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
Last updated at Sept. 4, 2021 by Teachoo
If x = a sec π, y = b tan π, then (d
^{
2
}
y)/(dx
^{
2
}
) at θ = π/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
Transcript
Question 22 If x = a sec π, y = b tan π, then (π^2 π¦)/(ππ₯^2 ) at π = π/6 is : (a) (β3β3 π)/π^2 (b) (β2β3 π)/π (c) (β3β3 π)/π (d) (βπ)/(3β3 π^2 )
Given
π₯=π sec β‘π, π¦=π tanβ‘π
ππ¦/ππ₯ = ππ¦/ππ₯ Γ ππ/ππ
ππ¦/ππ₯ = ππ¦/ππ Γ ππ/ππ₯
π
π/π
π = (π
π/π
π½)/(π
π/π
π½)
Calculating π
π/π
π½
π¦ = π tanβ‘π
ππ¦/ππ = π(π tanβ‘π )/ππ
ππ¦/ππ = π π(tanβ‘π )/ππ
π
π/π
π½ = π .sec^2β‘π
Calculating π
π/π
π½
π₯=π sec β‘π
ππ₯/ππ = π(π sec β‘π)/ππ
ππ₯/ππ = π π(sec β‘π)/ππ
π
π/π
π½ = π (secβ‘π.tanβ‘π )
ππ¦/ππ₯ = (π secβ‘π)/(π tanβ‘π )
ππ¦/ππ₯ = (π . 1/cosβ‘π )/(π (sinβ‘π/cosβ‘π ) )
ππ¦/ππ₯ = π Γ 1/cosβ‘π Γ cosβ‘π/γa sinγβ‘π
ππ¦/ππ₯ = π/π (1/sinβ‘π )
π
π/π
π = π/π πππππ π½
Now,
(π
^π π)/(π
π^π ) = π/π (π(πππππ π½))/π
π
(π^2 π¦)/(ππ₯^2 ) = π/π (d(πππ ππ π))/πΞΈ ΓπΞΈ/ππ₯
(π^2 π¦)/(ππ₯^2 ) = π/π (d(πππ ππ π))/πΞΈ Γπ/(π
π/π
π½)
(π^2 π¦)/(ππ₯^2 ) = π/π (βπππ ππ ΞΈ cotβ‘ΞΈ) Γπ/(π πππ ΞΈ πππ ΞΈ)
(π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 1/sinβ‘γΞΈ γ cotβ‘ΞΈ Γ πππ ΞΈ πππ ΞΈ
(π
^π π)/(π
π^π ) = (βπ)/π^π γπππγ^πβ‘π½
We need to find (π^2 π¦)/(ππ₯^2 ) ππ‘ π₯ = π/6
Putting π₯ = π
/π
(π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 γπππ‘γ^3β‘γπ/6γ
(π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 (β3)^3
(π^2 π¦)/(ππ₯^2 ) = (βπ)/π^2 3β3
(π
^π π)/(π
π^π ) = (βπβπ π)/π^π
So, the correct answer is (A)
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