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$8\pi\text{ sqp}.\text{units}$
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$4\pi\text{ sq}.\text{units}$
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$8\pi\text{ sq}.\text{units}$
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$3\pi\text{ sq}.\text{units}$
$8\pi\text{ sqp}.\text{units}$
$4\pi\text{ sq}.\text{units}$
$8\pi\text{ sq}.\text{units}$
$3\pi\text{ sq}.\text{units}$
$8\pi\text{ sq}.\text{units}$
Solution:
Inverse function is the mirror image with respect to y = x
Then area bounded by $\text{x}+\sin\text{x}$ and its inverse function is
$=4\int\limits^\pi_0(\text{x}+\sin\text{x}-\text{x})\text{ dx}=8$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $N ^{\top} M N$ is symmetric or skew symmetric, according as $M$ is symmetric or skew symmetric
$(B)$ $M N-N M$ is skew symmetric for all symmetric matrices $M$ and $N$
$(C)$ $M N$ is symetric for all symmetric matrices $M$ and $N$
$(D)$ $(\operatorname{adj} M)(\operatorname{adj} N)=\operatorname{adj}(M N)$ for all invertible matrices $M$ and $N$
If
$\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{3}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5},$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)$ is equas:$(A)$ $\hat{j}-\hat{k}$ $(B)$ $-\hat{i}+\hat{j}$ $(C)$ $\hat{i}-\hat{j}$ $(D)$ $-\hat{j}+\hat{k}$