# IB Math AA Question bank-Topic: SL 2.7 The quadratic equations SL Paper 1

### Question

The diagram shows the graph of the quadratic function $$f(x)=ax^2+bx+c$$Â , with vertex âˆ’2, 10) .

Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â

The equation f (x) = k has two solutions. One of these solutions is x = 2 .

1. Write down the other solution of f (x) = k . [2]

2. Complete the table below placing a tick (âœ”) to show whether the unknown parameters

a and b are positive, zero or negative. The row for c has been completed as an example. [2]

 Â positive zero negative a Â Â Â b Â Â Â c âœ” Â Â
3. State the values of x for which f (x) is decreasing. [2]

Â

Ans:Â

(a)

(x= (-2))-4 OR x= (-2)- (2-(-2))

X= -6

(b)

(c) x > âˆ’2Â  OR x â‰¥ âˆ’2

### Question

Let f (x)=- x2 +Â 4x +Â 5 and g (x) = -f (x) + k .

Find the values of k so that g (x) = 0Â has no real roots.

Ans:

Method 1(discriminant)

correct expression for g

eg $$-(-x^{2}+4x+5)+k$$ ,$$x^{2}$$-4x-5+k = 0

evidence of discriminant

eg $$b^{2}-4ac,\Delta$$

correct substitution into discriminant of g

eg $$(-4)^{2}-4(1)(-5+k)$$, $$16-4(k-5)$$

recognizing discriminant is negative

eg $$\Delta <0, (-4)^{2}-4(1)(-5+k)<0, 16-4 (-1)(5)<0$$

correct working (must be correct inequality)

eg -4k<-36, k-5>4,16+20-4k<0 k>9

### Question

Let f (x) = mx2 – 2mx , where x âˆˆÂ $$\mathbb{R}$$ and m âˆˆÂ Â $$\mathbb{R}$$Â The line y = mx – 9 meets the graph of f

at exactly one point.

(a)Â Â Â Â Â Â Â  Show that m = 4 .Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [6]

The function f can be expressed in the form f (x) = 4(x p)(x q) , where p , q âˆˆÂ Â $$\mathbb{R}$$.

(b) Find the value of p and the value of q .Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [2]

The function f can also be expressed in the form f (x) = 4(x h)2 + k , where h , k âˆˆÂ Â $$\mathbb{R}$$.

(c) Find the value of h and the value of k .Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [3]

(d)Â Â Â Â Â Â Â  Hence find the values of x where the graph of f is both negative and increasing.Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [3]

Ans

## Question

Let $$f(x) = 3{\tan ^4}x + 2k$$ and $$g(x) = Â – {\tan ^4}x + 8k{\tan ^2}x + k$$, for $$0 \leqslant x \leqslant 1$$, where $$0 < k < 1$$. The graphs of $$f$$ and $$g$$ intersect at exactly one point. Find the value of $$k$$.

## Markscheme

discriminant $$= 0$$Â (seen anywhere) Â  Â  M1

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$f = g,{\text{ }}3{\tan ^4}x + 2k = Â – {\tan ^4}x + 8k{\tan ^2}x + k$$

rearranging their equation (to equal zero) Â  Â  (M1)

eg$$\,\,\,\,\,$$$$4{\tan ^4}x – 8k{\tan ^2}x + k = 0,{\text{ }}4{\tan ^4}x – 8k{\tan ^2}x + k$$

recognizing LHS is quadratic Â  Â  (M1)

eg$$\,\,\,\,\,$$$$4{({\tan ^2}x)^2} – 8k{\tan ^2}x + k = 0,{\text{ }}4{m^2} – 8km + k$$

correct substitution into discriminant Â  Â  A1

eg$$\,\,\,\,\,$$$${( – 8k)^2} – 4(4)(k)$$

correct working to find discriminant or solve discriminant $$= 0$$Â Â  Â  (A1)

eg$$\,\,\,\,\,$$$$64{k^2} – 16k,{\text{ }}\frac{{ – ( – 16) \pm \sqrt {{{16}^2}} }}{{2 \times 64}}$$

correct simplification Â  Â  (A1)

egx$$\,\,\,\,\,$$$$16k(4k – 1),{\text{ }}\frac{{32}}{{2 \times 64}}$$

$$k = \frac{1}{4}$$Â Â  Â  A1 Â  Â  N2

[8 marks]

## Question

Three consecutive terms of a geometric sequence are $$x – 3$$, 6 and $$x + 2$$.

Find the possible values of $$x$$.

## Markscheme

METHOD 1

valid approach Â  Â  (M1)

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

correct equation in terms of $$x$$ only Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{6}{{x – 3}} = \frac{{x + 2}}{6},{\text{ }}(x – 3)(x + 2) = {6^2},{\text{ }}36 = {x^2} – x – 6$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$${x^2} – x – 42,{\text{ }}{x^2} – x = 42$$

valid attempt to solve their quadratic equation Â  Â  (M1)

eg$$\,\,\,\,\,$$factorizing, formula, completing the square

evidence of correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$(x – 7)(x + 6),{\text{ }}\frac{{1 \pm \sqrt {169} }}{2}$$

$$x = 7,{\text{ }}x = Â – 6$$ Â  Â Â A1 Â  Â  N4

METHOD 2 (finding r first)

valid approach Â  Â  (M1)

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

correct equation in terms of $$r$$ only Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{6}{r} + 3 = 6r – 2,{\text{ }}6 + 3r = 6{r^2} – 2r,{\text{ }}6{r^2} – 5r – 6 = 0$$

evidence of correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$(3r + 2)(2r – 3),{\text{ }}\frac{{5 \pm \sqrt {25 + 144} }}{{12}}$$

$$r = Â – \frac{2}{3},{\text{ }}r = \frac{3}{2}$$ Â  Â A1

substituting their values of $$r$$ to find $$x$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$(x – 3)\left( {\frac{2}{3}} \right) = 6,{\text{ }}x = 6\left( {\frac{3}{2}} \right) – 2$$

$$x = 7,{\text{ }}x = Â – 6$$ Â  Â A1 Â  Â  N4

[6 marks]

## Examiners report

Nearly all candidates attempted to set up an expression, or pair of expressions, for the common ratio of the geometric sequence. When done correctly, these expressions led to a quadratic equation which was solved correctly by many candidates.

