Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$-\frac{\pi }{2}<{\sin }^{-1}x\le \frac{\pi }{2}$ $-\frac{\pi }{2}<{\sin }^{-1}x\le \frac{\pi }{2}$

$$\begin{array}{rlr}& \therefore {\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}& \iff {\sin }^{-1}x={\sin }^{-1}y={\sin }^{-1}z=\frac{\pi }{2}\iff x=y=z=1\end{array}$$

Also $f(p+q)=f(p)\cdot f(q)\forall p\cdot q\in R$. Given $f(1)=2$. From (1), $f(1+1)=f(1)\cdot f(1)\Rightarrow f(2)=2^2=4$. From (2), $f(2+1)=f(2)\cdot f(1)=2^2\cdot 2=2^3=8$.

Now the given expression $=f(3)=8$. The given relation is possible when

$$a-\frac{a^2}{3}+\frac{a^3}{9}+\dots =1+b+b^2+\dots$$

Also

$$\begin{array}{rlr}& -1\le a-\frac{a^2}{3}+\frac{a^3}{9}+\dots .\le 1& \\ & -1\le 1+b+b^2+\dots .\le 1& \Rightarrow |b|<1\Rightarrow |a|<3\text{ and }\frac{a}{1+\frac{a}{3}}=\frac{1}{1-b}\end{array}$$

$\Rightarrow \frac{3a}{a+3}=\frac{1}{1-b}$, there are infinitely many solutions. But in the given options, it is satisfied only when $a=1$ and $b=\frac{1}{3}$.$b=\frac{1}{3}$