MCQ
$\int {\left( {{{\sin }^4}x - {{\cos }^4}x} \right)\,dx = } $
- A$ - \frac{{\cos 2x}}{2} + c$
- ✓$ - \frac{{\sin 2x}}{2} + c$
- C$\frac{{\sin 2x}}{2} + c$
- D$\frac{{\cos 2x}}{2} + c$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If $f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$
is a continuous function then the value of $k$ is
$f(x)=\min \{x-[x], 1+[x]-x\}$
where $[\mathrm{x}]$ is the greatest integer less than or equal to $\mathrm{x}$. Let $\mathrm{P}$ denote the set containing all $x \in[0,3]$ where $f$ is discontinuous, and $Q$ denote the set containing all $x \in(0,3)$ where $f$ is not differentiable. Then the sum of number of elements in $\mathrm{P}$ and $\mathrm{Q}$ is equal to $......$