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Date November 2020 Marks available 2 Reference code 20N.1.SL.TZ0.S_10
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number S_10 Adapted from N/A

Question

The following diagram shows part of the graph of fx=kx, for x>0, k>0.

Let Pp, kp be any point on the graph of f. Line L1 is the tangent to the graph of f at P.

Line L1 intersects the x-axis at point A2p, 0 and the y-axis at point B.

Find f'p in terms of k and p.

[2]
a.i.

Show that the equation of L1 is kx+p2y-2pk=0.

[2]
a.ii.

Find the area of triangle AOB in terms of k.

[5]
b.

The graph of f is translated by 43 to give the graph of g.
In the following diagram:

Line L2 is the tangent to the graph of g at Q, and passes through E and F.

Given that triangle EDF and rectangle CDFG have equal areas, find the gradient of L2 in terms of p.

[6]
c.

Markscheme

f'x=-kx-2        (A1)

f'p=-kp-2  =-kp2     A1     N2

[2 marks]

a.i.

attempt to use point and gradient to find equation of L1        M1

eg    y-kp=-kp-2x-p,  kp=-kp2p+b

correct working leading to answer        A1

eg    p2y-kp=-kx+kp,  y-kp=-kp2x+kp,  y=-kp2x+2kp

kx+p2y-2pk=0     AG     N0

[2 marks]

a.ii.

METHOD 1 – area of a triangle

recognizing x=0 at B       (M1)

correct working to find y-coordinate of null       (A1)

eg   p2y-2pk=0

y-coordinate of null at y=2kp (may be seen in area formula)        A1

correct substitution to find area of triangle       (A1)

eg   122p2kp,  p×2kp

area of triangle AOB=2k     A1     N3

 

METHOD 2 – integration

recognizing to integrate L1 between 0 and 2p       (M1)

eg   02pL1dx , 02p-kp2x+2kp

correct integration of both terms        A1

eg   -kx22p2+2kxp , -k2p2x2+2kpx+c , -k2p2x2+2kpx02p

substituting limits into their integrated function and subtracting (in either order)       (M1)

eg    -k2p22p2+2k2pp-0, -4kp22p2+4kpp

correct working       (A1)

eg    -2k+4k

area of triangle AOB=2k     A1     N3

 

[5 marks]

b.

Note: In this question, the second M mark may be awarded independently of the other marks, so it is possible to award (M0)(A0)M1(A0)(A0)A0.

 

recognizing use of transformation      (M1)

eg   area of triangle AOB = area of triangle DEFgx=kx-4+3, gradient of L2= gradient of L1, D4, 3, 2p+4,  one correct shift

correct working       (A1)

eg   area of triangle DEF=2k, CD=3, DF=2p, CG=2p, E4, 2kp+3, F2p+4, 3, Qp+4, kp+3, 

gradient of L2=-kp2, g'x=-kx-42, area of rectangle CDFG=2k

valid approach      (M1)

eg   ED×DF2=CD×DF, 2p·3=2k , ED=2CD , 42p+4L2dx=4k

correct working      (A1)

eg   ED=6, E4, 9, k=3p, gradient=3-2kp+32p+4-4, -62k3, -9k

correct expression for gradient (in terms of p)       (A1)

eg   -62p, 9-34-2p+4, -3pp2, 3-23pp+32p+4-4, -93p

gradient of L2 is -3p  =-3p-1     A1     N3

[6 marks]

c.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
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Topic 5—Calculus

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