Question
a.Write down the quadratic expression \(2{x^2} + x – 3\) as the product of two linear factors.[1]
b.Hence, or otherwise, find the coefficient of \(x\) in the expansion of \({\left( {2{x^2} + x – 3} \right)^8}\) .[4]
▶️Answer/Explanation
Markscheme
a.\(2{x^2} + x – 3 = \left( {2x + 3} \right)\left( {x – 1} \right)\) A1
Note: Accept \(2\left( {x + \frac{3}{2}} \right)\left( {x – 1} \right)\).
Note: Either of these may be seen in (b) and if so A1 should be awarded.
[1 mark]
EITHER
\({\left( {2{x^2} + x – 3} \right)^8} = {\left( {2x + 3} \right)^8}{\left( {x – 1} \right)^8}\) M1
\( = \left( {{3^8} + 8\left( {{3^7}} \right)\left( {2x} \right) + …} \right)\left( {{{\left( { – 1} \right)}^8} + 8{{\left( { – 1} \right)}^7}\left( x \right) + …} \right)\) (A1)
coefficient of \(x = {3^8} \times 8 \times {\left( { – 1} \right)^7} + {3^7} \times 8 \times 2 \times {\left( { – 1} \right)^8}\) M1
= −17 496 A1
Note: Under ft, final A1 can only be achieved for an integer answer.
OR
\({\left( {2{x^2} + x – 3} \right)^8} = {\left( {3 – \left( {x – 2{x^2}} \right)} \right)^8}\) M1
\( = {3^8} + 8\left( { – \left( {x – 2{x^2}} \right)\left( {{3^7}} \right) + …} \right)\) (A1)
coefficient of \(x = 8 \times \left( { – 1} \right) \times {3^7}\) M1
= −17 496 A1
Note: Under ft, final A1 can only be achieved for an integer answer.
[4 marks]
Question
a.Prove that the equation \(3{x^2} + 2kx + k – 1 = 0\) has two distinct real roots for all values of \(k \in \mathbb{R}\).[4]
b.Find the value of k for which the two roots of the equation are closest together.[3]
Answer/Explanation
Markscheme
\(\Delta = {b^2} – 4ac = 4{k^2} – 4 \times 3 \times (k – 1) = 4{k^2} – 12k + 12\) M1A1
Note: Award M1A1 if expression seen within quadratic formula.
EITHER
\(144 – 4 \times 4 \times 12 < 0\) M1
\(\Delta \) always positive, therefore the equation always has two distinct real roots R1
(and cannot be always negative as \(a > 0\))
OR
sketch of \(y = 4{k^2} – 12k + 12\) or \(y = {k^2} – 3k + 3\) not crossing the x-axis M1
\(\Delta \) always positive, therefore the equation always has two distinct real roots R1
OR
write \(\Delta \) as \(4{(k – 1.5)^2} + 3\) M1
\(\Delta \) always positive, therefore the equation always has two distinct real roots R1
[4 marks]
closest together when \(\Delta \) is least (M1)
minimum value occurs when k = 1.5 (M1)A1
[3 marks]
Examiners report
Most candidates were able to find the discriminant (sometimes only as part of the quadratic formula) but fewer were able to explain satisfactorily why there were two distinct roots.
Most candidates were able to find the discriminant (sometimes only as part of the quadratic formula) but fewer were able to explain satisfactorily why there were two distinct roots. Only the better candidates were able to give good answers to part (b).
Question
The vectors a and b are such that a \( = (3\cos \theta + 6)\)i \( + 7\) j and b \( = (\cos \theta – 2)\)i \( + (1 + \sin \theta )\)j.
Given that a and b are perpendicular,
a.show that \(3{\sin ^2}\theta – 7\sin \theta + 2 = 0\);[3]
b.find the smallest possible positive value of \(\theta \).[3]
Answer/Explanation
Markscheme
attempting to form \((3\cos \theta + 6)(\cos \theta – 2) + 7(1 + \sin \theta ) = 0\) M1
\(3{\cos ^2}\theta – 12 + 7\sin \theta + 7 = 0\) A1
\(3\left( {1 – {{\sin }^2}\theta } \right) + 7\sin \theta – 5 = 0\) M1
\(3{\sin ^2}\theta – 7\sin \theta + 2 = 0\) AG
[3 marks]
attempting to solve algebraically (including substitution) or graphically for \(\sin \theta \) (M1)
\(\sin \theta = \frac{1}{3}\) (A1)
\(\theta = 0.340{\text{ }}( = 19.5^\circ )\) A1
[3 marks]
Examiners report
Part (a) was very well done. Most candidates were able to use the scalar product and \({\cos ^2}\theta = 1 – {\sin ^2}\theta \) to show the required result.
Part (b) was reasonably well done. A few candidates confused ‘smallest possible positive value’ with a minimum function value. Some candidates gave \(\theta = 0.34\) as their final answer.
Question
Let \(z = r(\cos \alpha + {\text{i}}\sin \alpha )\), where \(\alpha \) is measured in degrees, be the solution of \({z^5} – 1 = 0\) which has the smallest positive argument.
a.(i) Use the binomial theorem to expand \({(\cos \theta + {\text{i}}\sin \theta )^5}\).
