Q48 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
$6 \int_{0}^{\pi}|(\sin 3 x+\sin 2 x+\sin x)| d x$ is equal to $\_\_\_\_$.
Correct answer: 17
Solution
Using $\sin 3x + \sin x = 2\sin 2x \cos x$:
$\sin 3x + \sin 2x + \sin x = \sin 2x(2\cos x + 1) = 2\sin x \cos x(2\cos x + 1)$
Zeros in $[0, \pi]$: $x = 0, \pi/2, 2\pi/3, \pi$
Expression is positive on $(0, \pi/2)$, negative on $(\pi/2, 2\pi/3)$, positive on $(2\pi/3, \pi)$
Let $F(x) = -\frac{\cos 3x}{3} - \frac{\cos 2x}{2} - \cos x$
$F(0) = -\frac{11}{6}$, $F(\pi/2) = \frac{1}{2}$, $F(2\pi/3) = \frac{5}{12}$, $F(\pi) = \frac{5}{6}$
$\int_0^{\pi/2} = \frac{7}{3}$, $\left|\int_{\pi/2}^{2\pi/3}\right| = \frac{1}{12}$, $\int_{2\pi/3}^{\pi} = \frac{5}{12}$
Total $= \frac{7}{3} + \frac{1}{12} + \frac{5}{12} = \frac{17}{6}$
$6 \times \frac{17}{6} = 17$
$\sin 3x + \sin 2x + \sin x = \sin 2x(2\cos x + 1) = 2\sin x \cos x(2\cos x + 1)$
Zeros in $[0, \pi]$: $x = 0, \pi/2, 2\pi/3, \pi$
Expression is positive on $(0, \pi/2)$, negative on $(\pi/2, 2\pi/3)$, positive on $(2\pi/3, \pi)$
Let $F(x) = -\frac{\cos 3x}{3} - \frac{\cos 2x}{2} - \cos x$
$F(0) = -\frac{11}{6}$, $F(\pi/2) = \frac{1}{2}$, $F(2\pi/3) = \frac{5}{12}$, $F(\pi) = \frac{5}{6}$
$\int_0^{\pi/2} = \frac{7}{3}$, $\left|\int_{\pi/2}^{2\pi/3}\right| = \frac{1}{12}$, $\int_{2\pi/3}^{\pi} = \frac{5}{12}$
Total $= \frac{7}{3} + \frac{1}{12} + \frac{5}{12} = \frac{17}{6}$
$6 \times \frac{17}{6} = 17$
Hence, the correct answer is 17.