IB DP Maths Topic 2.7 The discriminant Δ=b2−4ac and the nature of the roots SL Paper 1

Question

Consider \(f(x) = 2k{x^2} – 4kx + 1\) , for \(k \ne 0\) . The equation \(f(x) = 0\) has two equal roots.

Find the value of k .

[5]
a.

The line \(y = p\) intersects the graph of f . Find all possible values of p .

[2]
b.
Answer/Explanation

Markscheme

valid approach     (M1)

e.g. \({b^2} – 4ac\) , \(\Delta = 0\) , \({( – 4k)^2} – 4(2k)(1)\)

correct equation     A1

e.g. \({( – 4k)^2} – 4(2k)(1) = 0\) , \(16{k^2} = 8k\) , \(2{k^2} – k = 0\)

correct manipulation     A1

e.g. \(8k(2k – 1)\) , \(\frac{{8 \pm \sqrt {64} }}{{32}}\)

\(k = \frac{1}{2}\)     A2     N3

[5 marks]

a.

recognizing vertex is on the x-axis     M1

e.g. (1, 0) , sketch of parabola opening upward from the x-axis

\(p \ge 0\)     A1     N1

[2 marks]

b.

Question

Let \(f(x) = \frac{1}{2}{x^2} + kx + 8\) , where \(k \in \mathbb{Z}\) .

Find the values of k such that \(f(x) = 0\) has two equal roots.

[4]
a.

Each value of k is equally likely for \( – 5 \le k \le 5\) . Find the probability that \(f(x) = 0\) has no roots.

[4]
b.
Answer/Explanation

Markscheme

METHOD 1

evidence of discriminant     (M1)

e.g. \({b^2} – 4ac\) , discriminant = 0

correct substitution into discriminant     A1

e.g. \({k^2} – 4 \times \frac{1}{2} \times 8\) , \({k^2} – 16 = 0\)

\(k = \pm 4\)     A1A1     N3

METHOD 2

recognizing that equal roots means perfect square     (R1)

e.g. attempt to complete the square, \(\frac{1}{2}({x^2} + 2kx + 16)\)

correct working

e.g. \(\frac{1}{2}{(x + k)^2}\) ,  \(\frac{1}{2}{k^2} = 8\)     A1

\(k = \pm 4\)     A1A1     N3

[4 marks]

a.

evidence of appropriate approach     (M1)

e.g. \({b^2} – 4ac < 0\)

correct working for k     A1

e.g. \( – 4 < k < 4\) , \({k^2} < 16\) , list all correct values of k

\(p = \frac{7}{{11}}\)     A2     N3

[4 marks]

b.

Question

Consider the equation \({x^2} + (k – 1)x + 1 = 0\) , where k is a real number.

Find the values of k for which the equation has two equal real solutions.

Answer/Explanation

Markscheme

METHOD 1

evidence of valid approach     (M1)

e.g. \({b^2} – 4ac\) , quadratic formula

correct substitution into \({b^2} – 4ac\) (may be seen in formula)     (A1)

e.g. \({(k – 1)^2} – 4 \times 1 \times 1\) , \({(k – 1)^2} – 4\) , \({k^2} – 2k – 3\)

setting their discriminant equal to zero     M1

e.g. \(\Delta  = 0,{(k – 1)^2} – 4 = 0\)

attempt to solve the quadratic     (M1)

e.g. \({(k – 1)^2} = 4\) , factorizing

correct working     A1

e.g. \((k – 1) = \pm 2\) , \((k – 3)(k + 1)\)

\(k = – 1\) , \(k = 3\) (do not accept inequalities)     A1A1     N2

[7 marks]

METHOD 2

recognizing perfect square     (M1)

e.g. \({(x + 1)^2} = 0\) , \({(x – 1)^2}\)

correct expansion     (A1)(A1)

e.g. \({x^2} + 2x + 1 = 0\) , \({x^2} – 2x + 1\)

equating coefficients of x     A1A1

e.g. \(k – 1 = – 2\) , \(k – 1 = 2\)

\(k = – 1\) , \(k = 3\)     A1A1     N2

[7 marks]

Question

The equation \({x^2} – 3x + {k^2} = 4\) has two distinct real roots. Find the possible values of k .

