Assertion (A) If $y=f(x)=(|x|-|x-1|)^2$, then $\left(\frac{d y}{d x}\right)_{x=1}=1$
Reason (R) $\mathop {\lim }\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$ exist, then it is called derivative of $f(x)$ at $x=a$.
If $y=|\cos x-\sin x|+|\tan x-\cot x|$, then
$$ \left(\frac{d y}{d x}\right)_{x=\frac{\pi}{3}}+\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}}= $$
If the tangent drawn at the point $(\alpha, \beta)$ on the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$ is parallel to the line $\sqrt{3 x}+y=1$, then $\alpha^2+\beta^2=$
The displacement $S$ of a particle measured from a fixed point $O$ on a line is given by $S=t^3-16 t^2+64 t-16$. Then, the time at which displacement of the particle is maximum is
AP EAPCET Papers
All year-wise previous year question papers