The area of the region bounded by the curve y = x − x² and the line y = mx, where m > 0, is 9/2 sq units. The value of m is:

The area of the region bounded by the curve y = x − x² and the line y = mx, where m > 0, is 9/2 sq units. The value of m is:

• A. 4
• B. 3
• C. 2
• D. 1

Answer: A) 4

Explanation: Intersection: x − x² = mx → x(1 − m − x) = 0 → x = 0, 1 − m. Since m > 0 and area positive, we need 1−m > 0 → m < 1, but area is symmetric? Actually curve is y = x − x², line y = mx. Area = ∫[0 to 1−m] (x − x² − mx) dx = ∫[0 to 1−m] ((1−m)x − x²) dx = [(1−m)x²/2 − x³/3]₀^(1−m) = (1−m)³/2 − (1−m)³/3 = (1−m)³/6 = 9/2 → (1−m)³ = 27 → 1−m = 3 → m = −2 (not possible). If m > 1, parabola is below line? Actually area = ∫[0 to a] (mx − (x − x²)) dx where a solves x − x² = mx → x = 0, 1−m (negative if m>1). So region is different. Actually for m>1, line is above parabola from x=0 to some positive? No, x − x² is positive only in [0,1]. If m>1, mx > x − x² on [0,1]. The intersection x = 1−m is negative. So bounded region is in x<0. But area positive, limits from 1−m to 0. Area = ∫[1−m to 0] (x − x² − mx) dx = ∫[1−m to 0] ((1−m)x − x²) dx. Let u = 1−m, u is negative. Area = [u x²/2 − x³/3]_u⁰ = 0 − (u³/2 − u³/3) = −u³/6 = |u|³/6 = 9/2 → |1−m|³ = 27 → |1−m| = 3 → m−1 = 3 → m = 4.