Question

The number of solutions of
$si{n}^{-1}(1+b+b^2+....\infty )+co{s}^{-1}(a-\frac{a^2}{3}+\frac{a^3}{9}...\infty )$
$=\frac{\pi }{2}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $\infty$

Answer 1

Answer: (D)

The given equation will be valid if
$-1\le 1+b+b^2+...\infty \le 1$,
and $a-\frac{a^2}{3}+\frac{a^3}{9}-....\infty \in [-1,1]$
Also, $1+b+b^2+.....\infty =a-\frac{a^2}{3}+\frac{a^3}{9}-....\infty$
(Infinite G.P. series)
or $\frac{1}{1-b}=\frac{a}{1+\frac{a}{3}}$ or $\frac{1}{1-b}=\frac{3a}{a+3}$
which is true for number of values $a$ and $b$ so $\exists$ infinite solutions of the equation
$si{n}^{-1}(\frac{1}{1-b})+co{s}^{-1}(\frac{3a}{a+3})=\pi /2$