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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Area Under Curves

Let f(α) denote the area of the region in the first quadrant...

Let f(α)f(\alpha) denote the area of the region in the first quadrant bounded by x=0,x=1,y2=xx=0, x=1, y^{2}=x and y=∣αx−5∣−∣1−αx∣+αx2y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}. Then (f(0)+f(1))(f(0)+f(1)) is equal to

Options

  1. A12
  2. B14
  3. C7
  4. D9

Correct answer

C. 7

Step-by-step solution

For α=0\alpha = 0: curve is y=5−1=4y = 5 - 1 = 4 (horizontal line), area =∫01(4−x)dx=4−23=103= \int_0^1 (4 - \sqrt{x}) dx = 4 - \frac{2}{3} = \frac{10}{3}
For α=1\alpha = 1 and 0≤x≤10 \leq x \leq 1: ∣x−5∣=5−x|x-5| = 5-x, ∣1−x∣=1−x|1-x| = 1-x, so y=4+x2y = 4 + x^2
Area =∫01(4+x2−x)dx=4+13−23=113= \int_0^1 (4 + x^2 - \sqrt{x}) dx = 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}
f(0)+f(1)=103+113=7f(0) + f(1) = \frac{10}{3} + \frac{11}{3} = 7
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