Past Exam Questions

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.

Answer: \(\mathrm f^{-1}(x)=-\sqrt{\ln x-3}\), domain \(x\gt \mathrm e^3\), range \(\mathrm f^{-1}(x)\lt 0\), and \(\mathrm g(x)=-\sqrt{2x-3}\).

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

The function \(\mathrm f\) is defined by

\(\mathrm f(x)=\mathrm e^{x^2+3}\quad\text{for }x\lt 0.\)

On \(x\lt 0\), as \(x\) increases, \(x^2\) decreases. Hence \(\mathrm f\) is one-one, so \(\mathrm f^{-1}\) exists.

To find the inverse, let

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

Then

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

So

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

Since the original domain is \(x\lt 0\), take the negative square root:

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

Therefore

\(\mathrm f^{-1}(x)=-\sqrt{\ln x-3}.\)

The domain of \(\mathrm f^{-1}\) is the range of \(\mathrm f\). Since \(x\lt 0\), \(x^2\gt 0\), so \(x^2+3\gt 3\), and therefore

\(x\gt \mathrm e^3.\)

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

\(\mathrm f^{-1}(x)\lt 0.\)

Now use \(\mathrm{fg}(x)=\mathrm e^{2x}\). This means

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

Apply \(\mathrm f^{-1}\) to both sides:

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

Hence

\(\mathrm g(x)=-\sqrt{\ln(\mathrm e^{2x})-3}=-\sqrt{2x-3}.\)

This completes the solution and gives the required result.

Answer: (a) \(x=\ln\left(\frac3{\sqrt2}-1\right)\); (b) \(\mathrm f^{-1}\) does not exist; \(\mathrm g^{-1}(x)=\ln(x-1)\), domain \(x\gt1\).

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

(a) Since \(\mathrm{fg}(x)=\mathrm f(\mathrm g(x))\),

\(\mathrm{fg}(x)=2(\mathrm e^x+1)^2-1.\)

Set this equal to \(8\):

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

Then

\(2(\mathrm e^x+1)^2=9,\)

so

\((\mathrm e^x+1)^2=\frac92.\)

Because \(\mathrm e^x+1\gt0\),

\(\mathrm e^x+1=\frac3{\sqrt2}.\)

Therefore

\(\mathrm e^x=\frac3{\sqrt2}-1.\)

So

\(x=\ln\left(\frac3{\sqrt2}-1\right).\)

(b) The function \(\mathrm f(x)=2x^2-1\) is not one-to-one for all real values of \(x\). For example, \(\mathrm f(1)=\mathrm f(-1)\).

Therefore \(\mathrm f^{-1}\) does not exist on the stated domain.

For \(\mathrm g\), let

\(y=\mathrm e^x+1.\)

Then

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

Taking natural logarithms gives

\(x=\ln(y-1).\)

So

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

Since the input to the logarithm must be positive, the domain is

\(x\gt1.\)

This completes the solution and gives the required result.

Answer: domain of \(f^{-1}\): \(0\leqslant x\leqslant\dfrac{2\sqrt2}{3}\); range of \(f^{-1}\): \(\dfrac12\leqslant f^{-1}(x)\leqslant\dfrac32\); \(f^{-1}(x)=\sqrt{\dfrac1{4-4x^2}}\); \(gf(x)=e^{1-\frac1{4x^2}}\).

(a)(i) The domain of \(f\) is

\(\frac12\leqslant x\leqslant\frac32.\)

From the graph and the formula, \(f\) is increasing on this interval.

At \(x=\frac12\),

\(f\left(\frac12\right)=0.\)

At \(x=\frac32\),

So the range of \(f\), and hence the domain of \(f^{-1}\), is

\(0\leqslant x\leqslant\frac{2\sqrt2}{3}.\)

The range of \(f^{-1}\) is the domain of \(f\):

\(\frac12\leqslant f^{-1}(x)\leqslant\frac32.\)

(a)(ii) Let

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

Then

\(2xy=\sqrt{4x^2-1}.\)

Squaring both sides,

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

Rearrange:

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

Hence

\(x^2=\frac1{4(1-y^2)}=\frac1{4-4y^2}.\)

Since \(x\gt 0\),

\(x=\sqrt{\frac1{4-4y^2}}.\)

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

\(f^{-1}(x)=\sqrt{\frac1{4-4x^2}}.\)

(b) Since \(g(x)=e^{x^2}\),

\(gf(x)=g(f(x))=e^{(f(x))^2}.\)

Now

So

\((f(x))^2=\frac{4x^2-1}{4x^2} =1-\frac1{4x^2}.\)

Therefore

\(gf(x)=e^{1-\frac1{4x^2}},\)

which is of the required form with \(a=1\) and \(b=4\).

