JEE Main 2026 Jan 24 Shift 2, Mathematics Q12: Area as Definite Integral

JEE Main2026Jan 24, Shift 2Mathematics

Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0$ , $x = 1$ , $y^{2} = x$ and $y = ∣ α x - 5∣ - ∣1 - α x ∣ + α x^{2}$ . Then $f (0) + f (1)$ is equal to

A
9
B
14
C
7
D
12

Solution

Case $α = 0$ : $y = ∣0 - 5∣ - ∣1 - 0∣ + 0 = 5 - 1 = 4$ . Region bounded by $y = x$ below and $y = 4$ above.

$f (0) = \int_{0}^{1} (4 - x) d x = [4 x - \frac{2}{3} x^{3/2}]_{0}^{1} = 4 - \frac{2}{3} = \frac{10}{3}$ .

Case $α = 1$ : For $x \in (0, 1)$ , $∣ x - 5∣ = 5 - x$ and $∣1 - x ∣ = 1 - x$ , so $y = (5 - x) - (1 - x) + x^{2} = 4 + x^{2}$ .

$f (1) = \int_{0}^{1} ((4 + x^{2}) - x) d x = [4 x + \frac{x ^{3}}{3} - \frac{2}{3} x^{3/2}]_{0}^{1} = 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}$ .

$f (0) + f (1) = \frac{10}{3} + \frac{11}{3} = \frac{21}{3} = 7$ .

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JEE Main 2026 Jan 24 Shift 2, Mathematics Q12: Area as Definite Integral

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