## 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 Â  Â  (M1)

eg$$\,\,\,\,\,$$$$x – 3 = 0$$,

valid approach Â  Â  (M1)

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

$$p = 2$$ Â  Â  A1 Â  Â  N2

METHOD 2 (expanding f (x))Â

correct expansion (accept absence of $$a$$) Â  Â  (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 Â  Â  (M1)

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

$$p = 2$$ Â  Â  A1 Â  Â  N2

METHOD 3 (using derivative)

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

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

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$fâ€™(2.5) = 0$$

$$p = 2$$ Â  Â  A1 Â  Â  N2

[3 marks]

a.

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

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

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$– 6 = 6a$$

$$a =Â – 1$$ Â  Â  A1 Â  Â  N2

[3 marks]

b.

METHOD 1 (using discriminant)

recognizing tangent intersects curve once Â  Â  (M1)

recognizing one solution when discriminant = 0 Â  Â  M1

attempt to set up equation Â  Â  (M1)

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

rearranging their equation to equal zero Â  Â  (M1)

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

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

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

correct working Â  Â  (A1)

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

$$k = 3,{\text{ }}7$$ Â  Â  A1A1 Â  Â  N0

METHOD 2 (using derivatives)

attempt to set up equation Â  Â  (M1)

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

recognizing derivative/slope are equal Â  Â  (M1)

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

correct derivative of $$f$$ Â  Â  (A1)

eg$$\,\,\,\,\,$$$$– 2x + 5$$

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

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

rearranging their equation to equal zero Â  Â  (M1)

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

correct working Â  Â  (A1)

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

$$k = 3,{\text{ }}7$$ Â  Â  A1A1 Â  Â  N0

[8 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

The following table shows the probability distribution of a discrete random variable $$A$$, in terms of an angle $$\theta$$.

Show that $$\cos \theta Â = \frac{3}{4}$$.

[6]
a.

Given that $$\tan \thetaÂ > 0$$, find $$\tan \theta$$.

[3]
b.

Let $$y = \frac{1}{{\cos x}}$$, for $$0 < x < \frac{\pi }{2}$$. The graph of $$y$$between $$x = \theta$$ andÂ $$x = \frac{\pi }{4}$$ is rotated 360Â° about the $$x$$-axis. Find the volume of the solid formed.

[6]
c.

## Markscheme

evidence of summing to 1 Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\sum {p = 1}$$

correct equation Â  Â  A1

eg$$\,\,\,\,\,$$$$\cos \thetaÂ + 2\cos 2\thetaÂ = 1$$

correct equation in $$\cos \theta$$ Â  Â  A1

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

evidence of valid approach to solve quadratic Â  Â  (M1)

eg$$\,\,\,\,\,$$factorizing equation set equal to $$0,{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4 \times 4 \times ( – 3)} }}{8}$$

eg$$\,\,\,\,\,$$$$(4\cos \thetaÂ – 3)(\cos \thetaÂ + 1),{\text{ }}\frac{{ – 1 \pm 7}}{8}$$

correct reason for rejecting $$\cos \thetaÂ \neÂ – 1$$ Â  Â  R1

eg$$\,\,\,\,\,$$$$\cos \theta$$ is a probability (value must lie between 0 and 1), $$\cos \thetaÂ > 0$$

Note: Â  Â  Award R0 for $$\cos \thetaÂ \neÂ – 1$$ without a reason.

$$\cos \thetaÂ = \frac{3}{4}$$ Â  Â AG Â N0

a.

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$sketch of right triangle with sides 3 and 4, $${\sin ^2}x + {\cos ^2}x = 1$$

correct workingÂ  Â  Â

(A1)

eg$$\,\,\,\,\,$$missing side $$= \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}$$

$$\tan \thetaÂ = \frac{{\sqrt 7 }}{3}$$ Â  Â  A1 Â  Â  N2

[3 marks]

b.

attempt to substitute either limits or the function into formula involving $${f^2}$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} }$$

correct substitution of both limits and function Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x}$$

correct integration Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\tan x$$

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

eg$$\,\,\,\,\,$$$$\tan \frac{\pi }{4} – \tan \theta$$

Note: Â  Â  Award M0 if they substitute into original or differentiated function.

$$\tan \frac{\pi }{4} = 1$$Â  Â  (A1)

eg$$\,\,\,\,\,$$$$1 – \tan \theta$$

$$V = \piÂ – \frac{{\pi \sqrt 7 }}{3}$$ Â  Â  A1 Â  Â  N3

[6 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

Let $$f(x) = \frac{{{{(\ln x)}^2}}}{2}$$, for $$x > 0$$.

Let $$g(x) = \frac{1}{x}$$. The following diagram shows parts of the graphs of $$f’$$ and g.

The graph of $$f’$$ has an x-intercept at $$x = p$$.

Show that $$f'(x) = \frac{{\ln x}}{x}$$.

[2]
a.

There is a minimum on the graph of $$f$$. Find the $$x$$-coordinate of this minimum.

[3]
b.

Write down the value of $$p$$.

[2]
c.

The graph of $$g$$ intersects the graph of $$f’$$ when $$x = q$$.

Find the value of $$q$$.

[3]
d.

The graph of $$g$$ intersects the graph of $$f’$$ when $$x = q$$.

Let $$R$$ be the region enclosed by the graph of $$f’$$, the graph of $$g$$ and the line $$x = p$$.

Show that the area of $$R$$ is $$\frac{1}{2}$$.

[5]
e.

## Markscheme

METHOD 1

correct use of chain rule Â  Â  A1A1

eg Â  Â  $$\frac{{2\ln x}}{2} \times \frac{1}{x},{\text{ }}\frac{{2\ln x}}{{2x}}$$

Note:Â Award A1 for $$\frac{{2\ln x}}{{2x}}$$, A1 for $$\times \frac{1}{x}$$.

$$f'(x) = \frac{{\ln x}}{x}$$ Â  Â  AG Â  Â  N0

[2 marks]

METHOD 2

correct substitution into quotient rule, with derivatives seen Â  Â  A1

eg Â  Â  $$\frac{{2 \times 2\ln x \times \frac{1}{x} – 0 \times {{(\ln x)}^2}}}{4}$$

correct working Â  Â  A1

eg Â  Â  $$\frac{{4\ln x \times \frac{1}{x}}}{4}$$

$$f'(x) = \frac{{\ln x}}{x}$$ Â  Â  AG Â  Â  N0

[2 marks]

a.

