19M.2.SL.TZ2.T_5

pestleMathematicsAISLPaper 219M· sl-5-1-introduction-of-differential-calculussource ↗

Consider the function $f\left( x \right) = \frac{1}{3}{x^3} + \frac{3}{4}{x^2} - x - 1$ .

Find $f'\left( x \right)$ .

[3]

Find the gradient of the graph of $y = f\left( x \right)$ at $x = 2$ .

[2]

Find the equation of the tangent line to the graph of $y = f\left( x \right)$ at $x = 2$ . Give the equation in the form $ax + by + d = 0$ where, $a$ , $b$ , and $d \in \mathbb{Z}$ .

[2]

Markscheme / solution

${x^2} + \frac{3}{2}x - 1$ (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if there are extra terms.

[3 marks]

${2^2} + \frac{3}{2} \times 2 - 1$ (M1)

Note: Award (M1) for correct substitution of 2 in their derivative of the function.

6 (A1)(ft)(G2)

Note: Follow through from part (d).

[2 marks]

$\frac{8}{3} = 6\left( 2 \right) + c$ (M1)

Note: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.

$c = - \frac{{28}}{3}$

$\left( {y - \frac{8}{3}} \right) = 6\left( {x - 2} \right)$ (M1)

Note: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.

$y = 6x - \frac{{28}}{3}\,\,\left( {y = 6x - 9.33333 \ldots } \right)$ (M1)

Note: Award (M1) for their answer to (e) and intercept $- \frac{{28}}{3}$ substituted in the gradient-intercept line equation.

$- 18x + 3y + 28 = 0$ (accept integer multiples) (A1)(ft)(G2)

Note: Follow through from parts (a) and (e).

[2 marks]

Examiners’ report

[N/A]