Past Exam Questions

Answer: \(2\left(x+\dfrac34\right)^2-\dfrac{41}{8}\); stationary point \(\left(-\dfrac34,-\dfrac{41}{8}\right)\); \(k=\dfrac{41}{8}\).

(a) Complete the square:

\(2x^2+3x-4 =2\left(x^2+\frac32x\right)-4.\)

Now

\(x^2+\frac32x =\left(x+\frac34\right)^2-\frac9{16}.\)

Therefore

\(2x^2+3x-4 =2\left[\left(x+\frac34\right)^2-\frac9{16}\right]-4.\)

So

\(2x^2+3x-4 =2\left(x+\frac34\right)^2-\frac98-4.\)

Hence

\(2x^2+3x-4 =2\left(x+\frac34\right)^2-\frac{41}{8}.\)

(b) From the completed-square form, the stationary point of \(y=2x^2+3x-4\) is

\(\left(-\frac34,-\frac{41}{8}\right).\)

(c) The \(y\)-intercept is found by putting \(x=0\):

\(y=| -4 |=4.\)

So the \(y\)-intercept is \((0,4)\).

The \(x\)-intercepts occur when

\(2x^2+3x-4=0.\)

Using the quadratic formula,

\(x=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)} =\frac{-3\pm\sqrt{41}}{4}.\)

So the graph meets the \(x\)-axis at

\(\left(\frac{-3-\sqrt{41}}4,0\right) \quad\text{and}\quad \left(\frac{-3+\sqrt{41}}4,0\right).\)

The graph of \(y=|2x^2+3x-4|\) is obtained by reflecting the part of \(y=2x^2+3x-4\) below the \(x\)-axis above the \(x\)-axis. So it has cusps at the two \(x\)-intercepts and a maximum between them at

\(\left(-\frac34,\frac{41}{8}\right).\)

(d) The equation

\(|2x^2+3x-4|=k\)

has exactly 3 solutions when the horizontal line \(y=k\) just touches the reflected middle part at its maximum, while still meeting the two outer branches.

Therefore

\(k=\frac{41}{8}.\)

Answer: \(x\)-intercepts \(\left(-\dfrac12,0\right)\), \((3,0)\); \(y\)-intercept \((0,3)\).

First factorise the quadratic inside the modulus:

\(2x^2-5x-3=(2x+1)(x-3).\)

Therefore the graph meets the \(x\)-axis at

\(x=-\frac12\quad\text{and}\quad x=3.\)

So the \(x\)-intercepts are

\(\left(-\frac12,0\right)\quad\text{and}\quad(3,0).\)

For the \(y\)-intercept, put \(x=0\):

\(y=|-3|=3.\)

So the \(y\)-intercept is \((0,3)\).

The graph is the parabola \(y=2x^2-5x-3\) with the part below the \(x\)-axis reflected above the \(x\)-axis. Hence it touches the \(x\)-axis at the two roots and is never negative.

Answer: Intercepts \(\left(-\frac13,0\right)\), \((5,0)\), \((0,5)\); \(k=\frac{64}{3}\).

First find the zeros of \(3x^2-14x-5\):

\(3x^2-14x-5=0.\)

Using the quadratic formula,

\(x=\frac{14\pm\sqrt{196+60}}{6} =\frac{14\pm16}{6}.\)

So

\(x=-\frac13\quad\text{or}\quad x=5.\)

Therefore the graph meets the \(x\)-axis at

\(\left(-\frac13,0\right)\quad\text{and}\quad (5,0).\)

When \(x=0\),

\(y=|-5|=5,\)

so the graph meets the \(y\)-axis at \((0,5)\).

The quadratic \(3x^2-14x-5\) is negative between its roots, so the modulus reflects that part above the \(x\)-axis.

The turning point of \(3x^2-14x-5\) occurs at

\(x=\frac{-(-14)}{2(3)}=\frac73.\)

At this value,

\(3\left(\frac73\right)^2-14\left(\frac73\right)-5 =-\frac{64}{3}.\)

After reflection by the modulus, the maximum point of the middle section has height \(\frac{64}{3}\).

The equation \(\left|3x^2-14x-5\right|=k\) has exactly 3 solutions when the horizontal line touches this maximum point and also intersects the two outer branches. Hence

\(\boxed{k=\frac{64}{3}}.\)

Answer: Intercepts \(\left(-\frac12,0\right)\), \((5,0)\), \((0,5)\); exactly two solutions when \(k=0\) or \(k\gt \frac{121}{8}\).

