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Question

# Let f(x)=xex−1+x2+1 and g is either odd or even function. If fog is defined, then

A
fog is odd function
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B
fog is even function
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C
f(x) is even function
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D
f(x) is odd function
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Solution

## The correct option is C f(x) is even functionf(x)=xex−1+x2+1 ⇒f(−x)=−xe−x−1−x2+1 ⇒f(−x)=−xex1−ex−x2+1 ⇒f(−x)=−2xex−x+xex2(1−ex)+1 ⇒f(−x)=x(1−ex)−2x2(1−ex)+1 ⇒f(−x)=xex−1+x2+1 ⇒f(−x)=f(x) ∴f(x) is even function. As we know composite of two functions (which are either odd or even) will be even if atleast one of them is even, so fog is even function.

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