Calculus PYQ — Page 42

If $y=y(x)$ is the solution of the differential equation, $x\frac{dy}{dx}+2y={x}^{2}$ satisfying $y(1)=1,$ then $y(\frac{1}{2})$ is equal to

The area of the region $A={(x, y): 0\leq y\leq x|x|+1\mathrm{and}-1\leq x\leq 1}$ in sq. units, is

If the function $f(x)={\begin{matrix}a|\pi -x|+1, x\leq 5 \\ b|x-\pi |+3, x>5\end{matrix}$ is continuous at $x=5,$ then the value of $a-b$ is:

Let $f:[0, 2]\rightarrow R$ be a twice differentiable function such that ${f}^{''}(x)>0,$ for all $x\in [0, 2].$ If $\phi (x)= f(x)+ f(2–x),$ then $\phi$ is

Let $f(x)=15–|x –10|;x\in R.$ Then the set of all values of $x$, at which the function $g(x)=f(f(x))$ is not differentiable, is:

The maximum area (in sq. units) of a rectangle having its base on the $x-$ axis and its other two vertices on the parabola, $y=12-{x}^{2}$ such that the rectangle lies inside the parabola, is :

The maximum volume $(in cu.m)$ of the right circular cone having slant height $3 m$ is:

For, ${x}^{2}\neq n\pi +1,n\in N$ (the set of natural numbers), the integral $\int x\sqrt{\frac{2\mathrm{sin}({x}^{2}-1)-\mathrm{sin}2({x}^{2}-1)}{2\mathrm{sin}({x}^{2}-1)+\mathrm{sin}2({x}^{2}-1)}}dx$, is equal to (where $c$ is a constant of integration).

Let $f$ and $g$ be continuous functions on $[0,a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$, then ${\int }_{0}^{a}f(x)g(x)dx$ is equal to

Let $f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}, x \in \mathbb{R}$ wherea, b and d are non-zero real constants. Then :

Let $S(\alpha )={(x,y):{y}^{2}\leq x, 0\leq x\leq \alpha }$ and $A(\alpha )$ is area of the region $S(\alpha ).$ If for a $\lambda ,0<\lambda <4, A(\lambda ):A(4)=2:5,$then $\lambda$ equals:

For $x>1$, if ${(2x)}^{2y}=4{e}^{2x-2y}$, then ${(1+{\mathrm{log}}_{e}⁡2x)}^{2} \frac{dy}{dx}$ is equal to

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

If $\frac{dy}{dx}+\frac{3}{{\mathrm{cos}}^{2}x}y=\frac{1}{{\mathrm{cos}}^{2}x}$ , $x\in (-\frac{\pi }{3},\frac{\pi }{3}),$ and $y(\frac{\pi }{4})=\frac{4}{3},$ then $y(-\frac{\pi }{4})$ equals

The area (in sq. units) bounded by the parabola $y={x}^{2}-1,$ the tangent at the point $(2,3)$ to it and the $y$-axis is

The area (in sq. units) of the region bounded by the curve $x^{2}=4 y$ and the straight line $x=4 y-2$ is :

A curve amongst the family of curves represented by the differential equation, $({x}^{2}-{y}^{2}) dx+2xy dy=0$ which passes through $(1,1)$, is

If $f(x)=\frac{2-xcosx}{2+xcosx}$ and $g(x)={\mathrm{log}}_{e}⁡x,$ then the value of the integral $\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}g(f(x))dx$ is

$\underset{x\rightarrow 0}{lim} \frac{{sin}^{2}x}{\sqrt{2}-\sqrt{1+cosx}}$ equals

The value of $\int _{-\pi /2}^{\pi /2}\frac{dx}{[x]+[\mathrm{sin}⁡x] + 4},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is

The area (in sq. units) of the region $A={(x,y)\in R\times R|0\leq x\leq 3, 0\leq y\leq 4,y\leq {x}^{2}+3x}$ is

Let $f:[-1,3]\rightarrow R$ be defined as $f(x)={\begin{matrix}|x|+[x], \\ x+|x|, \\ x+[x],\end{matrix}\begin{matrix}-1\leq x<1 \\ 1\leq x<2 \\ 2\leq x\leq 3,\end{matrix}$ Where $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is discontinuous at:

$\int \frac{sin\frac{5x}{2}}{sin\frac{x}{2}}dx$, is equal to

$\underset{x\rightarrow {1}^{-}}{\mathrm{lim}}\frac{\sqrt{\pi }-\sqrt{2{\mathrm{sin}}^{-1}x}}{\sqrt{1-x}}$ is equal to