(ii) Hence use De Moivre’s theorem to prove
\[\sin 5\theta = 5{\cos ^4}\theta \sin \theta – 10{\cos ^2}\theta {\sin ^3}\theta + {\sin ^5}\theta .\]
(iii) State a similar expression for \(\cos 5\theta \) in terms of \(\cos \theta \) and \(\sin \theta \).[6]
c.Using (a) (ii) and your answer from (b) show that \(16{\sin ^4}\alpha – 20{\sin ^2}\alpha + 5 = 0\).[4]
d.Hence express \(\sin 72^\circ \) in the form \(\frac{{\sqrt {a + b\sqrt c } }}{d}\) where \(a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}\).[5]
Answer/Explanation
Markscheme
(i) \({(\cos \theta + {\text{i}}\sin \theta )^5}\)
\( = {\cos ^5}\theta + 5{\text{i}}{\cos ^4}\theta \sin \theta + 10{{\text{i}}^2}{\cos ^3}\theta {\sin ^2}\theta + \)
\(10{{\text{i}}^3}{\cos ^2}\theta {\sin ^3}\theta + 5{{\text{i}}^4}\cos \theta {\sin ^4}\theta + {{\text{i}}^5}{\sin ^5}\theta \) A1A1
\(( = {\cos ^5}\theta + 5{\text{i}}{\cos ^4}\theta \sin \theta – 10{\cos ^3}\theta {\sin ^2}\theta – \)
\(10{\text{i}}{\cos ^2}\theta {\sin ^3}\theta + 5\cos \theta {\sin ^4}\theta + {\text{i}}{\sin ^5}\theta )\)
Note: Award first A1 for correct binomial coefficients.
(ii) \({({\text{cis}}\theta )^5} = {\text{cis}}5\theta = \cos 5\theta + {\text{i}}\sin 5\theta \) M1
\( = {\cos ^5}\theta + 5{\text{i}}{\cos ^4}\theta \sin \theta – 10{\cos ^3}\theta {\sin ^2}\theta – 10{\text{i}}{\cos ^2}\theta {\sin ^3}\theta + \)
\(5\cos \theta {\sin ^4}\theta + {\text{i}}{\sin ^5}\theta \) A1
Note: Previous line may be seen in (i)
equating imaginary terms M1
\(\sin 5\theta = 5{\cos ^4}\theta \sin \theta – 10{\cos ^2}\theta {\sin ^3}\theta + {\sin ^5}\theta \) AG
(iii) equating real terms
\(\cos 5\theta = {\cos ^5}\theta – 10{\cos ^3}\theta {\sin ^2}\theta + 5\cos \theta {\sin ^4}\theta \) A1
[6 marks]
\({(r{\text{cis}}\alpha )^5} = 1 \Rightarrow {r^5}{\text{cis}}5\alpha = 1{\text{cis}}0\) M1
\({r^5} = 1 \Rightarrow r = 1\) A1
\(5\alpha = 0 \pm 360k,{\text{ }}k \in \mathbb{Z} \Rightarrow a = 72k\) (M1)
\(\alpha = 72^\circ \) A1
Note: Award M1A0 if final answer is given in radians.
[4 marks]
use of \(\sin (5 \times 72) = 0\) OR the imaginary part of \(1\) is \(0\) (M1)
\(0 = 5{\cos ^4}\alpha \sin \alpha – 10{\cos ^2}\alpha {\sin ^3}\alpha + {\sin ^5}\alpha \) A1
\(\sin \alpha \ne 0 \Rightarrow 0 = 5{(1 – {\sin ^2}\alpha )^2} – 10(1 – {\sin ^2}\alpha ){\sin ^2}\alpha + {\sin ^4}\alpha \) M1
Note: Award M1 for replacing \({\cos ^2}\alpha \).
\(0 = 5(1 – 2{\sin ^2}\alpha + {\sin ^4}\alpha ) – 10{\sin ^2}\alpha + 10{\sin ^4}\alpha + {\sin ^4}\alpha \) A1
Note: Award A1 for any correct simplification.
so \(16{\sin ^4}\alpha – 20{\sin ^2}\alpha + 5 = 0\) AG
[4 marks]
\({\sin ^2}\alpha = \frac{{20 \pm \sqrt {400 – 320} }}{{32}}\) M1A1
\(\sin \alpha = \pm \sqrt {\frac{{20 \pm \sqrt {80} }}{{32}}} \)
\(\sin \alpha = \frac{{ \pm \sqrt {10 \pm 2\sqrt 5 } }}{4}\) A1
Note: Award A1 regardless of signs. Accept equivalent forms with integral denominator, simplification may be seen later.
as \(72 > 60\), \(\sin 72 > \frac{{\sqrt 3 }}{2} = 0.866 \ldots \) we have to take both positive signs (or equivalent argument) R1
Note: Allow verification of correct signs with calculator if clearly stated
\(\sin 72 = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4}\) A1
[5 marks]
Total [19 marks]