CBSE Class 12 Sample Paper for 2026 Boards
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CBSE Class 12 Sample Paper for 2026 Boards
Last updated at November 26, 2025 by Teachoo
This question is similar to Chapter 5 Class 12 Continuity and Differentiability - Ex 5.6
Please check the question here
![[SQP] If π₯=π(π−sin π), π¦=π(1−cos π) find (π^2 π¦)/(ππ₯^2) - CBSE Class 1 - CBSE Class 12 Sample Paper for 2026 Boards](https://cdn.teachoo.com/6832186e-7720-4fd9-a374-95cd2c4e3208/slide32.jpg)
Transcript
Question 26 (B) If π₯=π(πβsin π), π¦=π(1βcos π) find (π^2 π¦)/(ππ₯^2 ).Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Finding π π/π π½ and π π/π π½ separately Calculating π π/π π½ π¦ = π (1+cosβ‘π) ππ¦/ππ = π(π (1 β cosβ‘π))/ππ ππ¦/ππ = π (π(1 β cosβ‘π )/ππ) ππ¦/ππ = π (0β(βsinβ‘π )) ππ¦/ππ = π (sinβ‘π ) π π/π π½ = π πππβ‘π½ Calculating π π/π π½ π₯=π (π βsinβ‘π ) ππ₯/ππ = π(π π βπ sinβ‘π )/ππ ππ₯/ππ = π(π π)/ππ β π(π sinβ‘π )/ππ ππ₯/ππ = πβγπ cosγβ‘π π π/π π½ = π(γπβπππγβ‘π½ ) Therefore, ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (π sinβ‘π)/π(γ1 β cosγβ‘π ) ππ¦/ππ₯ = sinβ‘π/γ1 β cosγβ‘π ππ¦/ππ₯ = (2 γsin γβ‘γπ/2γ γcos γβ‘γπ/2γ)/(2 γsin^2 γβ‘γπ/2γ ) ππ¦/ππ₯ = γcos γβ‘γπ/2γ/(sin π/2) π π/π π = πππβ‘γπ½/πγ Rough We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 1 β 2sin2 ΞΈ Replacing ΞΈ by π/2 cos ΞΈ = 1 β 2sin2 π/2 1 β cos ΞΈ = 2sin2 π½/π Finding (π ^π π)/(π π^π ) π π/π π = πππβ‘γπ½/πγ Differentiating again (π^2 π¦)/(ππ₯^2 )=π(γπππ‘ γβ‘γπ/2γ )/ππ₯ (π^2 π¦)/(ππ₯^2 )=π(γπππ‘ γβ‘γπ/2γ )/ππ Γππ/ππ₯ (π^2 π¦)/(ππ₯^2 )=βγπππππγ^π π½/π Γπ/π Γππ/ππ₯ (π^2 π¦)/(ππ₯^2 )=β1/2 γπππ ππγ^2 π/2 Γπ/(π π/π π½) (π^2 π¦)/(ππ₯^2 )=β1/2 γπππ ππγ^2 π/2 Γ1/π(π β ππ¨π¬β‘π½ ) (π^2 π¦)/(ππ₯^2 )=β1/2 γπππ ππγ^2 π/2 Γ1/(π Γ 2 sin^2β‘γπ/2γ ) (π^2 π¦)/(ππ₯^2 )=β1/4π Γ γπππ ππγ^2 π/2 Γ1/γπ ππγ^2β‘γπ/2γ (π^2 π¦)/(ππ₯^2 )=β1/4π Γ γπππ ππγ^2 π/2 Γ γπππ ππγ^2 π/2 (π ^π π)/(π π^π )=βπ/ππ Γ γπππππγ^π π½/π
