If y = 5 cos x – 3 sin x, then (d 2 y)/(dx 2 ) is equal to:

(a) −y     (b) y
(c) 25y   (d) 9y

This question is inspired from Ex 5.7, 11 - Chapter 5 Class 12 - Continuity and Differentiability

Ques 14 (MCQ) - If y = 5 cos x – 3 sin x, then 𝑑2𝑦/𝑑π‘₯2 is equal to - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

part 2 - Question 14 - 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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Transcript

Question 14 If y = 5 cos x – 3 sin x, then (𝑑^2 𝑦)/(𝑑π‘₯^2 ) is equal to: (a) βˆ’y (b) y (c) 25y (d) 9y y = 5 cos⁑〖π‘₯βˆ’3 sin⁑π‘₯ γ€— Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑(5 cos⁑〖π‘₯βˆ’3 sin⁑π‘₯ γ€—))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑(5 cos⁑π‘₯))/𝑑π‘₯ βˆ’ (𝑑(3 sin⁑π‘₯))/𝑑π‘₯ π’…π’š/𝒅𝒙 = βˆ’ 5 π’”π’Šπ’β‘π’™ βˆ’ 3 𝒄𝒐𝒔⁑𝒙 Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = (𝑑 γ€–(βˆ’ 5 sin〗⁑π‘₯ γ€–βˆ’ 3cos〗⁑〖π‘₯)γ€—)/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’(𝑑(5 sin⁑π‘₯))/𝑑π‘₯ βˆ’ (𝑑(3 cos⁑π‘₯))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ 5 cos⁑π‘₯ βˆ’ 3 γ€–(βˆ’sin〗⁑〖π‘₯)γ€— (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ 5 cos⁑π‘₯ + 3 sin⁑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ (5 𝒄𝒐𝒔⁑𝒙 βˆ’ 3 π’”π’Šπ’β‘π’™) (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = βˆ’y So, the correct answer is (A)

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