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JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Area Under Curves

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✓ Correct
(D)12

Answer: (C)

Solution

With

α = 0

,

y = ∣ - 5∣ - ∣1∣ + 0 = 4

, so

f (0) = \int_{0}^{1} (4 - x) d x = 4 - \frac{2}{3} = \frac{10}{3}

. With

α = 1

and

0 < x < 1

,

∣ x - 5∣ = 5 - x

and

∣1 - x ∣ = 1 - x

, so

y = (5 - x) - (1 - x) + x^{2} = 4 + x^{2}

, giving

f (1) = \int_{0}^{1} (4 + x^{2} - x) d x = 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}

. Therefore

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

.

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option

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