Statement-1 : $ sin^{-1}\begin{Bmatrix}x -\frac{x^2}{2}+\frac{x^3}{4}-.....\end{Bmatrix}=\frac{\pi}{2} -cos^{-1}\begin{Bmatrix}x^2 -\frac{x^4}{2}+\frac{x^6}{4}.....\end{Bmatrix}$ for $ 0< |x| < \sqrt{2}$ has a uniquee solution.
Statement-2: $ tan^{-1}\sqrt{x(x+1)} +sin^{-1}\sqrt{x^2+x+1} =\frac{\pi}{2}$ has no solution for $-\sqrt{2} < x< 0. $
Answer & explanation
Correct answer: option 3
Using $sin^{-1} x + cos^{-1} x =\frac{\pi}{2}$ in statement-1, we get
$x-\frac{x^2}{2}+\frac{x^3}{4} -........=x^2 -\frac{x^4}{2}+\frac{x^6}{4}-.....$
$⇒ \frac{x}{1+\frac{x}{2}}=\frac{x^2}{1+\frac{x^2}{2}}$
$⇒ x(2 +x^2) = x^2 (2 + x)$
$⇒x = 0, x= 1 ⇒ x = 1 $ $[∵ 0 < |x| < \sqrt{2}]$
So, statement-1 is true.
LHS of statement-2 is meaningful, if
$x^2 + x ≥ 0, x^2 + x + 1 ≥ 0 $ and $0 ≤ \sqrt{x^2+x+1} ≤1$
$⇒ x^2 + x ≥ 0 $ and $ x^2 + x ≤ 0 ⇒ x^2 + x = 0 ⇒ x =0, -1$
$⇒ x = - 1 $ $[∵ -\sqrt{2} < x < 0]$
Clearly,, x = -1 satisfies the satement-1.
So, statement-2 is not true.