setting derivative $$= 0$$ Â  Â  (M1)

eg Â  Â  $$f'(x) = 0,{\text{ }}\frac{{\ln x}}{x} = 0$$

correct working Â  Â  (A1)

eg Â  Â  $$\ln x = 0,{\text{ }}x = {{\text{e}}^0}$$

$$x = 1$$ Â  Â  A1 Â  Â  N2

[3 marks]Â

b.

intercept when $$f'(x) = 0$$ Â  Â  (M1)

$$p = 1$$ Â  Â  A1 Â  Â  N2

[2 marks]

c.

equating functions Â  Â  (M1)

eg Â  Â  $$f’ = g,{\text{ }}\frac{{\ln x}}{x} = \frac{1}{x}$$

correct working Â  Â  (A1)

eg Â  Â  $$\ln x = 1$$

$$q = {\text{e Â (accept }}x = {\text{e)}}$$ Â  Â  A1 Â  Â  N2

[3 marks]

d.

evidence of integrating and subtracting functions (in any order, seen anywhere) Â  Â  (M1)

eg Â  Â  $$\int_q^e {\left( {\frac{1}{x} – \frac{{\ln x}}{x}} \right){\text{d}}x{\text{, }}\int {f’ – g} }$$

correct integration $$\ln x – \frac{{{{(\ln x)}^2}}}{2}$$ Â  Â  A2

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

eg Â  Â  $$(\ln {\text{e}} – \ln 1) – \left( {\frac{{{{(\ln {\text{e}})}^2}}}{2} – \frac{{{{(\ln 1)}^2}}}{2}} \right)$$

Note:Â Do not award M1 if the integrated function has only one term.

correct working Â  Â  A1

eg Â  Â  $$(1 – 0) – \left( {\frac{1}{2} – 0} \right),{\text{ }}1 – \frac{1}{2}$$

$${\text{area}} = \frac{1}{2}$$ Â  Â  AG Â  Â  N0

Notes:Â Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the markscheme.

[5 marks]

e.

## Question

Write down the value of

(i) Â  Â  $${\log _3}27$$;

[1]
a(i).

(ii) Â  Â  $${\log _8}\frac{1}{8}$$;

[1]
a(ii).

(iii) Â  Â  $${\log _{16}}4$$.

[1]
a(iii).

Hence, solve $${\log _3}27 + {\log _8}\frac{1}{8} – {\log _{16}}4 = {\log _4}x$$.

[3]
b.

## Markscheme

(i) Â  Â  $${\log _3}27 = 3$$ Â  Â Â A1 Â  Â  N1

[1 mark]

a(i).

(ii) Â  Â  $${\log _8}\frac{1}{8} = Â – 1$$ Â  Â Â A1 Â  Â  N1

[1 mark]

a(ii).

(iii) Â  Â  $${\log _{16}}4 = \frac{1}{2}$$ Â  Â Â A1 Â  Â  N1

[1 mark]

a(iii).

correct equation with their three values Â  Â  (A1)

egÂ  Â  Â $$\frac{3}{2} = {\log _4}x{\text{, }}3 + ( – 1) – \frac{1}{2} = {\log _4}x$$

correct working involving powers Â  Â  (A1)

egÂ  Â  Â $$x = {4^{\frac{3}{2}}}{\text{, }}{4^{\frac{3}{2}}} = {4^{{{\log }_4}x}}$$

$$x = 8$$ Â  Â Â A1 Â  Â  N2

[3 marks]

b.

[N/A]

a(i).

[N/A]

a(ii).

[N/A]

a(iii).

[N/A]

b.

## Question

Let $$f(x) = 3{x^2} – 6x + p$$. The equation $$f(x) = 0$$ has two equal roots.

Write down the value of the discriminant.

[2]
a(i).

Hence, show that $$p = 3$$.

[1]
a(ii).

The graph of $$f$$has its vertex on the $$x$$-axis.

Find the coordinates of the vertex of the graph of $$f$$.

[4]
b.

The graph of $$f$$ has its vertex on the $$x$$-axis.

Write down the solution of $$f(x) = 0$$.

[1]
c.

The graph of $$f$$Â has its vertex on the $$x$$-axis.

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

[1]
d(i).

The graph of $$f$$ has its vertex on the $$x$$-axis.

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

[1]
d(ii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

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

[1]
d(iii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The graph of a function $$g$$ is obtained from the graph of $$f$$ by a reflection of $$f$$ in the $$x$$-axis, followed by a translation by the vector $$\left( \begin{array}{c}0\\6\end{array} \right)$$. Find $$g$$, giving your answer in the form $$g(x) = A{x^2} + Bx + C$$.

[4]
e.

## Markscheme

correct value $$0$$, or $$36 – 12p$$ Â  Â  A2 Â  Â  N2

[2 marks]

a(i).

correct equation which clearly leads to $$p = 3$$ Â  Â  A1

eg Â  Â  $$36 – 12p = 0,{\text{ }}36 = 12p$$

$$p = 3$$ Â  Â  AG Â  Â  N0

[1 mark]

a(ii).

METHOD 1

valid approach Â  Â  (M1)

eg Â  Â  $$x =Â – \frac{b}{{2a}}$$

correct working Â  Â  A1

eg Â  Â  $$– \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 2

valid approach Â  Â  (M1)

eg Â  Â  $$f(x) = 0$$, factorisation, completing the square

correct working Â  Â  A1

eg Â  Â  $${x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 3

valid approach using derivative Â  Â  (M1)

eg Â  Â  $$f'(x) = 0,{\text{ }}6x – 6$$

correct equation Â  Â  A1

eg Â  Â  $$6x – 6 = 0$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

[4 marks]

b.

$$x = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

c.

$$a = 3$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(i).

$$h = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(ii).

$$k = 0$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(iii).

attempt to apply vertical reflection Â  Â  (M1)

eg Â  Â Â $$– f(x),{\text{ }} – 3{(x – 1)^2}$$, sketch

attempt to apply vertical shift 6 units up Â  Â  (M1)

eg Â  Â Â $$– f(x) + 6$$, vertex $$(1, 6)$$

transformations performed correctly (in correct order) Â  Â  (A1)

eg Â  Â Â $$– 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6$$

$$g(x) =Â – 3{x^2} + 6x + 3$$ Â  Â  A1 Â  Â  N3

[4 marks]

e.

## Question

Write the expression $$3\ln 2 – \ln 4$$ in the form $$\ln k$$, where $$k \in \mathbb{Z}$$.

[3]
a.

Hence or otherwise, solve $$3\ln 2 – \ln 4 =Â – \ln x$$.

