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Question 1 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Misc 13 Find ππ¦/ππ₯ , if π¦=γπ ππγ^(βπ) π₯+γπ ππγ^(β1) β(1βπ₯2), β 1 β€ π₯ β€ 1 π¦=γπ ππγ^(βπ) π₯+γπ ππγ^(β1) β(1βπ₯^2 ) , β 1 β€ π₯ β€ 1 Putting π = πππβ‘π½ π¦=γπ ππγ^(βπ) (sinβ‘π)+γπ ππγ^(β1) β(1βsin^2 π ) π¦=π½+γπ ππγ^(β1) β(γππ¨π¬γ^π π ) π¦=π+γπ ππγ^(β1) (cos π) π¦=π+γπ ππγ^(β1) (sinβ‘(π /π βπ½) ) π¦=π+ (π/2 βπ) π¦=πβπ + π/2 π= π /π Differentiating π€.π.π‘.π₯. ππ¦/ππ₯ = π(π/2)/ππ₯ π π/π π = 0