if x = a (cos 2θ + 2θ sin 2θ) and y = a (sin 2θ − 2θ cos 2θ),
find (d ^{ 2 } y)/(dx ^{ 2 } ) at θ = π/8 .
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CBSE Class 12 Sample Paper for 2020 Boards
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Question 12
Question 13
Question 14 (OR 1st Question)
Question 14 (OR 2nd Question) Important
Question 15 (OR 1st Question)
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Question 16
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Question 28 (OR 1st Question) Important
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Question 31 (OR 1st Question) Important
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Question 33 (OR 1st Question) Important
Question 33 (OR 2nd Question) Important
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Question 35 (OR 1st Question) Important
Question 35 (OR 2nd Question) Important
Question 36 Important
CBSE Class 12 Sample Paper for 2020 Boards
Last updated at March 16, 2023 by Teachoo
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Question 28 (OR 2nd Question) if x = a (cos 2𝜃 + 2𝜃 sin 2𝜃) and y = a (sin 2𝜃 − 2𝜃 cos 2𝜃), find (𝑑^2 𝑦)/(𝑑𝑥^2 ) at 𝜃 = 𝜋/8 . Now, x = a (cos 2𝜃 + 2𝜃 sin 2𝜃) 𝑑𝑥/𝑑𝜃 = a (–2 sin 2𝜃 + 2 sin 2𝜃 + 2𝜃 × 2 cos 2𝜃) 𝑑𝑥/𝑑𝜃 = a (4𝜃 cos 2𝜃) And, y = a (sin 2𝜃 − 2𝜃 cos 2𝜃) 𝑑𝑦/𝑑𝜃 = a (2 cos 2𝜃 – 2 cos 2𝜃 + 2𝜃 × 2sin 2𝜃) 𝑑𝑦/𝑑𝜃 = a (4𝜃 sin 2𝜃) Therefore, Dividing (2) and (1) 𝑑𝑦/𝑑𝑥=(𝑎(4𝜃 sin2𝜃))/(𝑎(4𝜃 cos2𝜃)) 𝑑𝑦/𝑑𝑥=sin2𝜃/cos2𝜃 𝑑𝑦/𝑑𝑥=tan2𝜃 Now, differentiation again with respect to x (𝑑^2 𝑦)/(𝑑𝑥^2 )=(𝑑(tan2𝜃))/𝑑𝑥 = (𝑑(tan2𝜃))/𝑑𝑥×𝑑𝜃/𝑑𝜃 = (𝑑(tan2𝜃))/𝑑𝜃×𝑑𝜃/𝑑𝑥 = 2 sec2 2θ ×𝑑𝜃/𝑑𝑥 = 2 sec2 2θ ×1/(𝑑𝑥/𝑑𝜃) = 2 sec2 2θ ×1/(𝑎(4𝜃 cos2𝜃)) We need to find value of (𝑑^2 𝑦)/(𝑑𝑥^2 ) at 𝜃 = 𝜋/8 . Putting 𝜃 = 𝜋/8 in the equation (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 2 sec2 (2×𝜋/8) ×1/(𝑎(4 × 𝜋/8 × cos〖(2×𝜋/8)〗)) = 2 sec2 (𝜋/4) ×1/(𝑎(𝜋/2 × 〖cos 〗〖𝜋/4〗)) = 2 × 1/cos^2(𝜋/4) ×2/(𝑎𝜋 × 〖cos 〗〖𝜋/4〗 ) = 2 × 1/(1/√2)^2 ×2/(𝑎𝜋 × (1/√2) ) = 2 × 1/((1/2) ) ×2/(𝑎𝜋 × (1/√2) ) = 2 × 2 ×(2√2)/𝑎𝜋 = (𝟖√𝟐)/𝒂𝝅