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Question : 125 of 150
Marks:
+1,
-0
Solution:
We have,
sin‌x+2‌sin‌2‌x+3‌sin‌3‌x= ⇒−sin‌x−2‌sin‌2‌x−3‌sin‌3‌x=0 Let,
f(x)=x+cos‌x+cos‌2‌x+cos‌3‌x f(0)=0+1+1+1=3 f()=4+0−1+0=3 ∴
f(0)=f() Now,
f′(x)=−sin‌x−2‌sin‌2‌x−3‌sin‌3‌x By Rolles theorem,
at least one root of
f(x) lies between
(0,)≤[0,] i.e.
−sin‌x−2‌sin‌2‌x−3‌sin‌3‌x=0 for some
x∈[0,] ⇒sin‌x+2‌sin‌2‌x+3‌sin‌3‌x= for some
x∈[0,]
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