$$ \int_0^1 \frac{x^4+1}{x^6+1} d x= $$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$
$\frac{\pi}{2}$
The area of the region (in sq units) bounded by the curves $x^2+y^2=16$ and $y^2=6 x$ is
$4 \pi+4 \sqrt{3}$
$\frac{2}{3}(4 \pi+\sqrt{3})$
$\frac{4}{3}(4 \pi+\sqrt{3})$
$\frac{4 \pi+\sqrt{3}}{3}$
If $a$ and $b$ are arbitrary constants, then the differential equation corresponding to the family of curves $y=\tan (a x+b)$ is
$\left(1+x^2\right) y_2-2 y y_1+y=0$
$\left(1+y^2\right) y_2-2 y y_1^2=0$
$\left(1+x^2\right) y_2+2 y y_1^2=0$
$\left(1+y^2\right) y_2-2 y y_1^2+y=0$
The general solution of the differential equation $x y(y+2) d y+\left(y^3-1\right) d x=0$ is
$\log |x+2 y|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{y-x}{\sqrt{3} x}\right)=C$
$\log |2 x-y|+\frac{2}{3} \tan ^{-1}\left(\frac{x-y}{\sqrt{3} x}\right)=C$
$\log |x y-x|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right)=C$
$\log |x+y|+\frac{2}{3} \tan ^{-1}\left(\frac{x-2 y}{\sqrt{3 x}}\right)=C$
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