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Question : 108 of 150
Marks:
+1,
-0
Solution:
Since,
f(x)=cos‌x+‌cos‌2‌x−‌cos‌3‌x f′(x)=−sin‌x−(2)‌sin‌2‌x+(3)‌sin‌3‌x f′(x)=−sin‌x−sin‌2‌x+sin‌3‌x f′(x)=−(sin‌x+sin‌2‌x)+sin‌3‌x f′(x)=−(2‌sin‌.cos‌)+sin‌(3x) {∵sin‌C+sin‌D=2‌sin‌‌cos‌} f′(x)=−2‌sin‌.cos‌+2‌sin‌.cos‌ [∵sin‌2‌θ=2‌sin‌θ‌cos‌θ] f′(x)=2‌sin‌[cos‌−cos‌] f′(x)=2‌sin‌[−2‌sin‌x‌sin‌] f′(x)=−4‌sin‌‌sin‌x‌sin‌ For greatest and least value,
f′(x)=0 4‌sin‌‌sin‌x.sin‌=0 sin‌x=0 x=nπ,n∈Z sin‌=0 =nπ x=2nπ,n∈Z; sin‌=0 =nπ x=,n∈Z Now, For
x=0,π, ∵f(x)=cos‌x+‌cos‌2‌x−‌cos‌3‌x Now, at
x=0 f(0)=cos‌0+‌cos‌2(0)−‌cos‌3(0) =1+×1−= f(π)=cos‌π+‌cos‌2‌π−‌cos‌3‌π f(π)=−1+×1−(−1)=−1∕6 and
f()=cos‌+‌cos‌2() −‌cos‌3() =−+(−)−=− Hence, greatest value of
f(x)=7∕6 Least value of
f(x)=− Difference between the greatest and least value
=−(−)=+==
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