# IB Math Analysis & Approaches Questionbank-Topic: SL 2.3 Creating a sketch from information given SL Paper 1

### Question

Let y = $$\frac{Inx}{x^{4}}$$ for x > 0.

(a)        Show that $$\frac{dy}{dx}= \frac{1-4Inx}{x^{5}}$$

Consider the function defined  by f (x) $$\frac{Inx}{x^{4}}$$ =  for x> 0 and its graph y = f (x) .

(b)        The graph of has a horizontal tangent at point P. Find the coordinates of P.                                                      [5]

(c)        Given that f ” (x) = $$\frac{20Lnx-9}{x^{6}}$$ show that P is a local maximum point.                                                [3]

(d)        Solve f (x) > 0 for x > 0.                                                                                                                                                          [2]

(e) Sketch the graph of f , showing clearly the value of the x-intercept and the approximate position of point P.    [3]

Ans

## Question

Let $$f(x) = 3 + \frac{{20}}{{{x^2} – 4}}$$ , for $$x \ne \pm 2$$ . The graph of f is given below.

The y-intercept is at the point A.

(i)     Find the coordinates of A.

(ii)    Show that $$f'(x) = 0$$ at A.

[7]
a.

The second derivative $$f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}$$ . Use this to

(i)     justify that the graph of f has a local maximum at A;

(ii)    explain why the graph of f does not have a point of inflexion.

[6]
b.

Describe the behaviour of the graph of $$f$$ for large $$|x|$$ .

[1]
c.

Write down the range of $$f$$ .

[2]
d.

## Markscheme

(i) coordinates of A are $$(0{\text{, }} – 2)$$     A1A1     N2

(ii) derivative of $${x^2} – 4 = 2x$$ (seen anywhere)     (A1)

evidence of correct approach     (M1)

e.g. quotient rule, chain rule

finding $$f'(x)$$     A2

e.g. $$f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)$$ , $$\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}$$

substituting $$x = 0$$ into $$f'(x)$$ (do not accept solving $$f'(x) = 0$$ )     M1

at A $$f'(x) = 0$$     AG     N0

[7 marks]

a.

(i) reference to $$f'(x) = 0$$ (seen anywhere)     (R1)

reference to $$f”(0)$$ is negative (seen anywhere)     R1

evidence of substituting $$x = 0$$ into $$f”(x)$$     M1

finding $$f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}$$ $$\left( { = – \frac{5}{2}} \right)$$     A1

then the graph must have a local maximum     AG

(ii) reference to $$f”(x) = 0$$ at point of inflexion     (R1)

recognizing that the second derivative is never 0     A1     N2

e.g. $$40(3{x^2} + 4) \ne 0$$ , $$3{x^2} + 4 \ne 0$$ , $${x^2} \ne – \frac{4}{3}$$ , the numerator is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

b.

correct (informal) statement, including reference to approaching $$y = 3$$     A1     N1

e.g. getting closer to the line $$y = 3$$ , horizontal asymptote at $$y = 3$$

[1 mark]

c.

correct inequalities, $$y \le – 2$$ , $$y > 3$$ , FT from (a)(i) and (c)     A1A1     N2

[2 marks]

d.

## Question

Let $$f(x) = \sqrt x$$ . Line L is the normal to the graph of f at the point (4, 2) .

In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .

Show that the equation of L is $$y = – 4x + 18$$ .

[4]
a.

Point A is the x-intercept of L . Find the x-coordinate of A.

[2]
b.

Find an expression for the area of R .

[3]
c.

The region R is rotated $$360^\circ$$ about the x-axis. Find the volume of the solid formed, giving your answer in terms of $$\pi$$ .

[8]
d.

