Question

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

• A. $-\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. none

Answer 1

Answer: (A)

$-\frac{1}{3}$

$f\left(x\right)=\left|\begin{array}{ccc}\cos x& 1& 0\\ 1& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|=\cos x\left(4{\cos }^{2}x-1\right)-1\left(2{\cos }^{2}x\right)+0=4{\cos }^{3}x-\cos x-2\cos x=4{\cos }^{3}x-3\cos x=\cos 3x$
$\therefore {\int }_0^{\frac{\pi }{2}}f\left(x\right)={\int }_0^{\frac{\pi }{2}}\cos 3x={\left(\frac{\sin 3x}{3}\right)}_0^{\frac{\pi }{2}}=-\frac{1}{3}$