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Question

If k121e(x3+x)dxk2, then

A
k1=e3
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B
k2=e10
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C
k1=e2
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D
k2=e11
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Solution

The correct option is C k1=e2
e(x3+x) is increasing function in (1,2)
Minimum value =e2
Maximum value =e10
k1=e2(21)=e2
k2=e10(21)=e10

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