Answer: \(\mathrm{fg}(x)=\frac{2}{3x}-\frac{x}{3}\); range of \(\mathrm f^{-1}\) is \(x\gt 0\); \(\mathrm f^{-1}(x)=\frac{3x+\sqrt{9x^2+8}}{4}\).

(a) Since \(\mathrm g(x)=\frac1x\),

Substitute \(\frac1x\) into \(\mathrm f\):

So

Therefore

\(\mathrm{fg}(x)=\frac{2}{3x}-\frac{x}{3}.\)

(b)(i) The domain of \(\mathrm f\) is \(x\gt 0\). Therefore the range of \(\mathrm f^{-1}\) is

\(x\gt 0.\)

(b)(ii) Let

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

Then

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

Rearrange as a quadratic in \(x\):

\(2x^2-3yx-1=0.\)

Using the quadratic formula,

\(x=\frac{3y\pm\sqrt{9y^2+8}}{4}.\)

Since \(x\gt 0\), take the positive square-root branch:

\(x=\frac{3y+\sqrt{9y^2+8}}{4}.\)

Replacing \(y\) by \(x\),

\(\mathrm f^{-1}(x)=\frac{3x+\sqrt{9x^2+8}}{4}.\)

Answer: \(\mathrm f(x)=15-2(x+1)^2\), range \(\mathrm f\leq15\); domain of \(\mathrm g^{-1}\): \(x\geq\sqrt2\), range: \(x\geq1\); \(\mathrm g^{-1}(x)=-1+\sqrt{x^2+2}\).

(a)(i) Complete the square:

\(\mathrm f(x)=13-4x-2x^2.\)

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

\(\mathrm f(x)=-2(x^2+2x)+13.\)

Since

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

we get

\(\mathrm f(x)=-2\left((x+1)^2-1\right)+13.\)

So

\(\mathrm f(x)=15-2(x+1)^2.\)

(a)(ii) Since \((x+1)^2\geq0\),

\(-2(x+1)^2\leq0.\)

Therefore

\(\mathrm f(x)\leq15.\)

(b)(i) The domain of \(\mathrm g\) is \(x\geq1\). Hence the range of \(\mathrm g^{-1}\) is

\(x\geq1.\)

Also

\(\mathrm g(1)=\sqrt{1+2-1}=\sqrt2,\)

and \(\mathrm g\) is increasing for \(x\geq1\). So the range of \(\mathrm g\), and hence the domain of \(\mathrm g^{-1}\), is

\(x\geq\sqrt2.\)

(b)(ii) Let

\(y=\sqrt{x^2+2x-1}.\)

Then

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

Rearrange as a quadratic in \(x\):

\(x^2+2x-(1+y^2)=0.\)

Using the quadratic formula,

\(x=\frac{-2\pm\sqrt{4+4(1+y^2)}}{2}.\)

So

\(x=-1\pm\sqrt{y^2+2}.\)

Since the range of \(\mathrm g^{-1}\) is \(x\geq1\), choose the positive branch:

\(x=-1+\sqrt{y^2+2}.\)

Replacing \(y\) by \(x\),

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

Answer: \(2(x+\frac54)^2-\frac18\), stationary point \((-\frac54,-\frac18)\), least \(p=-\frac54\).

(a) Complete the square:

\(2x^2+5x+3=2\left(x^2+\dfrac52x\right)+3.\)

Inside the bracket,

\(x^2+\dfrac52x=\left(x+\dfrac54\right)^2-\left(\dfrac54\right)^2.\)

Therefore

\(2x^2+5x+3=2\left[\left(x+\dfrac54\right)^2-\dfrac{25}{16}\right]+3.\)

Simplifying,

\(2x^2+5x+3=2\left(x+\dfrac54\right)^2-\dfrac{25}{8}+3 =2\left(x+\dfrac54\right)^2-\dfrac18.\)

Thus \(a=\dfrac54\) and \(b=-\dfrac18\).

