If the normal to the curve y(x) = ∫₀ˣ (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of | a + 6b | is equal to ___________.

Solution

Maths sir · 4 steps

The integral hands you y'(x) for free, so match it to the tangent slope and solve for a first.

1. Slope from integral. By the fundamental theorem of calculus, y'(x) = 2x² - 15x + 10. You do not need to integrate to find the slope.
2. Tangent slope. The line x + 3y = -5 has slope -1/3, so the normal at (a, b) has slope -1/3. The tangent is perpendicular to it, so its slope is 3.
3. Find a. Set 2a² - 15a + 10 = 3, which gives 2a² - 15a + 7 = 0, that is (2a - 1)(a - 7) = 0. Since a > 1, take a = 7.

   b = ∫₀⁷ (2t² - 15t + 10) dt = (2 × 343)/3 - (15 × 49)/2 + 70 = (1372 - 2205 + 420)/6 = -413/6

4. So. 6b = -413, so a + 6b = 7 - 413 = -406, and |a + 6b| = 406.

The common slip: -406. Dropping the modulus leaves a + 6b = -406. The question asks for the absolute value, which is 406.

The idea to keep. When a curve is defined as ∫₀ˣ f(t) dt, its slope is just f(x). Turn the normal's direction into the tangent slope with m₁ m₂ = -1, then integrate only to get the point's y-value.

Answer from the NTA final answer key. Solution by PrepWiser.