Date | May Specimen paper | Marks available | 1 | Reference code | SPM.1.SL.TZ0.8 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Hence and Explain | Question number | 8 | Adapted from | N/A |
Question
Let f(x)=13x3+x2−15x+17.
The graph of f has horizontal tangents at the points where x = a and x = b, a < b.
Find f′(x).
Find the value of a and the value of b.
Sketch the graph of y=f′(x).
Hence explain why the graph of f has a local maximum point at x=a.
Find f″(b).
Hence, use your answer to part (d)(i) to show that the graph of f has a local minimum point at x=b.
The normal to the graph of f at x=a and the tangent to the graph of f at x=b intersect at the point (p, q) .
Find the value of p and the value of q.
Markscheme
f′(x)=x2+2x−15 (M1)A1
[2 marks]
correct reasoning that f′(x)=0 (seen anywhere) (M1)
x2+2x−15=0
valid approach to solve quadratic M1
(x−3)(x+5), quadratic formula
correct values for x
3, −5
correct values for a and b
a = −5 and b = 3 A1
[3 marks]
A1
[1 mark]
first derivative changes from positive to negative at x=a A1
so local maximum at x=a AG
[1 mark]
f″(x)=2x+2 A1
substituting their b into their second derivative (M1)
f″(3)=2×3+2
f″(b)=8 (A1)
[3 marks]
f″(b) is positive so graph is concave up R1
so local minimum at x=b AG
[1 mark]
normal to f at x=a is x = −5 (seen anywhere) (A1)
attempt to find y-coordinate at their value of b (M1)
f(3)= −10 (A1)
tangent at x=b has equation y = −10 (seen anywhere) A1
intersection at (−5, −10)
p = −5 and q = −10 A1
[5 marks]