(b) The completed square form shows that the minimum point of the parabola is

\(\left(-\dfrac54,-\dfrac18\right).\)

(c)(i) For \(\mathrm f^{-1}\) to exist, \(\mathrm f\) must be one-one on its domain. The parabola is one-one on either side of its vertex. Since the domain is \(x\geqslant p\), the least value of \(p\) that keeps only the right-hand branch is

\(p=-\dfrac54.\)

(c)(ii) With \(p=-\dfrac54\), the graph of \(y=\mathrm f(x)\) is the right-hand branch of the parabola with vertex

\(\left(-\dfrac54,-\dfrac18\right).\)

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

\(y=3,\)

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

For the \(x\)-intercepts, solve

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

Factorising,

\((2x+3)(x+1)=0,\)

so \(x=-\dfrac32\) or \(x=-1\). The value \(-\dfrac32\) is not in the restricted domain \(x\geqslant-\dfrac54\), so the only \(x\)-intercept of \(\mathrm f\) is

\((-1,0).\)

The graph of \(y=\mathrm f^{-1}(x)\) is the reflection of \(y=\mathrm f(x)\) in the line \(y=x\). Therefore the intercept \((-1,0)\) reflects to \((0,-1)\), and the intercept \((0,3)\) reflects to \((3,0)\). The vertex reflects to

\(\left(-\dfrac18,-\dfrac54\right).\)

Answer: range \(\mathrm f(x)\gt2\); \(\mathrm f^{-1}(x)=-\frac13\ln(x-2)\); asymptotes \(y=2\) and \(x=2\); \(\mathrm{gf}(x)=12\) gives \(x=-\frac13\ln2\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use the fact that the graph of an inverse function is the reflection in the line \(y=x\).

(a) Since \(e^{-3x}\gt0\) for all real \(x\),

\(\mathrm f(x)=2+e^{-3x}\gt2.\)

So the range is

\(\mathrm f(x)\gt2.\)

(b) Let

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

Then

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

Taking natural logarithms gives

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

So

\(x=-\frac13\ln(y-2).\)

Therefore

\(\mathrm f^{-1}(x)=-\frac13\ln(x-2),\quad x\gt2.\)

(c) The graph of \(y=\mathrm f(x)\) has \(y\)-intercept

\(2+e^0=3,\)

so it passes through \((0,3)\). Its horizontal asymptote is

\(y=2.\)

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

\(x=2.\)

(d) The equation \(\mathrm{gf}(x)=12\) means

\((\mathrm f(x))^{3/2}+4=12.\)

So

\((2+e^{-3x})^{3/2}=8.\)

Raise both sides to the power \(\frac23\):

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

Thus

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

Taking natural logarithms,

\(-3x=\ln2,\)

so

\(x=-\frac13\ln2.\)

This completes the solution and gives the required result.

Answer: (a)(i) domain \(0\leq x\leq\frac94\), range \(0\leq f^{-1}(x)\leq3\); (a)(ii) \(x=0\) or \(x=\frac85\); (b)(i) \(g^{-1}(x)=\sqrt[3]{\frac{x^3-3}{8}}\); (b)(ii) \(k=0\); (b)(iii) \(gh(x)=\sqrt[3]{8e^{12x}+3}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) The domain of \(f^{-1}\) is the range of \(f\). Since

\(f(0)=0\)

and

\(f(3)=\frac{9}{\sqrt{16}}=\frac94,\)

the domain of \(f^{-1}\) is

\(0\leq x\leq\frac94.\)

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

\(0\leq f^{-1}(x)\leq3.\)

(a)(ii) Solve

\(\frac{3x}{\sqrt{5x+1}}=x.\)

One solution is \(x=0\). If \(x\ne0\), divide by \(x\):

\(\frac{3}{\sqrt{5x+1}}=1.\)

Hence

\(\sqrt{5x+1}=3.\)

Squaring gives

\(5x+1=9,\)

so

\(x=\frac85.\)

Thus

\(x=0\quad\text{or}\quad x=\frac85.\)

(a)(iii) The graph of \(y=f^{-1}(x)\) is the reflection of the graph of \(y=f(x)\) in the line \(y=x\). It passes through \((0,0)\) and \(\left(\frac94,3\right)\), and it intersects the graph of \(y=f(x)\) at \((0,0)\) and \(\left(\frac85,\frac85\right)\).