[3]
b.

## Markscheme

correct application of $$\ln {a^b} = b\ln a$$ (seen anywhere) Â  Â  (A1)

eg$$\;\;\;\ln 4 = 2\ln 2,{\text{ }}3\ln 2 = \ln {2^3},{\text{ }}3\log 2 = \log 8$$

correct working Â  Â  (A1)

eg$$\;\;\;3\ln 2 – 2\ln 2,{\text{ }}\ln 8 – \ln 4$$

$$\ln 2\;\;\;{\text{(accept }}k = 2{\text{)}}$$ Â  Â  A1 Â  Â  N2

[3 marks]

a.

METHOD 1

attempt to substitute their answer into the equation Â  Â  (M1)

eg$$\;\;\;\ln 2 =Â – \ln x$$

correct application of a log rule Â  Â  (A1)

eg$$\;\;\;\ln \frac{1}{x},{\text{ }}\ln \frac{1}{2} = \ln x,{\text{ }}\ln 2 + \ln x = \ln 2x\;\;\;( = 0)$$

$$x = \frac{1}{2}$$ Â  Â  A1 Â  Â  N2

METHOD 2

attempt to rearrange equation, withÂ  $$3\ln 2$$ written as $$\ln {2^3}$$ or $$\ln 8$$ Â  Â  (M1)

eg$$\;\;\;\ln x = \ln 4 – \ln {2^3},{\text{ }}\ln 8 + \ln x = \ln 4,{\text{ }}\ln {2^3} = \ln 4 – \ln x$$

correct working applying $$\ln a \pm \ln b$$ Â  Â  (A1)

eg$$\;\;\;\frac{4}{8},{\text{ }}8x = 4,{\text{ }}\ln {2^3} = \ln \frac{4}{x}$$

$$x = \frac{1}{2}$$ Â  Â  A1 Â  Â  N2

[3 marks]

Total [6 marks]

b.

## Question

Three consecutive terms of a geometric sequence are $$x – 3$$, 6 and $$x + 2$$.

Find the possible values of $$x$$.

## Markscheme

METHOD 1

valid approach Â  Â  (M1)

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

correct equation in terms of $$x$$ only Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{6}{{x – 3}} = \frac{{x + 2}}{6},{\text{ }}(x – 3)(x + 2) = {6^2},{\text{ }}36 = {x^2} – x – 6$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$${x^2} – x – 42,{\text{ }}{x^2} – x = 42$$

valid attempt to solve their quadratic equation Â  Â  (M1)

eg$$\,\,\,\,\,$$factorizing, formula, completing the square

evidence of correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$(x – 7)(x + 6),{\text{ }}\frac{{1 \pm \sqrt {169} }}{2}$$

$$x = 7,{\text{ }}x = Â – 6$$ Â  Â Â A1 Â  Â  N4

METHOD 2 (finding r first)

valid approach Â  Â  (M1)

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

correct equation in terms of $$r$$ only Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{6}{r} + 3 = 6r – 2,{\text{ }}6 + 3r = 6{r^2} – 2r,{\text{ }}6{r^2} – 5r – 6 = 0$$

evidence of correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$(3r + 2)(2r – 3),{\text{ }}\frac{{5 \pm \sqrt {25 + 144} }}{{12}}$$

$$r = Â – \frac{2}{3},{\text{ }}r = \frac{3}{2}$$ Â  Â A1

substituting their values of $$r$$ to find $$x$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$(x – 3)\left( {\frac{2}{3}} \right) = 6,{\text{ }}x = 6\left( {\frac{3}{2}} \right) – 2$$

$$x = 7,{\text{ }}x = Â – 6$$ Â  Â A1 Â  Â  N4

[6 marks]

## Question

Let $$f(x) = {x^2} – x$$, for $$x \in \mathbb{R}$$. The following diagram shows part of the graph of $$f$$.

The graph of $$f$$ crosses the $$x$$-axis at the origin and at the point $${\text{P}}(1,{\text{ }}0)$$.

The line L is the normal to the graph of f at P.

The line $$L$$ intersects the graph of $$f$$ at another point Q, as shown in the following diagram.

Show that $$fâ€™(1) = 1$$.

[3]
a.

Find the equation of $$L$$ in the form $$y = ax + b$$.

[3]
b.

Find the $$x$$-coordinate of Q.

[4]
c.

Find the area of the region enclosed by the graph of $$f$$ and the line $$L$$.

[6]
d.

## Markscheme

$$fâ€™(x) = 2x – 1$$ Â  Â  A1A1

correct substitution Â  Â  A1

eg$$\,\,\,\,\,$$$$2(1) – 1,{\text{ }}2 – 1$$

$$fâ€™(1) = 1$$ Â  Â  AG Â  Â  N0

[3 marks]

a.

correct approach to find the gradient of the normal Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\frac{{ – 1}}{{f'(1)}},{\text{ }}{m_1}{m_2} =Â – 1,{\text{ slope}} =Â – 1$$

attempt to substitute correct normal gradient and coordinates into equation of a line Â  Â  (M1)

eg$$\,\,\,\,\,$$$$y – 0 =Â – 1(x – 1),{\text{ }}0 =Â – 1 + b,{\text{ }}b = 1,{\text{ }}L =Â – x + 1$$

$$y =Â – x + 1$$ Â  Â  A1 Â  Â  N2

[3 marks]

b.

equating expressions Â  Â  (M1)

eg$$\,\,\,\,\,$$$$f(x) = L,{\text{ }} – x + 1 = {x^2} – x$$

correct working (must involve combining terms) Â  Â  (A1)

eg$$\,\,\,\,\,$$$${x^2} – 1 = 0,{\text{ }}{x^2} = 1,{\text{ }}x = 1$$

$$x =Â – 1\,\,\,\,\,\left( {{\text{accept }}Q( – 1,{\text{ }}2)} \right)$$ Â  Â  A2 Â  Â  N3

[4 marks]

c.

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\int {L – f,{\text{ }}\int_{ – 1}^1 {(1 – {x^2}){\text{d}}x} }$$, splitting area into triangles and integrals

correct integration Â  Â  (A1)(A1)

eg$$\,\,\,\,\,$$$$\left[ {x – \frac{{{x^3}}}{3}} \right]_{ – 1}^1,{\text{ }} – \frac{{{x^3}}}{3} – \frac{{{x^2}}}{2} + \frac{{{x^2}}}{2} + x$$

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

eg$$\,\,\,\,\,$$$$1 – \frac{1}{3} – \left( { – 1 – \frac{{ – 1}}{3}} \right)$$

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

area $$= \frac{4}{3}$$ Â  Â  A2 Â  Â  N3

[6 marks]

d.

