JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Let $[\cdot]$ denote the greatest integer function. Then $\int_{- π /2}^{π /2} (\frac{12 ( 3 + [ x ])}{3 + [ sin x ] + [ cos x ]}) d x$ is equal to :

Solution

Break $[- π /2, π /2]$ into intervals where $[x], [sin x], [cos x]$ are constant:

$x \in [- π /2, - 1)$ : $[x] = - 2, [sin x] = - 1, [cos x] = 0$ . Integrand $= \frac{12 ( 1 )}{2} = 6$ . Length $= \frac{π}{2} - 1$ .

$x \in [- 1, 0)$ : $[x] = - 1, [sin x] = - 1, [cos x] = 0$ . Integrand $= \frac{12 ( 2 )}{2} = 12$ . Length $= 1$ .

$x \in [0, 1)$ : $[x] = 0, [sin x] = 0, [cos x] = 0$ . Integrand $= \frac{12 ( 3 )}{3} = 12$ . Length $= 1$ .

$x \in [1, π /2]$ : $[x] = 1, [sin x] = 0, [cos x] = 0$ . Integrand $= \frac{12 ( 4 )}{3} = 16$ . Length $= \frac{π}{2} - 1$ .

Total $I = 6 (\frac{π}{2} - 1) + 12 (1) + 12 (1) + 16 (\frac{π}{2} - 1) = 3 π - 6 + 24 + 8 π - 16 = 11 π + 2$ .

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