Past Exam Questions

Answer: \(4-\sqrt6\).

Answer: \(4-\sqrt6\).

Substitute \(x=6\) into the expression for \(y\):

\(\displaystyle y=\frac{6+\sqrt6}{3+\sqrt6}\).

Rationalise the denominator by multiplying by \(3-\sqrt6\):

\(\displaystyle y=\frac{6+\sqrt6}{3+\sqrt6}\times\frac{3-\sqrt6}{3-\sqrt6}\).

The denominator is

\((3+\sqrt6)(3-\sqrt6)=9-6=3\).

The numerator is

\((6+\sqrt6)(3-\sqrt6)=18-6\sqrt6+3\sqrt6-6=12-3\sqrt6\).

Therefore

\(\displaystyle y=\frac{12-3\sqrt6}{3}=4-\sqrt6\).

Answer: \(k=7-5\sqrt2\).

Answer: \(k=7-5\sqrt2\).

Since \(y=kx^2\),

\(\displaystyle k=\frac{y}{x^2}\).

Here \(x=1+\sqrt2\) and \(y=1-\sqrt2\), so

\(\displaystyle k=\frac{1-\sqrt2}{(1+\sqrt2)^2}\).

Now

\((1+\sqrt2)^2=1+2\sqrt2+2=3+2\sqrt2\).

Therefore

\(\displaystyle k=\frac{1-\sqrt2}{3+2\sqrt2}\).

Rationalise the denominator using \(3-2\sqrt2\):

\(\displaystyle k=\frac{(1-\sqrt2)(3-2\sqrt2)}{(3+2\sqrt2)(3-2\sqrt2)}\).

The denominator is \(9-8=1\). The numerator is

\(3-2\sqrt2-3\sqrt2+4=7-5\sqrt2\).

Hence \(k=7-5\sqrt2\).

Answer: \(a=-\frac52\), range \(\mathbb{R}\), and \(x=\pm\sqrt{\frac{\mathrm{e}^4-7}{2}}\).

For \(\ln(2x+5)\) to be defined, \(2x+5\gt0\), so \(x\gt-\frac52\). Hence the least possible value of \(a\) is \(-\frac52\).

With this domain, \(2x+5\) can take every positive value, so \(\ln(2x+5)\) can take every real value. Thus the range is \(\mathbb{R}\).

Now

\(\mathrm{fg}(x)=\mathrm{f}(\mathrm{g}(x))=\ln(2(x^2+1)+5)=\ln(2x^2+7).\)

Set this equal to \(4\):

\(\ln(2x^2+7)=4.\)

Therefore \(2x^2+7=\mathrm{e}^4\), so

\(x^2=\frac{\mathrm{e}^4-7}{2}.\)

Hence

\(x=\pm\sqrt{\frac{\mathrm{e}^4-7}{2}}.\)

Answer: \(a=150\), range \(0\leqslant \mathrm{g}\mathrm{f}(x)\leqslant\frac1{\sqrt2}\); \(\mathrm{g}^2\) does not exist.

(b) Use the horizontal line test to decide whether a relation is one-one or many-one. A relation is one-one if no horizontal line cuts its graph more than once; otherwise it is many-one.

From the given diagrams, the relation with repeated horizontal values is many-one. The straight line \(y=4-x\) is one-one, and it is its own inverse because reflecting it in the line \(y=x\) gives the same line. The multi-turn curve is many-one. The increasing curve is one-one.

(c) The composite function \(\mathrm g\mathrm f\) exists only if every output of \(\mathrm f\) is allowed as an input of \(\mathrm g\).

The domain of \(\mathrm g\) is

\(x\geqslant\dfrac12.\)

Since \(\mathrm f(x)=\sin x\), we need

\(\sin x\geqslant\dfrac12\)

for every \(x\) in the interval \(30^\circ\leqslant x\leqslant a^\circ\).