Answer/Explanation

Markscheme

evidence of rearranged quadratic equation (may be seen in working)     A1

e.g. \({x^2} – 3x + {k^2} – 4 = 0\) , \({k^2} – 4\) 

evidence of discriminant (must be seen explicitly, not in quadratic formula)     (M1)

e.g. \({b^2} – 4ac\) , \(\Delta  = {( – 3)^2} – 4(1)({k^2} – 4)\)

recognizing that discriminant is greater than zero (seen anywhere, including answer)     R1

e.g. \({b^2} – 4ac > 0\) , \(9 + 16 – 4{k^2} > 0\)

correct working (accept equality)     A1

e.g. \(25 – 4{k^2} > 0\) , \(4{k^2} < 25\) , \({k^2} = \frac{{25}}{4}\)

both correct values (even if inequality never seen)     (A1)

e.g. \(\pm \sqrt{{\frac{{25}}{4}}}\) , \( \pm 2.5\)

correct interval     A1     N3

e.g. \( – \frac{5}{2} < k < \frac{5}{2}\) , \( – 2.5 < k < 2.5\)

Note: Do not award the final mark for unfinished values, or for incorrect or reversed inequalities, including \( \le \) , \(k > – 2.5\) , \(k < 2.5\) .

Special cases:

If working shown, and candidates attempt to rearrange the quadratic equation to equal zero, but find an incorrect value of c, award A1M1R1A0A0A0.

If working shown, and candidates do not rearrange the quadratic equation to equal zero, but find \(c = {k^2}\) or \(c = \pm 4\) , award A0M1R1A0A0A0.

[6 marks]

Question

The equation \({x^2} + (k + 2)x + 2k = 0\) has two distinct real roots.

Find the possible values of \(k\).

Answer/Explanation

Markscheme

evidence of discriminant     (M1)

eg     \({b^2} – 4ac,{\text{ }}\Delta  = 0\)

correct substitution into discriminant     (A1)

eg     \({(k + 2)^2} – 4(2k),{\text{ }}{k^2} + 4k + 4 – 8k\)

correct discriminant     A1

eg     \({k^2} – 4k + 4,{\text{ }}{(k – 2)^2}\)

recognizing discriminant is positive     R1

eg     \(\Delta  > 0,{\text{ }}{(k + 2)^2} – 4(2k) > 0\)

attempt to solve their quadratic in \(k\)     (M1)

eg     factorizing, \(k = \frac{{4 \pm \sqrt {16 – 16} }}{2}\)

correct working     A1

eg     \({(k – 2)^2} > 0,{\text{ }}k = 2\), sketch of positive parabola on the x-axis

correct values     A2     N4

eg     \(k \in \mathbb{R}{\text{ and }}k \ne 2,{\text{ }}\mathbb{R}\backslash 2,{\text{ }}\left] { – \infty ,{\text{ }}2} \right[ \cup \left] {2,{\text{ }}\infty } \right[\)

[8 marks]

Question

Let \(f(x) = p{x^3} + p{x^2} + qx\).

Find \(f'(x)\).

[2]
a.

Given that \(f'(x) \geqslant 0\), show that \({p^2} \leqslant 3pq\).

[5]
b.
Answer/Explanation

Markscheme

\(f'(x) = 3p{x^2} + 2px + q\)     A2     N2

 

Note:     Award A1 if only 1 error.

 

[2 marks]

a.

evidence of discriminant (must be seen explicitly, not in quadratic formula)     (M1)

eg     \({b^2} – 4ac\)

correct substitution into discriminant (may be seen in inequality)     A1

eg     \({(2p)^2} – 4 \times 3p \times q,{\text{ }}4{p^2} – 12pq\)

\(f'(x) \geqslant 0\) then \(f’\) has two equal roots or no roots     (R1)

recognizing discriminant less or equal than zero     R1

eg     \(\Delta  \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0\)

correct working that clearly leads to the required answer     A1

eg     \({p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq\)

\({p^2} \leqslant 3pq\)     AG     N0

[5 marks] 

b.

Question

Let \(f(x) = 3{x^2} – 6x + p\). The equation \(f(x) = 0\) has two equal roots.