The graph meets the \(x\)-axis when

\(2x^2-9x-5=0.\)

Factorising gives

\((2x+1)(x-5)=0.\)

So the \(x\)-intercepts are

\(\left(-\frac12,0\right)\quad\text{and}\quad(5,0).\)

The \(y\)-intercept is

\(\left|2(0)^2-9(0)-5\right|=5,\)

so the graph meets the y-axis at \((0,5)\).

The original quadratic \(2x^2-9x-5\) is negative between the roots, so that part is reflected above the \(x\)-axis by the modulus.

The maximum of this reflected middle section occurs at

\(x=\frac{-(-9)}{2(2)}=\frac94.\)

At this value,

\(2\left(\frac94\right)^2-9\left(\frac94\right)-5=-\frac{121}{8}.\)

So the reflected maximum is

\(\frac{121}{8}.\)

For \(\left|2x^2-9x-5\right|=k\) to have exactly two solutions, either the horizontal line is the \(x\)-axis, giving the two roots, or it lies above the reflected maximum.

Therefore

\(\boxed{k=0\quad\text{or}\quad k\gt \frac{121}{8}}.\)

Answer: (i) \(5\left(x-\dfrac75\right)^2-\dfrac{64}{5}\). (ii) Intercepts: \((-0.2,0)\), \((3,0)\), \((0,3)\); reflected vertex \(\left(\dfrac75,\dfrac{64}{5}\right)\). (iii) \(0\lt k\lt\dfrac{64}{5}\).

(i) Complete the square:

\(5x^2-14x-3 =5\left(x^2-\frac{14}{5}x\right)-3.\)

Now

\(x^2-\frac{14}{5}x =\left(x-\frac75\right)^2-\left(\frac75\right)^2.\)

Therefore

\(5x^2-14x-3 = 5\left(x-\frac75\right)^2 -5\left(\frac{49}{25}\right)-3.\)

So

\(5x^2-14x-3 = 5\left(x-\frac75\right)^2-\frac{64}{5}.\)

(ii) The \(x\)-intercepts are found from

\(5x^2-14x-3=0.\)

Using the quadratic formula,

\(x=\frac{14\pm\sqrt{196+60}}{10} =\frac{14\pm16}{10}.\)

So

\(x=3\quad\text{or}\quad x=-0.2.\)

The \(y\)-intercept is

\(y=|-3|=3.\)

The original quadratic has minimum value \(-\frac{64}{5}\) at \(x=\frac75\). For \(y=\left|5x^2-14x-3\right|\), the part below the \(x\)-axis is reflected above it, giving a highest point on the reflected middle section at

\(\left(\frac75,\frac{64}{5}\right).\)

(iii) A horizontal line \(y=k\) cuts the modulus graph in exactly four points only when it lies above the \(x\)-axis and below the reflected maximum \(\frac{64}{5}\). Therefore

\(0\lt k\lt\frac{64}{5}.\)

Answer: (i) \(5\left(x-\dfrac75\right)^2-\dfrac{64}{5}\). (ii) Intercepts: \((-0.2,0)\), \((3,0)\), \((0,3)\); reflected vertex \(\left(\dfrac75,\dfrac{64}{5}\right)\). (iii) \(0\lt k\lt\dfrac{64}{5}\).

(i) Complete the square:

\(5x^2-14x-3 =5\left(x^2-\frac{14}{5}x\right)-3.\)

Now

\(x^2-\frac{14}{5}x =\left(x-\frac75\right)^2-\left(\frac75\right)^2.\)

Therefore

\(5x^2-14x-3 = 5\left(x-\frac75\right)^2 -5\left(\frac{49}{25}\right)-3.\)

So

\(5x^2-14x-3 = 5\left(x-\frac75\right)^2-\frac{64}{5}.\)

(ii) The \(x\)-intercepts are found from

\(5x^2-14x-3=0.\)

Using the quadratic formula,

\(x=\frac{14\pm\sqrt{196+60}}{10} =\frac{14\pm16}{10}.\)

So

\(x=3\quad\text{or}\quad x=-0.2.\)

The \(y\)-intercept is

\(y=|-3|=3.\)

The original quadratic has minimum value \(-\frac{64}{5}\) at \(x=\frac75\). For \(y=\left|5x^2-14x-3\right|\), the part below the \(x\)-axis is reflected above it, giving a highest point on the reflected middle section at

\(\left(\frac75,\frac{64}{5}\right).\)

(iii) A horizontal line \(y=k\) cuts the modulus graph in exactly four points only when it lies above the \(x\)-axis and below the reflected maximum \(\frac{64}{5}\). Therefore

\(0\lt k\lt\frac{64}{5}.\)

Answer: \(\left(x-\dfrac92\right)^2-\dfrac{49}{4}\); minimum \(\left(\dfrac92,-\dfrac{49}{4}\right)\); \(k=\dfrac{49}{4}\).