Starting at \(30^\circ\), the largest interval for which \(\sin x\geqslant\dfrac12\) is

\(30^\circ\leqslant x\leqslant150^\circ.\)

Hence the largest possible value is

\(a=150.\)

Now

\(\mathrm g\mathrm f(x)=\sqrt{\sin x-\dfrac12}.\)

On \(30^\circ\leqslant x\leqslant150^\circ\), the smallest value of \(\sin x\) is \(\dfrac12\), and the largest value is \(1\). Therefore

\(0\leqslant \sin x-\dfrac12\leqslant\dfrac12.\)

Taking square roots, the range of \(\mathrm g\mathrm f\) is

\(0\leqslant \mathrm g\mathrm f(x)\leqslant\dfrac1{\sqrt2}.\)

For \(\mathrm g^2\) to exist, the range of \(\mathrm g\) must be a subset of the domain of \(\mathrm g\). The range of \(\mathrm g\) is \([0,\infty)\), while the domain is \([\frac12,\infty)\). Since values such as \(0\) are outputs of \(\mathrm g\) but are not in the domain of \(\mathrm g\), \(\mathrm g^2\) does not exist.

Answer: \(x=2\).

The notation \(\mathrm{fg}(x)\) means \(\mathrm{f}(\mathrm{g}(x))\).

Since

\(\mathrm{g}(x)=\sqrt{x+2},\)

we have

\(\mathrm{fg}(x)=\mathrm{f}(\sqrt{x+2}).\)

Using \(\mathrm{f}(x)=\dfrac{3x}{x+4}\),

\(\mathrm{fg}(x)=\dfrac{3\sqrt{x+2}}{\sqrt{x+2}+4}.\)

Now solve \(\mathrm{fg}(x)=1\):

\(\dfrac{3\sqrt{x+2}}{\sqrt{x+2}+4}=1.\)

Multiply by \(\sqrt{x+2}+4\):

\(3\sqrt{x+2}=\sqrt{x+2}+4.\)

So

\(2\sqrt{x+2}=4,\)

and hence

\(\sqrt{x+2}=2.\)

Squaring both sides gives

\(x+2=4,\)

so

\(x=2.\)

This satisfies the domain condition \(x\gt -2\), so the solution is valid.

Answer: \(a=-1\), and \(x=0.916\).

Rewrite the exponential terms in a common form, or introduce a substitution, so that the equation becomes algebraic.

For \(\mathrm{gf}(x)=\mathrm g(\mathrm f(x))\) to exist, the output of \(\mathrm f\) must be at least \(1\), because the domain of \(\mathrm g\) is \(x\geqslant1\).

Since \(x\geqslant0\), the least value of \(2\mathrm e^x+a\) is \(2+a\). Therefore

\(2+a\geq1.\)

The least integer value is

\(a=-1.\)

When \(a=5\),

\(\mathrm{gf}(x)=\sqrt{(2\mathrm e^x+5)-1}=\sqrt{2\mathrm e^x+4}.\)

Solve \(\mathrm{gf}(x)=3\):

\(\sqrt{2\mathrm e^x+4}=3.\)

Squaring gives

\(2\mathrm e^x+4=9.\)

So

\(\mathrm e^x=\frac52.\)

Hence

\(x=\ln\left(\frac52\right)=0.916\ldots.\)

Therefore \(x=0.916\) to 3 decimal places.

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

Answer: (a) \(3\); (b) \(\mathrm{gg}(x)=\mathrm e^{3\mathrm e^{3x}-3}-1\); (c) \(x=4\).

We start with the main method. Substitute one function into the other carefully and respect the stated domains.