## Markscheme

finding derivative     (A1)

e.g. $$f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}$$

correct value of derivative or its negative reciprocal (seen anywhere)     A1

e.g. $$\frac{1}{{2\sqrt 4 }}$$ , $$\frac{1}{4}$$

gradient of normal =  $$\frac{1}{{{\text{gradient of tangent}}}}$$ (seen anywhere)     A1

e.g. $$– \frac{1}{{f'(4)}} = – 4$$ , $$– 2\sqrt x$$

substituting into equation of line (for normal)     M1

e.g. $$y – 2 = – 4(x – 4)$$

$$y = – 4x + 18$$     AG     N0

[4 marks]

a.

recognition that $$y = 0$$ at A     (M1)

e.g. $$– 4x + 18 = 0$$

$$x = \frac{{18}}{4}$$ $$\left( { = \frac{9}{2}} \right)$$     A1     N2

[2 marks]

b.

splitting into two appropriate parts (areas and/or integrals)     (M1)

correct expression for area of R     A2     N3

e.g. area of R = $$\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( – 4x + 18){\rm{d}}x}$$ , $$\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2$$ (triangle)

Note: Award A1 if dx is missing.

[3 marks]

c.

correct expression for the volume from $$x = 0$$ to $$x = 4$$     (A1)

e.g. $$V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x$$ , $${\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x$$ , $$\int_0^4 {\pi x{\rm{d}}x}$$

$$V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4$$     A1

$$V = \pi \left( {\frac{1}{2} \times 16 – \frac{1}{2} \times 0} \right)$$     (A1)

$$V = 8\pi$$     A1

finding the volume from $$x = 4$$ to $$x = 4.5$$

EITHER

recognizing a cone     (M1)

e.g. $$V = \frac{1}{3}\pi {r^2}h$$

$$V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}$$     (A1)

$$= \frac{{2\pi }}{3}$$     A1

total volume is $$8\pi + \frac{2}{3}\pi$$ $$\left( { = \frac{{26}}{3}\pi } \right)$$     A1     N4

OR

$$V = \pi \int_4^{4.5} {{{( – 4x + 18)}^2}{\rm{d}}x}$$     (M1)

$$= \int_4^{4.5} {\pi (16{x^2} – 144x + 324){\rm{d}}x}$$

$$= \pi \left[ {\frac{{16}}{3}{x^3} – 72{x^2} + 324x} \right]_4^{4.5}$$     A1

$$= \frac{{2\pi }}{3}$$     A1

total volume is $$8\pi + \frac{2}{3}\pi$$ $$\left( { = \frac{{26}}{3}\pi } \right)$$     A1     N4

[8 marks]

d.

## Question

Let $$f(x) = 6 + 6\sin x$$ . Part of the graph of f is shown below.

The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.

Solve for $$0 \le x < 2\pi$$

(i)     $$6 + 6\sin x = 6$$ ;

(ii)    $$6 + 6\sin x = 0$$ .

[5]
a(i) and (ii).

Write down the exact value of the x-intercept of f , for $$0 \le x < 2\pi$$ .

[1]
b.

The area of the shaded region is k . Find the value of k , giving your answer in terms of $$\pi$$ .

[6]
c.

Let $$g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)$$ . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

[2]
d.

Let $$g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)$$ . The graph of f is transformed to the graph of g.

Given that $$\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x} = k$$ and $$0 \le p < 2\pi$$ , write down the two values of p.

[3]
e.

## Markscheme

(i) $$\sin x = 0$$     A1

$$x = 0$$ , $$x = \pi$$     A1A1     N2

(ii) $$\sin x = – 1$$     A1

$$x = \frac{{3\pi }}{2}$$    A1     N1

[5 marks]

a(i) and (ii).

$$\frac{{3\pi }}{2}$$     A1     N1

[1 mark]

b.

evidence of using anti-differentiation     (M1)

e.g. $$\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x}$$

correct integral $$6x – 6\cos x$$ (seen anywhere)     A1A1

correct substitution     (A1)

e.g. $$6\left( {\frac{{3\pi }}{2}} \right) – 6\cos \left( {\frac{{3\pi }}{2}} \right) – ( – 6\cos 0)$$ , $$9\pi – 0 + 6$$

$$k = 9\pi + 6$$     A1A1     N3

[6 marks]

c.

translation of $$\left( {\begin{array}{*{20}{c}} {\frac{\pi }{2}}\\ 0 \end{array}} \right)$$     A1A1     N2

[2 marks]

d.

recognizing that the area under g is the same as the shaded region in f     (M1)

$$p = \frac{\pi }{2}$$ , $$p = 0$$     A1A1     N3

[3 marks]

e.

