Inverse Circular Functions Ques 6
- The number of real solutions of the equation $\sin ^{-1} \sum _{i=1}^{\infty} x^{i+1}-x \sum _{i=1}^{\infty} \frac{x}{2}$
$=\frac{\pi}{2}-\cos ^{-1} \sum _{i=1}^{\infty}-\frac{x^{i}}{2}-\sum _{i=1}^{\infty}(-x)^{i}$ lying in the interval
$ -\frac{1}{2}, \frac{1}{2} \text { is } $
(Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\cos ^{-1} x$ assume values in $-\frac{\pi}{2}, \frac{\pi}{2}$ and $[0, \pi]$, respectively.)
(2018 Adv.)
Show Answer
Answer:
Correct Answer: 6.(2)
Solution:
- We have,
$ \begin{gathered} \sin ^{-1} \sum _{i=1}^{\infty} x^{i+1}-x \sum _{i=1}^{\infty} \frac{x^{i}}{2} \\ =\frac{\pi}{2}-\cos ^{-1} \sum _{i=1}^{\infty} \frac{-x^{i}}{2}-\sum _{i=1}^{\infty}(-x)^{i} \\ \Rightarrow \quad \sin ^{-1} [\frac{x^{2}}{1-x}-\frac{x \cdot \frac{x}{2}}{1-\frac{x}{2}}] \\ =\frac{\pi}{2}-\cos ^{-1} [\frac{\frac{-x}{2}}{1+\frac{x}{2}}-\frac{(-x)}{1+x}] \\ [\because \sum _{i=1}^{\infty} x^{i+1}=x^{2}+x^{3}+x^{4}+\ldots=\frac{x^{2}}{1-x}] \end{gathered} $
using sum of infinite terms of GP
$\Rightarrow \quad \sin ^{-1} [\frac{x^{2}}{1-x}-\frac{x^{2}}{2-x}]=\frac{\pi}{2}-\cos ^{-1} [\frac{x}{1+x}-\frac{x}{2+x}]$
$\Rightarrow \sin ^{-1} [\frac{x^{2}}{1-x}-\frac{x^{2}}{2-x}]=\sin ^{-1} (\frac{x}{1+x}-\frac{x}{2+x})$
$ \begin{aligned} & \qquad \quad [\because \sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x] \\ & \Rightarrow \quad x^{2} (\frac{2-x-1+x}{(1-x)(2-x)})=x \frac{x^{2}}{(1+x)(2+x)}=\frac{x}{1+x}-\frac{x}{2+x} \\ & \Rightarrow \quad \frac{x}{2-3 x+x^{2}}=\frac{1}{2+3 x+x^{2}} \quad \text { or } x=0 \\ & \Rightarrow \quad x^{3}+3 x^{2}+2 x=x^{2}-3 x+2 \\ & \Rightarrow \quad x^{3}+2 x^{2}+5 x-2=0 \text { or } x=0 \end{aligned} $
Let $\quad f(x)=x^{3}+2 x^{2}+5 x-2$
$ f^{\prime}(x)=3 x^{2}+4 x+5 $
$ f^{\prime}(x)>0, \forall x \in R $
$\therefore \quad x^{3}+2 x^{2}+5 x-2$ has only one real roots
Therefore, total number of real solution is 2 .
BETA
