Past Exam Questions

Answer: (a)(i) The graph has roots \((-3,0)\) and \((5,0)\). (a)(ii) For example, \(x\geq5\). (b)(i) \(hg(x)=\dfrac4{3x-1}\). (b)(ii) \((hg)^{-1}(x)=\dfrac{x+4}{3x}\). (c) \(a\).

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

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

So

\((x+3)(x-5)=0,\)

giving \(x=-3\) or \(x=5\). Therefore the intercepts are

\((-3,0)\quad\text{and}\quad(5,0).\)

The graph of \(y=(x+3)(x-5)\) is reflected above the \(x\)-axis wherever it would be negative, giving a modulus graph with sharp points at the two roots.

(a)(ii) One suitable domain is

\(x\geq5.\)

On this interval the function is one-to-one, so it has an inverse.

(b)(i) Since \(hg(x)=h(g(x))\),

\(hg(x)=h(3x-1)=\frac4{3x-1}.\)

(b)(ii) Let

\(y=\frac4{3x-1}.\)

Then

\(y(3x-1)=4.\)

So

\(3xy-y=4, \qquad 3xy=y+4, \qquad x=\frac{y+4}{3y}.\)

Therefore

\((hg)^{-1}(x)=\frac{x+4}{3x}.\)

(c) If \(p(a)=b\), then applying the inverse function gives

\(p^{-1}(b)=a.\)

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: Vertex \(\left(\frac75,0\right)\), y-intercept \((0,7)\); \(x=\frac45\) or \(x=2\).

For \(y=|5x-7|\), the graph is V-shaped.

The vertex occurs when

\(5x-7=0,\)

so

\(x=\frac75.\)

Thus the vertex is

\(\left(\frac75,0\right).\)

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

\(y=|-7|=7.\)

So the graph meets the y-axis at \((0,7)\).

For part (b),

\(5|5x-7|-1=14.\)

Hence

\(5|5x-7|=15,\)

so

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

Therefore

\(5x-7=3\quad\text{or}\quad 5x-7=-3.\)

This gives

\(x=2\quad\text{or}\quad x=\frac45.\)

So the solutions are

\(\boxed{x=\frac45\text{ or }x=2}.\)

Answer: Intercepts \(\left(\frac35,0\right)\), \((0,3)\); solutions \(x=\frac14,\frac56\).

The graph \(y=|5x-3|\) is a V-shaped graph.

For the \(x\)-intercept, set \(5x-3=0\):

\(x=\frac35.\)

So the graph meets the \(x\)-axis at \(\left(\frac35,0\right)\).

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

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

So the graph meets the \(y\)-axis at \((0,3)\).

To solve \(|5x-3|=2-x\), consider the two cases.

If \(5x-3\geq0\), then

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

so \(6x=5\), giving \(x=\frac56\).

If \(5x-3\lt 0\), then

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

Thus

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

so \(1=4x\), giving \(x=\frac14\).

Both values satisfy the original equation, so

\(\boxed{x=\frac14,\ \frac56}.\)

Answer: Vertex \(\left(\frac32,0\right)\), y-intercept \((0,3)\); \(x=-2\) or \(x=5\).

The graph of \(y=|2x-3|\) is V-shaped. Its vertex occurs when

\(2x-3=0,\)

so

\(x=\frac32.\)

At this point, \(y=0\), so the vertex is

\(\left(\frac32,0\right).\)

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

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

For the equation,

\(7-|2x-3|=0\)

gives

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

Therefore

\(2x-3=7\quad\text{or}\quad 2x-3=-7.\)

So

\(\boxed{x=5\quad\text{or}\quad x=-2}.\)

Answer: vertex \((2,0)\), \(y\)-intercept \((0,6)\); \(x=\dfrac43,\dfrac83\); \(x\lt \dfrac43\) or \(x\gt \dfrac83\).

The graph of \(y=\left|6-3x\right|\) is a V-shaped graph. The expression inside the modulus is zero when

\(6-3x=0,\qquad x=2.\)

So the vertex is \((2,0)\). When \(x=0\),

\(y=|6|=6,\)

so the graph meets the \(y\)-axis at \((0,6)\).

