The area (in sq. units) bounded by the curves $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

If $\int \frac{d x}{\sqrt[3]{\sin ^{11} x \cos x}}=-\left(\frac{3}{8} f(x)+\frac{3}{2} g(x)\right)+c$ then

A random variable X takes values $-1,0,1,2$ with probabilities $\frac{1+3 \mathrm{p}}{4}, \frac{1-\mathrm{p}}{4}, \frac{1+2 \mathrm{p}}{4}, \frac{1-4 \mathrm{p}}{4}$ respectively, where p varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of X are respectively.

Let C be a curve given by $y(x)=1+\sqrt{4 x-3}$, $x>\frac{3}{4}$. If P is a point on C , such that the tangent at P has slope $\frac{2}{3}$, then a point through which the normal at P passes, is