Calculus PYQ — Page 40

The solution of the differential equation $\frac{dy}{dx}-\frac{y+3x}{{\mathrm{log}}_{e}(y+3x)}+3=0$ is (where $C$ is a constant of integration)

Let $y=y(x)$ be the solution of the differential equation, $x{y}^{'}-y={x}^{2}(x\mathrm{cos}x+\mathrm{sin}x)$,$x>0$. If $y(\pi )=\pi ,$ then $y''(\frac{\pi }{2})+y(\frac{\pi }{2})$ is equal to :

Let $y=y(x)$ be a solution of the differential equation, $\sqrt{1-{x}^{2}}\frac{dy}{dx}+\sqrt{1-{y}^{2}}=0,|x|<1.$ If $y(\frac{1}{2})=\frac{\sqrt{3}}{2},$ then $y(\frac{-1}{\sqrt{2}})$ is equal to

Let $[t]$ denote the greatest integer less than or equal to $t$. Then the value of ${\int }_{1}^{2}|2x-[3x]|dx$ is

The derivative of ${\mathrm{tan}}^{-1}(\frac{\sqrt{1+{x}^{2}}-1}{x})$ with respect to ${\mathrm{tan}}^{-1}(\frac{2x\sqrt{1-{x}^{2}}}{1-2{x}^{2}})$ at $x=\frac{1}{2}$ is :

If $p(x)$ be a polynomial of degree three that has a local maximum value $8$ at $x=1$ and a local minimum value $4$ at $x=2$ then $p(0)$ is equal to

If ${\theta }_{1}$ and ${\theta }_{2}$ be respectively the smallest and the largest values of $\theta$ in $(0,2\pi )-{\pi }$ which satisfy the equation, $2{\mathrm{cot}}^{2}⁡\theta -\frac{5}{\mathrm{sin}⁡\theta }+4=0$ , then $\int _{{\theta }_{1}}^{{\theta }_{2}}{\mathrm{cos}}^{2}⁡3\theta d\theta$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $\mathrm{cos}x\frac{dy}{dx}+2y\mathrm{sin}x=\mathrm{sin}2x,x\in (0,\frac{\pi }{2})$ If $y(\pi /3)=0,$ then $y(\pi /4)$ is equal to :

The integral $\int \frac{dx}{{(x+4)}^{\frac{8}{7}}{(x-3)}^{\frac{6}{7}}}$ is equal to: (where $C$ is a constant of integration)

Let $y=y(x)$ be the solution of the differential equation, $\frac{2+\mathrm{sin}x}{y+1}.\frac{dy}{dx}=-\mathrm{cos}x$, $y>0,y(0)=1$. If $y(\pi )=a$ and $\frac{dy}{dx}$ at $x=\pi$ is $b$, then the ordered pair $(a,b)$ is equal to

The general solution of the differential equation $\sqrt{1+{x}^{2}+{y}^{2}+{x}^{2}{y}^{2}}+xy\frac{dy}{dx}=0$ (where C is a constant of integration)

The function $f(x)={\begin{matrix}\frac{\pi }{4}+{\mathrm{tan}}^{-1}x, & |x|\leq 1 \\ \frac{1}{2}(|x|-1), & |x|>1\end{matrix}$ is :

If ${f}^{'}(x)={\mathrm{tan}}^{-1}⁡(\mathrm{sec}⁡x+\mathrm{tan}⁡x),-\frac{\pi }{2}<x<\frac{\pi }{2}$ and $f(0)=0$ , then $f(1)$ is equal to:

The integral $\int {(\frac{x}{x\mathrm{sin}x+\mathrm{cos}x})}^{2}dx$ is equal to, (where $C$ is a constant of integration):

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

$\underset{x\rightarrow 0}{lim}\frac{{\int }_{0}^{x}tsin(10t)dt}{x}$, is equal to

The area (in sq. units) of the region $A={(x,y):(x-1)[x]\leq y\leq 2\sqrt{x},0\leq x\leq 2},$ where $[t]$ denotes the greatest integer function, is :

Let $f(x)=x{\mathrm{cos}}^{-1}⁡(-\mathrm{sin}⁡|x|),x\in [-\frac{\pi }{2},\frac{\pi }{2}],$ then which of the following is true?

The integral ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\mathrm{tan}}^{3}x\cdot {\mathrm{sin}}^{2}3x(2{\mathrm{sec}}^{2}x\cdot {\mathrm{sin}}^{2}3x+3\mathrm{tan}x\cdot \mathrm{sin}6x)dx$ is equal to:

Given: $f(x)={\begin{matrix}\begin{matrix}x,0\leq x<\frac{1}{2} \\ \frac{1}{2},x=\frac{1}{2}\end{matrix} \\ 1-x,\frac{1}{2}<x\leq 1\end{matrix}$ and $g(x)={(x-\frac{1}{2})}^{2},x\in R.$ Then, the area (in sq. units) of the region bounded by the curves, $y=f(x)$ and $y=g(x)$ between the lines $2x=1$ and $2x=\sqrt{3},$ is:

If $\int \frac{d\theta }{{\mathrm{cos}}^{2}\theta (\mathrm{tan}2\theta +\mathrm{sec}2\theta )}=$ $\lambda \mathrm{tan}\theta +2{\mathrm{log}}_{e}|f(\theta )|+C$ where $C$ is a constant of integration, then the ordered pair $(\lambda ,f(\theta ))$ is equal to:

If $\frac{dy}{dx}=\frac{xy}{{x}^{2}+{y}^{2}};y(1)=1;$ then a value of $x$ satisfying $y(x)=e$ is:

Let $f:R\rightarrow R$ be a function defined by $f(x)=\mathrm{max}{x,{x}^{2}}$.Let $S$ denote the set of all points in $R$,where $f$ is not differentiable.Then :

Let $f(x)=|x-2|$ and $g(x)=f(f(x)),x\in [0,4]$. Then ${\int }_{0}^{3}(g(x)-f(x))dx$ is equal to