To solve \(\left|6-3x\right|=2\), use

\(6-3x=2\quad\text{or}\quad 6-3x=-2.\)

These give

\(x=\frac43\quad\text{or}\quad x=\frac83.\)

From the graph, \(\left|6-3x\right|\) is greater than \(2\) outside these two values. Therefore

\(\boxed{x\lt \frac43\quad\text{or}\quad x\gt \frac83}.\)

Answer: \(|2x-5|=4\) gives \(x=\dfrac12,\dfrac92\). The inequality holds for \(\dfrac12\le x\le\dfrac92\).

The graph \(y=|2x-5|\) is a V-shaped graph with vertex where \(2x-5=0\), so its vertex is at

\(x=\frac52,\qquad y=0.\)

Its \(y\)-intercept is \(5\).

The equation \(9y=80x-16x^2\) gives

\(y=\frac{80x-16x^2}{9}.\)

This is a downward-opening quadratic. Its \(x\)-intercepts are found from

\(80x-16x^2=0,\)

so

\(16x(5-x)=0.\)

Thus its \(x\)-intercepts are \(0\) and \(5\).

For part (ii),

\(|2x-5|=4\)

means

\(2x-5=4 \quad\text{or}\quad 2x-5=-4.\)

So

\(x=\frac92 \quad\text{or}\quad x=\frac12.\)

At both these \(x\)-values, \(y=|2x-5|=4\).

Check the quadratic at these values:

\(\frac{80(1/2)-16(1/2)^2}{9}=4\)

and

\(\frac{80(9/2)-16(9/2)^2}{9}=4.\)

Hence the two graphs intersect at the points where \(y=4\).

The inequality

\(9|2x-5|\le 80x-16x^2\)

is equivalent to saying that the quadratic graph is on or above the modulus graph.

From the sketch, this happens between the two intersection points, inclusive.

Therefore

\(\boxed{\frac12\le x\le\frac92}.\)

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

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

Answer: \(a=\frac43\), range \(\mathbb{R}\), and \(\mathrm{f}^{-1}(x)=\frac{4+\mathrm{e}^{x/2}}{3}\).

The logarithm requires

\(3x-4\gt 0,\)

so

\(x\gt \frac43.\)

Therefore the least possible value is

\(a=\frac43.\)

As \(x\to\frac43^+\), \(2\ln(3x-4)\to-\infty\). As \(x\to\infty\), \(2\ln(3x-4)\to\infty\). Hence the range is

\(\mathbb{R}.\)

To find the inverse, let

\(y=2\ln(3x-4).\)

Then

\(\mathrm{e}^{y/2}=3x-4.\)

So

\(x=\frac{4+\mathrm{e}^{y/2}}{3}.\)

Therefore

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

For \(y=\mathrm{f}(x)\), the vertical asymptote is \(x=\frac43\), and the \(x\)-intercept is found from

\(2\ln(3x-4)=0,\)

so \(3x-4=1\) and \(x=\frac53\). There is no \(y\)-intercept because the domain is \(x\gt \frac43\).

For \(y=\mathrm{f}^{-1}(x)\), the horizontal asymptote is \(y=\frac43\). The \(y\)-intercept is

\(\frac{4+\mathrm{e}^{0}}{3}=\frac53,\)

and there is no \(x\)-intercept.

The two graphs are reflections in the line \(y=x\), so they should be sketched symmetrically about that line.

Answer: \(\mathrm{f}(x)=\frac18-2\left(x-\frac94\right)^2\); \(\mathrm{f}^{-1}\) does not exist. \(\mathrm{g}^{-1}(x)=\frac12\left(5-\mathrm{e}^{x/3}\right)\), domain \(x\leqslant3\ln5\), range \(0\leqslant\mathrm{g}^{-1}(x)\lt \frac52\).

Complete the square:

\(\mathrm{f}(x)=-2x^2+9x-10=-2\left(x-\frac94\right)^2+\frac18.\)

So

\(\mathrm{f}(x)=\frac18-2\left(x-\frac94\right)^2.\)

The turning point is \(\left(\frac94,\frac18\right)\). Since the given domain \(0\leqslant x\leqslant3\) includes this turning point, \(\mathrm{f}\) is not one-to-one on its domain. Therefore \(\mathrm{f}^{-1}\) does not exist.

