Q115 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

If $f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|$ then $\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x=$

  • A
    $\frac{-1}{6}$
  • Correct answer
    B
    $\frac{-1}{9}$
  • C
    $\frac{-2}{9}$
  • D
    $\frac{1}{27}$

Solution

Given that, $f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|$ $C_1 \rightarrow C_1-2 C_2-3 C_3$ $f(x)=\left|\begin{array}{ccc}\sin x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 0 & \frac{3}{2} & \frac{4}{3} \sin x \\ 0 & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|$ On expanding, we get $f(x)=\sin x\left|\begin{array}{cc}\frac{3}{2} & \frac{4}{3} \sin x \\ \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|+0+0$ $=\sin x\left(\frac{3}{2} \times \frac{1}{3}-\frac{4}{3} \sin x \times \frac{1}{2} \sin x\right)$ $=\sin x\left(\frac{1}{2}-\frac{2}{3} \sin ^2 x\right)$ $f(x)=\frac{1}{6}\left(3 \sin x-4 \sin ^3 x\right)$ $f(x)=\frac{\sin 3 x}{6} \quad\left[\because \sin 3 x=3 \sin x-4 \sin ^3 x\right]$ Differentiating on both sides w.r.t. ' $x$ ', we get $f^{\prime}(x)=\frac{d}{d x}\left(\frac{\sin 3 x}{6}\right)=\frac{1}{6} \cdot \cos 3 x \cdot 3$ $f^{\prime}(x)=\frac{\cos 3 x}{2}$ Let $\quad I=\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x$ $I=\int_0^{\pi / 2}\left(\frac{\sin 3 x}{6}+\frac{\cos 3 x}{2}\right) d x$ $\ldots$ (i) or $\quad I=\int_0^{\pi / 2}\left(\frac{\sin 3\left(\frac{\pi}{2}-x\right)}{6}+\frac{\cos 3\left(\frac{\pi}{2}-x\right)}{2}\right) d x$ $=\int_0^{\pi / 2}\left(\frac{\sin \left(\frac{3 \pi}{2}-3 x\right)}{6}+\frac{\cos \left(\frac{3 \pi}{2}-3 x\right)}{2}\right) d x$ $I=\int_0^{\pi / 2}\left(-\frac{\cos 3 x}{6}-\frac{\sin 3 x}{2}\right) d x$ $\ldots$ (ii) Adding Eqs. (i) and (ii), we get $2 I=\int_0^{\pi / 2}\left(\frac{\cos 3 x}{2}-\frac{\cos 3 x}{6}+\frac{\sin 3 x}{6}-\frac{\sin 3 x}{2}\right) d x$ $2 I=\frac{1}{3} \int_0^{\pi / 2}(\cos 3 x-\sin 3 x) d x$ or $2 I=\frac{1}{3}\left[\frac{\sin 3 x}{3}+\frac{\cos 3 x}{3}\right]_0^{\pi / 2}$ $=\frac{1}{9}\left[\sin 3 \times \frac{\pi}{2}+\cos 3 \times \frac{\pi}{2}-\sin 3 \times 0-\cos 3 \times 0\right]$ $=\frac{1}{9}[-1+0-0-1]=-\frac{2}{9}$ or $I=-\frac{1}{9}$

Hence, the correct answer is option: 'B'.

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