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 .
The line \(y = p\) intersects the graph of f . Find all possible values of p .
Answer/Explanation
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]
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]
Question
Let \(f(x) = \frac{1}{2}{x^2} + kx + 8\) , where \(k \in \mathbb{Z}\) .
Find the values of k such that \(f(x) = 0\) has two equal roots.
Each value of k is equally likely for \( – 5 \le k \le 5\) . Find the probability that \(f(x) = 0\) has no roots.
Answer/Explanation
Markscheme
METHOD 1
evidence of discriminant (M1)
e.g. \({b^2} – 4ac\) , discriminant = 0
correct substitution into discriminant A1
e.g. \({k^2} – 4 \times \frac{1}{2} \times 8\) , \({k^2} – 16 = 0\)
\(k = \pm 4\) A1A1 N3
METHOD 2
recognizing that equal roots means perfect square (R1)
e.g. attempt to complete the square, \(\frac{1}{2}({x^2} + 2kx + 16)\)
correct working
e.g. \(\frac{1}{2}{(x + k)^2}\) , \(\frac{1}{2}{k^2} = 8\) A1
\(k = \pm 4\) A1A1 N3
[4 marks]
evidence of appropriate approach (M1)
e.g. \({b^2} – 4ac < 0\)
correct working for k A1
e.g. \( – 4 < k < 4\) , \({k^2} < 16\) , list all correct values of k
\(p = \frac{7}{{11}}\) A2 N3
[4 marks]
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.
Answer/Explanation
Markscheme
METHOD 1
evidence of valid approach (M1)
e.g. \({b^2} – 4ac\) , quadratic formula
correct substitution into \({b^2} – 4ac\) (may be seen in formula) (A1)
e.g. \({(k – 1)^2} – 4 \times 1 \times 1\) , \({(k – 1)^2} – 4\) , \({k^2} – 2k – 3\)
setting their discriminant equal to zero M1
e.g. \(\Delta = 0,{(k – 1)^2} – 4 = 0\)
attempt to solve the quadratic (M1)
e.g. \({(k – 1)^2} = 4\) , factorizing
correct working A1
e.g. \((k – 1) = \pm 2\) , \((k – 3)(k + 1)\)
\(k = – 1\) , \(k = 3\) (do not accept inequalities) A1A1 N2
[7 marks]
METHOD 2
recognizing perfect square (M1)
e.g. \({(x + 1)^2} = 0\) , \({(x – 1)^2}\)
correct expansion (A1)(A1)
e.g. \({x^2} + 2x + 1 = 0\) , \({x^2} – 2x + 1\)
equating coefficients of x A1A1
e.g. \(k – 1 = – 2\) , \(k – 1 = 2\)
\(k = – 1\) , \(k = 3\) A1A1 N2
[7 marks]
Question
The equation \({x^2} – 3x + {k^2} = 4\) has two distinct real roots. Find the possible values of k .
Answer/Explanation
Markscheme
evidence of rearranged quadratic equation (may be seen in working) A1
e.g. \({x^2} – 3x + {k^2} – 4 = 0\) , \({k^2} – 4\)
evidence of discriminant (must be seen explicitly, not in quadratic formula) (M1)
e.g. \({b^2} – 4ac\) , \(\Delta = {( – 3)^2} – 4(1)({k^2} – 4)\)
recognizing that discriminant is greater than zero (seen anywhere, including answer) R1
e.g. \({b^2} – 4ac > 0\) , \(9 + 16 – 4{k^2} > 0\)
correct working (accept equality) A1
e.g. \(25 – 4{k^2} > 0\) , \(4{k^2} < 25\) , \({k^2} = \frac{{25}}{4}\)
both correct values (even if inequality never seen) (A1)
e.g. \(\pm \sqrt{{\frac{{25}}{4}}}\) , \( \pm 2.5\)
correct interval A1 N3
e.g. \( – \frac{5}{2} < k < \frac{5}{2}\) , \( – 2.5 < k < 2.5\)
Note: Do not award the final mark for unfinished values, or for incorrect or reversed inequalities, including \( \le \) , \(k > – 2.5\) , \(k < 2.5\) .
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 find \(c = {k^2}\) or \(c = \pm 4\) , award A0M1R1A0A0A0.
[6 marks]
Question
The equation \({x^2} + (k + 2)x + 2k = 0\) has two distinct real roots.
Find the possible values of \(k\).
