Calculus PYQ — Page 39

The solution curve of the differential equation, $(1+{e}^{-x})(1+{y}^{2})\frac{dy}{dx}={y}^{2}$ which passes through the point $(0,1)$, is

The set of all real values $\lambda$ for which the function $f(x)=(1-{\mathrm{cos}}^{2}x).(\lambda +\mathrm{sin}x),x\epsilon (-\frac{\pi }{2},\frac{\pi }{2})$, has exactly one maxima and exactly one minima, is :

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:

If for $x\geq 0,y=y(x)$ is the solution of the differential equation, $(x+1)dy=({(x+1)}^{2}+y-3)dx,y(2)=0$ then $y(3)$ is equal to ________

If $x=1$ is a critical point of the function $f(x)=(3{x}^{2}+ax-2-a){e}^{x},$ then

Let $y=y(x)$ be the solution curve of the differential equation, $({y}^{2}-x)\frac{dy}{dx}=1$ , satisfying $y(0)=1$ . This curve intersects the $X-$axis at a point whose abscissa is

Let $f(x)=x\cdot [\frac{x}{2}],$ for $-10<x<10,$ where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f(x)$ is equal to

Let $[t]$ denote the greatest integer $\leq t$ and $\underset{x\rightarrow 0}{\mathrm{lim}}⁡x[\frac{4}{x}]=A.$ Then the function, $f(x)=[{x}^{2}]\mathrm{sin}⁡(\pi x)$ is discontinuous, when $x$ is equal to:

If $y(\alpha )=\sqrt{2(\frac{tan\alpha +cot\alpha }{1+ta{n}^{2}\alpha })+\frac{1}{si{n}^{2}\alpha }},\alpha \in (\frac{3\pi }{4},\pi )$, then $\frac{dy}{d\alpha }$ at $\alpha =\frac{5\pi }{6}$ is

Let ${x}^{k}+{y}^{k}={a}^{k},(a,k>0)$ and $\frac{dy}{dx}+{(\frac{y}{x})}^{\frac{1}{3}}=0,$ then $k$ is

If the surface area of a cube is increasing at a rate of $3.6c{m}^{2}/sec$, retaining its shape; then the rate of change of its volume (in $c{m}^{3}/sec$), when the length of a side of the cube is $10cm$, is:

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

$\underset{x\rightarrow 0}{\mathrm{lim}}{(\mathrm{tan}(\frac{\pi }{4}+x))}^{1/x}$ is equal to

Let $f(x)$, be a polynomial of degree $3$, such that $f(-1)=10, f(1)=-6, f(x)$, has a critical point at $x=-1$ and $f'(x)$, has a critical point at $x=1.$ Then $f(x)$, has local minima at $x=$

Let $f:(0,\infty )\rightarrow (0,\infty )$ be a differentiable function such that $f(1)=e$ and $\underset{t\rightarrow x}{\mathrm{lim}}\frac{{t}^{2}{f}^{2}(x)-{x}^{2}{f}^{2}(t)}{t-x}=0.$ If$f(x)=1$, then $x$ is equal to:

If$\int \frac{\mathrm{cos}\theta }{5+7\mathrm{sin}\theta -2{\mathrm{cos}}^{2}\theta }d\theta ={Alog}_{e}|B(\theta )|+C,$ where $C$ is a constant of integration, then $\frac{B(\theta )}{A}$ can be:

If $\int ({e}^{2x}+2{e}^{x}-{e}^{-x}-1){e}^{({e}^{x}+{e}^{-x})}dx=g(x){e}^{({e}^{x}+{e}^{-x})}+c,$ where $c$ is a constant of integration, then $g(0)$ is

If $\int {\mathrm{sin}}^{-1}(\frac{\sqrt{x}}{1+x})dx=A(x){\mathrm{tan}}^{-1}(\sqrt{x})+B(x)+C$, where $C$ is a constant of integration, then the ordered pair $(A(x),B(x))$ can be :

Suppose $f(x)$ is a polynomial of degree four having critical points at $-1,0,1$. If $T={x\in R|f(x)=f(0)}$, then the sum of squares of all the elements of $T$ is :

Let ${x}$ and $[x]$ denote the fractional part of $x$ and the greatest integer $\leq x$ respectively of a real number $x$. if ${\int }_{0}^{n}{x}dx,{\int }_{0}^{n}[x]dx$ and $10({n}^{2}-n),(n\in N,n>1)$ are three consecutive terms of a G.P. then $n$ is equal to__

The area (in sq. units) of the region ${(x,y):0\leq y\leq {x}^{2}+1,0\leq y\leq x+1,\frac{1}{2}\leq x\leq 2}$ is

If $(a+\sqrt{2}b\mathrm{cos}x)(a-\sqrt{2}b\mathrm{cos}y)={a}^{2}-{b}^{2}$, where $a>b>0,$ then$\frac{dx}{dy}$ at $(\frac{\pi }{4},\frac{\pi }{4})$ is:

If $f(a+b+1-x)=f(x),$ for all $x,$ where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{a+b}\int _{a}^{b}x(f(x)+f(x+1))dx$ is equal to

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 :