For \(\mathrm{g}(x)=3\ln(5-2x)\), the \(y\)-intercept is

\(\mathrm{g}(0)=3\ln5.\)

The \(x\)-intercept is found from \(3\ln(5-2x)=0\), so \(5-2x=1\), giving \(x=2\).

The vertical asymptote is

\(x=\frac52.\)

To find the inverse, write

\(y=3\ln(5-2x).\)

Then

\(\mathrm{e}^{y/3}=5-2x.\)

Hence

\(x=\frac12\left(5-\mathrm{e}^{y/3}\right).\)

Therefore

\(\mathrm{g}^{-1}(x)=\frac12\left(5-\mathrm{e}^{x/3}\right).\)

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

\(x\leqslant3\ln5.\)

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

\(0\leqslant\mathrm{g}^{-1}(x)\lt \frac52.\)

Answer: \(\mathrm{g}^{-1}(x)=\ln\left(\frac{3}{2-x}-2\right)\), domain \(1\leqslant x\lt 2\).

For \(\mathrm{f}(x)=2\mathrm{e}^{-x}+3\), the \(y\)-intercept is

\(\mathrm{f}(0)=5,\)

and the horizontal asymptote is \(y=3\). Since \(2\mathrm{e}^{-x}\gt 0\), the graph does not meet the \(x\)-axis.

The graph of \(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in the line \(y=x\). It has vertical asymptote \(x=3\) and \(x\)-intercept \((5,0)\).

For \(\mathrm{g}^{-1}\), write

\(y=2-\frac{3}{\mathrm{e}^x+2}.\)

Then

\(2-y=\frac{3}{\mathrm{e}^x+2}.\)

So

\(\mathrm{e}^x+2=\frac{3}{2-y},\)

and hence

\(x=\ln\left(\frac{3}{2-y}-2\right).\)

Interchanging \(x\) and \(y\),

\(\mathrm{g}^{-1}(x)=\ln\left(\frac{3}{2-x}-2\right).\)

The domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\). Since \(x\geqslant0\), \(\mathrm{g}(0)=1\), and as \(x\to\infty\), \(\mathrm{g}(x)\to2\) from below.

Therefore the domain is

\(1\leqslant x\lt 2.\)

Answer: \(2\left(x-\frac12\right)^2+\frac52\), \(p=\frac12\), range \(\mathrm{f}(x)\geqslant\frac52\), and \(\mathrm{f}^{-1}(x)=\frac12-\sqrt{\frac{x-\frac52}{2}}\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Complete the square:

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

Since \(x^2-x=\left(x-\frac12\right)^2-\frac14\),

\(2x^2-2x+3=2\left(x-\frac12\right)^2-\frac12+3.\)

So

\(2x^2-2x+3=2\left(x-\frac12\right)^2+\frac52.\)

The vertex is at \(x=\frac12\). For the restricted domain \(x\leqslant p\), the greatest value of \(p\) for which \(\mathrm{f}\) is one-one is

\(p=\frac12.\)

The minimum value is \(\frac52\), so the range is

\(\mathrm{f}(x)\geqslant\frac52.\)

Let \(y=2\left(x-\frac12\right)^2+\frac52\). Then

\(\frac{y-\frac52}{2}=\left(x-\frac12\right)^2.\)

Since the domain is \(x\leqslant\frac12\), take the negative square root:

\(x-\frac12=-\sqrt{\frac{y-\frac52}{2}}.\)

Hence

\(\mathrm{f}^{-1}(x)=\frac12-\sqrt{\frac{x-\frac52}{2}}.\)

This gives the required answer: \(2\left(x-\frac12\right)^2+\frac52\), \(p=\frac12\), range \(\mathrm{f}(x)\geqslant\frac52\), and \(\mathrm{f}^{-1}(x)=\frac12-\sqrt{\frac{x-\frac52}{2}}\).

Answer: \(\mathrm{ff}(x)=x\); \(\mathrm f\) is self-inverse; range of \(\mathrm g\) is \(1\lt\mathrm g(x)\leqslant2\); domain of \(\mathrm h\) is \(x\ne-\frac13\).