## Question

Solve $${\log _2}x + {\log _2}(x – 2) = 3$$ , for $$x > 2$$ .

## Markscheme

recognizing $$\log a + \log b = \log ab$$ (seen anywhere)Â Â Â Â  (A1)

e.g. $${\log _2}(x(x – 2))$$ , $${x^2} – 2x$$

recognizing $${\log _a}b = x \Leftrightarrow {a^x} = b$$Â Â Â Â  (A1)

e.g. $${2^3} = 8$$

correct simplificationÂ Â Â Â  A1

e.g. $$x(x – 2) = {2^3}$$ , $${x^2} – 2x – 8$$

evidence of correct approach to solveÂ Â Â Â  (M1)

correct workingÂ Â Â Â  A1

e.g. $$(x – 4)(x + 2)$$ , $$\frac{{2 \pm \sqrt {36} }}{2}$$

$$x = 4$$Â Â Â Â  A2Â Â Â Â  N3

[7 marks]

## Question

Let $$f(x) = {x^2} + x – 6$$.

Write down the $$y$$-intercept of the graph of $$f$$.

[1]
a.

Solve $$f(x) = 0$$.

[3]
b.

On the following grid, sketch the graph of $$f$$, for $$– 4 \le x \le 3$$.

[3]
c.

## Markscheme

$$y$$-intercept is $$– 6,{\text{ }}(0,{\text{ }} – 6),{\text{ }}y =Â – 6$$ Â  Â  A1

[1 mark]

a.

valid attempt to solve Â  Â  (M1)

eg$$\;\;\;(x – 2)(x + 3) = 0,{\text{ }}x = \frac{{ – 1 \pm \sqrt {1 + 24} }}{2}$$, one correct answer

$$x = 2,{\text{ }}x =Â – 3$$ Â  Â  A1A1 Â  Â  N3

[3 marks]

b.

Â  Â Â A1A1A1

Note: Â  Â  The shape must be an approximately correct concave up parabola. Only if the shape is correct, award the following:

A1 for the $$y$$-intercept in circle and the vertex approximately on $$x =Â – \frac{1}{2}$$, below $$y =Â – 6$$,

A1 for both theÂ $$x$$-intercepts in circles,

A1 for both end points in ovals.

[3 marks]

Total [7 marks]

c.

## Examiners report

Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).

a.

Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).

b.

In part (c), although most candidates were familiar with the general parabolic shape of the graph, many placed the vertex at the $$y$$-intercept $$(0,{\text{ }} – 6)$$, and very few candidates considered the endpoints of the function with the given domain.

c.

## 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 approachÂ Â Â Â  (M1)

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

correct equationÂ Â Â Â  A1

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

correct manipulationÂ Â Â Â  A1

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

$$k = \frac{1}{2}$$Â Â Â Â Â A2Â Â Â Â  N3

[5 marks]

a.

recognizing vertex is on the x-axisÂ Â Â Â  M1

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

$$p \ge 0$$Â Â Â Â Â A1Â Â Â Â  N1

[2 marks]

b.

## Question

LetÂ $$f(x) = \frac{1}{2}{x^2} + kx + 8$$ , whereÂ $$k \in \mathbb{Z}$$ .

Find the values of k such that $$f(x) = 0$$ has two equal roots.

[4]
a.

Each value of k is equally likely for $$– 5 \le k \le 5$$ . Find the probability that $$f(x) = 0$$ has no roots.

[4]
b.

## Markscheme

METHOD 1

evidence of discriminantÂ Â Â Â  (M1)

e.g. $${b^2} – 4ac$$ , discriminant = 0

correct substitution into discriminantÂ Â Â Â  A1

e.g. $${k^2} – 4 \times \frac{1}{2} \times 8$$ , $${k^2} – 16 = 0$$

$$k = \pm 4$$Â Â Â Â  A1A1Â Â Â Â  N3

METHOD 2

recognizing that equal roots means perfect squareÂ Â Â Â  (R1)

e.g. attempt to complete the square, $$\frac{1}{2}({x^2} + 2kx + 16)$$

correct working

e.g. $$\frac{1}{2}{(x + k)^2}$$ , Â $$\frac{1}{2}{k^2} = 8$$Â Â Â Â  A1

$$k = \pm 4$$Â Â Â Â  A1A1 Â  Â  N3

[4 marks]

a.

evidence of appropriate approachÂ Â Â Â  (M1)

e.g. $${b^2} – 4ac < 0$$

correct working for kÂ Â Â Â  A1

e.g. $$– 4 < k < 4$$ , $${k^2} < 16$$ , list all correct values of k

$$p = \frac{7}{{11}}$$Â Â Â Â Â A2Â Â Â Â  N3

[4 marks]

b.

## Question

Consider the equation $${x^2} + (k – 1)x + 1 = 0$$ , where k is a real number.

Find the values of k for which the equation has two equal real solutions.

## Markscheme

METHOD 1

evidence of valid approachÂ Â Â Â  (M1)

e.g. $${b^2} – 4ac$$Â , quadratic formula

correct substitution into $${b^2} – 4ac$$Â (may be seen in formula)Â Â Â Â  (A1)

e.g. $${(k – 1)^2} – 4 \times 1 \times 1$$ , $${(k – 1)^2} – 4$$ , $${k^2} – 2k – 3$$

setting their discriminant equal to zeroÂ Â Â Â  M1

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

attempt to solve the quadraticÂ Â Â Â  (M1)

e.g. $${(k – 1)^2} = 4$$Â , factorizing

correct workingÂ Â Â Â  A1

e.g. $$(k – 1) = \pm 2$$ , $$(k – 3)(k + 1)$$

$$k = – 1$$ , $$k = 3$$Â (do not accept inequalities)Â Â Â Â  A1A1 Â  Â  N2

[7 marks]

METHOD 2

recognizing perfect squareÂ Â Â Â  (M1)

e.g. $${(x + 1)^2} = 0$$ , $${(x – 1)^2}$$

correct expansionÂ Â Â  Â (A1)(A1)

e.g. $${x^2} + 2x + 1 = 0$$ , $${x^2} – 2x + 1$$

equating coefficients of xÂ Â Â Â  A1A1

e.g. $$k – 1 = – 2$$ , $$k – 1 = 2$$

$$k = – 1$$ , $$k = 3$$Â Â Â Â Â A1A1Â Â Â Â  N2

[7 marks]

## Question

The equation $${x^2} – 3x + {k^2} = 4$$ has two distinct real roots. Find the possible values of k .

