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Mathematics Question on Functions

f(x)=xex−1+x2+2cos⁡3x2f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} on R-\left\\{0\right\\} is

A

One one function

B

Bisection

C

Algebraic function

D

Even function

Answer

Even function

Explanation

Solution

Given function
f(x)=xex−1+x2+2cos⁡3x2f(x)=\frac{x}{e^{x}-1}+\frac{x}{2}+2 \cos ^{3} \frac{x}{2} on
R−0R-\\{0\\}.
∵f(−x)=−xe−x−1−x2+2cos⁡3(−x2)\because f(-x) =\frac{-x}{e^{-x}-1}-\frac{x}{2}+2 \cos ^{3}\left(-\frac{x}{2}\right)
=−xex1−ex−x2+2cos⁡3x2=\frac{-x e^{x}}{1-e^{x}}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}
=xexex−1−x2+2cos⁡3x2=\frac{x e^{x}}{e^{x}-1}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}
=x(ex−1+1)ex−1−x2+2cos⁡3x2=\frac{x\left(e^{x}-1+1\right)}{e^{x}-1}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}
=x+xex−1−x2+2cos⁡3x2=x+\frac{x}{e^{x}-1}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}
=xex−1+x2+2cos⁡3x2=f(x)=\frac{x}{e^{x}-1}+\frac{x}{2}+2 \cos ^{3} \frac{x}{2}=f(x)
∵f(−x)=f(x),∀x∈R−0\because f(-x)=f(x), \forall x \in R-\\{0\\}
∴f(x)\therefore f(x) is an even function.