Home / IBDP Maths AA: Topic: SL 2.7: Solution of quadratic equations: IB style Questions SL Paper 2

IBDP Maths AA: Topic: SL 2.7: Solution of quadratic equations: IB style Questions SL Paper 2

Question

The quadratic equation \(k{x^2} + (k – 3)x + 1 = 0\) has two equal real roots.

Find the possible values of k.[5]

a.

Write down the values of k for which \({x^2} + (k – 3)x + k = 0\) has two equal real roots.[2]

b.
Answer/Explanation

Markscheme

attempt to use discriminant     (M1)

correct substitution, \({(k – 3)^2} – 4 \times k \times 1\)    (A1)

setting their discriminant equal to zero     M1

e.g. \({(k – 3)^2} – 4 \times k \times 1 = 0\) , \({k^2} – 10k + 9 = 0\)

\(k = 1\) , \(k = 9\)     A1A1     N3

[5 marks]

a.

\(k = 1\) , \(k = 9\)     A2     N2

[2 marks]

b.

Question

Let \(f(x) = k{x^2} + kx\) and \(g(x) = x – 0.8\). The graphs of \(f\) and \(g\) intersect at two distinct points.

Find the possible values of \(k\).

Answer/Explanation

Markscheme

attempt to set up equation     (M1)

eg\(\;\;\;f = g,{\text{ }}k{x^2} + kx = x – 0.8\)

rearranging their equation to equal zero     M1

eg\(\;\;\;k{x^2} + kx – x + 0.8 = 0,{\text{ }}k{x^2} + x(k – 1) + 0.8 = 0\)

evidence of discriminant (if seen explicitly, not just in quadratic formula)     (M1)

eg\(\;\;\;{b^2} – 4ac,{\text{ }}\Delta  = {(k – 1)^2} – 4k \times 0.8,{\text{ }}D = 0\)

correct discriminant     (A1)

eg\(\;\;\;{(k – 1)^2} – 4k \times 0.8,{\text{ }}{k^2} – 5.2k + 1\)

evidence of correct discriminant greater than zero     R1

eg\(\;\;\;{k^2} – 5.2k + 1 > 0,{\text{ }}{(k – 1)^2} – 4k \times 0.8 > 0\), correct answer

both correct values     (A1)

eg\(\;\;\;0.2,{\text{ }}5\)

correct answer     A2     N3

eg\(\;\;\;k < 0.2,{\text{ }}k \ne 0,{\text{ }}k > 5\)

[8 marks]

Question

Consider an infinite geometric sequence with \({u_1} = 40\) and \(r = \frac{1}{2}\) .

(i)     Find \({u_4}\) .

(ii)    Find the sum of the infinite sequence.

[4]
a(i) and (ii).

Consider an arithmetic sequence with n terms, with first term (\( – 36\)) and eighth term (\( – 8\)) .

(i)     Find the common difference.

(ii)    Show that \({S_n} = 2{n^2} – 38n\) .

[5]
b(i) and (ii).

The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n .

[5]
c.
Answer/Explanation

Markscheme

(i) correct approach     (A1)

e.g. \({u_4} = (40){\frac{1}{2}^{(4 – 1)}}\) , listing terms

\({u_4} = 5\)     A1     N2

(ii) correct substitution into formula for infinite sum     (A1)

e.g. \({S_\infty } = \frac{{40}}{{1 – 0.5}}\) , \({S_\infty } = \frac{{40}}{{0.5}}\)

\({S_\infty } = 80\)     A1     N2

[4 marks]

a(i) and (ii).

(i) attempt to set up expression for \({u_8}\)     (M1)

e.g. \( – 36 + (8 – 1)d\)

correct working     A1

e.g. \( – 8 = – 36 + (8 – 1)d\) , \(\frac{{ – 8 – ( – 36)}}{7}\)

\(d = 4\)     A1     N2

(ii) correct substitution into formula for sum     (A1)

e.g. \({S_n} = \frac{n}{2}(2( – 36) + (n – 1)4)\)

correct working     A1

e.g. \({S_n} = \frac{n}{2}(4n – 76)\) , \( – 36n + 2{n^2} – 2n\)

\({S_n} = 2{n^2} – 38n\)     AG     N0

[5 marks]

b(i) and (ii).

multiplying \({S_n}\) (AP) by 2 or dividing S (infinite GP) by 2     (M1)

e.g. \(2{S_n}\) , \(\frac{{{S_\infty }}}{2}\) , 40

evidence of substituting into \(2{S_n} = {S_\infty }\)     A1

e.g. \(2{n^2} – 38n = 40\) , \(4{n^2} – 76n – 80\) (\( = 0\))

attempt to solve their quadratic (equation)     (M1)

e.g. intersection of graphs, formula

\(n = 20\)     A2     N3

[5 marks]

c.

Question

[Maximum mark: 6] [without GDC]
Consider the function \(f(x)=2x^{2}-8x+5\).

(a) Express \(f (x)\) in the form \(a(x-p)^{2}+q\), where \(a,p,q \in \mathbb{Z}\) .
(b) Find the minimum value of \(f (x)\).

Answer/Explanation

Ans.

(a) \(2x^{2}-8x+5=2(x^{2}-4x+4)+5-8=2(x-2)^{2}-3\)
OR vertex at (2,-3)\(\Rightarrow y=2(x-2)^{2}-3\)

=> \(a\) = 2, \(p\) = 2, \(q\) = –3

(b) Minimum value of \(f (x) = –3\)

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