Past Exam Questions

Answer: \(x=6\sqrt7-7\), \(y=6-2\sqrt7\).

Answer: \(x=6\sqrt7-7\), \(y=6-2\sqrt7\).

From the first equation,

\(x=11-3y.\)

Substitute this into \(x-\sqrt7y=7\):

\(11-3y-\sqrt7y=7.\)

So

\((3+\sqrt7)y=4.\)

Hence

\(y=\frac4{3+\sqrt7}.\)

Rationalising the denominator,

\(y=\frac{4(3-\sqrt7)}{9-7}=2(3-\sqrt7)=6-2\sqrt7.\)

Now substitute into \(x=11-3y\):

\(x=11-3(6-2\sqrt7)=6\sqrt7-7.\)

Answer: \((x,y)=(6,-2)\) or \(\left(-7,\frac35\right)\).

Answer: \((x,y)=(6,-2)\) or \(\left(-7,\frac35\right)\).

From \(x+5y=-4\),

\(x=-4-5y.\)

Substitute this into \(3y-xy=6\):

\(3y-(-4-5y)y=6.\)

So

\(3y+4y+5y^2=6.\)

Hence

\(5y^2+7y-6=0.\)

Factorising,

\((5y-3)(y+2)=0.\)

Therefore

\(y=\frac35\quad\text{or}\quad y=-2.\)

If \(y=\frac35\), then

\(x=-4-5\left(\frac35\right)=-7.\)

If \(y=-2\), then

\(x=-4-5(-2)=6.\)

So the solutions are

\((x,y)=\left(-7,\frac35\right)\quad\text{or}\quad (6,-2).\)

Answer: \(3\sqrt5\).

From the line

\(x+2y=0,\)

we have

\(x=-2y.\)

Substitute this into the curve equation:

\(\frac4{x^2}+\frac5{4y^2}=1.\)

Since \(x^2=4y^2\),

\(\frac4{4y^2}+\frac5{4y^2}=1.\)

So

\(\frac9{4y^2}=1.\)

Hence

\(4y^2=9,\qquad y=\pm\frac32.\)

When \(y=\frac32\), \(x=-3\). When \(y=-\frac32\), \(x=3\).

The two points are therefore

The distance between them is

Thus

Answer: \((1,-4)\), \((-1,4)\), \((2,-2)\), \((-2,2)\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

From

\(xy+4=0,\)

we have

\(xy=-4.\)

Substitute this into the first equation:

\(4x^2+3xy+y^2=8.\)

This gives

\(4x^2+3(-4)+y^2=8.\)

So

\(4x^2+y^2=20. \tag{1}\)

Since \(y=-\frac4x\), substitute into (1):

\(4x^2+\frac{16}{x^2}=20.\)

Multiply by \(x^2\):

\(4x^4-20x^2+16=0.\)

Divide by \(4\):

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

Factorise:

\((x^2-1)(x^2-4)=0.\)

Thus

\(x=\pm1,\quad \pm2.\)

Using \(y=-\frac4x\), the solutions are

\((1,-4),\quad(-1,4),\quad(2,-2),\quad(-2,2).\)

Answer: \((x,y)=\left(\frac52,\frac72\right)\) or \((3,2)\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

From

\(y+3x=11,\)

we have

\(y=11-3x.\)

Substitute this into \(xy+x^2=15\):

\(x(11-3x)+x^2=15.\)

Hence

\(11x-2x^2=15,\)

so

\(2x^2-11x+15=0.\)

Factorise:

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

Therefore

\(x=\frac52\quad\text{or}\quad x=3.\)

If \(x=\frac52\), then

\(y=11-3\left(\frac52\right)=\frac72.\)

If \(x=3\), then

\(y=11-9=2.\)

So the solutions are

Answer: \(x=1+\sqrt3,\ y=2-\sqrt3\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

From

\(x+y=3,\)

we have

\(y=3-x.\)

Substitute this into the second equation:

\(2x-\sqrt3(3-x)=5.\)

Expand:

\(2x-3\sqrt3+\sqrt3x=5.\)

So

\(x(2+\sqrt3)=5+3\sqrt3.\)

Hence

\(x=\frac{5+3\sqrt3}{2+\sqrt3}.\)

Rationalise the denominator:

The denominator is \(1\), and the numerator is

\(10-5\sqrt3+6\sqrt3-9=1+\sqrt3.\)

Therefore

\(x=1+\sqrt3.\)

Then

\(y=3-(1+\sqrt3)=2-\sqrt3.\)

Answer: Stationary point \((-1,-4)\). Intercepts: \((-3,0)\), \((1,0)\), \((0,-3)\).

