Question

Assertion :If $\sum _{i=1}^{2n}{\sin }^{-1}{x}_i=n\pi \forall nϵN$ then $\sum _{i=1}^{2n}{x}_i=\sum _{i=1}^{2n}{x}_i^2=\sum _{i=1}^{2n}{x}_i^n=2n$ Reason: $-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}\forall xϵ[-1,1]$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion false but Reason is true

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$\sum _{i=1}^{2n}{sin}^{-1}{x}_i=n\pi =2n\left(\frac{\pi }{2}\right)\forall ϵ(i=1,2,\mathrm{3...2}n)$

$\Rightarrow {sin}^{-1}{x}_i=\left(\frac{\pi }{2}\right)\forall ϵ(i=1,2,\mathrm{3...2}n)$

$\Rightarrow x_i=1\forall ϵ(i=1,2,\mathrm{3...2}n)$

$\sum _{i=1}^{2n}{x}_i=1+1+...1\left(2ntimes\right)=2n$

$\sum _{i=1}^{2n}{x_i}^2=1^2+1^2+....1^2\left(2ntimes\right)=2n$

$\sum _{i=1}^{2n}{x_i}^n=1^n+1^n+....1^n\left(2ntimes\right)=2n$