Question

Let $f\left(x\right)=(x^3\sec x+3)\sqrt{16-x^2},$ then $f\left(x\right)$ is

• A. an odd function
• B. an even function
• C. neither odd nor even function
• D. both even and odd function

Answer 1

The correct option is C neither odd nor even function

$f\left(x\right)=(x^3\sec x+3)\sqrt{16-x^2}=x^3\sec x\sqrt{16-x^2}+3\sqrt{16-x^2}=g\left(x\right)+h\left(x\right)$

$g\left(x\right)=x^3\sec x\sqrt{16-x^2}$
$\Rightarrow g(-x)=-g\left(x\right)$
$\Rightarrow g\left(x\right)$ is an odd function.

$h\left(x\right)=3\sqrt{16-x^2}$
$\Rightarrow h(-x)=h\left(x\right)$
$\Rightarrow h\left(x\right)$ is an even function.

$\therefore f\left(x\right)$ is neither odd nor even function