If function f(x)=λsinx+cosx has two extremum points in [0,2π], then the value of λ can be
A
−313
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B
0
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C
27
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D
1
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Solution
The correct option is D1 Given : f(x)=λsinx+cosx ⇒f′(x)=λcosx−sinx
for critical points : f′(x)=0 ⇒λcosx−sinx=0⇒tanx=λ
Plotting the graph :
Clearly, for λ>0 or λ<0, we get two solutions
but for λ=0 gives three solutions