(a) First find \(\mathrm g(0)\):

\(\mathrm g(0)=\mathrm e^0-1=0.\)

Therefore

\(\mathrm{fg}(0)=\mathrm f(\mathrm g(0))=\mathrm f(0).\)

So

\(\mathrm{fg}(0)=2\sin0+3\cos0=3.\)

(b) Since \(\mathrm g(x)=\mathrm e^{3x}-1\),

\(\mathrm{gg}(x)=\mathrm g(\mathrm g(x))=\mathrm e^{3(\mathrm e^{3x}-1)}-1.\)

Hence

\(\mathrm{gg}(x)=\mathrm e^{3\mathrm e^{3x}-3}-1.\)

(c) To find \(\mathrm g^{-1}\), let

\(y=\mathrm e^{3x}-1.\)

Then

\(y+1=\mathrm e^{3x},\)

so

\(\ln(y+1)=3x.\)

Thus

\(\mathrm g^{-1}(x)=\frac13\ln(x+1).\)

Now solve

\(\frac13\ln(x+1)=\frac13\ln5.\)

Multiplying by \(3\),

\(\ln(x+1)=\ln5,\)

so

\(x+1=5.\)

Therefore

\(x=4.\)

This completes the solution and gives the required result.

Answer:

(b) \(a=1\), range \(\mathrm h(x)\ge3\), \(\mathrm h^{-1}(x)=1+\sqrt{x-3}\), domain \(x\ge3\).

Answer: \(a=1\), \(\mathrm h(x)\ge3\), \(\mathrm h^{-1}(x)=1+\sqrt{x-3}\), with domain \(x\ge3\).

For \(\mathrm f(x)=2x^2\),

\(\mathrm f'(x)=4x.\)

Also,

\(\mathrm{fg}(x)=\mathrm f(\mathrm g(x))=2(2x+1)^2=8x^2+8x+2.\)

Next,

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

Finally, from \(y=2x+1\),

\(x=\frac{y-1}{2}.\)

So

\(\mathrm g^{-1}(x)=\frac{x-1}{2}.\)

For part (b),

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

The turning point is at \(x=1\). For \(\mathrm h^{-1}\) to exist with \(a\) as small as possible, use the domain

\(x\ge1.\)

So \(a=1\). The least value of \(\mathrm h(x)\) is \(3\), hence the range is

\(\mathrm h(x)\ge3.\)

To find the inverse, write

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

Then

\(x-1=\sqrt{y-3}\)

because \(x\ge1\). Thus

\(x=1+\sqrt{y-3}.\)

Therefore

\(\mathrm h^{-1}(x)=1+\sqrt{x-3}.\)

The domain of \(\mathrm h^{-1}\) is the range of \(\mathrm h\), so

\(x\ge3.\)

Answer: (a) \((\ln(3x+2))^2+1\); (b) \(x=\frac{e^2-2}{3}\); (c) \(x=-0.243\) approximately.

Answer: (a) \((\ln(3x+2))^2+1\); (b) \(x=\frac{e^2-2}{3}\); (c) \(x=-0.243\) approximately.

(a) Since \(f(x)=x^2+1\) and \(g(x)=\ln(3x+2)\),

\(fg(x)=f(g(x)).\)

Therefore

\(fg(x)=(\ln(3x+2))^2+1.\)

(b) Solve

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

Then

\((\ln(3x+2))^2=4,\)

so

\(\ln(3x+2)=\pm2.\)

Because the composite function \(fg(x)\) requires \(g(x)\) to be in the domain of \(f\), we need \(g(x)\gt -\frac23\). The value \(\ln(3x+2)=-2\) is not allowed.

Hence

\(\ln(3x+2)=2.\)

So

\(3x+2=e^2, \qquad x=\frac{e^2-2}{3}.\)

(c) Now

\(gg(x)=g(g(x))=\ln(3\ln(3x+2)+2).\)

Set this equal to \(1\):

\(\ln(3\ln(3x+2)+2)=1.\)

Therefore

\(3\ln(3x+2)+2=e.\)

So

\(\ln(3x+2)=\frac{e-2}{3}.\)

Exponentiating gives

\(3x+2=e^{(e-2)/3}.\)

Hence

\(x=\frac{e^{(e-2)/3}-2}{3}.\)

Numerically,

\(x=-0.243\)

to 3 significant figures.

