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TMUA 2020 · Paper 1 · Question 11 of 20

TMUA 2020 Paper 1 Question 11

Differentiation and integration — Areas above and below the axis · equal regions. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 1Differentiation and integrationAreas above and below the axis · equal regions6 options

The original question includes a diagram: A upward parabola crossing the x-axis at x = 2 and x = q; region R is between the y-axis, the x-axis and the curve for 0 ≤ x ≤ 2, and region S is the region below the axis between x = 2 and x = q.

The diagram shows a quadratic function passing through (2,  0) and (q,  0), where q > 2. The curve opens upwards. Region R is enclosed by the y-axis, the x-axis and the curve between x = 0 and x = 2; region S is enclosed by the curve and the x-axis between x = 2 and x = q.

What is the value of q such that the area of region R equals the area of region S?

  1. A√6
  2. B3
  3. C185
  4. D4
  5. E6
  6. F335
Show the answer and worked solution
answer · E
  1. A√6
  2. B3
  3. C185
  4. D4
  5. E6
  6. F335
The leading coefficient does not matter, since both areas scale by it, so take f(x) = (x−2)(x−q). Equal areas means the signed integral over the whole stretch is zero: ∫0q (x−2)(x−q) dx = 0, because the part above the axis counts positive and the part below counts negative. Expanding gives ∫0q (x2 − (2+q)x + 2q)dx = q33 − (2+q)q22 + 2q2 = −q36 + q2. Setting this to zero gives q2(1 − q6) = 0, so q = 6.