The value of $\cos(2\cos^{-1}x + \sin^{-1}x)$ at $x =\frac{1}{5}$ is

$-\sqrt{\frac{24}{25}}$

The correct answer is Option (2) → $-\sqrt{\frac{24}{25}}$

Let $\theta = \cos^{-1}x$.

Then $\sin^{-1}x = \frac{\pi}{2} - \theta$.

Expression: $\cos(2\cos^{-1}x + \sin^{-1}x)$

$= \cos(2\theta + \frac{\pi}{2} - \theta) = \cos(\theta + \frac{\pi}{2}) = -\sin\theta$

$\sin\theta = \sqrt{1 - \cos^{2}\theta} = \sqrt{1 - x^{2}}$

Therefore, $\cos(2\cos^{-1}x + \sin^{-1}x) = -\sqrt{1 - x^{2}}$

At $x = \frac{1}{5}$:

$\cos(2\cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5}) = -\sqrt{1 - \left(\frac{1}{5}\right)^{2}} = -\sqrt{\frac{24}{25}} = -\frac{2\sqrt{6}}{5}$