Answer: range of \(\mathrm f\): \(\mathrm f(x)\geq -4\); range of \(\mathrm g\): \(\mathrm g(x)\gt1\); \(x=\ln\sqrt2\).

(a) Complete the square:

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

Since \((x+2)^2\geq0\), the least value of \(\mathrm f(x)\) is \(-4\). Hence the range of \(\mathrm f\) is

\(\mathrm f(x)\geq -4.\)

(b) Since \(\mathrm e^{2x}\gt0\) for all real \(x\),

\(\mathrm g(x)=1+\mathrm e^{2x}\gt1.\)

So the range of \(\mathrm g\) is

\(\mathrm g(x)\gt1.\)

(c) The equation \(\mathrm{fg}(x)=21\) means

\(\mathrm f(\mathrm g(x))=21.\)

Substitute \(\mathrm g(x)=1+\mathrm e^{2x}\) into \(\mathrm f(u)=u^2+4u\):

\((1+\mathrm e^{2x})^2+4(1+\mathrm e^{2x})=21.\)

Expand:

So

\(\mathrm e^{4x}+6\mathrm e^{2x}-16=0.\)

Let \(u=\mathrm e^{2x}\). Then

\(u^2+6u-16=0.\)

Factorise:

\((u+8)(u-2)=0.\)

Since \(u=\mathrm e^{2x}\gt0\), take \(u=2\). Hence

\(\mathrm e^{2x}=2.\)

Taking logarithms,

\(2x=\ln2,\)

so

\(x=\frac12\ln2=\ln\sqrt2.\)

Answer: \(\mathrm{fg}(x)=\frac{x^2+4}{x^2}\); \(\mathrm g^{-1}(x)=\sqrt{x-1}\); \(x=2\).

(a) Here \(\mathrm{fg}(x)\) means \(\mathrm f(\mathrm g(x))\). Since \(\mathrm g(x)=1+x^2\),

Therefore

\(\mathrm{fg}(x)=\frac{x^2+4}{x^2}.\)

(b) Let

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

Then

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

Since the domain has \(x\gt 1\), use the positive square root:

\(x=\sqrt{y-1}.\)

Hence

\(\mathrm g^{-1}(x)=\sqrt{x-1}.\)

(c) Solve

\(\frac{x+3}{x-1}=1+x^2.\)

Multiplying by \(x-1\),

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

Expanding,

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

Cancel \(x\) from both sides and rearrange:

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

Since \(x=2\) is a root, factorise:

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

The quadratic \(x^2+x+2\) has discriminant

\(1-8=-7,\)

so it has no real roots. Therefore the only solution is

\(x=2.\)

Answer: \(\mathrm{f}^2(x)=4x-3\), intercepts \((\frac34,0)\) and \((0,3)\), \(a=\frac12\), \(b=\frac{15}{2}\).

(a) The range of \(\mathrm{f}\) is all real numbers, and this is the same as the domain of \(\mathrm{f}\). Therefore \(\mathrm{f}^2=\mathrm{f}\mathrm{f}\) can be formed.

(b) First find \(\mathrm{f}^2(x)\):

\(\mathrm{f}^2(x)=\mathrm{f}(\mathrm{f}(x))=2(2x-1)-1=4x-3.\)

So

\(y=\left|\mathrm{f}^2(x)\right|=|4x-3|.\)

This is a V-shaped graph with vertex where \(4x-3=0\), so the \(x\)-intercept is

\(\left(\frac34,0\right).\)

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

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

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

(c) The inequality holds exactly for \(-1\leqslant x\leqslant3\), so the line \(y=ax+b\) passes through the corresponding end points on \(y=|4x-3|\).

When \(x=-1\),

\(|4(-1)-3|=7.\)

When \(x=3\),

\(|4(3)-3|=9.\)

So the line passes through \((-1,7)\) and \((3,9)\). Its gradient is

\(a=\frac{9-7}{3-(-1)}=\frac12.\)

Using \((-1,7)\),

\(7=-\frac12+b,\)

so

\(b=\frac{15}{2}.\)

Answer: vertex \(\left(\dfrac32,0\right)\), intercepts \(\left(\dfrac32,0\right)\) and \((0,6)\); \(x=1\) or \(x=3\).