Complete the square:

\(x^2-9x+8=\left(x-\frac92\right)^2-\frac{81}{4}+8 =\left(x-\frac92\right)^2-\frac{49}{4}.\)

So the minimum point on \(y=x^2-9x+8\) is

\(\left(\frac92,-\frac{49}{4}\right).\)

For the modulus graph, first find the intercepts of \(x^2-9x+8\):

\(x^2-9x+8=(x-1)(x-8).\)

So the \(x\)-intercepts are \((1,0)\) and \((8,0)\). The \(y\)-intercept is

\(|8|=8,\)

so it is \((0,8)\).

The part of the parabola below the \(x\)-axis is reflected above the \(x\)-axis. This gives cusps at \((1,0)\) and \((8,0)\), and the reflected maximum point

\(\left(\frac92,\frac{49}{4}\right).\)

The horizontal line \(y=k\) gives exactly 3 intersections when it passes through this reflected maximum. Therefore

\(\boxed{k=\frac{49}{4}}.\)

Answer: \(-1\leqslant x\leqslant3\).

First expand the left-hand side carefully:

\((3-x)(5x+8)=15x+24-5x^2-8x=-5x^2+7x+24.\)

So the inequality becomes

\(-5x^2+7x+24\geqslant9-3x.\)

Bring all terms to the left-hand side:

\(-5x^2+10x+15\geqslant0.\)

It is easier to factorise with a positive coefficient of \(x^2\). Multiplying by \(-1\) reverses the inequality:

\(5x^2-10x-15\leqslant0.\)

Divide by \(5\):

\(x^2-2x-3\leqslant0.\)

Now factorise:

\((x-3)(x+1)\leqslant0.\)

The critical values are \(x=-1\) and \(x=3\). Since the quadratic opens upwards, it is non-positive between the roots, including the roots themselves.

Therefore

\(-1\leqslant x\leqslant3.\)

Answer: \(-2\leqslant x\leqslant\frac54\).

The critical values occur when one of the factors is zero:

\((x+2)(4x-5)=0.\)

So

\(x=-2\quad\text{or}\quad x=\frac54.\)

The product \((x+2)(4x-5)\) is a quadratic with positive leading coefficient, so it is positive outside the roots and non-positive between the roots.

Because the inequality is

\((x+2)(4x-5)\leqslant0,\)

we take the interval between the two critical values, including both endpoints.

Therefore

\(-2\leqslant x\leqslant\frac54.\)

Answer: (a) \(x\lt 1\) or \(x\gt 4\); (b)(i) \(3(x-2)^2+4\); (b)(ii) \(y=4\).

For a quadratic expression, first find its roots or complete the square. The sign of the quadratic then follows from its shape and the position of the roots.

(a) Start with

\(3x^2-12x+16\gt 3x+4.\)

Bring all terms to one side:

\(3x^2-15x+12\gt 0.\)

Factorise:

\(3(x^2-5x+4)\gt 0,\)

so

\(3(x-1)(x-4)\gt 0.\)

The quadratic is positive outside its roots, so

\(x\lt 1\quad\text{or}\quad x\gt 4.\)

(b)(i) Complete the square:

\(3x^2-12x+16=3(x^2-4x)+16.\)

Since

\(x^2-4x=(x-2)^2-4,\)

we get

\(3x^2-12x+16=3\bigl((x-2)^2-4\bigr)+16.\)

Hence

\(3x^2-12x+16=3(x-2)^2+4.\)

(b)(ii) The minimum point is \((2,4)\). At the minimum point, the tangent is horizontal, so its equation is

\(y=4.\)

Answer: \(-4\lt x\lt -2\).

Answer: \(-4\lt x\lt -2\).

Expand both sides:

\((2x+3)(x-4)=2x^2-5x-12,\)

and

\((3x+4)(x-1)=3x^2+x-4.\)

So

\(2x^2-5x-12\gt 3x^2+x-4.\)

Bring all terms to one side:

\(0\gt x^2+6x+8.\)

Hence

\(x^2+6x+8\lt 0.\)

Factorise:

\((x+2)(x+4)\lt 0.\)

The product is negative between the two roots, so

\(-4\lt x\lt -2.\)

Answer: \(x\lt -4\) or \(x\gt 4\).

Expand the left side:

\((x+5)(x-2)=x^2+3x-10.\)

So

\(x^2+3x-10\gt 3x+6.\)

Subtracting \(3x+6\) from both sides gives

\(x^2-16\gt 0.\)

Factorise:

\((x-4)(x+4)\gt 0.\)

The product is positive outside the two roots, so

\(x\lt -4\quad\text{or}\quad x\gt 4.\)

Answer: \(\frac{2}{5} \lt x \lt \frac{3}{2}\).

