Calculus PYQ — Page 38

Let $y=y(x)$ be a function of $x$ satisfying $y\sqrt{1-{x}^{2}}=k-x\sqrt{1-{y}^{2}}$ where $k$ is a constant and $y(\frac{1}{2})=-\frac{1}{4}$.Then $\frac{dy}{dx}$ at $x=\frac{1}{2}$ , is equal to

Area (in sq. units) of the region outside $\frac{|x|}{2}+\frac{|y|}{3}=1$ and inside the ellipse $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$ is

The area (in sq. units) of the region $A={(x,y):|x|+|y|\leq 1,2{y}^{2}\geq |x|}$

If $y=y(x)$ is the solution of the differential equation $\frac{5+{e}^{x}}{2+y}\cdot \frac{dy}{dx}+{e}^{x}=0$ satisfying $y(0)=1$ then value of $y({\mathrm{log}}_{e}13)$ is

Let $[t]$ denote the greatest integer $\leq t$. If $\lambda \epsilon R-{0,1},\underset{x\rightarrow 0}{\mathrm{lim}}|\frac{1-x+|x|}{\lambda -x+[x]}|=L$, then $L$ is equal to

The area of the region (in sq. units), enclosed by the circle ${x}^{2}+{y}^{2}=2$ which is not common to the region bounded by the parabola ${y}^{2}=x$ and the straight line $y=x$, 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 :

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:

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 y = x³ + x², then dy/dx at x = 1 is:

The area bounded by y = x², x-axis, and x = 2 is:

Let $S$, be the set of all functions $f:[0,1]\rightarrow R$, which are continuous on $[0,1]$, and differentiable on $(0,1)$. Then for every $f$ in $S$, there exists $c\in (0,1)$, depending on $f$, such that.

If a function $f(x)$ defined by $f(x)={\begin{matrix}a{e}^{x}+b{e}^{-x}, & -1\leq x<1 \\ c{x}^{2}, & 1\leq x\leq 3 \\ a{x}^{2}+2cx, & 3<x\leq 4\end{matrix}$be continuous for some $a,b,c\in R$ and ${f}^{'}(0)+{f}^{'}(2)=e$, then the value of $a$ is

The area (in sq. units) of the region enclosed by the curves $y={x}^{2}-1$ and $y=1-{x}^{2}$ is equal to:

Let $f(x)=\int \frac{\sqrt{x}}{{(1+x)}^{2}}dx(x\geq 0)$. Then $f(3)-f(1)$ is equal to :

If the function $f(x)={\begin{matrix}{k}_{1}(x-\pi {)}^{2}-1, & x\leq \pi \\ {k}_{2}\mathrm{cos}x, & x>\pi \end{matrix}$ is twice differentiable, then the ordered pair $({k}_{1},{k}_{2})$ is equal to:

Consider a region $R={(x,y)\in {R}^{2}:{x}^{2}\leq y\leq 2x}$. If a line $y=\alpha$ divides the area of region $R$ into two equal parts, then which of the following is true ?

The position of a moving car at time $t$ is given by $f(t)=a{t}^{2}+bt+c,t>0$, where $a,b\text{and}c$ are real numbers greater than $1$. Then the average speed of the car over the time interval $[{t}_{1},{t}_{2}]$ is attained at the point:

If the function $f$ defined on $(-\frac{1}{3},1/3)$ by $f(x)={\begin{matrix}\frac{1}{x}{log}_{e}(\frac{1+3x}{1-2x}), & \text{when }x\neq 0 \\ k & ,\text{when }x=0\end{matrix}$, is continuous, then $k$ is equal to.

If $f(x)={\begin{matrix}\frac{sin(a+2)x+sinx}{x} & ;x<0 \\ b & ;x=0 \\ \frac{{(x+3{x}^{2})}^{1/3}-{x}^{1/3}}{{x}^{1/3}} & ;x>0\end{matrix}$ is continuous at $x=0$ , then $a+2b$ is equal to:

If $\alpha$ is the positive root of the equation, $p(x)={x}^{2}-x-2=0$, then $\underset{x\rightarrow {\alpha }^{+}}{\mathrm{lim}}\frac{\sqrt{1-\mathrm{cos}p(x)}}{x+\alpha -4}$is equal to

$\underset{x\rightarrow a}{\mathrm{lim}}\frac{{(a+2x)}^{\frac{1}{3}}-{(3x)}^{\frac{1}{3}}}{{(3a+x)}^{\frac{1}{3}}-{(4x)}^{\frac{1}{3}}}(a\neq 0)$ 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

If for all real triplets $(a,b,c),f(x)=a+bx+c{x}^{2};$ then $\int _{0}^{1}f(x)dx$ is equal to: