JAMB Mathematics 1999

9 reviewed questions with answers and explanations.

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Question 2

Solve √2/(x + √2) = 1/(x − √2), excluding zero denominators.

  1. 3√2 + 4
  2. 3√2 − 4
  3. 3 − 2√2
  4. 4 + 2√2
Answer and explanation

A: 3√2 + 4

Cross-multiplying gives√2x − 2 = x + √2, so x = (2 + √2)/(√2 − 1). Rationalising gives 4 + 3√2.

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Question 8

If log₈10 = x, express log₈5 in terms of x.

  1. x/2
  2. x − 1/4
  3. x − 1/3
  4. x − 1/2
Answer and explanation

C: x − 1/3

Since 5 = 10/2, log₈5 = log₈10 − log₈2. Because 8^(1/3) = 2, the result is x − 1/3. The logarithm of a quotient is the difference of its logarithms.

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Question 12

A convergent geometric progression has first term twice its common ratio and infinite sum 8. Find the sum of its first two terms.

  1. 8/5
  2. 8/3
  3. 72/25
  4. 56/9
Answer and explanation

C: 72/25

Writing a = 2r gives 2r/(1 − r) = 8, so r = 4/5 and a = 8/5. The first two terms sum to a(1 + r) = 72/25.

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Question 16

Divide 4x³ − 3x + 1 by 2x − 1.

  1. 2x² − x + 1
  2. 2x² − x − 1
  3. 2x² + x + 1
  4. 2x² + x − 1
Answer and explanation

D: 2x² + x − 1

The quotient is 2x² + x − 1. Multiplying by 2x − 1 gives 4x³ − 3x + 1, with no remainder. Each coefficient agrees with the original cubic polynomial.

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Question 20

Two real numbers have positive difference p and sum 2p. Find the larger number.

  1. p/2
  2. 3p/2
  3. 5p/2
  4. 3p
Answer and explanation

B: 3p/2

If the larger number is L and the smaller is S, then L − S = p and L + S = 2p. Adding gives 2L = 3p, so L = 3p/2.

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Question 23

Find the locus of P(x,y) such that PV = PW, where V = (1,1) and W = (3,5).

  1. 2x + 2y = 9
  2. 2x + 3y = 8
  3. 2x + y = 9
  4. x + 2y = 8
Answer and explanation

D: x + 2y = 8

Equating squared distances gives (x − 1)² + (y − 1)² = (x − 3)² + (y − 5)². Cancelling and simplifying gives 4x + 8y = 32, hence x + 2y = 8.

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Question 29

Find the tangent of the acute angle between 2x + y = 3 and 3x − 2y = 5.

  1. −7/4
  2. 7/8
  3. 7/4
  4. 7/2
Answer and explanation

C: 7/4

The slopes are −2 and 3/2. For the acute angle, tan θ = |(m₂ − m₁)/(1 + m₁m₂)| = |(7/2)/(−2)| = 7/4.

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Question 30

From P, the bearings of Q and R are N67°W and N23°E. From Q, R is on bearing N68°E. If PQ = 150 m, find PR.

  1. 120 m
  2. 140 m
  3. 150 m
  4. 160 m
Answer and explanation

C: 150 m

The angle at P is 67° + 23° = 90°. The reverse bearing QP is 113°, so angle PQR is 113° − 68° = 45°. The remaining angle is also 45°, giving PR = PQ = 150 m.

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Question 40

At what value of x does y = x³ − x have a local minimum?

  1. −√3
  2. −√3/2
  3. √3/3
  4. √3
Answer and explanation

C: √3/3

Stationary points satisfy 3x² − 1 = 0. The second derivative is 6x, which is positive at x = 1/√3. Thus the local minimum occurs at √3/3.

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