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Date May 2016 Marks available 1 Reference code 16M.1.sl.TZ1.11
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 11 Adapted from N/A

Question

Consider the function \(f(x) = a{x^2} + c\).

Find \(f'(x)\)

 

[1]
a.

Point \({\text{A}}( - 2,\,5)\)  lies on the graph of \(y = f(x)\) . The gradient of the tangent to this graph at \({\text{A}}\) is \( - 6\) .

Find the value of \(a\) .

[3]
b.

Find the value of \(c\) .

[2]
c.

Markscheme

\(2ax\)      (A1)   (C1)

Note: Award (A1) for \(2ax\).  Award (A0) if other terms are seen.

a.

\(2a( - 2) =  - 6\)       (M1)(M1)

Note: Award (M1) for correct substitution of \(x =  - 2\)  in their gradient function, (M1) for equating their gradient function to \( - 6\) . Follow through from part (a).

\((a = )1.5\,\,\,\left( {\frac{3}{2}} \right)\)       (A1)(ft) (C3)

b.

\({\text{their }}1.5 \times {( - 2)^2} + c = 5\)         (M1)

Note: Award (M1) for correct substitution of their \(a\) and point \({\text{A}}\). Follow through from part (b).

\((c = ) - 1\)         (A1)(ft) (C2)

c.

Examiners report

Question 11: Equation of tangent
Part (a) was generally well answered.

a.

In part (b), many candidates substituted the value of the function, rather than its gradient; this was usually correctly followed through into part (c).

b.

In part (b), many candidates substituted the value of the function, rather than its gradient; this was usually correctly followed through into part (c).

c.

Syllabus sections

Topic 7 - Introduction to differential calculus » 7.3 » Gradients of curves for given values of \(x\).
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