Expand the right hand side first:

\((2x+1)(x-1)=2x^2-x-1.\)

So the inequality becomes

\(12x^2-20x+5 \lt 2x^2-x-1.\)

Bring all terms to the left:

\(10x^2-19x+6 \lt 0.\)

Factorise:

\(10x^2-19x+6=(5x-2)(2x-3).\)

Hence

\((5x-2)(2x-3)\lt 0.\)

The critical values are \(x=\frac25\) and \(x=\frac32\). Since the quadratic has a positive coefficient of \(x^2\), it is negative between the roots.

Therefore

\(\boxed{\frac25\lt x\lt \frac32}.\)

Answer: \(x\lt-\frac52\) or \(x\gt3\).

Consider the two possible cases for the modulus.

If

\(3x+2\gt 8+x,\)

then

\(2x\gt6,\)

so

\(x\gt3.\)

If

\(-(3x+2)\gt 8+x,\)

then

\(-3x-2\gt 8+x.\)

Hence

\(-4x\gt10.\)

Dividing by a negative number reverses the inequality, so

\(x\lt-\frac52.\)

Therefore the solution is

\(x\lt-\frac52\quad\text{or}\quad x\gt3.\)

Answer: \(x\lt3\) or \(x\gt15\).

Expand the left-hand side:

\((x-8)(x-10)=x^2-18x+80.\)

So

\(x^2-18x+80\gt35.\)

Hence

\(x^2-18x+45\gt0.\)

Factorise:

\((x-15)(x-3)\gt0.\)

The quadratic is positive outside its two roots, so

\(x\lt3\quad\text{or}\quad x\gt15.\)

Answer: \(-1\lt x\lt \dfrac34\); discriminant \(=-1\), so no real roots.

Expand and rearrange:

\((2x+1)^2\lt 3x+4\)

gives

\(4x^2+4x+1\lt 3x+4,\)

so

\(4x^2+x-3\lt 0.\)

Factorise:

\(4x^2+x-3=(4x-3)(x+1).\)

The critical values are \(x=\frac34\) and \(x=-1\). Since the quadratic opens upwards, it is negative between the roots:

\(\boxed{-1\lt x\lt \frac34}.\)

For the equation

\(\frac{x^2}{4}+kx+k^2+1=0,\)

the discriminant is

\(k^2-4\left(\frac14\right)(k^2+1)=k^2-k^2-1=-1.\)

The discriminant is always negative, so the equation has no real roots for any value of \(k\).

Answer: \(x\leq-\frac52\) or \(x\geq\frac43\).

Start by bringing all terms to one side:

\(x(6x+7)\geq20\)

\(6x^2+7x-20\geq0.\)

Factorise:

\(6x^2+7x-20=(3x-4)(2x+5).\)

The critical values are

\(x=\frac43,\qquad x=-\frac52.\)

The quadratic opens upwards, so it is non-negative outside the two roots. Therefore

\(\boxed{x\leq-\frac52\quad\text{or}\quad x\geq\frac43}.\)

Answer: \(-2\lt x\lt \dfrac19\).

Bring all terms to the left-hand side:

\(9x^2+18x-1\lt x+1.\)

This gives

\(9x^2+17x-2\lt 0.\)

Factorise:

\(9x^2+17x-2=(9x-1)(x+2).\)

The critical values are \(x=\frac19\) and \(x=-2\). Since the quadratic opens upwards, it is negative between the roots. Therefore

\(\boxed{-2\lt x\lt \frac19}.\)

Answer: \(x\lt -8\) or \(x\gt 1\).

Expand the left-hand side:

\((2-x)(x+9)=2x+18-x^2-9x=-x^2-7x+18.\)

So

\(-x^2-7x+18\lt 10.\)

Bring everything to one side:

\(-x^2-7x+8\lt 0.\)

Multiplying by \(-1\) reverses the inequality:

\(x^2+7x-8\gt 0.\)

Factorise:

\((x+8)(x-1)\gt 0.\)

The product is positive outside the two roots, so

\(\boxed{x\lt -8\quad\text{or}\quad x\gt 1}.\)

Answer: \(x\lt -5\) or \(x\gt 5\).

Expand the left-hand side:

\((x-3)(x+4)=x^2+x-12.\)

So the inequality becomes

\(x^2+x-12\gt x+13.\)

Subtract \(x+13\) from both sides:

\(x^2-25\gt 0.\)

Factorise:

\((x-5)(x+5)\gt 0.\)

The product is positive outside the two roots, so

\(\boxed{x\lt -5\quad\text{or}\quad x\gt 5}.\)