CUET UG 2025 — Mathematics Calculus

For the function $y = \sin^{-1}(x-1) + \cos^{-1}\sqrt{x-1}$y=sin−1(x−1)+cos−1x−1​ to be defined, all parts must work.

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The function $\sin^{-1}$sin−1 only accepts inputs between $-1$−1 and $1$1 (inclusive).

For $\sin^{-1}(x-1)$sin−1(x−1):

$$-1 \leq (x-1) \leq 1$$−1≤(x−1)≤1

$$-1 + 1 \leq x - 1 + 1 \leq 1 + 1$$−1+1≤x−1+1≤1+1

$$0 \leq x \leq 2$$0≤x≤2

From the first term: $x \in [0, 2]$x∈[0,2]

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For $\cos^{-1}\sqrt{x-1}$cos−1x−1​, there are two conditions:

The square root must be real (non-negative radicand):

$$x - 1 \geq 0$$x−1≥0

$$x \geq 1$$x≥1

The function $\cos^{-1}$cos−1 only accepts inputs between $-1$−1 and $1$1.

Since square roots are always non-negative:

$$0 \leq \sqrt{x-1} \leq 1$$0≤x−1​≤1

$$\sqrt{x-1} \leq 1$$x−1​≤1

$$x - 1 \leq 1$$x−1≤1

$$x \leq 2$$x≤2

Combining both conditions: $1 \leq x \leq 2$1≤x≤2

From the second term: $x \in [1, 2]$x∈[1,2]

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For the entire function to be defined:

First term requires: $[0, 2]$[0,2]

Second term requires: $[1, 2]$[1,2]

The intersection is: $[1, 2]$[1,2]

Therefore, the domain is $[1, 2]$[1,2].