(a) Complete the square:

\(y=x^2+2x-3.\)

Since

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

we get

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

This is a positive quadratic, so the stationary point is a minimum. The minimum occurs when

\(x+1=0,\)

so \(x=-1\), and then \(y=-4\).

Therefore the stationary point is

\((-1,-4).\)

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

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

Factorise:

\((x+3)(x-1)=0.\)

So the \(x\)-intercepts are

\((-3,0)\quad\text{and}\quad(1,0).\)

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

\(y=-3.\)

So the \(y\)-intercept is

\((0,-3).\)

The sketch should be an upward-opening parabola passing through these intercepts with minimum point \((-1,-4)\).

Answer: \(3+4x-2x^2=5-2(x-1)^2\), and \(f(x)\leqslant5\).

We start with the main method. Complete the square first, because this immediately shows the turning value and the value of the variable at which it occurs.

First rewrite the quadratic by completing the square:

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

Factor out \(-2\) from the quadratic terms:

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

Since \(x^2-2x=(x-1)^2-1\),

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

Therefore

\(3+4x-2x^2=5-2(x-1)^2.\)

Because \((x-1)^2\geqslant0\), the greatest possible value of \(5-2(x-1)^2\) is \(5\).

Hence the range is

\(f(x)\leqslant5.\)

This completes the solution and gives the required result.

Answer: (a) \(-3(x+2)^2+31\); (b) maximum \(31\) when \(x=-2\); (c) \(u=\left(-2+\sqrt{\frac{31}{3}}\right)^2\approx1.48\).

We start with the main method. Complete the square first, because this immediately shows the turning value and the value of the variable at which it occurs.

(a) Start by factorising the quadratic terms:

\(19-12x-3x^2=-3(x^2+4x)+19.\)

Complete the square inside the bracket:

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

So

\(19-12x-3x^2=-3\big((x+2)^2-4\big)+19.\)

Hence

\(19-12x-3x^2=-3(x+2)^2+12+19.\)

Therefore

\(19-12x-3x^2=-3(x+2)^2+31.\)

(b) Since \((x+2)^2\geq0\), the term \(-3(x+2)^2\leq0\).

So the greatest possible value occurs when

\((x+2)^2=0,\)

which gives \(x=-2\).

The maximum value is then

\(31.\)

(c) Let \(x=\sqrt u\). Then \(u=x^2\), and the equation becomes

\(19-12x-3x^2=0.\)

Using part (a),

\(-3(x+2)^2+31=0.\)

So

\((x+2)^2=\frac{31}{3}.\)

Thus

\(x=-2\pm\sqrt{\frac{31}{3}}.\)

But \(x=\sqrt u\geq0\), so only the positive value is possible:

\(x=-2+\sqrt{\frac{31}{3}}.\)

Therefore

\(u=x^2=\left(-2+\sqrt{\frac{31}{3}}\right)^2.\)

Numerically,

\(u\approx1.48.\)

This completes the solution and gives the required result.

Answer: (a) \(2\left(x+\frac54\right)^2-\frac18\); (b) \(\left(-\frac54,-\frac18\right)\); (c) \(-\frac94\lt x\lt-\frac14\).

