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Question

If function f(x)=λcos2x+2sinx has only one extremum point in [0,π], then the value of λ can not be

A
14
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B
25
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C
37
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D
78
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Solution

The correct option is D 78
Given : f(x)=λcos2x+2sinx
f(x)=2λsin2x+2cosx =2cosx(12λsinx)
for critical points : f(x)=0
cosx=0 or sinx=12λ
cosx=0 gives x=π2
So, there should not be any solution for sinx=12λin [0,π], as f(x) has only one extremum point in [0,π].
12λ>1 or 12λ<0
0<λ<12 or λ<0

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