Inverse Circular Functions 2 Question 7
7. If $\sin ^{-1} x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\ldots+\cos ^{-1} x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}-\ldots$ $=\frac{\pi}{2}$, for $0<|x|<\sqrt{2}$, then $x$ equals
(2001, 1M)
(a) $1 / 2$
(b) 1
(c) $-1 / 2$
(d) -1
Show Answer
Answer:
Correct Answer: 7. $P \rightarrow 4 ; Q \rightarrow 3 ; R \rightarrow 2 ; S \rightarrow 1$
Solution:
- We know that, $\sin ^{-1}(\alpha)+\cos ^{-1}(\alpha)=\frac{\pi}{2}$
Therefore, $\alpha$ should be equal in both functions.
$$ \begin{aligned} & \therefore \quad x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\ldots=x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}-\ldots \\ & \Rightarrow \quad \frac{x}{1+\frac{x}{2}}=\frac{x^{2}}{1+\frac{x^{2}}{2}} \Rightarrow \frac{x}{\frac{2+x}{2}}=\frac{x^{2}}{\frac{2+x^{2}}{2}} \\ & \Rightarrow \quad \frac{2 x}{2+x}=\frac{2 x^{2}}{2+x^{2}} \\ & \Rightarrow \quad 2 x\left(2+x^{2}\right)=2 x^{2}(2+x) \\ & \Rightarrow \quad 4 x+2 x^{3}=4 x^{2}+2 x^{3} \\ & \Rightarrow \quad x\left(4+2 x^{2}-4 x-2 x^{2}\right)=0 \\ & \Rightarrow \quad \text { Either } x=0 \text { or } 4-4 x=0 \\ & \Rightarrow \quad x=0 \text { or } x=1 \\ & \because \quad 0<|x|<\sqrt{2} \\ & \therefore \quad x=1 \text { and } x \neq 0 \end{aligned} $$
