Question

If $f\left(x\right)=\left|\begin{array}{ccc}x& \cos x& e^{x^2}\\ \sin x& x^2& secx\\ \tan x& 1& 2\end{array}\right|$ then the value of ${\int }_{-\pi /2}^{\pi /2}$ f(x)dx is equal to.

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

Answer: (A)

$0$

Given : $f\left(x\right)=\left|\begin{array}{ccc}x& \cos x& x^{x^2}\\ \sin x& x^2& \sec x\\ \tan x& 1& 2\end{array}\right|$

$f(-x)=\left|\begin{array}{ccc}-x& \cos x& x^{x^2}\\ -\sin x& x^2& \sec x\\ -\tan x& 1& 2\end{array}\right|=-\left|\begin{array}{ccc}x& \cos x& x^{x^2}\\ \sin x& x^2& \sec x\\ \tan x& 1& 2\end{array}\right|f(-x)=-f\left(x\right)$

Since $f\left(x\right)$ is an odd function

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\left(x\right)dxI={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f(0-x)dxI={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f(-x)dx-{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\left(x\right)dxI=0$

Hence the correct answer is $0$