Write down the value of the discriminant.

[2]
a(i).

Hence, show that \(p = 3\).

[1]
a(ii).

The graph of \(f\)has its vertex on the \(x\)-axis.

Find the coordinates of the vertex of the graph of \(f\).

[4]
b.

The graph of \(f\) has its vertex on the \(x\)-axis.

Write down the solution of \(f(x) = 0\).

[1]
c.

The graph of \(f\) has its vertex on the \(x\)-axis.

The function can be written in the form \(f(x) = a{(x – h)^2} + k\). Write down the value of \(a\).

[1]
d(i).

The graph of \(f\) has its vertex on the \(x\)-axis.

The function can be written in the form \(f(x) = a{(x – h)^2} + k\). Write down the value of \(h\).

[1]
d(ii).

The graph of \(f\) has its vertex on the \(x\)-axis.

The function can be written in the form \(f(x) = a{(x – h)^2} + k\). Write down the value of \(k\).

[1]
d(iii).

The graph of \(f\) has its vertex on the \(x\)-axis.

The graph of a function \(g\) is obtained from the graph of \(f\) by a reflection of \(f\) in the \(x\)-axis, followed by a translation by the vector \(\left( \begin{array}{c}0\\6\end{array} \right)\). Find \(g\), giving your answer in the form \(g(x) = A{x^2} + Bx + C\).

[4]
e.
Answer/Explanation

Markscheme

correct value \(0\), or \(36 – 12p\)     A2     N2

[2 marks]

a(i).

correct equation which clearly leads to \(p = 3\)     A1

eg     \(36 – 12p = 0,{\text{ }}36 = 12p\)

\(p = 3\)     AG     N0

[1 mark]

a(ii).

METHOD 1

valid approach     (M1)

eg     \(x =  – \frac{b}{{2a}}\)

correct working     A1

eg     \( – \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

METHOD 2

valid approach     (M1)

eg     \(f(x) = 0\), factorisation, completing the square

correct working     A1

eg     \({x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

METHOD 3

valid approach using derivative     (M1)

eg     \(f'(x) = 0,{\text{ }}6x – 6\)

correct equation     A1

eg     \(6x – 6 = 0\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

[4 marks]

b.

\(x = 1\)     A1     N1

[1 mark]

c.

\(a = 3\)     A1     N1

[1 mark]

d(i).

\(h = 1\)     A1     N1

[1 mark]

d(ii).

\(k = 0\)     A1     N1

[1 mark]

d(iii).

attempt to apply vertical reflection     (M1)

eg     \( – f(x),{\text{ }} – 3{(x – 1)^2}\), sketch

attempt to apply vertical shift 6 units up     (M1)

eg     \( – f(x) + 6\), vertex \((1, 6)\)

transformations performed correctly (in correct order)     (A1)

eg     \( – 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6\)

\(g(x) =  – 3{x^2} + 6x + 3\)     A1     N3

[4 marks]

e.

Examiners report

[N/A]

a(i).

[N/A]

a(ii).

[N/A]

b.

[N/A]

c.

[N/A]

d(i).

[N/A]

d(ii).

[N/A]

d(iii).

[N/A]

e.

Question

Let \(f(x) = p{x^2} + (10 – p)x + \frac{5}{4}p – 5\).

Show that the discriminant of \(f(x)\) is \(100 – 4{p^2}\).

[3]
a.

Find the values of \(p\) so that \(f(x) = 0\) has two equal roots.

[3]
b.
Answer/Explanation

Markscheme

correct substitution into \({b^2} – 4ac\)     A1

eg\(\;\;\;{(10 – p)^2} – 4(p)\left( {\frac{5}{4}p – 5} \right)\)

correct expansion of each term     A1A1

eg\(\;\;\;100 – 20p + {p^2} – 5{p^2} + 20p,{\text{ }}100 – 20p + {p^2} – (5{p^2} – 20p)\)

\(100 – 4{p^2}\)     AG     N0

[3 marks]

a.

recognizing discriminant is zero for equal roots     (R1)

eg\(\;\;\;D = 0,{\text{ }}4{p^2} = 100\)

correct working     (A1)

eg\(\;\;\;{p^2} = 25\), \(1\) correct value of \(p\)

both correct values \(p =  \pm 5\)     A1     N2

[3 marks]

Total [6 marks]

b.