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) Complete the square:

\(2x^2+5x+3=2\left(x^2+\frac52x\right)+3.\)

Half of \(\frac52\) is \(\frac54\), so

\(2x^2+5x+3 =2\left[\left(x+\frac54\right)^2-\frac{25}{16}\right]+3.\)

Hence

(b) The stationary point is therefore

\(\left(-\frac54,-\frac18\right).\)

(c) Solve

\(2\left(x+\frac54\right)^2-\frac18\lt\frac{15}{8}.\)

Then

\(2\left(x+\frac54\right)^2\lt2,\)

so

\(\left(x+\frac54\right)^2\lt1.\)

Thus

\(-1\lt x+\frac54\lt1.\)

Subtracting \(\frac54\),

\(-\frac94\lt x\lt-\frac14.\)

Answer: (a) \(3\left(x+\frac52\right)^2-\frac{155}{4}\); (b) minimum value \(-\frac{155}{4}\) when \(x=-\frac52\); (c) \(y=1.31\) or \(y=-226\).

Answer: (a) \(3\left(x+\frac52\right)^2-\frac{155}{4}\); (b) minimum value \(-\frac{155}{4}\) when \(x=-\frac52\); (c) \(y=1.31\) or \(y=-226\).

(a) Complete the square:

\(3x^2+15x-20=3(x^2+5x)-20.\)

Now

\(x^2+5x=\left(x+\frac52\right)^2-\frac{25}{4}.\)

Therefore

\(3x^2+15x-20=3\left(x+\frac52\right)^2-\frac{75}{4}-20.\)

So

\(3x^2+15x-20=3\left(x+\frac52\right)^2-\frac{155}{4}.\)

(b) Since \(3\left(x+\frac52\right)^2\geq0\), the minimum value is

\(-\frac{155}{4},\)

and it occurs when

\(x=-\frac52.\)

(c) Let

\(u=y^{1/3}.\)

Then \(y^{2/3}=u^2\), so the equation becomes

\(3u^2+15u-20=0.\)

Using the completed square form,

\(3\left(u+\frac52\right)^2-\frac{155}{4}=0.\)

Thus

\(3\left(u+\frac52\right)^2=\frac{155}{4},\)

so

\(u=-\frac52\pm\sqrt{\frac{155}{12}}.\)

Since \(u=y^{1/3}\),

Therefore

\(y=1.31\quad\text{or}\quad y=-226\)

to 3 significant figures.

Answer: \((x-3)^2-8\), minimum point \((3,-8)\).

(a) Complete the square:

\(x^2-6x+1=(x-3)^2-9+1.\)

Therefore

\(x^2-6x+1=(x-3)^2-8.\)

(b) Since \((x-3)^2\geq0\), the minimum value of \(y\) is \(-8\), which occurs when \(x=3\).

So the minimum point is

\((3,-8).\)

Answer: \(x^2-x-6=\left(x-\frac12\right)^2-\frac{25}{4}\), the stationary point is \(\left(\frac12,-\frac{25}{4}\right)\), and \(\left|x^2-x-6\right|\lt4\) for \(\dfrac{1-\sqrt{41}}{2}\lt x\lt-1\) or \(2\lt x\lt\dfrac{1+\sqrt{41}}{2}\).

(a)(i) Complete the square:

\(x^2-x-6=x^2-x+\frac14-\frac14-6\).

Hence

\(x^2-x-6=\left(x-\frac12\right)^2-\frac{25}{4}\).

(a)(ii) From the completed-square form, the stationary point of

\(y=x^2-x-6\)

is

\(\left(\frac12,-\frac{25}{4}\right)\).

(b) First consider the graph of

\(y=x^2-x-6\).

Its roots are found from

\(x^2-x-6=(x-3)(x+2)=0\),

so the roots are \(x=-2\) and \(x=3\).

Therefore the graph of \(y=\left|x^2-x-6\right|\) touches the \(x\)-axis at

\((-2,0)\quad\text{and}\quad(3,0)\).

The part of the quadratic below the \(x\)-axis is reflected above the \(x\)-axis. Thus the stationary point \(\left(\frac12,-\frac{25}{4}\right)\) becomes

\(\left(\frac12,\frac{25}{4}\right)\)

on the modulus graph.

Useful points for the sketch include

\((-4,14),\quad(-2,0),\quad\left(\frac12,\frac{25}{4}\right),\quad(3,0),\quad(4,6)\).

(c) To solve

\(\left|x^2-x-6\right|\lt4\),

use the equivalent double inequality

\(-4\lt x^2-x-6\lt4\).

First solve

\(x^2-x-6\lt4\).

