(a) Given: \( g(x) = x^2 – \frac{3}{x} \). \[ g(x) = x^2 – 3x^{-1} \] \[ \begin{aligned} g'(x) &= \frac{d}{dx}(x^2) – \frac{d}{dx}(3x^{-1}) \\ &= 2x – 3 \times (-1)x^{-2} \\ &= 2x + \frac{3}{x^2} \end{aligned} \] Answer: \( g'(x) = 2x + \frac{3}{x^2} \) A1 A1 [2 marks]
(b) From part (a): \( g'(x) = 2x + \frac{3}{x^2} \). \[ g'(1) = 2 \times 1 + \frac{3}{1^2} = 2 + 3 = 5 \] Answer: Gradient of L = 5 M1 A1 [2 marks]
(c) EITHER
Solve for points where the tangent is parallel to \( L \): \( g'(x) = 5 \). \[ 2x + \frac{3}{x^2} = 5 \] M1
\[ 2x \times x^2 + 3 = 5x^2 \] \[ 2x^3 – 5x^2 + 3 = 0 \] Solutions: \( x \approx -0.686, 1, 2.19 \). A1
OR
Sketch of \( y = g'(x) \) with \( y = 5 \).
M1
Three points of intersection at \( x \approx -0.686, 1, 2.19 \). A1
THEN there are two other tangent lines to \( g(x) \) parallel to \( L \). A1
[3 marks]
