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Question 1 Deleted for CBSE Board 2025 Exams

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Last updated at April 16, 2024 by Teachoo

Example 34 (Method 1) Find ππ¦/ππ₯ , if π₯^(2/3) + π¦^(2/3) = π^(2/3) . π₯^(2/3) + π¦^(2/3) = π^(2/3) Differentiating w.r.t. x (π(π₯^(2/3)))/ππ₯ + (π(π¦^(2/3)))/ππ₯ = (π(π^(2/3)))/ππ₯ 2/3 π₯^(2/3 β 1) + (π(π¦^(2/3)))/ππ₯ Γ ππ¦/ππ¦ = 0 2/3 π₯^((β1)/3) + (π(π¦^(2/3)))/ππ¦ Γ ππ¦/ππ₯ = 0 2/3 π₯^((β1)/3) + 2/3 π^(π/π β π) Γ ππ¦/ππ₯ = 0 2/3 1/π₯^(1/3) + 2/3 π^((βπ)/π) Γ ππ¦/ππ₯ = 0 2/3 1/π₯^(1/3) + 2/3 π/π^(π/π) Γ ππ¦/ππ₯ = 0 2/3 1/π¦^(1/3) Γ ππ¦/ππ₯ = βπ/π 1/π₯^(1/3) 1/π¦^(1/3) Γ ππ¦/ππ₯ = (β1)/π₯^(1/3) ππ¦/ππ₯ = β π¦^(1/3)/π₯^(1/3) π π/π π = β (π/π)^(π/π) Example 34 (Method 2) Find ππ¦/ππ₯ , if π₯^(2/3) + π¦^(2/3) = π^(2/3) . Let π = π γπππγ^πβ‘π½ & π = π γπ¬π’π§γ^πβ‘π½ Putting values in equation π₯^(2/3) + π¦^(2/3) = π^(2/3) γ(π γππ¨π¬γ^πβ‘π½)γ^(π/π) + γ(π γπ¬π’πγ^πβ‘π½)γ^(π/π) = π^(π/π) π^(2/3) γ(cos^3β‘π)γ^(2/3) + π^(2/3) γ(γsiπγ^3β‘π)γ^(2/3) = π^(2/3) π^(2/3) . (sin^2β‘π+cos^2β‘π )= π^(2/3) π^(2/3) = π^(2/3) So, our assumed values of x and y is correct Now, ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ π¦ = a sin^3β‘π ππ¦/ππ = π(a sin^3β‘π )/ππ ππ¦/ππ = a (3 sin^2β‘π) π (πππβ‘π½ )/π π½ ππ¦/ππ = 3π sin2 π½ cos π½ Calculating π π/π π½ π₯ = a cos^3β‘π ππ₯/ππ = (π (γa cosγ^3β‘γπ)γ)/ππ ππ₯/ππ = a(3 cos^2β‘π) (π (πππ π½) )/π π½ π π/π π½ = ππ γπππγ^πβ‘π½ . (βπππβ‘π½ ) Therefore π π/π π = ππ¦/ππ Γ· ππ₯/ππ = (3a sin^2β‘π cosβ‘π)/(β3a cos^2β‘π sinβ‘π ) = (βsinβ‘π)/cosβ‘π = βtan π½ Finding value of tan π½ π¦/π₯ = (π sin^3β‘π)/(π cos^3β‘π ) sin^3β‘π/cos^3β‘π = π¦/π₯ (sinβ‘π/cosβ‘π )^3 = π¦/π₯ sinβ‘π/cosβ‘π = (π¦/π₯)^(1/3) πππβ‘π½ = β(π&π/π) Thus, ππ¦/ππ₯ = βtan π π π/π π = ββ(π&π/π)