JEE Main 2011 — Calculus

The value of $\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x$ is

The shortest distance between line $y-x=1$ and curve $x=y^2$ is

Let I be the purchase value of an equipment and $\mathrm{V}(\mathrm{t})$ be the value after it has been used for t years. The value $\mathrm{V}(\mathrm{t})$ depreciates at a rate given by differential equation $\frac{\mathrm{dV}(\mathrm{t})}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{t})$, where $\mathrm{k}>0$ is a constant and $\mathrm{T}$ is the total life in years of the equipment. Then the scrap value $\mathrm{V}(\mathrm{T})$ of the equipment is

$$ \lim _{x \rightarrow 2}\left(\frac{\sqrt{1-\cos \{2(x-2)\}}}{x-2}\right)

The value of $p$ and $q$ for which the function $f(x)=\left\{\begin{array}{cl}\frac{\sin (p+1) x+\sin x}{x} & x < 0 \\ q & , x=0 \\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} & , x>0\end{array}\right.$ is continuous for all $\mathrm{x}$ in $\mathrm{R}$, is

$\frac{d^2 x}{d y^2}$ equals

For $x \in\left(0, \frac{5 \pi}{2}\right)$, define $f(x)=\int_0^x \sqrt{t} \sin t d t$. Then $f$ has

If $\frac{d y}{d x}=y+3>0$ and $y(0)=2$, then $y(\ln 2)$ is equal to

The area of the region enclosed by the curves $y=x, x=e, y=\frac{1}{x}$ and the positive $x$-axis is