- A$\frac{{10}}{{11}}.\frac{8}{9}.\frac{6}{7}.\frac{4}{5}.\frac{2}{3}$
- B$\frac{{10}}{{11}}.\frac{8}{9}.\frac{6}{7}.\frac{4}{5}.\frac{2}{3}.\frac{\pi }{2}$
- C$1$
- ✓$0$
therefore $\int_{ - 1}^1 {{{\sin }^{11}}x\,\,dx} = 0$.
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If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then A - B is:
$\text{I}$
$0$
$2\text{I}$
$\frac{1}{2}\text{I}$
$f(x)=\left\{\begin{array}{l}\max \left\{t^{3}-3 t\right\} ; x \leq 2 \\ t \leq x \\ x^{2}+2 x-6 ; 2 < x < 3 \\ {[x-3]+9 ; 3 \leq x \leq 5} \\ 2 x+1 \quad ; \quad x > 5\end{array}\right\}$
Where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I =\int\limits_{-2}^{2} f( x ) dx$. Then the ordered pair $( m , I )$ is equal to