Answer/Explanation
Markscheme
evidence of discriminant (M1)
eg \({b^2} – 4ac,{\text{ }}\Delta = 0\)
correct substitution into discriminant (A1)
eg \({(k + 2)^2} – 4(2k),{\text{ }}{k^2} + 4k + 4 – 8k\)
correct discriminant A1
eg \({k^2} – 4k + 4,{\text{ }}{(k – 2)^2}\)
recognizing discriminant is positive R1
eg \(\Delta > 0,{\text{ }}{(k + 2)^2} – 4(2k) > 0\)
attempt to solve their quadratic in \(k\) (M1)
eg factorizing, \(k = \frac{{4 \pm \sqrt {16 – 16} }}{2}\)
correct working A1
eg \({(k – 2)^2} > 0,{\text{ }}k = 2\), sketch of positive parabola on the x-axis
correct values A2 N4
eg \(k \in \mathbb{R}{\text{ and }}k \ne 2,{\text{ }}\mathbb{R}\backslash 2,{\text{ }}\left] { – \infty ,{\text{ }}2} \right[ \cup \left] {2,{\text{ }}\infty } \right[\)
[8 marks]
Question
Let \(f(x) = p{x^3} + p{x^2} + qx\).
Find \(f'(x)\).
Given that \(f'(x) \geqslant 0\), show that \({p^2} \leqslant 3pq\).
Answer/Explanation
Markscheme
\(f'(x) = 3p{x^2} + 2px + q\) A2 N2
Note: Award A1 if only 1 error.
[2 marks]
evidence of discriminant (must be seen explicitly, not in quadratic formula) (M1)
eg \({b^2} – 4ac\)
correct substitution into discriminant (may be seen in inequality) A1
eg \({(2p)^2} – 4 \times 3p \times q,{\text{ }}4{p^2} – 12pq\)
\(f'(x) \geqslant 0\) then \(f’\) has two equal roots or no roots (R1)
recognizing discriminant less or equal than zero R1
eg \(\Delta \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0\)
correct working that clearly leads to the required answer A1
eg \({p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq\)
\({p^2} \leqslant 3pq\) AG N0
[5 marks]
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.
Hence, show that \(p = 3\).
The graph of \(f\)has its vertex on the \(x\)-axis.
Find the coordinates of the vertex of the graph of \(f\).
The graph of \(f\) has its vertex on the \(x\)-axis.
Write down the solution of \(f(x) = 0\).
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 \(a\).
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 \(h\).
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 \(k\).
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 \(g(x) = A{x^2} + Bx + C\).
Answer/Explanation
Markscheme
correct value \(0\), or \(36 – 12p\) A2 N2
[2 marks]
correct equation which clearly leads to \(p = 3\) A1
eg \(36 – 12p = 0,{\text{ }}36 = 12p\)
\(p = 3\) AG N0
[1 mark]
METHOD 1
valid approach (M1)
eg \(x = – \frac{b}{{2a}}\)
correct working A1
eg \( – \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}\)
correct answers A1A1 N2
eg \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)
METHOD 2
valid approach (M1)
eg \(f(x) = 0\), factorisation, completing the square
correct working A1
eg \({x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}\)
correct answers A1A1 N2
eg \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)
METHOD 3
valid approach using derivative (M1)
eg \(f'(x) = 0,{\text{ }}6x – 6\)
correct equation A1
eg \(6x – 6 = 0\)
correct answers A1A1 N2
eg \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)
[4 marks]
\(x = 1\) A1 N1
[1 mark]
\(a = 3\) A1 N1
[1 mark]
\(h = 1\) A1 N1
[1 mark]
\(k = 0\) A1 N1
[1 mark]
attempt to apply vertical reflection (M1)
eg \( – f(x),{\text{ }} – 3{(x – 1)^2}\), sketch
attempt to apply vertical shift 6 units up (M1)
eg \( – f(x) + 6\), vertex \((1, 6)\)
transformations performed correctly (in correct order) (A1)
eg \( – 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6\)
\(g(x) = – 3{x^2} + 6x + 3\) A1 N3
[4 marks]
Examiners report
[N/A]
[N/A]
[N/A]
[N/A]
[N/A]
[N/A]
[N/A]
[N/A]
Question
Let \(f(x) = p{x^2} + (10 – p)x + \frac{5}{4}p – 5\).
Show that the discriminant of \(f(x)\) is \(100 – 4{p^2}\).
Find the values of \(p\) so that \(f(x) = 0\) has two equal roots.
