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

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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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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.