(b)(i) Let

\(y=\sqrt[3]{8x^3+3}.\)

Then

\(y^3=8x^3+3.\)

So

\(x^3=\frac{y^3-3}{8}.\)

Therefore

\(x=\sqrt[3]{\frac{y^3-3}{8}}.\)

Replacing \(y\) by \(x\),

\(g^{-1}(x)=\sqrt[3]{\frac{x^3-3}{8}}.\)

(b)(ii) For \(gh(x)\) to be formed, the output of \(h\) must be in the domain of \(g\). Since the domain of \(g\) is \(x\geq1\), we need

\(e^{4x}\geq1.\)

This is true when \(x\geq0\), so the least possible value is

\(k=0.\)

(b)(iii) Now

\(gh(x)=g(h(x))=g(e^{4x}).\)

Therefore

\(gh(x)=\sqrt[3]{8(e^{4x})^3+3} =\sqrt[3]{8e^{12x}+3}.\)

Answer: (a)(i) \(f(x)\leq -1\); (a)(ii) \(x=-2\); (a)(iii) \(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}\); (a)(iv) \(x=\frac{5\pi}{6},\frac{7\pi}{6}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) For

\(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2},\)

we have \(\cos x\lt 0\). Therefore

\(\operatorname{sec} x=\frac{1}{\cos x}\leq -1.\)

So the range of \(f\) is

\(f(x)\leq -1.\)

(a)(ii) If

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

then

\(x=f\left(\frac{2\pi}{3}\right).\)

Hence

\(x=\operatorname{sec}\frac{2\pi}{3}.\)

Since

\(\cos\frac{2\pi}{3}=-\frac12,\)

we get

\(x=-2.\)

(a)(iii) Since \(g\) is defined for all real inputs, the domain of \(gf\) is just the domain of \(f\):

\(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}.\)

(a)(iv) Now

\(gf(x)=g(f(x))=3\bigl((\operatorname{sec} x)^2-1\bigr).\)

Using \(\operatorname{sec}^2x-1=\tan^2x\),

\(gf(x)=3\tan^2x.\)

So \(gf(x)=1\) gives

\(3\tan^2x=1.\)

Therefore

\(\tan^2x=\frac13.\)

In the interval \(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}\), the solutions are

\(x=\frac{5\pi}{6} \quad\text{and}\quad x=\frac{7\pi}{6}.\)

(b) For

\(h(x)=\ln(4-x),\)

the vertical asymptote is

\(x=4.\)

The intercepts of \(y=h(x)\) are:

\(y\text{-intercept: }(0,\ln4),\)

and

\(x\text{-intercept: }(3,0).\)

To find the inverse, let

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

Then

\(e^y=4-x,\)

so

\(x=4-e^y.\)

Thus

\(h^{-1}(x)=4-e^x.\)

The graph of \(y=h^{-1}(x)\) has horizontal asymptote

\(y=4.\)

Its intercepts are

\((0,3) \quad\text{and}\quad (\ln4,0).\)

The two graphs are reflections of each other in the line \(y=x\).

Answer: (a)(i) \(a=-\frac13\), (ii) \(f\geq-4\); (b) \(x=\sqrt{\frac{e^2-5}{2}}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) The function is

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

This is a quadratic with vertex at

\(3x+1=0\),

so

\(x=-\frac13\).

For \(f^{-1}\) to exist on the domain \(x\geq a\), the function must be one-to-one on that domain. The least possible value is therefore the \(x\)-coordinate of the vertex:

\(a=-\frac13\).

(a)(ii) At the vertex,

\(f\left(-\frac13\right)=-4\).

Since the parabola opens upwards and the domain starts at the vertex, the range is

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

(a)(iii) The graph of \(y=f(x)\) is the right-hand half of the parabola with vertex \(\left(-\frac13,-4\right)\).

It passes through the \(y\)-axis at

\(f(0)=1-4=-3\),

so it passes through \((0,-3)\).

It meets the \(x\)-axis when

\((3x+1)^2-4=0\).

So \((3x+1)^2=4\), giving \(3x+1=2\) on the chosen domain, and hence \(x=\frac13\).

Thus \(y=f(x)\) passes through \(\left(\frac13,0\right)\).

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

So it passes through \((-3,0)\) and \(\left(0,\frac13\right)\), and its vertex is \(\left(-4,-\frac13\right)\).

(b) Since \(h(x)=3x-2\),

\(hg(x)=h(g(x))=3\ln(2x^2+5)-2\).

Set this equal to 4:

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

So

\(3\ln(2x^2+5)=6\),

and hence

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

Therefore

\(2x^2+5=e^2\).

So

\(x^2=\frac{e^2-5}{2}\).

Since \(x\geq0\),

\(x=\sqrt{\frac{e^2-5}{2}}\).

Answer: (a)(i) \(a=-\frac25\). (ii) Range: \(\mathbb{R}\). (iii) \(\displaystyle f^{-1}(x)=\frac{e^x-2}{5}\), domain \(\mathbb{R}\). (b) \(x=64\).