The graph of \(y=|4x-6|\) is a V-shaped graph.

The vertex occurs where \(4x-6=0\), so \(x=\frac32\). Hence the vertex is \(\left(\frac32,0\right)\).

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

\(y=|4(0)-6|=6.\)

So the graph meets the axes at \(\left(\frac32,0\right)\) and \((0,6)\).

For part (b), solve the absolute value equation by using the two possible signs:

\(4x-6=2x\quad\text{or}\quad 4x-6=-2x.\)

From \(4x-6=2x\),

\(2x=6\), so \(x=3\).

From \(4x-6=-2x\),

\(6x=6\), so \(x=1\).

Therefore the solutions are

\(x=1\quad\text{or}\quad x=3.\)

This completes the solution and gives the required result.

Answer: \(2\le x\le4\).

Answer: \(2\le x\le4\).

From the graph, the straight line \(f(x)\) has intercept 5 and gradient \(-2\), so

\(f(x)=-2x+5\).

The line \(g(x)\) has intercept \(-1\) and gradient 1, so

\(g(x)=x-1\).

The graph of \(|f(x)|\) is formed by reflecting the part of \(f(x)\) below the \(x\)-axis above the \(x\)-axis. Hence

\(|f(x)|=|-2x+5|\).

The critical points occur when

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

For \(x\le2.5\), \(|-2x+5|=-2x+5\). Then

\(-2x+5=x-1\),

so \(x=2\).

For \(x\ge2.5\), \(|-2x+5|=2x-5\). Then

\(2x-5=x-1\),

so \(x=4\).

Between these two intersection points, the graph of \(|f(x)|\) lies below or on the graph of \(g(x)\). Therefore

\(2\le x\le4\).

Answer: (a) Intercepts: \((0,1)\), \(\left(-\frac12,0\right)\), \((0,5)\), and \(\left(\frac53,0\right)\); (b) \(\displaystyle x=\frac45\) or \(x=6\).

Answer: (a) Intercepts: \((0,1)\), \(\left(-\frac12,0\right)\), \((0,5)\), and \(\left(\frac53,0\right)\); (b) \(\displaystyle x=\frac45\) or \(x=6\).

(a) For \(y=\left|2x+1\right|\), the vertex is where \(2x+1=0\), so

\(x=-\frac12.\)

Therefore this graph meets the \(x\)-axis at \(\left(-\frac12,0\right)\). It meets the \(y\)-axis when \(x=0\), giving \(y=1\), so the \(y\)-intercept is \((0,1)\).

For \(y=\left|5-3x\right|\), the vertex is where \(5-3x=0\), so

\(x=\frac53.\)

Therefore this graph meets the \(x\)-axis at \(\left(\frac53,0\right)\). It meets the \(y\)-axis when \(x=0\), giving \(y=5\), so the \(y\)-intercept is \((0,5)\).

(b) To solve

\(\left|2x+1\right|=\left|5-3x\right|,\)

use the two possible cases:

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

or

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

The first case gives

\(5x=4,\)

so

\(x=\frac45.\)

The second case gives

\(2x+1=-5+3x,\)

so

\(x=6.\)

Answer: \(x\lt -2\) or \(x\gt 2\).

For

\(y=5+|3x-2|,\)

the vertex occurs when \(3x-2=0\), so

\(x=\frac23.\)

At this point,

\(y=5.\)

So the vertex is \(\left(\frac23,5\right)\). Also, when \(x=0\),

\(y=5+|-2|=7,\)

so the \(y\)-intercept is \((0,7)\). The graph is a V-shape.

The straight line \(y=11-x\) has \(y\)-intercept \((0,11)\) and gradient \(-1\).

To solve the inequality algebraically, first find the intersection points.

