Limits; derivative as gradient and rate of change practice questions.

Determine the value of lim(x→3) (x² − 9)/(x − 3).

• 0
• 6
• 9
• The limit does not exist

Answer: 6

Factorise the numerator: (x − 3)(x + 3)/(x − 3) = x + 3 for x ≠ 3, so the limit is 3 + 3 = 6. The gap at x = 3 does not stop the limit existing.

Let f(x) = 3x². Calculate f'(2).

• 6
• 24
• 12
• 36

Answer: 12

f'(x) = 3 × 2x = 6x, so f'(2) = 6 × 2 = 12. Answering 6 means stopping at the coefficient 3 × 2 without substituting x = 2; 24 comes from doubling f(2) = 12 instead of differentiating.

Determine the gradient of the tangent to the curve y = x³ − 4x at the point where x = 1.

• −1
• −3
• 3
• 7

Answer: −1

dy/dx = 3x² − 4, so at x = 1 the gradient is 3 − 4 = −1. Note that −3 is the y-coordinate of the point, not the gradient.

A particle moves in a straight line with displacement s(t) = t² − 6t + 8 metres after t seconds. Calculate its velocity at t = 4.

• 0 m s⁻¹
• 8 m s⁻¹
• −6 m s⁻¹
• 2 m s⁻¹

Answer: 2 m s⁻¹

Velocity is the rate of change of displacement: v(t) = s'(t) = 2t − 6, so v(4) = 2. Evaluating s(4) = 0 gives the displacement, not the velocity.

Which expression defines the derivative f'(a) from first principles?

• lim(h→0) [f(a + h) − f(a)] / h
• lim(h→0) [f(a + h) + f(a)] / h
• lim(h→a) [f(h) − f(a)] / h
• [f(a + h) − f(a)] / h with h = 0

Answer: lim(h→0) [f(a + h) − f(a)] / h

The derivative is the limit of the difference quotient [f(a + h) − f(a)]/h as h → 0. Setting h = 0 directly gives 0/0, which is undefined.

The tangent to a curve at point P has gradient 2. Determine the gradient of the normal to the curve at P.

• 2
• −2
• −1/2
• 1/2

Answer: −1/2

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: −1/2. Taking only the negative (−2) or only the reciprocal (1/2) are classic errors.