CUET UG 2025 — Mathematics Calculus

$$y = \sin^{-1} \sqrt{\frac{x}{x+1}} + \sec^{-1} \sqrt{\frac{x+1}{x}}$$y=sin−1x+1x​​+sec−1xx+1​​

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Using the identity $\sec^{-1}(u) = \cos^{-1}\left(\frac{1}{u}\right)$sec−1(u)=cos−1(u1​) with $u = \sqrt{\dfrac{x+1}{x}}$u=xx+1​​:

$$\sec^{-1}\sqrt{\frac{x+1}{x}} = \cos^{-1}\left(\frac{1}{\sqrt{\frac{x+1}{x}}}\right)$$sec−1xx+1​​=cos−1​xx+1​​1​​

$$= \cos^{-1}\sqrt{\frac{x}{x+1}}$$=cos−1x+1x​​

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$$y = \sin^{-1}\sqrt{\frac{x}{x+1}} + \cos^{-1}\sqrt{\frac{x}{x+1}}$$y=sin−1x+1x​​+cos−1x+1x​​

Both terms have the same argument $t = \sqrt{\dfrac{x}{x+1}}$t=x+1x​​.

Using the identity $\sin^{-1}(t) + \cos^{-1}(t) = \dfrac{\pi}{2}$sin−1(t)+cos−1(t)=2π​, for $t \in [-1, 1]$t∈[−1,1]:

$$y = \dfrac{\pi}{2}$$y=2π​

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Since $y = \dfrac{\pi}{2}$y=2π​ is a constant:

$$\dfrac{dy}{dx} = \boxed{0}$$dxdy​=0​