## Markscheme

evidence of rearranged quadratic equation (may be seen in working)Â Â Â Â  A1

e.g.Â $${x^2} – 3x + {k^2} – 4 = 0$$ , $${k^2} – 4$$Â

evidence of discriminant (must be seen explicitly, not in quadratic formula)Â Â Â Â  (M1)

e.g.Â $${b^2} – 4ac$$ ,Â $$\DeltaÂ = {( – 3)^2} – 4(1)({k^2} – 4)$$

recognizing that discriminant is greater than zero (seen anywhere, including answer)Â Â Â Â  R1

e.g.Â $${b^2} – 4ac > 0$$ ,Â $$9 + 16 – 4{k^2} > 0$$

correct working (accept equality)Â Â Â Â  A1

e.g.Â $$25 – 4{k^2} > 0$$ , $$4{k^2} < 25$$Â ,Â $${k^2} = \frac{{25}}{4}$$

both correct values (even if inequality never seen)Â Â Â Â  (A1)

e.g.Â $$\pm \sqrt{{\frac{{25}}{4}}}$$ , $$\pm 2.5$$

correct intervalÂ Â Â Â  A1Â Â Â Â  N3

e.g.Â $$– \frac{5}{2} < k < \frac{5}{2}$$ , $$– 2.5 < k < 2.5$$

Note: Do not award the final mark for unfinished values, or for incorrect or reversed inequalities, including $$\le$$Â ,Â $$k > – 2.5$$ , $$k < 2.5$$Â .

Special cases:

If working shown, and candidates attempt to rearrange the quadratic equation to equal zero, but find an incorrect value of c, award A1M1R1A0A0A0.

If working shown, and candidates do not rearrange the quadratic equation to equal zero, but findÂ $$c = {k^2}$$ or $$c = \pm 4$$Â , award A0M1R1A0A0A0.

[6 marks]

## Question

The equation $${x^2} + (k + 2)x + 2k = 0$$ has two distinct real roots.

Find the possible values of $$k$$.

## Markscheme

evidence of discriminant Â  Â  (M1)

eg Â  Â  $${b^2} – 4ac,{\text{ }}\DeltaÂ = 0$$

correct substitution into discriminant Â  Â  (A1)

eg Â  Â  $${(k + 2)^2} – 4(2k),{\text{ }}{k^2} + 4k + 4 – 8k$$

correct discriminant Â  Â  A1

eg Â  Â  $${k^2} – 4k + 4,{\text{ }}{(k – 2)^2}$$

recognizing discriminant is positive Â  Â  R1

eg Â  Â  $$\DeltaÂ > 0,{\text{ }}{(k + 2)^2} – 4(2k) > 0$$

attempt to solve their quadratic in $$k$$ Â  Â  (M1)

eg Â  Â  factorizing, $$k = \frac{{4 \pm \sqrt {16 – 16} }}{2}$$

correct working Â  Â  A1

eg Â  Â  $${(k – 2)^2} > 0,{\text{ }}k = 2$$, sketch of positive parabola on the x-axis

correct values Â  Â  A2 Â  Â  N4

eg Â  Â  $$k \in \mathbb{R}{\text{ and }}k \ne 2,{\text{ }}\mathbb{R}\backslash 2,{\text{ }}\left] { – \infty ,{\text{ }}2} \right[ \cup \left] {2,{\text{ }}\infty } \right[$$

[8 marks]

## Question

Let $$f(x) = p{x^3} + p{x^2} + qx$$.

Find $$f'(x)$$.

[2]
a.

Given that $$f'(x) \geqslant 0$$, show that $${p^2} \leqslant 3pq$$.

[5]
b.

## Markscheme

$$f'(x) = 3p{x^2} + 2px + q$$ Â  Â Â A2 Â  Â  N2

Â

Note: Â  Â  Award A1 if only 1 error.

Â

[2 marks]

a.

evidence of discriminant (must be seen explicitly, not in quadratic formula) Â  Â Â (M1)

egÂ  Â  Â $${b^2} – 4ac$$

correct substitution into discriminant (may be seen in inequality) Â  Â Â A1

egÂ  Â  Â $${(2p)^2} – 4 \times 3p \times q,{\text{ }}4{p^2} – 12pq$$

$$f'(x) \geqslant 0$$ then $$f’$$Â has two equal roots or no roots Â  Â Â (R1)

recognizing discriminant less or equal than zero Â  Â Â R1

egÂ  Â  Â $$\Delta Â \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0$$

correct working that clearly leads to the required answer Â  Â Â A1

egÂ Â  Â  $${p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq$$

$${p^2} \leqslant 3pq$$ Â  Â Â AG Â  Â  N0

[5 marks]Â

b.

## Question

Let $$f(x) = 3{x^2} – 6x + p$$. The equation $$f(x) = 0$$ has two equal roots.

Write down the value of the discriminant.

[2]
a(i).

Hence, show that $$p = 3$$.

[1]
a(ii).

The graph of $$f$$has its vertex on the $$x$$-axis.

Find the coordinates of the vertex of the graph of $$f$$.

[4]
b.

The graph of $$f$$ has its vertex on the $$x$$-axis.

Write down the solution of $$f(x) = 0$$.

[1]
c.

The graph of $$f$$Â has its vertex on the $$x$$-axis.

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

[1]
d(i).

The graph of $$f$$ has its vertex on the $$x$$-axis.

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

[1]
d(ii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

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

[1]
d(iii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The graph of a function $$g$$ is obtained from the graph of $$f$$ by a reflection of $$f$$ in the $$x$$-axis, followed by a translation by the vector $$\left( \begin{array}{c}0\\6\end{array} \right)$$. Find $$g$$, giving your answer in the form $$g(x) = A{x^2} + Bx + C$$.

[4]
e.

## Markscheme

correct value $$0$$, or $$36 – 12p$$ Â  Â  A2 Â  Â  N2

[2 marks]

a(i).

correct equation which clearly leads to $$p = 3$$ Â  Â  A1

eg Â  Â  $$36 – 12p = 0,{\text{ }}36 = 12p$$

$$p = 3$$ Â  Â  AG Â  Â  N0

[1 mark]

a(ii).

