The derivative of sinβˆ’1 (2x √(1 - x 2 )) w.r.t sinβˆ’1 x, 1/√2Β  < π‘₯ < 1, is:

(a) 2Β  Β  Β  Β  Β  Β  Β  Β  Β (b) Ο€/2 βˆ’ 2
(c) Ο€/2Β  Β  Β  Β  Β  Β  Β  (d) βˆ’2

Ques 24 (MCQ) - The derivative of sin-1 (2π‘₯√1 βˆ’ 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo