MCQ
$\int \limits_{-\pi}^{\pi}|\pi-| x || d x$ is equal to :
- ✓$\pi^{2}$
- B$2 \pi^{2}$
- C$\sqrt{2} \pi^{2}$
- D$\frac{\pi^{2}}{2}$
$=2 \int_{0}^{\pi}(\pi- x ) d x$
$=2\left[\pi x -\frac{ x ^{2}}{2}\right]_{0}^{\pi}=\pi^{2}$
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$g(x)=\left\{\begin{array}{cl}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\1, & x=-1\end{array} \text { and } h(x)=2[x]-f(x),\right.$
where $[x]$ is the greatest integer $\leq x$. Then the value of $\lim _{x \rightarrow 1} g(h(x-1))$ is
Statement-$2$ : ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right]$ = ${\tan ^{ - 1}}\,1 + \,{\tan ^{ - 1}}\left( {\log {x^2}} \right)$