CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## (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

### Transcript

Question 22 If x = a sec π, y = b tan π, then (π^2 π¦)/(ππ₯^2 ) at π = π/6 is : (a) (β3β3 π)/π^2 (b) (β2β3 π)/π (c) (β3β3 π)/π (d) (βπ)/(3β3 π^2 ) Given π₯=π sec β‘π, π¦=π tanβ‘π ππ¦/ππ₯ = ππ¦/ππ₯ Γ ππ/ππ ππ¦/ππ₯ = ππ¦/ππ Γ ππ/ππ₯ ππ/ππ = (ππ/ππ½)/(ππ/ππ½) Calculating ππ/ππ½ π¦ = π tanβ‘π ππ¦/ππ = π(π tanβ‘π )/ππ ππ¦/ππ = π π(tanβ‘π )/ππ ππ/ππ½ = π .sec^2β‘π Calculating ππ/ππ½ π₯=π sec β‘π ππ₯/ππ = π(π sec β‘π)/ππ ππ₯/ππ = π π(sec β‘π)/ππ ππ/ππ½ = π (secβ‘π.tanβ‘π ) ππ¦/ππ₯ = (π secβ‘π)/(π tanβ‘π ) ππ¦/ππ₯ = (π . 1/cosβ‘π )/(π (sinβ‘π/cosβ‘π ) ) ππ¦/ππ₯ = π Γ 1/cosβ‘π Γ cosβ‘π/γa sinγβ‘π ππ¦/ππ₯ = π/π (1/sinβ‘π ) ππ/ππ = π/π πππππ π½ Now, (π^π π)/(ππ^π ) = π/π (π(πππππ π½))/ππ (π^2 π¦)/(ππ₯^2 ) = π/π (d(πππ ππ π))/πΞΈ ΓπΞΈ/ππ₯ (π^2 π¦)/(ππ₯^2 ) = π/π (d(πππ ππ π))/πΞΈ Γπ/(ππ/ππ½) (π^2 π¦)/(ππ₯^2 ) = π/π (βπππ ππ ΞΈ cotβ‘ΞΈ) Γπ/(π πππ ΞΈ πππ ΞΈ) (π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 1/sinβ‘γΞΈ γ cotβ‘ΞΈ Γ πππ ΞΈ πππ ΞΈ (π^π π)/(ππ^π ) = (βπ)/π^π γπππγ^πβ‘π½ We need to find (π^2 π¦)/(ππ₯^2 ) ππ‘ π₯ = π/6 Putting π₯ = π/π (π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 γπππ‘γ^3β‘γπ/6γ (π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 (β3)^3 (π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 3β3 (π^π π)/(ππ^π ) = (βπβπ π)/π^π So, the correct answer is (A)