CBSE Class 12 Sample Paper for 2021 Boards

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

## If π₯ = π π ππ π , π¦ = π Β π‘πππΒ  findΒ  d 2 y / dx 2 Β  Β ππ‘ π₯ = Ο/6

Β

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Question 31 (Choice 2) If π₯ = π π ππ π , π¦ = π π‘ππ π ππππ (π^2 π¦)/(ππ₯^2 ) ππ‘ π₯ = π/6 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)/(ππ₯/ππ) 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 = 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 = π (secβ‘π.tanβ‘π ) Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (π .γ secγ^2β‘π)/(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