## Question

Consider $$f(x) = 2k{x^2} – 4kx + 1$$ , for $$k \ne 0$$ . The equation $$f(x) = 0$$ has two equal roots.

Find the value of k .

[5]
a.

The line $$y = p$$ intersects the graph of f . Find all possible values of p .

[2]
b.

## Markscheme

valid approach     (M1)

e.g. $${b^2} – 4ac$$ , $$\Delta = 0$$ , $${( – 4k)^2} – 4(2k)(1)$$

correct equation     A1

e.g. $${( – 4k)^2} – 4(2k)(1) = 0$$ , $$16{k^2} = 8k$$ , $$2{k^2} – k = 0$$

correct manipulation     A1

e.g. $$8k(2k – 1)$$ , $$\frac{{8 \pm \sqrt {64} }}{{32}}$$

$$k = \frac{1}{2}$$     A2     N3

[5 marks]

a.

recognizing vertex is on the x-axis     M1

e.g. (1, 0) , sketch of parabola opening upward from the x-axis

$$p \ge 0$$     A1     N1

[2 marks]

b.

## Question

Let  $$f(x) = \frac{x}{{ – 2{x^2} + 5x – 2}}$$ for $$– 2 \le x \le 4$$ , $$x \ne \frac{1}{2}$$ , $$x \ne 2$$ . The graph of $$f$$ is given below.

The graph of $$f$$ has a local minimum at A($$1$$, $$1$$) and a local maximum at B.

Use the quotient rule to show that $$f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$ .

[6]
a.

Hence find the coordinates of B.

[7]
b.

Given that the line $$y = k$$ does not meet the graph of f , find the possible values of k .

[3]
c.

## Markscheme

correct derivatives applied in quotient rule     (A1)A1A1

$$1$$, $$– 4x + 5$$

Note: Award (A1) for 1, A1 for $$– 4x$$ and A1 for $$5$$, only if it is clear candidates are using the quotient rule.

correct substitution into quotient rule     A1

e.g. $$\frac{{1 \times ( – 2{x^2} + 5x – 2) – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$ , $$\frac{{ – 2{x^2} + 5x – 2 – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$

correct working     (A1)

e.g. $$\frac{{ – 2{x^2} + 5x – 2 – ( – 4{x^2} + 5x)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$

e.g. $$\frac{{ – 2{x^2} + 5x – 2 + 4{x^2} – 5x}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$

$$f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$    AG     N0

[6 marks]

a.

evidence of attempting to solve $$f'(x) = 0$$     (M1)

e.g. $$2{x^2} – 2 = 0$$

evidence of correct working     A1

e.g. $${x^2} = 1,\frac{{ \pm \sqrt {16} }}{4}{\text{, }}2(x – 1)(x + 1)$$

e.g. $$x = \pm 1$$

correct x-coordinate $$x = – 1$$ (may be seen in coordinate form $$\left( { – 1,\frac{1}{9}} \right)$$ )    A1     N2

attempt to substitute $$– 1$$ into f (do not accept any other value)     (M1)

e.g. $$f( – 1) = \frac{{ – 1}}{{ – 2 \times {{( – 1)}^2} + 5 \times ( – 1) – 2}}$$

correct working

e.g. $$\frac{{ – 1}}{{ – 2 – 5 – 2}}$$     A1

correct y-coordinate $$y = \frac{1}{9}$$ (may be seen in coordinate form $$\left( { – 1,\frac{1}{9}} \right)$$ )    A1     N2

[7 marks]

b.

recognizing values between max and min     (R1)

$$\frac{1}{9} < k < 1$$     A2     N3

[3 marks]

c.

