The derivative of sin−1 (2x √(1 - x 2 )) w.r.t sin−1 x, 1/√2  < π‘₯ < 1, is:

(a) 2                 (b) π/2 − 2
(c) π/2              (d) −2

Ques 24 (MCQ) - The derivative of sin-1 (2π‘₯√1 − π‘₯2) w.r.t sin-1 x, - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

part 2 - Question 24 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 24 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 24 The derivative of sinβˆ’1 ("2π‘₯" √("1 βˆ’ " π‘₯^2 )) w.r.t sinβˆ’1 x, 1/√2 < π‘₯ < 1, is: (a) 2 (b) πœ‹/2 βˆ’ 2 (c) πœ‹/2 (d) βˆ’2 Let y = sinβˆ’1 x sin y = x x = sin y We need to find the derivative of sinβˆ’1 ("2π‘₯" √("1 βˆ’ " π‘₯^2 )) w.r.t sinβˆ’1 x i.e. sinβˆ’1 ("2π‘₯" √("1 βˆ’ " π‘₯^2 )) w.r.t y i.e. (𝒅(〖𝐬𝐒𝐧〗^(βˆ’πŸ)⁑〖(πŸπ’™βˆš(𝟏 βˆ’ 𝒙^𝟐 ))) γ€—)/π’…π’š Now, (𝑑(sin^(βˆ’1)⁑〖(2π‘₯√(1 βˆ’ π‘₯^2 ))) γ€—)/𝑑𝑦 Putting x = sin y = (𝑑(sin^(βˆ’1)⁑〖(2 sin⁑𝑦 √(1 βˆ’ 〖𝑠𝑖𝑛〗^2 𝑦))) γ€—)/𝑑𝑦 = (𝑑(sin^(βˆ’1)⁑〖(2 sin⁑𝑦 √(cos^2⁑𝑦 )))γ€—)/𝑑𝑦 = (𝑑(sin^(βˆ’1)⁑〖(2 sin⁑𝑦 cos⁑𝑦 ))γ€—)/𝑑𝑦 = (𝑑(sin^(βˆ’1)⁑〖(sin⁑2𝑦 ))γ€—)/𝑑𝑦 = 𝑑(2𝑦)/𝑑𝑦 = 2 So, the correct answer is (A)

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