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
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
When \({x^2} + 4x – b\) is divided by \(x – a\) the remainder is 2.
Given that \(a,{\text{ }}b \in \mathbb{R}\), find the smallest possible value for \(b\).
▶️Answer/Explanation
Markscheme
\({a^2} + 4a – b = 2\) M1A1
EITHER
\({a^2} + 4a – (b + 2) = 0\)
as \(a\) is real \( \Rightarrow 16 + 4(b + 2) \geqslant 0\) M1A1
OR
\(b = {a^2} + 4a – 2\) M1
\( = {(a + 2)^2} – 6\) (A1)
THEN
\(b \geqslant – 6\)
hence smallest possible value for \(b\) is \( – 6\) A1
[5 marks]
Examiners report
For quite a difficult question, there were many good solutions for this, including many different methods. It was disturbing to see how many students did not seem to be aware of the remainder theorem, instead choosing to divide the polynomial.
Question
The polynomial \({x^4} + p{x^3} + q{x^2} + rx + 6\) is exactly divisible by each of \(\left( {x – 1} \right)\), \(\left( {x – 2} \right)\) and \(\left( {x – 3} \right)\).
Find the values of \(p\), \(q\) and \(r\).
▶️Answer/Explanation
Markscheme
METHOD 1
substitute each of \(x\) = 1,2 and 3 into the quartic and equate to zero (M1)
\(p + q + r = – 7\)
\(4p + 2q + r = – 11\) or equivalent (A2)
\(9p + 3q + r = – 29\)
Note: Award A2 for all three equations correct, A1 for two correct.
attempting to solve the system of equations (M1)
\(p\) = −7, \(q\) = 17, \(r\) = −17 A1
Note: Only award M1 when some numerical values are found when solving algebraically or using GDC.
METHOD 2
attempt to find fourth factor (M1)
\(\left( {x – 1} \right)\) A1
attempt to expand \({\left( {x – 1} \right)^2}\left( {x – 2} \right)\left( {x – 3} \right)\) M1
\({x^4} – 7{x^3} + 17{x^2} – 17x + 6\) (\(p\) = −7, \(q\) = 17, \(r\) = −17) A2
Note: Award A2 for all three values correct, A1 for two correct.
Note: Accept long / synthetic division.
[5 marks]
Examiners report
Question
Given that (x − 2) is a factor of \(f(x) = {x^3} + a{x^2} + bx – 4\) and that division \(f(x)\) by (x − 1) leaves a remainder of −6 , find the value of a and the value of b .
▶️Answer/Explanation
Markscheme
\(f(2) = 8 + 4a + 2b – 4 = 0\) M1
\( \Rightarrow 4a + 2b = – 4\) A1
\(f(1) = 1 + a + b – 4 = – 6\) M1
\( \Rightarrow a + b = – 3\) A1
solving, \(a = 1,{\text{ }}b = – 4\) A1A1
[6 marks]