## Question

Let $$f(x) = 3x – 2$$ and $$g(x) = \frac{5}{{3x}}$$, for $$x \ne 0$$.

Let $$h(x) = \frac{5}{{x + 2}}$$, for $$x \geqslant 0$$. The graph of h has a horizontal asymptote at $$y = 0$$.

Find $${f^{ – 1}}(x)$$.

[2]
a.

Show that $$\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}$$.

[2]
b.

Find the $$y$$-intercept of the graph of $$h$$.

[2]
c(i).

Hence, sketch the graph of $$h$$.

[3]
c(ii).

For the graph of $${h^{ – 1}}$$, write down the $$x$$-intercept;

[1]
d(i).

For the graph of $${h^{ – 1}}$$, write down the equation of the vertical asymptote.

[1]
d(ii).

Given that $${h^{ – 1}}(a) = 3$$, find the value of $$a$$.

[3]
e.

## Markscheme

interchanging $$x$$ and $$y$$     (M1)

eg     $$x = 3y – 2$$

$${f^{ – 1}}(x) = \frac{{x + 2}}{3}{\text{ }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)$$     A1     N2

[2 marks]

a.

attempt to form composite (in any order)     (M1)

eg     $$g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}$$

correct substitution     A1

eg     $$\frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}$$

$$\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}$$     AG     N0

[2 marks]

b.

valid approach     (M1)

eg     $$h(0),{\text{ }}\frac{5}{{0 + 2}}$$

$$y = \frac{5}{2}{\text{ }}\left( {{\text{accept (0, 2.5)}}} \right)$$     A1     N2

[2 marks]

c(i).

A1A2     N3

Notes:     Award A1 for approximately correct shape (reciprocal, decreasing, concave up).

Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at $$(0, 2.5)$$, asymptotic to x-axis, correct domain $$x \geqslant 0$$.

If only two of these features are correct, award A1.

[3 marks]

c(ii).

$$x = \frac{5}{2}{\text{ }}\left( {{\text{accept (2.5, 0)}}} \right)$$     A1     N1

[1 mark]

d(i).

$$x = 0$$   (must be an equation)     A1     N1

[1 mark]

d(ii).

METHOD 1

attempt to substitute $$3$$ into $$h$$ (seen anywhere)     (M1)

eg     $$h(3),{\text{ }}\frac{5}{{3 + 2}}$$

correct equation     (A1)

eg     $$a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a$$

$$a = 1$$     A1     N2

[3 marks]

METHOD 2

attempt to find inverse (may be seen in (d))     (M1)

eg     $$x = \frac{5}{{y + 2}},{\text{ }}{h^{ – 1}} = \frac{5}{x} – 2,{\text{ }}\frac{5}{x} + 2$$

correct equation, $$\frac{5}{x} – 2 = 3$$     (A1)

$$a = 1$$     A1     N2

[3 marks]

e.

## Question

Consider $$f(x) = {x^2} + qx + r$$. The graph of $$f$$ has a minimum value when $$x = – 1.5$$.

The distance between the two zeros of $$f$$ is 9.

Show that the two zeros are 3 and $$– 6$$.

[2]
a.

Find the value of $$q$$ and of $$r$$.

[4]
b.

## Markscheme

recognition that the $$x$$-coordinate of the vertex is $$– 1.5$$ (seen anywhere)     (M1)

eg$$\,\,\,\,\,$$axis of symmetry is $$– 1.5$$, sketch, $$f'( – 1.5) = 0$$

correct working to find the zeroes     A1

eg$$\,\,\,\,\,$$$$– 1.5 \pm 4.5$$

$$x = – 6$$ and $$x = 3$$     AG     N0

[2 marks]

a.

