If x = a sec πœƒ, y = b tan πœƒ, then (d 2 y)/(dx 2 ) at ΞΈ = Ο€/6 is :

(a) (-3√3 b)/a 2  (b) (-2√3 b)/a
(c) (-3√3 b)/a  (d) (-b)/(3√3 a 2 )

This question is inspired from Question 31 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards

Ques 22 (MCQ) - If x = a sec πœƒ, y = b tan πœƒ, then 𝑑2𝑦/𝑑π‘₯2 at πœƒ - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

part 2 - Question 22 - 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 22 - 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 4 - Question 22 - 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 5 - Question 22 - 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 22 If x = a sec πœƒ, y = b tan πœƒ, then (𝑑^2 𝑦)/(𝑑π‘₯^2 ) at πœƒ = πœ‹/6 is : (a) (βˆ’3√3 𝑏)/π‘Ž^2 (b) (βˆ’2√3 𝑏)/π‘Ž (c) (βˆ’3√3 𝑏)/π‘Ž (d) (βˆ’π‘)/(3√3 π‘Ž^2 ) Given π‘₯=π‘Ž sec β‘πœƒ, 𝑦=𝑏 tanβ‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/𝑑π‘₯ Γ— π‘‘πœƒ/π‘‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/π‘‘πœƒ Γ— π‘‘πœƒ/𝑑π‘₯ π’…π’š/𝒅𝒙 = (π’…π’š/π’…πœ½)/(𝒅𝒙/π’…πœ½) Calculating π’…π’š/π’…πœ½ 𝑦 = 𝑏 tanβ‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑑(𝑏 tanβ‘πœƒ )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑏 𝑑(tanβ‘πœƒ )/π‘‘πœƒ π’…π’š/π’…πœ½ = 𝑏 .sec^2β‘πœƒ Calculating 𝒅𝒙/π’…πœ½ π‘₯=π‘Ž sec β‘πœƒ 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž sec β‘πœƒ)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž 𝑑(sec β‘πœƒ)/π‘‘πœƒ 𝒅𝒙/π’…πœ½ = π‘Ž (secβ‘πœƒ.tanβ‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = (𝑏 secβ‘πœƒ)/(π‘Ž tanβ‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = (𝑏 . 1/cosβ‘πœƒ )/(π‘Ž (sinβ‘πœƒ/cosβ‘πœƒ ) ) 𝑑𝑦/𝑑π‘₯ = 𝑏 Γ— 1/cosβ‘πœƒ Γ— cosβ‘πœƒ/γ€–a sinγ€—β‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑏/π‘Ž (1/sinβ‘πœƒ ) π’…π’š/𝒅𝒙 = 𝒃/𝒂 𝒄𝒐𝒔𝒆𝒄 𝜽 Now, (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = 𝒃/𝒂 (𝐝(𝒄𝒐𝒔𝒆𝒄 𝜽))/𝒅𝒙 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑏/π‘Ž (d(π‘π‘œπ‘ π‘’π‘ πœƒ))/𝑑θ ×𝑑θ/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑏/π‘Ž (d(π‘π‘œπ‘ π‘’π‘ πœƒ))/𝑑θ Γ—πŸ/(𝒅𝒙/π’…πœ½) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑏/π‘Ž (βˆ’π‘π‘œπ‘ π‘’π‘ ΞΈ cot⁑θ) Γ—πŸ/(𝒂 𝒔𝒆𝒄 ΞΈ 𝒕𝒂𝒏 ΞΈ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 1/sin⁑〖θ γ€— cot⁑θ Γ— 𝒄𝒐𝒔 ΞΈ 𝒄𝒐𝒕 ΞΈ (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (βˆ’π’ƒ)/𝒂^𝟐 〖𝒄𝒐𝒕〗^πŸ‘β‘πœ½ We need to find (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘Žπ‘‘ π‘₯ = πœ‹/6 Putting π‘₯ = 𝝅/πŸ” (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 γ€–π‘π‘œπ‘‘γ€—^3β‘γ€–πœ‹/6γ€— (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 (√3)^3 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 3√3 (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (βˆ’πŸ‘βˆšπŸ‘ 𝒃)/𝒂^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