If $f(t)=\int_0^t \tan ^{(2 n-1)} x d x, n \in N$, then $f(t+\pi)=$
$f(t) f(\pi)$
$f(t)-f(\pi)$
$f(t)+f(\pi)$
$\frac{f(t)}{f(\pi)}$
$$ \int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{\frac{5}{2}} d x= $$
$\frac{2^{15}}{63}$
$\frac{2^{16}}{315}$
$\frac{2^{16}}{189}$
$\frac{2^{10}}{63}$
The area (in sq. units) of the region bounded by the curves $y=x^2$ and $y=8-x^2$ is
$\frac{32}{3}$
$\frac{16}{3}$
$\frac{64}{3}$
$\frac{128}{3}$
The solution of the differential equation $x^2(y+1) \frac{d y}{d x}+y^2(x+1)^2=0$, when $y(1)=2$, is
$\log \left|x^2 y\right|=\frac{2}{x}+\frac{1}{y}+x-1$
$\log \left|\frac{1}{4} x^2 y\right|=\frac{1}{x}+\frac{2}{y}+x-1$
$\log \left|\frac{1}{2} x^2 y\right|=\frac{1}{x}+\frac{1}{y}-x-\frac{1}{2}$
$\log \left|\frac{1}{3} x^2 y\right|=\frac{1}{x}+\frac{1}{y}-x+\frac{1}{2}$
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