METHOD 1

valid approach Â  Â  (M1)

eg Â  Â  $$x =Â – \frac{b}{{2a}}$$

correct working Â  Â  A1

eg Â  Â  $$– \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 2

valid approach Â  Â  (M1)

eg Â  Â  $$f(x) = 0$$, factorisation, completing the square

correct working Â  Â  A1

eg Â  Â  $${x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 3

valid approach using derivative Â  Â  (M1)

eg Â  Â  $$f'(x) = 0,{\text{ }}6x – 6$$

correct equation Â  Â  A1

eg Â  Â  $$6x – 6 = 0$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

[4 marks]

b.

$$x = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

c.

$$a = 3$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(i).

$$h = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(ii).

$$k = 0$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(iii).

attempt to apply vertical reflection Â  Â  (M1)

eg Â  Â Â $$– f(x),{\text{ }} – 3{(x – 1)^2}$$, sketch

attempt to apply vertical shift 6 units up Â  Â  (M1)

eg Â  Â Â $$– f(x) + 6$$, vertex $$(1, 6)$$

transformations performed correctly (in correct order) Â  Â  (A1)

eg Â  Â Â $$– 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6$$

$$g(x) =Â – 3{x^2} + 6x + 3$$ Â  Â  A1 Â  Â  N3

[4 marks]

e.

[N/A]

a(i).

[N/A]

a(ii).

[N/A]

b.

[N/A]

c.

[N/A]

d(i).

[N/A]

d(ii).

[N/A]

d(iii).

[N/A]

e.

## Question

Let $$f(x) = p{x^2} + (10 – p)x + \frac{5}{4}p – 5$$.

Show that the discriminant of $$f(x)$$ is $$100 – 4{p^2}$$.

[3]
a.

Find the values of $$p$$ so that $$f(x) = 0$$ has two equal roots.

[3]
b.

## Markscheme

correct substitution into $${b^2} – 4ac$$ Â  Â  A1

eg$$\;\;\;{(10 – p)^2} – 4(p)\left( {\frac{5}{4}p – 5} \right)$$

correct expansion of each term Â  Â  A1A1

eg$$\;\;\;100 – 20p + {p^2} – 5{p^2} + 20p,{\text{ }}100 – 20p + {p^2} – (5{p^2} – 20p)$$

$$100 – 4{p^2}$$ Â  Â  AG Â  Â  N0

[3 marks]

a.

recognizing discriminant is zero for equal roots Â  Â  (R1)

eg$$\;\;\;D = 0,{\text{ }}4{p^2} = 100$$

correct working Â  Â  (A1)

eg$$\;\;\;{p^2} = 25$$,Â $$1$$ correct value of $$p$$

both correct values $$p =Â \pm 5$$ Â  Â  A1 Â  Â  N2

[3 marks]

Total [6 marks]

b.

## Examiners report

Many candidates were able to identify the discriminant correctly and continued with good algebraic manipulation. A commonly seen mistake was identifying the constant as $$\frac{5}{4}p$$ instead of $$\frac{5}{4}p – 5$$. Mostly a correct approach to part b) was seen $$(\DeltaÂ = 0)$$, with the common error being only one answer given for $$p$$, even though the question said values (plural).

a.

Many candidates were able to identify the discriminant correctly and continued with good algebraic manipulation. A commonly seen mistake was identifying the constant as $$\frac{5}{4}p$$ instead of $$\frac{5}{4}p – 5$$. Mostly a correct approach to part b) was seen $$(\DeltaÂ = 0)$$, with the common error being only one answer given for $$p$$, even though the question said values (plural).

b.

## Question

Let $$f(x) = 3{\tan ^4}x + 2k$$ and $$g(x) = Â – {\tan ^4}x + 8k{\tan ^2}x + k$$, for $$0 \leqslant x \leqslant 1$$, where $$0 < k < 1$$. The graphs of $$f$$ and $$g$$ intersect at exactly one point. Find the value of $$k$$.

## Markscheme

discriminant $$= 0$$Â (seen anywhere) Â  Â  M1

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$f = g,{\text{ }}3{\tan ^4}x + 2k = Â – {\tan ^4}x + 8k{\tan ^2}x + k$$

rearranging their equation (to equal zero) Â  Â  (M1)

eg$$\,\,\,\,\,$$$$4{\tan ^4}x – 8k{\tan ^2}x + k = 0,{\text{ }}4{\tan ^4}x – 8k{\tan ^2}x + k$$

recognizing LHS is quadratic Â  Â  (M1)

eg$$\,\,\,\,\,$$$$4{({\tan ^2}x)^2} – 8k{\tan ^2}x + k = 0,{\text{ }}4{m^2} – 8km + k$$

correct substitution into discriminant Â  Â  A1

eg$$\,\,\,\,\,$$$${( – 8k)^2} – 4(4)(k)$$

correct working to find discriminant or solve discriminant $$= 0$$Â Â  Â  (A1)

eg$$\,\,\,\,\,$$$$64{k^2} – 16k,{\text{ }}\frac{{ – ( – 16) \pm \sqrt {{{16}^2}} }}{{2 \times 64}}$$

correct simplification Â  Â  (A1)

egx$$\,\,\,\,\,$$$$16k(4k – 1),{\text{ }}\frac{{32}}{{2 \times 64}}$$

$$k = \frac{1}{4}$$Â Â  Â  A1 Â  Â  N2

[8 marks]

## Question

Let $$f(x) = m – \frac{1}{x}$$, for $$x \ne 0$$. The line $$y = x – m$$ intersects the graph of $$f$$ in two distinct points. Find the possible values of $$m$$.

