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. $ |
Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1 Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1 Statement 1 is True, Statement 2 is False. Statement 1 is False, Statement 2 is True. |
Statement 1 is True, Statement 2 is False. |
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. |