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.

The graph represents a function because each permitted vertical line \(x=k\) meets the graph in at most one point.

Now

\(\mathrm f(\mathrm f(x))=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}.\)

The denominator is

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

Therefore

\(\mathrm{ff}(x)=\frac{\frac{x}{x-1}}{\frac{1}{x-1}}=x.\)

So applying \(\mathrm f\) twice gives the identity function. Hence \(\mathrm f\) is its own inverse, so \(\mathrm f=\mathrm f^{-1}\).

On the diagram, this is shown by the graph being symmetrical about the line \(y=x\).

For \(\mathrm g(x)=\frac{x}{x-1}\), where \(x\geqslant2\), we have \(\mathrm g(2)=2\). As \(x\) increases, \(\frac{x}{x-1}=1+\frac{1}{x-1}\), so the values approach \(1\) from above.

Thus the range is \(1\lt\mathrm g(x)\leqslant2\).

For \(\mathrm h(x)=\frac{2x}{3x+1}\), the denominator must not be zero. Therefore \(3x+1\ne0\), so the domain is \(x\ne-\frac13\).

This gives the required answer: \(\mathrm{ff}(x)=x\); \(\mathrm f\) is self-inverse; range of \(\mathrm g\) is \(1\lt\mathrm g(x)\leqslant2\); domain of \(\mathrm h\) is \(x\ne-\frac13\).

Answer: \(5-(x+2)^2\); range \(\mathrm{f}(x)\leqslant5\); \(k=-2\); \(\mathrm{g}^{-1}(x)=-2+\sqrt{5-x}\), with domain \(x\leqslant5\) and range \(\mathrm{g}^{-1}(x)\geqslant-2\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Complete the square:

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

Since \((x+2)^2\geqslant0\), the greatest value of \(\mathrm{f}(x)\) is \(5\). Hence the range is \(\mathrm{f}(x)\leqslant5\).

For \(\mathrm{g}\) to have an inverse, its domain must be restricted to one side of the turning point. The least possible value is therefore \(k=-2\).

Now write \(y=5-(x+2)^2\), with \(x\geqslant-2\). Then

\((x+2)^2=5-y.\)

Since \(x+2\geqslant0\), \(x+2=\sqrt{5-y}\), so \(x=-2+\sqrt{5-y}\).

Interchanging \(x\) and \(y\),

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

The domain of \(\mathrm{g}^{-1}\) is \(x\leqslant5\), and the range is \(\mathrm{g}^{-1}(x)\geqslant-2\).

This gives the required answer: \(5-(x+2)^2\); range \(\mathrm{f}(x)\leqslant5\); \(k=-2\); \(\mathrm{g}^{-1}(x)=-2+\sqrt{5-x}\), with domain \(x\leqslant5\) and range \(\mathrm{g}^{-1}(x)\geqslant-2\).

Answer: (b) \(\mathrm{gf}(x)=\dfrac{(4x-1)^2}{9x^4}\); (c) \(\mathrm f^{-1}(x)=\dfrac{2x-\sqrt{x(4x-3)}}{3}\).

We start with the main method. Check whether the function is one-to-one on the stated domain before finding an inverse.

The function \(\mathrm g\) has domain \(x\lt 0\), and

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

For \(x\lt 0\), \(\mathrm g(x)\gt 0\). However, \(\mathrm f\) is defined only for inputs less than \(0\).

Therefore \(\mathrm{fg}\) does not exist, because the range of \(\mathrm g\) is not inside the domain of \(\mathrm f\).

For part (b),

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

Since

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

we get

\(\mathrm{gf}(x)=\frac{1}{\left(\frac{3x^2}{4x-1}\right)^2} =\frac{(4x-1)^2}{9x^4}.\)

For part (c), let

\(y=\frac{3x^2}{4x-1}.\)

Then

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

So

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

Now treat this as a quadratic equation in \(x\):

\(x=\frac{4y\pm\sqrt{16y^2-12y}}{6}.\)

Hence

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

Since the original domain of \(\mathrm f\) is \(x\lt 0\), we choose the negative square-root branch.

Therefore

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

This completes the solution and gives the required result.