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Date May 2013 Marks available 3 Reference code 13M.2.hl.TZ1.9
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 9 Adapted from N/A

Question

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]
a.

Find the value of k for which the two roots of the equation are closest together.

[3]
b.

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]

a.

closest together when \(\Delta \) is least     (M1)

minimum value occurs when k = 1.5     (M1)A1

[3 marks]

b.

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.

a.

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).

b.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.6 » Solving quadratic equations using the quadratic formula.

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