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Question : 35 of 160
Marks:
+1,
-0
Solution:
‌f(x)=(cos‌x)(cos‌2‌x)...(cos‌n‌x)‌f′(x)=−sin‌x⋅cos‌2‌x...cos‌n‌x‌‌‌+cos‌x‌{cos‌2‌x⋅cos‌3‌x...cos‌n‌x}‌f′(x)=−sin‌x⋅cos‌2‌x...cos‌n‌x+cos‌x[−2sin‌2x⋅cos‌3‌x...cos‌n‌x+cos‌2‌x‌‌cos‌3‌x⋅cos‌4‌x...cos‌n‌x]‌f′(x)⇒−−(sin‌x⋅cos‌2‌x...cos‌n‌x)‌‌−(2‌cos‌x⋅sin‌2x⋅cos‌3‌x...cos‌n‌x)+cos‌x⋅cos‌2‌x‌(cos‌3‌x⋅cos‌4‌x⋅cos‌n‌x)‌f′(x)⇒−(sin‌x⋅cos‌2‌x...cos‌n‌x)‌‌−(2‌cos‌x⋅sin‌2x...cos‌n‌x)‌‌‌−(3‌cos‌x⋅cos‌2‌x⋅sin‌3x...cos‌n‌x)‌‌‌−(n‌cos‌x⋅cos‌2‌x...sin‌nx)So,
⇒f′(x)+(r‌tan‌r‌x)f(x) =f′(x)+{tan‌x+2‌tan‌2‌x+3‌tan‌3‌x ‌‌+...+n‌tan‌n‌x}‌f(x) =f′(x)+f(x)‌tan‌x+2f(x)‌tan‌2‌x ‌‌+...+nf(x)‌tan‌n‌x =f′(x)+[(sin‌x⋅cos‌2‌x...cos‌n‌x) ‌‌+(2‌cos‌x⋅sin‌2x...cos‌n‌x) ‌‌‌+...+(n‌cos‌x⋅cos‌2‌x...sin‌nx)]=‌f′(x)−f′(x)⇒0Hence,
f′(x)+(r‌tan‌r‌x)f(x)=0
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