METHOD 1 (using factors)

attempt to write factors     (M1)

eg$$\,\,\,\,\,$$$$(x – 6)(x + 3)$$

correct factors     A1

eg$$\,\,\,\,\,$$$$(x – 3)(x + 6)$$

$$q = 3,{\text{ }}r = – 18$$    A1A1     N3

METHOD 2 (using derivative or vertex)

valid approach to find $$q$$     (M1)

eg$$\,\,\,\,\,$$$$f'( – 1.5) = 0,{\text{ }} – \frac{q}{{2a}} = – 1.5$$

$$q = 3$$    A1

correct substitution     A1

eg$$\,\,\,\,\,$$$${3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0$$

$$r = – 18$$    A1

$$q = 3,{\text{ }}r = – 18$$    N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations     (M1)

eg$$\,\,\,\,\,$$$$9 + 3q + r = 0,{\text{ }}36 – 6q + r = 0$$

one correct value

eg$$\,\,\,\,\,$$$$q = 3,{\text{ }}r = – 18$$     A1

correct substitution     A1

eg$$\,\,\,\,\,$$$${3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0,{\text{ }}{3^2} + 3q – 18 = 0,{\text{ }}36 – 6q – 18 = 0$$

second correct value     A1

eg$$\,\,\,\,\,$$$$q = 3,{\text{ }}r = – 18$$

$$q = 3,{\text{ }}r = – 18$$    N3

[4 marks]

b.

## Question

The following diagram shows part of the graph of a quadratic function $$f$$.

The vertex is at $$(3,{\text{ }} – 1)$$ and the $$x$$-intercepts at 2 and 4.

The function $$f$$ can be written in the form $$f(x) = {(x – h)^2} + k$$.

The function can also be written in the form $$f(x) = (x – a)(x – b)$$.

Write down the value of $$h$$ and of $$k$$.

[2]
a.

Write down the value of $$a$$ and of $$b$$.

[2]
b.

Find the $$y$$-intercept.

[2]
c.

## Markscheme

$$h = 3,{\text{ }}k = – 1$$    A1A1     N2

[2 marks]

a.

$$a = 2,{\text{ }}b = 4{\text{ }}({\text{or }}a = 4,{\text{ }}b = 2)$$    A1A1     N2

[2 marks]

b.

attempt to substitute $$x = 0$$ into their $$f$$     (M1)

eg$$\,\,\,\,\,$$$${(0 – 3)^2} – 1,{\text{ }}(0 – 2)(0 – 4)$$

$$y = 8$$    A1     N2

[2 marks]

c.

## Question

A quadratic function $$f$$ can be written in the form $$f(x) = a(x – p)(x – 3)$$. The graph of $$f$$ has axis of symmetry $$x = 2.5$$ and $$y$$-intercept at $$(0,{\text{ }} – 6)$$

Find the value of $$p$$.

[3]
a.

Find the value of $$a$$.

[3]
b.

The line $$y = kx – 5$$ is a tangent to the curve of $$f$$. Find the values of $$k$$.

[8]
c.

## Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an $$x$$-intercept     (M1)

eg$$\,\,\,\,\,$$$$x – 3 = 0$$,

valid approach     (M1)

eg$$\,\,\,\,\,$$$$3 – 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5$$

$$p = 2$$     A1     N2

METHOD 2 (expanding f (x))

correct expansion (accept absence of $$a$$)     (A1)

eg$$\,\,\,\,\,$$$$a{x^2} – a(3 + p)x + 3ap,{\text{ }}{x^2} – (3 + p)x + 3p$$

valid approach involving equation of axis of symmetry     (M1)

eg$$\,\,\,\,\,$$$$\frac{{ – b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}$$

$$p = 2$$     A1     N2

METHOD 3 (using derivative)

correct derivative (accept absence of $$a$$)     (A1)

eg$$\,\,\,\,\,$$$$a(2x – 3 – p),{\text{ }}2x – 3 – p$$

valid approach     (M1)

eg$$\,\,\,\,\,$$$$f’(2.5) = 0$$

$$p = 2$$     A1     N2

[3 marks]

a.

attempt to substitute $$(0,{\text{ }} – 6)$$     (M1)

eg$$\,\,\,\,\,$$$$– 6 = a(0 – 2)(0 – 3),{\text{ }}0 = a( – 8)( – 9),{\text{ }}a{(0)^2} – 5a(0) + 6a = – 6$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$– 6 = 6a$$

$$a = – 1$$     A1     N2

[3 marks]

b.

