IIT JEE 1982 Maths - 'Adapted' - Subjective to Multi-Correct Q19

$\begin{aligned} I & = \int_{-1}^\frac 32 |x \sin (\pi x)| \ dx \\ & = {\color{#3D99F6}\int_{-1}^0 |x \sin (\pi x)| \ dx} + {\color{#D61F06}\int_0^\frac 32 |x \sin (\pi x)| \ dx} & \small \color{#3D99F6} \text{Note that } \sin (\pi x) < 0, \ -1< x < 0 \\ & = {\color{#3D99F6}\int_0^1 x \sin (\pi x) \ dx} + {\color{#D61F06}\int_0^1 |x \sin (\pi x)| \ dx + \int_1^\frac 32 |x \sin (\pi x)| \ dx} & \small \color{#D61F06} \text{Note that } \sin (\pi x) < 0, \ 1 < x \le \frac 32 \\ & = \int_0^1 x \sin (\pi x) \ dx + {\color{#D61F06}\int_0^1 x \sin (\pi x) \ dx - \int_1^\frac 32 x \sin (\pi x) \ dx} \\ & = 2 \int_0^1 x \sin (\pi x) \ dx - \int_1^\frac 32 x \sin (\pi x) \ dx & \small \color{#3D99F6}\text{By integration by parts} \\ & = -\frac {2x\cos (\pi x)}\pi \bigg|_0^1 + 2\int_0^1 \frac {\cos (\pi x)}\pi dx + \frac {x\cos (\pi x)}\pi \bigg|_1^\frac 32 - \int_1^\frac 32 \frac {\cos (\pi x)}\pi dx \\ & = \frac 2 \pi + \frac {2 \sin (\pi x)}{\pi^2} \bigg|_0^1 + \frac 1 \pi - \frac {\sin (\pi x)}{\pi^2} \bigg|_1^\frac 32 \\ & = \frac 2 \pi + 0 + \frac 1 \pi + \frac 1{\pi^2} \bigg|_1^\frac 32 \\ & = \frac 3 \pi + \frac 1{\pi^2} = \sum_{i=1}^2 \frac {a_i}{b_i\pi^i} \end{aligned}$

Let us check the statements:

• A: $n=2$ --- true (1)
• B: $a_1=3$ --- true (1)
• C: $b_1=1 \ne 3$ --- false (9)
• D: $a_2=1$ --- true (1)

Therefore, the answer is $\boxed{1191}$ .