Let f(x) = √(x - 1) + √(x + 24 - 10-1) be a real valued function for 1 < x < 26 . Then f'(x) is: - Mathematics MCQ Mathematics Mock Test - 8 2026 | ExamTest.live

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1 mark

Let f (x) = x - 1 + x + 24 - 10 x - 1 be a real valued function for 1 < x < 26 . Then f^{'} (x) is:

Solution & Step-by-step Explanation

Let u = x - 1 .The expression becomes: f (x) = u + u^{2} + 1 - 1 + 24 - 10 u = u + u^{2} - 10 u + 25 f (x) = u + (u - 5)^{2} = u + ∣ u - 5∣ For 1 < x < 26 : 0 < x - 1 < 25 ⟹ 0 < x - 1 < 5 .Thus u < 5, so ∣ u - 5∣ = - (u - 5) = 5 - u . f (x) = u + (5 - u) = 5 Since f (x) = 5 (a constant), the derivative f^{'} (x) = 0 .

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Let f (x) = x - 1 + x + 24 - 10 x - 1 be a real valued function for 1 < x < 26 . Then f^{'} (x) is:

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