Question

Assertion :$si{n}^{-1}[x-\frac{x^2}{2}+\frac{x^3}{4}....]=\pi /2-co{s}^{-1}[x^2-\frac{x^4}{2}+\frac{x^6}{4}....]$ for $0<|x|<\sqrt{2}$ has a unique solution. Reason: $ta{n}^{-1}\sqrt{x(x+1)}+si{n}^{-1}\sqrt{x^2+x+1}=\pi /2$ has no solution for $-\sqrt{2}<x<0$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Since ${\sin }^{-1}+{\cos }^{-1}x=\frac{\pi }{2}$, using this result we get
$x-\frac{x^2}{2}+\frac{x^3}{4}+.....=x^2-\frac{x^4}{2}+\frac{x^6}{4}+.....$
$\Rightarrow \frac{x}{1+\left(x/2\right)}=\frac{x^2}{1+\left(x^2/2\right)}$
$\Rightarrow 2x(2+x^2)=2x^2(2+x)$
$\Rightarrow x[4+2x^2-4x-2x^2]=0$
$\Rightarrow x=1$ as 0 < |x| < $\sqrt{2}$
So statement-1 is true.
For statement-2 $x(x+1)\ge 0$ and $0\le x^2+x+1\le 1$
$\Rightarrow x(x+1)=0\Rightarrow x=-1$ as $-\sqrt{2}<x<0$
and $x=-1$ satisfies the given equation
So statement-2 is false.