Question
The function \(f(x) = 4{x^3} + 2ax – 7a\) , \(a \in \mathbb{R}\), leaves a remainder of \(−10\) when divided by \(\left( {x – a} \right)\) .
a.Find the value of \(a\) .[3]
b.Show that for this value of \(a\) there is a unique real solution to the equation \(f (x) = 0\) .[2]
▶️Answer/Explanation
Markscheme
\(f(a) = 4{a^3} + 2{a^2} – 7a = – 10\) M1
\(4{a^3} + 2{a^2} – 7a + 10 = 0\)
\(\left( {a + 2} \right)\left( {4{a^2} – 6a + 5} \right) = 0\) or sketch or GDC (M1)
\(a = – 2\) A1
[3 marks]
substituting \(a = – 2\) into \(f (x)\)
\(f(x) = 4{x^3} – 4x + 14 = 0\) A1
EITHER
graph showing unique solution which is indicated (must include max and min) R1
OR
convincing argument that only one of the solutions is real R1
(−1.74, 0.868 ±1.12i)
[5 marks]
Question
Given that the graph of \(y = {x^3} – 6{x^2} + kx – 4\) has exactly one point at which the gradient is zero, find the value of k .
▶️Answer/Explanation
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 3{x^2} – 12x + k\) M1A1
For use of discriminant \({b^2} – 4ac = 0\) or completing the square \(3{(x – 2)^2} + k – 12\) (M1)
\(144 – 12k = 0\) (A1)
Note: Accept trial and error, sketches of parabolas with vertex (2,0) or use of second derivative.
\(k = 12\) A1
[5 marks]
Examiners report
Generally candidates answer this question well using a diversity of methods. Surprisingly, a small number of candidates were successful in answering this question using the discriminant of the quadratic and in many cases reverted to trial and error to obtain the correct answer.
</
Question
Show that the quadratic equation \({x^2} – (5 – k)x – (k + 2) = 0\) has two distinct real roots for all real values of k .
▶️Answer/Explanation
Markscheme
\(\Delta = {(5 – k)^2} + 4(k + 2)\) M1A1
\( = {k^2} – 6k + 33\) (A1)
\( = {(k – 3)^2} + 24\) which is positive for all k R1
Note: Accept analytical, graphical or other correct methods. In all cases only award R1 if a reason is given in words or graphically. Award M1A1A0R1 if mistakes are made in the simplification but the argument given is correct.
[4 marks]
Examiners report
Overall the question was pretty well answered but some candidates seemed to have mixed up the terms determinant with discriminant. In some cases a lack of quality mathematical reasoning and understanding of the discriminant was evident. Many worked with the quadratic formula rather than just the discriminant, conveying a lack of understanding of the strategy required. Errors in algebraic simplification (expanding terms involving negative signs) prevented many candidates from scoring well in this question. Many candidates were not able to give a clear reason why the quadratic has always two distinct real solutions; in some cases a vague explanation was given, often referring to a graph which was not sketched.
Question
Consider \(p(x) = 3{x^3} + ax + 5a,\;\;\;a \in \mathbb{R}\).
The polynomial \(p(x)\) leaves a remainder of \( – 7\) when divided by \((x – a)\).
Show that only one value of \(a\) satisfies the above condition and state its value.
▶️Answer/Explanation
Markscheme
using \(p(a) = – 7\) to obtain \(3{a^3} + {a^2} + 5a + 7 = 0\) M1A1
\((a + 1)(3{a^3} – 2a + 7) = 0\) (M1)(A1)
Note: Award M1 for a cubic graph with correct shape and A1 for clearly showing that the above cubic crosses the horizontal axis at \(( – 1,{\text{ }}0)\) only.
\(a = – 1\) A1
EITHER
showing that \(3{a^2} – 2a + 7 = 0\) has no real (two complex) solutions for \(a\) R1
OR
showing that \(3{a^3} + {a^2} + 5a + 7 = 0\) has one real (and two complex) solutions for \(a\) R1
Note: Award R1 for solutions that make specific reference to an appropriate graph.
[6 marks]
Examiners report
A large number of candidates, either by graphical (mostly) or algebraic or via use of a GDC solver, were able to readily obtain \(a = – 1\). Most candidates who were awarded full marks however, made specific reference to an appropriate graph. Only a small percentage of candidates used the discriminant to justify that only one value of \(a\) satisfied the required condition. A number of candidates erroneously obtained \(3{a^3} + {a^2} + 5a – 7 = 0\) or equivalent rather than \(3{a^3} + {a^2} + 5a + 7 = 0\).
Question
In the quadratic equation \(7{x^2} – 8x + p = 0,{\text{ }}(p \in \mathbb{Q})\), one root is three times the other root.
Find the value of \(p\).
▶️Answer/Explanation
Markscheme
METHOD 1
let roots be \(\alpha \) and \(3\alpha \) (M1)
sum of roots \((4\alpha ) = \frac{8}{7}\) M1
\( \Rightarrow \alpha = \frac{2}{7}\) A1
EITHER
product of roots \((3{\alpha ^2}) = \frac{p}{7}\) M1
\(p = 21{\alpha ^2} = 21 \times \frac{4}{{49}}\)
OR
\(7{\left( {\frac{2}{7}} \right)^2} – 8\left( {\frac{2}{7}} \right) + p = 0\) M1
\(\frac{4}{7} – \frac{{16}}{7} + p = 0\)
THEN
\( \Rightarrow p = \frac{{12}}{7}{\text{ }}( = 1.71)\) A1
METHOD 2
\(x = \frac{{8 \pm \sqrt {64 – 28p} }}{{14}}\) (M1)
\(\frac{{8 + \sqrt {64 – 28p} }}{{14}} = 3\left( {\frac{{8 – \sqrt {64 – 28p} }}{{14}}} \right)\) M1A1
\(8 + \sqrt {64 – 28p} = 24 – 3\sqrt {64 – 28p} \Rightarrow \sqrt {64 – 28p} = 4\) (M1)
\(p = \frac{{12}}{7}{\text{ }}( = 1.71)\) A1
[5 marks]