Examiners report

Many candidates were able to identify the discriminant correctly and continued with good algebraic manipulation. A commonly seen mistake was identifying the constant as \(\frac{5}{4}p\) instead of \(\frac{5}{4}p – 5\). Mostly a correct approach to part b) was seen \((\Delta  = 0)\), with the common error being only one answer given for \(p\), even though the question said values (plural).

a.

Many candidates were able to identify the discriminant correctly and continued with good algebraic manipulation. A commonly seen mistake was identifying the constant as \(\frac{5}{4}p\) instead of \(\frac{5}{4}p – 5\). Mostly a correct approach to part b) was seen \((\Delta  = 0)\), with the common error being only one answer given for \(p\), even though the question said values (plural).

b.

Question

Let \(f(x) = 3{\tan ^4}x + 2k\) and \(g(x) =  – {\tan ^4}x + 8k{\tan ^2}x + k\), for \(0 \leqslant x \leqslant 1\), where \(0 < k < 1\). The graphs of \(f\) and \(g\) intersect at exactly one point. Find the value of \(k\).

Answer/Explanation

Markscheme

discriminant \( = 0\) (seen anywhere)     M1

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f = g,{\text{ }}3{\tan ^4}x + 2k =  – {\tan ^4}x + 8k{\tan ^2}x + k\)

rearranging their equation (to equal zero)     (M1)

eg\(\,\,\,\,\,\)\(4{\tan ^4}x – 8k{\tan ^2}x + k = 0,{\text{ }}4{\tan ^4}x – 8k{\tan ^2}x + k\)

recognizing LHS is quadratic     (M1)

eg\(\,\,\,\,\,\)\(4{({\tan ^2}x)^2} – 8k{\tan ^2}x + k = 0,{\text{ }}4{m^2} – 8km + k\)

correct substitution into discriminant     A1

eg\(\,\,\,\,\,\)\({( – 8k)^2} – 4(4)(k)\)

correct working to find discriminant or solve discriminant \( = 0\)     (A1)

eg\(\,\,\,\,\,\)\(64{k^2} – 16k,{\text{ }}\frac{{ – ( – 16) \pm \sqrt {{{16}^2}} }}{{2 \times 64}}\)

correct simplification     (A1)

egx\(\,\,\,\,\,\)\(16k(4k – 1),{\text{ }}\frac{{32}}{{2 \times 64}}\)

\(k = \frac{1}{4}\)     A1     N2

[8 marks]

Question

Let \(f(x) = m – \frac{1}{x}\), for \(x \ne 0\). The line \(y = x – m\) intersects the graph of \(f\) in two distinct points. Find the possible values of \(m\).

Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f = y,{\text{ }}m – \frac{1}{x} = x – m\)

correct working to eliminate denominator     (A1)

eg\(\,\,\,\,\,\)\(mx – 1 = x(x – m),{\text{ }}mx – 1 = {x^2} – mx\)

correct quadratic equal to zero     A1

eg\(\,\,\,\,\,\)\({x^2} – 2mx + 1 = 0\)

correct reasoning     R1

eg\(\,\,\,\,\,\)for two solutions, \({b^2} – 4ac > 0\)

correct substitution into the discriminant formula     (A1)

eg\(\,\,\,\,\,\)\({( – 2m)^2} – 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(4{m^2} > 4,{\text{ }}{m^2} = 1\), sketch of positive parabola on the \(x\)-axis

correct interval     A1     N4

eg\(\,\,\,\,\,\)\(\left| m \right| > 1,{\text{ }}m <  – 1\) or \(m > 1\)

[7 marks]

Question

A quadratic function \(f\) can be written in the form \(f(x) = a(x – p)(x – 3)\). The graph of \(f\) has axis of symmetry \(x = 2.5\) and \(y\)-intercept at \((0,{\text{ }} – 6)\)

Find the value of \(p\).

[3]
a.

Find the value of \(a\).

[3]
b.

The line \(y = kx – 5\) is a tangent to the curve of \(f\). Find the values of \(k\).