Answer: (a)(i) \(a=-\frac25\). (ii) Range: \(\mathbb{R}\). (iii) \(\displaystyle f^{-1}(x)=\frac{e^x-2}{5}\), domain \(\mathbb{R}\). (b) \(x=64\).

(a)(i) For \(f(x)=\ln(5x+2)\), the argument must be positive:

\(5x+2\gt 0\).

Thus \(x\gt -\frac25\), so the smallest possible value of \(a\) is

\(a=-\frac25\).

(a)(ii) The logarithm function can take all real values, so the range of \(f\) is \(\mathbb{R}\).

(a)(iii) Let \(y=\ln(5x+2)\). Then

\(e^y=5x+2\),

so

\(\displaystyle x=\frac{e^y-2}{5}\).

Therefore

\(\displaystyle f^{-1}(x)=\frac{e^x-2}{5}\).

The domain of \(f^{-1}\) is the range of \(f\), so the domain is \(\mathbb{R}\).

(a)(iv) The graph of \(y=f(x)\) has vertical asymptote \(x=-\frac25\), \(x\)-intercept \(\left(-\frac15,0\right)\), and \(y\)-intercept \((0,\ln2)\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). Its intercepts are \((\ln2,0)\) and \(\left(0,-\frac15\right)\).

(b) Since \(g(x)=\sqrt{x}-4\),

\(g^2(x)=g(g(x))=\sqrt{\sqrt{x}-4}-4\).

Set this equal to \(-2\):

\(\sqrt{\sqrt{x}-4}-4=-2\).

Then

\(\sqrt{\sqrt{x}-4}=2\).

Squaring gives

\(\sqrt{x}-4=4\),

so \(\sqrt{x}=8\), and therefore

\(x=64\).

Answer: (a)(i) \(3\leq x\lt 5\); (a)(ii) \(x=4\); (b) \(\displaystyle f^{-1}g(x)=\ln\left(\frac{2-2x}{2-5x}\right)\).

Answer: (a)(i) \(3\leq x\lt 5\); (a)(ii) \(x=4\); (b) \(\displaystyle f^{-1}g(x)=\ln\left(\frac{2-2x}{2-5x}\right)\).

(a)(i) Since \(x\geq0\), we have \(0\lt \mathrm e^{-x}\leq1\).

Therefore

\(3\leq 5-2\mathrm e^{-x}\lt 5.\)

The range of \(f\) is \(3\leq y\lt 5\), so the domain of \(f^{-1}\) is

\(3\leq x\lt 5.\)

(a)(ii) If \(f^{-1}(x)=\sqrt{5x-4}\), apply \(f\) to both sides:

\(x=f(\sqrt{5x-4}).\)

It is quicker to note that \(f^{-1}(x)=u\) means \(x=f(u)\). The equation given also says \(u=\sqrt{5x-4}\). Thus the mark scheme equation reduces to

\(x=\sqrt{5x-4}.\)

Squaring gives

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

so

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

Factorising,

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

The domain of \(f^{-1}\) is \(3\leq x\lt 5\), so reject \(x=1\). Hence

\(x=4.\)

(a)(iii) The graph of \(y=f(x)\) starts at \((0,3)\), increases, and has horizontal asymptote \(y=5\).

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

(b) To find \(f^{-1}\), let

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

Then

\(2\mathrm e^{-x}=5-y,\)

so

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

Taking logarithms,

\(-x=\ln\left(\frac{5-y}{2}\right),\)

and hence

\(f^{-1}(y)=\ln\left(\frac{2}{5-y}\right).\)

Now substitute \(g(x)=\frac{3}{1-x}\):

\(f^{-1}g(x)=\ln\left(\frac{2}{5-\frac{3}{1-x}}\right).\)

Simplify the denominator:

Therefore

\(f^{-1}g(x)=\ln\left(\frac{2}{(2-5x)/(1-x)}\right) =\ln\left(\frac{2-2x}{2-5x}\right).\)

Start with the function \(y = h(x) = 4x^2 - 12x + 13\).

Complete the square for the quadratic expression:

\(4x^2 - 12x + 13 = 4(x^2 - 3x) + 13\).

\(= 4((x - \frac{3}{2})^2 - \frac{9}{4}) + 13\).

\(= 4(x - \frac{3}{2})^2 - 9 + 13\).

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

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

Rearrange to solve for \(x\):

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

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

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

Since \(x < 0\), choose the negative root:

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

Thus, \(h^{-1}(x) = \frac{3}{2} - \frac{\sqrt{x - 4}}{2}\).