Write down the equation of the vertical asymptote of the graph of y = g(x) .

Write down g′(x) .

The line T is the tangent to the graph of y = g(x) at the point where x = 1. The gradient of T is 3.

Show that b = 5.

The line T is the tangent to the graph of y = g(x) at the point where x = 1. The gradient of T is 3.

Find the equation of T.

Using your graphic display calculator find the coordinates of the point where the graph of y = g(x) intersects the x-axis.

(i) Sketch the graph of y = g(x) for −2 ≤ x ≤ 5 and −15 ≤ y ≤ 25, indicating clearly your answer to part (e).

(ii) Draw the line T on your sketch.

Using your graphic display calculator find the coordinates of the local minimum point of y = g(x) .

Write down the interval for which g(x) is increasing in the domain 0 < x < 5 .

x = 0     (A1)(A1)

Notes: Award (A1) for x=constant, (A1) for 0. Award (A0)(A0) if answer is not an equation.

[2 marks]

\(b - \frac{2}{{{x^3}}}\)     (A1)(A1)(A1)

Note: Award (A1) for b, (A1) for −2, (A1) for \(\frac{1}{{{x^3}}}\) (or x⁻³). Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

\(3 = b - \frac{2}{{{{(1)}^3}}}\)     (M1)(M1)

Note: Award (M1) for substituting 1 into their gradient function, (M1) for equating their gradient function to 3.

b = 5     (AG)

Note: Award at most (M1)(A0) if final line is not seen or b does not equal 5.

[2 marks]

g(1) = 3 or (1, 3) (seen or implied from the line below)     (A1)

3 = 3 × 1 + c     (M1)

Note: Award (M1) for correct substitution of their point (1, 3) and gradient 3 into equation y = mx + c. Follow through from their point of tangency.

y = 3x     (A1)(ft)(G2)

OR

y − 3 = 3(x − 1)     (M1)(A1)(ft)(G2) 

Note: Award (M1) for substitution of gradient 3 and their point (1, 3) into y − y₁ = m(x − x₁), (A1)(ft) for correct substitutions. Follow through from their point of tangency. Award at most (A1)(M1)(A0)(ft) if further incorrect working seen.

[3 marks]

(−0.439, 0)   ((−0.438785..., 0))     (G1)(G1)

Notes: If no parentheses award at most (G1)(G0). Accept x = 0.439, y = 0.

[2 marks]

(i)

Award (A1) for labels and some indication of scale in the stated window.

Award (A1) for correct general shape (curve must be smooth and must not cross the y-axis)

Award (A1)(ft) for x-intercept consistent with their part (e).

Award (A1) for local minimum in the first quadrant.     (A1)(A1)(A1)(ft)(A1)

(ii) Tangent to curve drawn at approximately x = 1     (A1)(A1) 

Note: Award (A1) for a line tangent to curve approximately at x = 1. Must be a straight line for the mark to be awarded. Award (A1)(ft) for line passing through the origin. Follow through from their answer to part (d).

[6 marks]

(0.737, 2.53) ((0.736806..., 2.52604...))     (G1)(G1)

Notes: Do not penalize for lack of parentheses if already penalized in (e). Accept x = 0.737, y = 2.53.

[2 marks]

0.737 < x < 5  OR  (0.737;5)     (A1)(A1)(ft)

Notes: Award (A1) for correct strict or weak inequalities with x seen if the interval is given as inequalities, (A1)(ft) for 0.737 and 5 or their value from part (g).

[2 marks]

This question was moderately well answered. The concept of vertical asymptote in part (a) seemed to be problematic for a great number of candidates. In many cases students showed partial understanding of the vertical asymptote but found it difficult to write a correct equation.

This question was moderately well answered. The concept of vertical asymptote in part (a) seemed to be problematic for a great number of candidates. In many cases students showed partial understanding of the vertical asymptote but found it difficult to write a correct equation. Finding the derivative in part (b) proved problematic as well. It seems that the presence of the parameter b in the function may have contributed to this.

This question was moderately well answered. In part (c) a great number of students substituted b = 5 in the equation of the function instead of substituting it in the equation of their derivative.

This question was moderately well answered. Very few students used the GDC to find the equation of the tangent at x = 1 in part (d).

This question was moderately well answered. Good use of the GDC was seen in part (e), although some students wrote the x-coordinates of the point of intersection and neglected to write the y-coordinate.

This question was moderately well answered. The sketch in part (f) was, for the most part, not well done. Often the axes labels were missing. Very few tangents to the curve at the correct point were seen. Often the intended tangent lines intersected the curve, which shows that candidates either did not know what a tangent is or did not make sense of the sketch.

This question was moderately well answered. Good use of the GDC was shown in part (g) for finding the coordinates of the minimum point.

This question was moderately well answered. Few acceptable answers were given in part (h).