Question

If $f\left(x\right)=\left|\begin{array}{ccc}\cos x& 1& 0\\ 1& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|$, then ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx$

• A. $\frac{1}{4}$
• B. $-\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is B $-\frac{1}{3}$

Solving determinant by expanding across row 1, we get

$f\left(x\right)=\cos x\left|\begin{array}{cc}2\cos x& 1\\ 1& 2\cos x\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 0& 2\cos x\end{array}\right|+0\left|\begin{array}{cc}1& 2\cos x\\ 0& 1\end{array}\right|$

$\Rightarrow f\left(x\right)=\cos x(4{\cos }^{2}x-1)-1(2\cos x-0)+0(1-0)$

$\Rightarrow f\left(x\right)=4{\cos }^{3}x-3\cos x$

$\Rightarrow f\left(x\right)=\cos 3x$

Now, ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx$

$={\int }_0^{\frac{\pi }{2}}\cos 3xdx$

$={\left(\frac{\sin 3x}{3}\right)}_0^{\frac{\pi }{2}}$

$=\frac{1}{3}(\sin \frac{3\pi }{2}-\sin 0)$

$=\frac{1}{3}(-1-0)$

$=-\frac{1}{3}$