This gives

\(x^2-x-10\lt0\).

The roots of \(x^2-x-10=0\) are

\(x=\dfrac{1\pm\sqrt{41}}{2}\).

Since the quadratic opens upwards,

\(\dfrac{1-\sqrt{41}}{2}\lt x\lt\dfrac{1+\sqrt{41}}{2}\).

Now solve

\(x^2-x-6\gt-4\).

This gives

\(x^2-x-2\gt0\),

so

\((x-2)(x+1)\gt0\).

Thus

\(x\lt-1\quad\text{or}\quad x\gt2\).

Combining both conditions gives

\(\dfrac{1-\sqrt{41}}{2}\lt x\lt-1\quad\text{or}\quad2\lt x\lt\dfrac{1+\sqrt{41}}{2}\).

Answer: Stationary point \(\left(\frac12,-\frac{49}{4}\right)\); intercepts \((-3,0)\), \((4,0)\), \((0,12)\); exactly 2 roots when \(k\gt\frac{49}{4}\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

Expand the quadratic:

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

The stationary point is found from

\(\frac{\mathrm dy}{\mathrm dx}=2x-1=0\), so \(x=\frac12\).

Then

\(y=\left(\frac12+3\right)\left(\frac12-4\right)=\frac72\cdot\left(-\frac72\right)=-\frac{49}{4}.\)

So the stationary point is \(\left(\frac12,-\frac{49}{4}\right)\).

The graph of \(y=|(x+3)(x-4)|\) is obtained by reflecting the part of the quadratic below the \(x\)-axis above the \(x\)-axis.

The \(x\)-intercepts are \(x=-3\) and \(x=4\), and the \(y\)-intercept is \(y=12\).

The reflected central part has maximum value \(\frac{49}{4}\). Therefore a horizontal line \(y=k\), with \(k\gt0\), cuts the graph in exactly 2 distinct points only when it is above this maximum.

Hence \(k\gt\frac{49}{4}\).

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.

Answer: (a) \(5\left(x-\frac75\right)^2-\frac95\); (b) \(\left(\frac75,-\frac95\right)\); (d) \(0\lt k\lt\frac95\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a) Complete the square:

\(5x^2-14x+8=5\left(x^2-\frac{14}{5}x\right)+8.\)

Half of \(-\frac{14}{5}\) is \(-\frac75\), so

\(5x^2-14x+8=5\left[\left(x-\frac75\right)^2-\frac{49}{25}\right]+8.\)

Hence

(b) The stationary point of \(y=5\left(x-\frac75\right)^2-\frac95\) is

\(\left(\frac75,-\frac95\right).\)

(c) First find where the original quadratic is zero:

\(5x^2-14x+8=0.\)

Factorising gives

\((5x-4)(x-2)=0,\)

so the \(x\)-intercepts are

\(\left(\frac45,0\right)\quad\text{and}\quad(2,0).\)

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

\(y=\lvert8\rvert=8.\)

So the curve passes through \((0,8)\). The part of \(5x^2-14x+8\) below the \(x\)-axis is reflected above the \(x\)-axis, giving a maximum point on the reflected section at

\(\left(\frac75,\frac95\right).\)

(d) The horizontal line \(y=k\) cuts the absolute value graph in four distinct points only when it lies strictly between the \(x\)-axis and the top of the reflected middle section. Therefore

\(0\lt k\lt \frac95.\)

Answer: (a) intercepts \((-5,0)\), \(\left(\frac23,0\right)\), \((0,10)\); (b) \(\displaystyle k=0\) or \(\displaystyle k\gt \frac{289}{12}\).

Answer: (a) intercepts \((-5,0)\), \(\left(\frac23,0\right)\), \((0,10)\); (b) \(\displaystyle k=0\) or \(\displaystyle k\gt \frac{289}{12}\).

(a) First find where the graph meets the axes.

