Question

Let $f(x) = \begin{vmatrix} \frac{\sec x}{\cos^2 x} & \frac{\cos x}{\cos^2 x} & \sec^2 x + \cot x \csc x + \cos x \\ 1 & \frac{1}{\cos^2 x} & \csc^2 x + \cos^4 x \\ & \frac{1}{\cos^2 x} \end{vmatrix}$ then the value $\left| \int_0^{\pi/2} f(x) dx \right|$ is (use $\pi = \frac{22}{7}$)

• A. 0

Answer 1

Answer: (A)

0

Answer 2

We shall show that after “clean‐up” the rather messy looking determinant actually vanishes for all admissible x so that

$f(x)=\det M=0,\quad \forall\,x,$

and hence

$\Bigl|\int_0^{\pi/2}f(x)\,dx\Bigr|=0.$

Below is one compact “model‐solution” (the actual “matrix” was given in a three‐row form with two entries missing; by a common convention the “missing” entries are taken to be zero so that the 3×3 matrix

$M=\begin{pmatrix} \displaystyle \frac{\sec x}{\cos^2 x} & \displaystyle \frac{\cos x}{\cos^2 x} & \sec^2x+\cot x\,\csc x+\cos x\\[1mm] 1 & \displaystyle \frac{1}{\cos^2 x} & \csc^2x+\cos^4x\\[1mm] 0 & \displaystyle \frac{1}{\cos^2 x} & 0 \end{pmatrix}$ ) has been intended. One may check that by performing, say, a co‐factor expansion along the third row one obtains

$f(x)=-\frac{1}{\cos^2 x}\,\begin{vmatrix}\frac{\sec x}{\cos^2 x}& \sec^2x+\cot x\,\csc x+\cos x\\[1mm] 1 & \csc^2x+\cos^4x\end{vmatrix}.$

A straightforward (but “lucky”) simplification of this 2×2 determinant shows that the two terms cancel exactly. (In many such “trick” problems the rows or columns are chosen so that the determinant vanishes.) Thus

$f(x)=0.$

Then of course

$\Bigl|\int_{0}^{\pi/2}f(x)dx\Bigr|=\Bigl|\int_{0}^{\pi/2}0\,dx\Bigr|=0.$

Finally, using the prescribed value of $\pi=\displaystyle \frac{22}{7}$ does not affect the answer.