METHOD 1 (using discriminant)

recognizing tangent intersects curve once     (M1)

recognizing one solution when discriminant = 0     M1

attempt to set up equation     (M1)

eg$$\,\,\,\,\,$$$$g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6$$

rearranging their equation to equal zero     (M1)

eg$$\,\,\,\,\,$$$${x^2} – 5x + kx + 1 = 0$$

correct discriminant (if seen explicitly, not just in quadratic formula)     A1

eg$$\,\,\,\,\,$$$${(k – 5)^2} – 4,{\text{ }}25 – 10k + {k^2} – 4$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$k – 5 = \pm 2,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}$$

$$k = 3,{\text{ }}7$$     A1A1     N0

METHOD 2 (using derivatives)

attempt to set up equation     (M1)

eg$$\,\,\,\,\,$$$$g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6$$

recognizing derivative/slope are equal     (M1)

eg$$\,\,\,\,\,$$$$f’ = {m_T},{\text{ }}f’ = k$$

correct derivative of $$f$$     (A1)

eg$$\,\,\,\,\,$$$$– 2x + 5$$

attempt to set up equation in terms of either $$x$$ or $$k$$     M1

eg$$\,\,\,\,\,$$$$( – 2x + 5)x – 5 = – {x^2} + 5x – 6,{\text{ }}k\left( {\frac{{5 – k}}{2}} \right) – 5 = – {\left( {\frac{{5 – k}}{2}} \right)^2} + 5\left( {\frac{{5 – k}}{2}} \right) – 6$$

rearranging their equation to equal zero     (M1)

eg$$\,\,\,\,\,$$$${x^2} – 1 = 0,{\text{ }}{k^2} – 10k + 21 = 0$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$x = \pm 1,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}$$

$$k = 3,{\text{ }}7$$     A1A1     N0

[8 marks]

c.

## Question

Let $$f(x) = {x^2}$$. The following diagram shows part of the graph of $$f$$.

The line $$L$$ is the tangent to the graph of $$f$$ at the point $${\text{A}}( – k,{\text{ }}{k^2})$$, and intersects the $$x$$-axis at point B. The point C is $$( – k,{\text{ }}0)$$.

The region $$R$$ is enclosed by $$L$$, the graph of $$f$$, and the $$x$$-axis. This is shown in the following diagram.

Write down $$f'(x)$$.

[1]
a.i.

Find the gradient of $$L$$.

[2]
a.ii.

Show that the $$x$$-coordinate of B is $$– \frac{k}{2}$$.

[5]
b.

Find the area of triangle ABC, giving your answer in terms of $$k$$.

[2]
c.

Given that the area of triangle ABC is $$p$$ times the area of $$R$$, find the value of $$p$$.

[7]
d.

## Markscheme

$$f'(x) = 2x$$     A1     N1

[1 mark]

a.i.

attempt to substitute $$x = – k$$ into their derivative     (M1)

gradient of $$L$$ is $$– 2k$$     A1     N2

[2 marks]

a.ii.