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

\(y=\left|-10\right|=10.\)

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

\((0,10).\)

For the \(x\)-intercepts, solve

\(3x^2+13x-10=0.\)

Factorising gives

\((3x-2)(x+5)=0.\)

Hence

\(x=\frac23\quad\text{or}\quad x=-5.\)

So the \(x\)-intercepts are

\(\left(\frac23,0\right)\quad\text{and}\quad(-5,0).\)

The graph is the quadratic \(y=3x^2+13x-10\), with the part below the \(x\)-axis reflected above the \(x\)-axis.

(b) The stationary point of

\(3x^2+13x-10\)

occurs when

\(x=-\frac{13}{2\cdot3}=-\frac{13}{6}.\)

At this point,

So the reflected maximum of the middle part of \(y=\left|3x^2+13x-10\right|\) is

\(\frac{289}{12}.\)

If \(k=0\), the equation has exactly the two intercept roots.

If \(0\lt k\lt \frac{289}{12}\), the horizontal line \(y=k\) cuts the graph in four distinct points.

If \(k=\frac{289}{12}\), it cuts the graph in three distinct points, because it is tangent at the reflected maximum.

If \(k\gt \frac{289}{12}\), it cuts the graph in exactly two distinct points.

Therefore

\(k=0\quad\text{or}\quad k\gt \frac{289}{12}.\)

Answer: (a) \(\displaystyle 2\left(x+\frac14\right)^2-\frac{121}{8}\); (b) \(\displaystyle \left(-\frac14,-\frac{121}{8}\right)\); (c) intercepts \((-3,0)\), \(\left(\frac52,0\right)\), \((0,15)\); (d) \(\displaystyle k=\frac{121}{8}\).

Answer: (a) \(\displaystyle 2\left(x+\frac14\right)^2-\frac{121}{8}\); (b) \(\displaystyle \left(-\frac14,-\frac{121}{8}\right)\); (c) intercepts \((-3,0)\), \(\left(\frac52,0\right)\), \((0,15)\); (d) \(\displaystyle k=\frac{121}{8}\).

(a) Complete the square:

\(2x^2+x-15=2\left(x^2+\frac12x\right)-15.\)

Inside the bracket,

Therefore

\(2x^2+x-15 =2\left(x+\frac14\right)^2-\frac18-15.\)

So

\(2x^2+x-15=2\left(x+\frac14\right)^2-\frac{121}{8}.\)

(b) The stationary point is therefore

\(\left(-\frac14,-\frac{121}{8}\right).\)

(c) To find the \(x\)-intercepts, solve

\(2x^2+x-15=0.\)

Factorising gives

\((2x-5)(x+3)=0.\)

Thus

\(x=\frac52\quad\text{or}\quad x=-3.\)

The \(y\)-intercept is

\(\left|2(0)^2+0-15\right|=15,\)

so it is \((0,15)\).

The graph is the quadratic \(y=2x^2+x-15\), with the part below the \(x\)-axis reflected above it.

(d) The equation

\(\left|2x^2+x-15\right|=k\)

has \(3\) distinct solutions when the horizontal line \(y=k\) touches the reflected maximum of the middle part and also meets the two outer branches.

This occurs when

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

Answer: \(2\left(x+\dfrac54\right)^2-\dfrac{49}{8}\); stationary point \(\left(-\dfrac54,-\dfrac{49}{8}\right)\); intercepts \((-3,0)\), \(\left(\dfrac12,0\right)\), \((0,3)\); \(k=\dfrac{49}{8}\).

(a) Complete the square:

\(2x^2+5x-3=2\left(x^2+\frac52x\right)-3.\)

Now

Therefore

(b) The stationary point of \(y=2x^2+5x-3\) is

\(\left(-\frac54,-\frac{49}{8}\right).\)

(c) The \(x\)-intercepts of \(y=|2x^2+5x-3|\) occur when

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

Factorising,

\((2x-1)(x+3)=0.\)

So the \(x\)-intercepts are

The \(y\)-intercept is

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

so it is \((0,3)\).

The part of the original quadratic below the \(x\)-axis is reflected above the \(x\)-axis. Hence the reflected central maximum is

\(\left(-\frac54,\frac{49}{8}\right).\)

(d) A horizontal line \(y=k\) cuts the modulus graph in exactly 3 distinct points when it passes through this central maximum. Therefore

\(k=\frac{49}{8}.\)

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

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