Question

Q.):- If ${\sin }^{-1}\left(x-\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$x^3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.-\raisebox{1ex}{$x^4$}\!\left/ \!\raisebox{-1ex}{$8$}\right.+----\infty \right)=\frac{\overline{\Lambda }}{6}$ where $\left|x\right|<2$ then value of x is
$1.\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.2.\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.3.\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.4.\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Answer 1

Dear student
$$\begin{gathered} Given: \\ {\sin }^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+...\infty \right)=\frac{\mathrm{\pi }}{6} \\ x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+...\infty =\sin \left(\frac{\mathrm{\pi }}{6}\right) \\ x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+...\infty =\frac{1}{2} \\ LHSformsaGPwithfirstterm,a=xandcommonratio=\frac{\frac{x^2}{2}}{x}=-\frac{x}{2} \\ andweknowthatsumofinfiniteGPisgivenby:\frac{a}{1-r} \\ \Rightarrow \frac{x}{1-\left(-\frac{x}{2}\right)}=\frac{1}{2} \\ \Rightarrow \frac{2x}{2+x}=\frac{1}{2} \\ \Rightarrow 4x=2+x \\ \Rightarrow 3x=2 \\ \Rightarrow x=\frac{2}{3} \end{gathered}$$
Regards