Answer/Explanation
Markscheme
correct substitution into \({b^2} – 4ac\) A1
eg\(\;\;\;{(10 – p)^2} – 4(p)\left( {\frac{5}{4}p – 5} \right)\)
correct expansion of each term A1A1
eg\(\;\;\;100 – 20p + {p^2} – 5{p^2} + 20p,{\text{ }}100 – 20p + {p^2} – (5{p^2} – 20p)\)
\(100 – 4{p^2}\) AG N0
[3 marks]
recognizing discriminant is zero for equal roots (R1)
eg\(\;\;\;D = 0,{\text{ }}4{p^2} = 100\)
correct working (A1)
eg\(\;\;\;{p^2} = 25\), \(1\) correct value of \(p\)
both correct values \(p = \pm 5\) A1 N2
[3 marks]
Total [6 marks]
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 \((\Delta = 0)\), with the common error being only one answer given for \(p\), even though the question said values (plural).
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 \((\Delta = 0)\), with the common error being only one answer given for \(p\), even though the question said values (plural).
Question
Let \(f(x) = 3{\tan ^4}x + 2k\) and \(g(x) = – {\tan ^4}x + 8k{\tan ^2}x + k\), 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 \(k\).
Answer/Explanation
Markscheme
discriminant \( = 0\) (seen anywhere) M1
valid approach (M1)
eg\(\,\,\,\,\,\)\(f = g,{\text{ }}3{\tan ^4}x + 2k = – {\tan ^4}x + 8k{\tan ^2}x + k\)
rearranging their equation (to equal zero) (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 (M1)
eg\(\,\,\,\,\,\)\(4{({\tan ^2}x)^2} – 8k{\tan ^2}x + k = 0,{\text{ }}4{m^2} – 8km + k\)
correct substitution into discriminant A1
eg\(\,\,\,\,\,\)\({( – 8k)^2} – 4(4)(k)\)
correct working to find discriminant or solve discriminant \( = 0\) (A1)
eg\(\,\,\,\,\,\)\(64{k^2} – 16k,{\text{ }}\frac{{ – ( – 16) \pm \sqrt {{{16}^2}} }}{{2 \times 64}}\)
correct simplification (A1)
egx\(\,\,\,\,\,\)\(16k(4k – 1),{\text{ }}\frac{{32}}{{2 \times 64}}\)
\(k = \frac{1}{4}\) A1 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 \(m\).
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)\(f = y,{\text{ }}m – \frac{1}{x} = x – m\)
correct working to eliminate denominator (A1)
eg\(\,\,\,\,\,\)\(mx – 1 = x(x – m),{\text{ }}mx – 1 = {x^2} – mx\)
correct quadratic equal to zero A1
eg\(\,\,\,\,\,\)\({x^2} – 2mx + 1 = 0\)
correct reasoning R1
eg\(\,\,\,\,\,\)for two solutions, \({b^2} – 4ac > 0\)
correct substitution into the discriminant formula (A1)
eg\(\,\,\,\,\,\)\({( – 2m)^2} – 4\)
correct working (A1)
eg\(\,\,\,\,\,\)\(4{m^2} > 4,{\text{ }}{m^2} = 1\), sketch of positive parabola on the \(x\)-axis
correct interval A1 N4
eg\(\,\,\,\,\,\)\(\left| m \right| > 1,{\text{ }}m < – 1\) or \(m > 1\)
[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 \(p\).
Find the value of \(a\).
The line \(y = kx – 5\) is a tangent to the curve of \(f\). Find the values of \(k\).
Answer/Explanation
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]
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]
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]
Question
Let \(f\left( x \right) = p{x^2} + qx – 4p\), where p ≠ 0. Find Find the number of roots for the equation \(f\left( x \right) = 0\).
Justify your answer.
Answer/Explanation
Markscheme
METHOD 1
evidence of discriminant (M1)
eg \({b^2} – 4ac,\,\,\Delta \)
correct substitution into discriminant (A1)
eg \({q^2} – 4p\left( { – 4p} \right)\)
correct discriminant A1
eg \({q^2} + 16{p^2}\)
\(16{p^2} > 0\,\,\,\,\left( {{\text{accept}}\,\,{p^2} > 0} \right)\) A1
\({q^2} \geqslant 0\,\,\,\,\left( {{\text{do not accept}}\,\,{q^2} > 0} \right)\) A1
\({q^2} + 16{p^2} > 0\) A1
\(f\) has 2 roots A1 N0
METHOD 2
y-intercept = −4p (seen anywhere) A1
if p is positive, then the y-intercept will be negative A1
an upward-opening parabola with a negative y-intercept R1
eg sketch that must indicate p > 0.
if p is negative, then the y-intercept will be positive A1
a downward-opening parabola with a positive y-intercept R1
eg sketch that must indicate p > 0.
\(f\) has 2 roots A2 N0
[7 marks]