(x)={\begin{matrix}\frac{1}{|x|}, & |x|\geq 2 \\ a{x}^{2}+2b, & |x|<2\end{matrix}$ is differentiable on $R$, then $48(a+b)$ is equal to _______.

Let $y={\mathrm{log}}_{e}(\frac{1-{x}^{2}}{1+{x}^{2}}),-1<x<1$. Then at $x=\frac{1}{2}$, the value of $225({y}^{'}-{y}^{"})$ is equal to

Let $f(x)=\int_0^x\left(t+\sin \left(1-e^{\prime}\right)\right) d t, x \in \mathbb{R}$. Then, $\lim _{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to

$|\frac{120}{{\pi }^{3}}\int _{0}^{\pi }\frac{{x}^{2}\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx|$ is equal to ______.

If ${\int }_{-\pi /2}^{\pi /2}\frac{8\sqrt{2}\mathrm{cos}xdx}{(1+{e}^{\mathrm{sin}x})(1+{\mathrm{sin}}^{4}x)}=\alpha \pi +\beta {\mathrm{log}}_{e}(3+2\sqrt{2})$, where $\alpha ,\beta$ are integers, then ${\alpha }^{2}+{\beta }^{2}$ equals __________

If $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{a{x}^{2}{e}^{x}-b{\mathrm{log}}_{e}(1+x)+cx{e}^{-x}}{{x}^{2}\mathrm{sin}x}=1$, then $16({a}^{2}+{b}^{2}+{c}^{2})$ is equal to ______.

If $x=x(t)$ is the solution of the differential equation $(t+1)dx=(2x+{(t+1)}^{4})dt,x(0)=2$, then $x(1)$ equals ________

Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y-y={Y}^{'}(x)(X-x)$ and the co-ordinate axes, where $(x,y)$ is any point on the curve, is always $\frac{-{y}^{2}}{2{Y}^{'}(x)}+1,{Y}^{'}(x)\neq 0$. If $Y(1)=1$, then $12Y(2)$ equals ________.

Let $f(x)={\int }_{0}^{x}g(t){\mathrm{log}}_{e}(\frac{1-t}{1+t})\mathrm{dt}$, where $g$ is a continuous odd function. If ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f(x)+\frac{{x}^{2}\mathrm{cosx}}{1+{e}^{x}})\mathrm{dx}={(\frac{\pi }{\alpha })}^{2}-\alpha$, then $\alpha$ is equal to _____.

The derivative of sin(x) with respect to x is:

$\underset{x\rightarrow \frac{\pi }{2}}{\mathrm{lim}}(\frac{1}{{(x-\frac{\pi }{2})}^{2}}{\int }_{{x}^{3}}^{{(\frac{\pi }{2})}^{3}}\mathrm{cos}(\frac{1}{{t}^{3}})dt)$ is equal to

Let $f(x)=\left\{\begin{array}{lr}-2, & -2 \leq x \leq 0 \\ x-2, & 0 < x \leq 2\end{array}\right.$ and $h(x)=f(|x|)+|f(x)|$. Then $\int_{-2}^2 h(x) \mathrm{d} x$ is equal to :

Let $f:R-{0}\rightarrow R$ be a function satisfying $f(\frac{x}{y})=\frac{f(x)}{f(y)}$ for all $x,y,f(y)\neq 0$. If ${f}^{'}(1)=2024$, then

Consider the function $f:(0,\infty )\rightarrow R$ defined by $f(x)={e}^{-|{\mathrm{log}}_{e}x|}$. If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m+n$ is

Let $\int_\alpha^{\log _e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}$. Then $\mathrm{e}^\alpha$ and $\mathrm{e}^{-\alpha}$ are the roots of the equation :

For $0<a<1$, the value of the integral ${\int }_{0}^{\pi }\frac{dx}{1-2a\mathrm{cos}x+{a}^{2}}$ is :

Let $f:\rightarrow R\rightarrow (0,\infty )$ be strictly increasing function such that $\underset{x\rightarrow \infty }{\mathrm{lim}}\frac{f(7x)}{f(x)}=1$. Then, the value of $\underset{x\rightarrow \infty }{\mathrm{lim}}[\frac{f(5x)}{f(x)}-1]$ is equal to