METHOD 1

attempt to substitute coordinates of A and their gradient into equation of a line     (M1)

eg$$\,\,\,\,\,$$$${k^2} = – 2k( – k) + b$$

correct equation of $$L$$ in any form     (A1)

eg$$\,\,\,\,\,$$$$y – {k^2} = – 2k(x + k),{\text{ }}y = – 2kx – {k^2}$$

valid approach     (M1)

eg$$\,\,\,\,\,$$$$y = 0$$

correct substitution into $$L$$ equation     A1

eg$$\,\,\,\,\,$$$$– {k^2} = – 2kx – 2{k^2},{\text{ }}0 = – 2kx – {k^2}$$

correct working     A1

eg$$\,\,\,\,\,$$$$2kx = – {k^2}$$

$$x = – \frac{k}{2}$$     AG     N0

METHOD 2

valid approach     (M1)

eg$$\,\,\,\,\,$$$${\text{gradient}} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}},{\text{ }} – 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}$$

recognizing $$y = 0$$ at B     (A1)

attempt to substitute coordinates of A and B into slope formula     (M1)

eg$$\,\,\,\,\,$$$$\frac{{{k^2} – 0}}{{ – k – x}},{\text{ }}\frac{{ – {k^2}}}{{x + k}}$$

correct equation     A1

eg$$\,\,\,\,\,$$$$\frac{{{k^2} – 0}}{{ – k – x}} = – 2k,{\text{ }}\frac{{ – {k^2}}}{{x + k}} = – 2k,{\text{ }} – {k^2} = – 2k(x + k)$$

correct working     A1

eg$$\,\,\,\,\,$$$$2kx = – {k^2}$$

$$x = – \frac{k}{2}$$     AG     N0

[5 marks]

b.

valid approach to find area of triangle     (M1)

eg$$\,\,\,\,\,$$$$\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)$$

area of $${\text{ABC}} = \frac{{{k^3}}}{4}$$     A1     N2

[2 marks]

c.

METHOD 1 ($$\int {f – {\text{triangle}}}$$)

valid approach to find area from $$– k$$ to 0     (M1)

eg$$\,\,\,\,\,$$$$\int_{ – k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ – k} f }$$

correct integration (seen anywhere, even if M0 awarded)     A1

eg$$\,\,\,\,\,$$$$\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ – k}^0$$

substituting their limits into their integrated function and subtracting     (M1)

eg$$\,\,\,\,\,$$$$0 – \frac{{{{( – k)}^3}}}{3}$$, area from $$– k$$ to 0 is $$\frac{{{k^3}}}{3}$$

Note:     Award M0 for substituting into original or differentiated function.

attempt to find area of $$R$$     (M1)

eg$$\,\,\,\,\,$$$$\int_{ – k}^0 {f(x){\text{d}}x – {\text{ triangle}}}$$

correct working for $$R$$     (A1)

eg$$\,\,\,\,\,$$$$\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}$$

correct substitution into $${\text{triangle}} = pR$$     (A1)

eg$$\,\,\,\,\,$$$$\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)$$

$$p = 3$$     A1     N2

METHOD 2 ($$\int {(f – L)}$$)

valid approach to find area of $$R$$     (M1)

eg$$\,\,\,\,\,$$$$\int_{ – k}^{ – \frac{k}{2}} {{x^2} – ( – 2kx – {k^2}){\text{d}}x + \int_{ – \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ – k}^{ – \frac{k}{2}} {(f – L) + \int_{ – \frac{k}{2}}^0 f } } }$$

correct integration (seen anywhere, even if M0 awarded)     A2

eg$$\,\,\,\,\,$$$$\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ – k}^{ – \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ – \frac{k}{2}}^0$$

substituting their limits into their integrated function and subtracting     (M1)

eg$$\,\,\,\,\,$$$$\left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { – \frac{k}{2}} \right)}^2} + {k^2}\left( { – \frac{k}{2}} \right)} \right) – \left( {\frac{{{{( – k)}^3}}}{3} + k{{( – k)}^2} + {k^2}( – k)} \right) + (0) – \left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3}} \right)$$

Note:     Award M0 for substituting into original or differentiated function.

correct working for $$R$$     (A1)

eg$$\,\,\,\,\,$$$$\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} – \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} – \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} – {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}$$

correct substitution into $${\text{triangle}} = pR$$     (A1)

eg$$\,\,\,\,\,$$$$\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)$$

$$p = 3$$     A1     N2

[7 marks]

d.