[8]
c.
Answer/Explanation

Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an \(x\)-intercept     (M1)

eg\(\,\,\,\,\,\)\(x – 3 = 0\), M17/5/MATME/SP1/ENG/TZ1/09.a/M

valid approach     (M1)

eg\(\,\,\,\,\,\)\(3 – 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5\)

\(p = 2\)     A1     N2

METHOD 2 (expanding f (x)) 

correct expansion (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a{x^2} – a(3 + p)x + 3ap,{\text{ }}{x^2} – (3 + p)x + 3p\)

valid approach involving equation of axis of symmetry     (M1)

eg\(\,\,\,\,\,\)\(\frac{{ – b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}\)

\(p = 2\)     A1     N2

METHOD 3 (using derivative)

correct derivative (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a(2x – 3 – p),{\text{ }}2x – 3 – p\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f’(2.5) = 0\)

\(p = 2\)     A1     N2

[3 marks]

a.

attempt to substitute \((0,{\text{ }} – 6)\)     (M1)

eg\(\,\,\,\,\,\)\( – 6 = a(0 – 2)(0 – 3),{\text{ }}0 = a( – 8)( – 9),{\text{ }}a{(0)^2} – 5a(0) + 6a =  – 6\)

correct working     (A1)

eg\(\,\,\,\,\,\)\( – 6 = 6a\)

\(a =  – 1\)     A1     N2

[3 marks]

b.

METHOD 1 (using discriminant)

recognizing tangent intersects curve once     (M1)

recognizing one solution when discriminant = 0     M1

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 =  – {x^2} + 5x – 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} – 5x + kx + 1 = 0\)

correct discriminant (if seen explicitly, not just in quadratic formula)     A1

eg\(\,\,\,\,\,\)\({(k – 5)^2} – 4,{\text{ }}25 – 10k + {k^2} – 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(k – 5 =  \pm 2,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

METHOD 2 (using derivatives)

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 =  – {x^2} + 5x – 6\)

recognizing derivative/slope are equal     (M1)

eg\(\,\,\,\,\,\)\(f’ = {m_T},{\text{ }}f’ = k\)

correct derivative of \(f\)     (A1)

eg\(\,\,\,\,\,\)\( – 2x + 5\)

attempt to set up equation in terms of either \(x\) or \(k\)     M1

eg\(\,\,\,\,\,\)\(( – 2x + 5)x – 5 =  – {x^2} + 5x – 6,{\text{ }}k\left( {\frac{{5 – k}}{2}} \right) – 5 =  – {\left( {\frac{{5 – k}}{2}} \right)^2} + 5\left( {\frac{{5 – k}}{2}} \right) – 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} – 1 = 0,{\text{ }}{k^2} – 10k + 21 = 0\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(x =  \pm 1,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

[8 marks]

c.

Question

Let \(f\left( x \right) = p{x^2} + qx – 4p\), where p ≠ 0. Find Find the number of roots for the equation \(f\left( x \right) = 0\).

Justify your answer.

Answer/Explanation

Markscheme

METHOD 1

evidence of discriminant      (M1)
eg  \({b^2} – 4ac,\,\,\Delta \)

correct substitution into discriminant      (A1)
eg  \({q^2} – 4p\left( { – 4p} \right)\)

correct discriminant       A1
eg  \({q^2} + 16{p^2}\)

\(16{p^2} > 0\,\,\,\,\left( {{\text{accept}}\,\,{p^2} > 0} \right)\)     A1

\({q^2} \geqslant 0\,\,\,\,\left( {{\text{do not accept}}\,\,{q^2} > 0} \right)\)     A1

\({q^2} + 16{p^2} > 0\)      A1

\(f\) has 2 roots     A1 N0

METHOD 2

y-intercept = −4p (seen anywhere)      A1

if p is positive, then the y-intercept will be negative      A1

an upward-opening parabola with a negative y-intercept      R1
eg  sketch that must indicate p > 0.

if p is negative, then the y-intercept will be positive      A1

a downward-opening parabola with a positive y-intercept      R1
eg  sketch that must indicate p > 0.

\(f\) has 2 roots     A2 N0

[7 marks]

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