If \(x\lt \frac23\), then

\(5+|3x-2|=5-(3x-2)=7-3x.\)

So

\(11-x=7-3x,\)

giving

\(x=-2.\)

If \(x\geq\frac23\), then

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

So

\(11-x=3x+3,\)

giving

\(x=2.\)

The inequality

\(11-x\lt 5+|3x-2|\)

means that the straight line is below the modulus graph. This happens outside the two intersection points, so

\(x\lt -2\quad\text{or}\quad x\gt 2.\)

Answer: \(x\lt 2\) or \(x\gt 6\).

The key point is to split the absolute-value expression into the correct cases before solving.

The graph of \(y=|x-5|\) is a V-shape with vertex \((5,0)\) and \(y\)-intercept \((0,5)\).

The graph of \(y=6-|2x-7|\) is an inverted V-shape. Its vertex occurs when \(2x-7=0\), so

\(x=\frac72.\)

The vertex is therefore

\(\left(\frac72,6\right).\)

When \(x=0\),

\(y=6-|-7|=-1,\)

so the \(y\)-intercept is \((0,-1)\).

The two graphs intersect at \(x=2\) and \(x=6\).

The inequality

\(|x-5|\gt 6-|2x-7|\)

means that the graph of \(y=|x-5|\) is above the graph of \(y=6-|2x-7|\).

This happens outside the two intersection points, so

\(x\lt 2\quad\text{or}\quad x\gt 6.\)

Answer: The graph of \(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in the line \(y=x\). Also \(\mathrm{g}^{-1}(x)=(\ln x)^2+2\), the range of \(\mathrm{g}^{-1}\) is \([2,\infty)\), and \(\mathrm{gh}(x)=\mathrm{e}^{1/x}\).

(a) To sketch \(y=\mathrm{f}^{-1}(x)\), interchange the coordinates of points on the graph of \(y=\mathrm{f}(x)\). For example, a point \((a,b)\) on \(y=\mathrm{f}(x)\) becomes \((b,a)\) on \(y=\mathrm{f}^{-1}(x)\).

This is exactly the same as reflecting the graph of \(y=\mathrm{f}(x)\) in the line

\(y=x\).

Therefore the relationship is:

\(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in \(y=x\).

(b)(i) Let

\(y=\mathrm{g}(x)=\mathrm{e}^{\sqrt{x-2}}\), where \(x\geqslant2\).

Take natural logarithms:

\(\ln y=\sqrt{x-2}\).

Now square both sides:

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

So

\(x=(\ln y)^2+2\).

Interchange \(x\) and \(y\) to obtain the inverse function:

\(\mathrm{g}^{-1}(x)=(\ln x)^2+2\).

Since \(\mathrm{g}(x)\geqslant1\), the domain of \(\mathrm{g}^{-1}\) is \(x\geqslant1\).

(b)(ii) The range of \(\mathrm{g}^{-1}\) is the original domain of \(\mathrm{g}\). Since the domain of \(\mathrm{g}\) is \(x\geqslant2\),

\(\text{range of }\mathrm{g}^{-1}=[2,\infty)\).

(b)(iii) The notation \(\mathrm{gh}(x)\) means

\(\mathrm{g}(\mathrm{h}(x))\).

Now

\(\mathrm{h}(x)=\frac1{x^2}+2\), where \(x\gt0\). Hence

\(\mathrm{gh}(x)=\mathrm{g}\left(\frac1{x^2}+2\right)\).

Using \(\mathrm{g}(x)=\mathrm{e}^{\sqrt{x-2}}\),

\(\mathrm{gh}(x)=\mathrm{e}^{\sqrt{\left(\frac1{x^2}+2\right)-2}}\).

This simplifies to

\(\mathrm{gh}(x)=\mathrm{e}^{\sqrt{1/x^2}}\).

Since \(x\gt0\), \(\sqrt{1/x^2}=1/x\). Therefore

\(\mathrm{gh}(x)=\mathrm{e}^{1/x}\).