## Markscheme

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$f = y,{\text{ }}m – \frac{1}{x} = x – m$$

correct working to eliminate denominator Â  Â  (A1)

eg$$\,\,\,\,\,$$$$mx – 1 = x(x – m),{\text{ }}mx – 1 = {x^2} – mx$$

correct quadratic equal to zero Â  Â  A1

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

correct reasoning Â  Â  R1

eg$$\,\,\,\,\,$$for two solutions,Â $${b^2} – 4ac > 0$$

correct substitution into the discriminant formula Â  Â  (A1)

eg$$\,\,\,\,\,$$$${( – 2m)^2} – 4$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$4{m^2} > 4,{\text{ }}{m^2} = 1$$, sketch of positive parabola on the $$x$$-axis

correct interval Â  Â  A1 Â  Â  N4

eg$$\,\,\,\,\,$$$$\left| m \right| > 1,{\text{ }}m < Â – 1$$Â orÂ $$m > 1$$

[7 marks]

## 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 Â  Â  (M1)

eg$$\,\,\,\,\,$$$$x – 3 = 0$$,

valid approach Â  Â  (M1)

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

$$p = 2$$ Â  Â  A1 Â  Â  N2

METHOD 2 (expanding f (x))Â

correct expansion (accept absence of $$a$$) Â  Â  (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 Â  Â  (M1)

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

$$p = 2$$ Â  Â  A1 Â  Â  N2

METHOD 3 (using derivative)

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

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

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$fâ€™(2.5) = 0$$

$$p = 2$$ Â  Â  A1 Â  Â  N2

[3 marks]

a.

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

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

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$– 6 = 6a$$

$$a =Â – 1$$ Â  Â  A1 Â  Â  N2

[3 marks]

b.

METHOD 1 (using discriminant)

recognizing tangent intersects curve once Â  Â  (M1)

recognizing one solution when discriminant = 0 Â  Â  M1

attempt to set up equation Â  Â  (M1)

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

rearranging their equation to equal zero Â  Â  (M1)

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

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

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

correct working Â  Â  (A1)

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

$$k = 3,{\text{ }}7$$ Â  Â  A1A1 Â  Â  N0

METHOD 2 (using derivatives)

attempt to set up equation Â  Â  (M1)

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

recognizing derivative/slope are equal Â  Â  (M1)

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

correct derivative of $$f$$ Â  Â  (A1)

eg$$\,\,\,\,\,$$$$– 2x + 5$$

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

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

rearranging their equation to equal zero Â  Â  (M1)

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

correct working Â  Â  (A1)

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

$$k = 3,{\text{ }}7$$ Â  Â  A1A1 Â  Â  N0

[8 marks]

c.

## Question

Let $$f\left( x \right) = p{x^2} + qx – 4p$$, where pÂ â‰  0. FindÂ Find the number of roots for the equation $$f\left( x \right) = 0$$.

## Markscheme

METHOD 1

evidence of discriminantÂ  Â  Â  (M1)
egÂ Â $${b^2} – 4ac,\,\,\Delta$$

correct substitution into discriminantÂ  Â  Â  (A1)
egÂ Â $${q^2} – 4p\left( { – 4p} \right)$$

correct discriminantÂ  Â  Â  Â A1
egÂ Â $${q^2}Â + 16{p^2}$$

$$16{p^2} > 0\,\,\,\,\left( {{\text{accept}}\,\,{p^2} > 0} \right)$$Â  Â  Â A1

$${q^2} \geqslant 0\,\,\,\,\left( {{\text{do not accept}}\,\,{q^2} > 0} \right)$$Â  Â  Â A1

$${q^2} + 16{p^2} > 0$$Â  Â  Â  A1

$$f$$ has 2 rootsÂ  Â  Â A1 N0

METHOD 2

y-intercept = âˆ’4p (seen anywhere)Â  Â  Â  A1

if p is positive, then the y-intercept will be negativeÂ  Â  Â  A1

an upward-opening parabola with a negative y-interceptÂ  Â  Â  R1
egÂ  sketch that must indicate p > 0.

if p is negative, then the y-intercept will be positiveÂ  Â  Â  A1

a downward-opening parabola with a positive y-interceptÂ  Â  Â  R1
egÂ  sketch that must indicate pÂ >Â 0.

$$f$$ has 2 rootsÂ  Â  Â A2 N0

[7 marks]

## Question

Given that $${2^m} = 8$$ and $${2^n} = 16$$, write down the value of $$m$$ and of $$n$$.

[2]
a.

Hence or otherwise solve $${8^{2x + 1}} = {16^{2x – 3}}$$.

[4]
b.

## Markscheme

$$m = 3,{\text{ }}n = 4$$ Â  Â  A1A1 Â  Â  N2

[2 marks]

a.

attempt to apply $${({2^a})^b} = {2^{ab}}$$ Â  Â  (M1)

eg$$\;\;\;6x + 3,{\text{ }}4(2x – 3)$$

equating their powers ofÂ $$2$$ (seen anywhere) Â  Â  M1

eg$$\;\;\;3(2x + 1) = 8x – 12$$

correct working Â  Â  A1

eg$$\;\;\;8x – 12 = 6x + 3,{\text{ }}2x = 15$$

$$x = \frac{{15}}{2}\;\;\;(7.5)$$ Â  Â  A1 Â  Â  N2

[4 marks]

Total [6 marks]

b.

## Question

Let $$f(x) = 3{\tan ^4}x + 2k$$ and $$g(x) = Â – {\tan ^4}x + 8k{\tan ^2}x + k$$, for $$0 \leqslant x \leqslant 1$$, where $$0 < k < 1$$. The graphs of $$f$$ and $$g$$ intersect at exactly one point. Find the value of $$k$$.

## Markscheme

discriminant $$= 0$$Â (seen anywhere) Â  Â  M1

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$f = g,{\text{ }}3{\tan ^4}x + 2k = Â – {\tan ^4}x + 8k{\tan ^2}x + k$$

rearranging their equation (to equal zero) Â  Â  (M1)

eg$$\,\,\,\,\,$$$$4{\tan ^4}x – 8k{\tan ^2}x + k = 0,{\text{ }}4{\tan ^4}x – 8k{\tan ^2}x + k$$

recognizing LHS is quadratic Â  Â  (M1)

eg$$\,\,\,\,\,$$$$4{({\tan ^2}x)^2} – 8k{\tan ^2}x + k = 0,{\text{ }}4{m^2} – 8km + k$$

correct substitution into discriminant Â  Â  A1

eg$$\,\,\,\,\,$$$${( – 8k)^2} – 4(4)(k)$$

correct working to find discriminant or solve discriminant $$= 0$$Â Â  Â  (A1)

eg$$\,\,\,\,\,$$$$64{k^2} – 16k,{\text{ }}\frac{{ – ( – 16) \pm \sqrt {{{16}^2}} }}{{2 \times 64}}$$

correct simplification Â  Â  (A1)

egx$$\,\,\,\,\,$$$$16k(4k – 1),{\text{ }}\frac{{32}}{{2 \times 64}}$$

$$k = \frac{1}{4}$$Â Â  Â  A1 Â  Â  N2

[8 marks]