Past Exam Questions

Answer: \(X\) and \(Y\) must not overlap; \(C\) must be outside both \(A\) and \(B\).

The condition \(n(X\cap Y)=0\) means that there are no elements in both \(X\) and \(Y\). So the two circles \(X\) and \(Y\) should be drawn with no overlap.

The condition \(C\subset(A\cup B)'\) means that every element of \(C\) is outside \(A\cup B\).

So \(C\) should be drawn completely outside both \(A\) and \(B\), with no intersection with either circle.

Answer: \(n(P')=18\), \(n((Q\cup R)\cap P)=11\), \(n(Q'\cup P)=29\).

For \(n(P')\), count all regions outside \(P\):

\(4+3+6+5=18.\)

So \(n(P')=18\).

For \(n((Q\cup R)\cap P)\), we need the elements that are in \(P\) and also in either \(Q\) or \(R\).

Those regions contain \(2\), \(1\), and \(8\), so

\(n((Q\cup R)\cap P)=2+1+8=11.\)

For \(n(Q'\cup P)\), it is easier to count the complement. The only elements not in \(Q'\cup P\) are those in \(Q\) but not in \(P\).

These regions contain \(4\) and \(3\), so there are \(7\) elements outside \(Q'\cup P\).

The total number of elements is

\(7+2+4+1+8+3+6+5=36.\)

Therefore

\(n(Q'\cup P)=36-7=29.\)

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: \(p=5+2\sqrt5\).

Since the point lies on the curve, substitute \(x=1-\sqrt5\):

\(p=\frac{10+2\sqrt5}{(1-\sqrt5)^2}.\)

Now

\((1-\sqrt5)^2=1-2\sqrt5+5=6-2\sqrt5.\)

So

\(p=\frac{10+2\sqrt5}{6-2\sqrt5}=\frac{5+\sqrt5}{3-\sqrt5}.\)

Rationalise the denominator:

\(p=\frac{(5+\sqrt5)(3+\sqrt5)}{(3-\sqrt5)(3+\sqrt5)}.\)

The denominator is \(9-5=4\), and the numerator is

\(15+5\sqrt5+3\sqrt5+5=20+8\sqrt5.\)

Hence

\(p=\frac{20+8\sqrt5}{4}=5+2\sqrt5.\)

Therefore

\(\boxed{p=5+2\sqrt5}.\)

Answer: \(A\): a function with no inverse. \(B\): not a function. \(C\): one-one function and its own inverse. \(D\): one-one function.

Use the vertical line test to decide whether a graph represents a function. A vertical line cuts graph \(B\) more than once, so \(B\) is not a function.

Graph \(A\) is a function, but it is not one-one because horizontal lines can cut it twice. Therefore \(A\) is a function with no inverse.

Graph \(C\) is one-one, and because it is symmetric in the line \(y=x\), it is its own inverse.

Graph \(D\) is one-one, since each horizontal line cuts it at most once.

So the ticks are:

Answer: \(A\): a function with no inverse. \(B\): not a function. \(C\): one-one function and its own inverse. \(D\): one-one function.

Use the vertical line test to decide whether a graph represents a function. A vertical line cuts graph \(B\) more than once, so \(B\) is not a function.

Graph \(A\) is a function, but it is not one-one because horizontal lines can cut it twice. Therefore \(A\) is a function with no inverse.

Graph \(C\) is one-one, and because it is symmetric in the line \(y=x\), it is its own inverse.

Graph \(D\) is one-one, since each horizontal line cuts it at most once.

So the ticks are:

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: range of \(f\): \(f\gt3\); range of \(g\): \(\mathbb R\); \(x=e^9+3\); \(x=2\).

(a) Since \(e^x\gt0\) for all real \(x\),

\(f(x)=3+e^x\gt3.\)

So the range of \(f\) is

\(f\gt3.\)

The function \(g(x)=9x-5\) is linear with non-zero gradient and domain \(\mathbb R\). Therefore its range is

\(\mathbb R.\)

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

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

Then

\(e^x=y-3,\)

so

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

Hence

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

Also

\(g'(x)=9.\)

So

\(\ln(x-3)=9.\)

Therefore

\(x-3=e^9,\)

and hence

\(x=e^9+3.\)

(c) Here \(g^2(x)\) means \(g(g(x))\). Since \(g(x)=9x-5\),

\(g(g(x))=9(9x-5)-5.\)

So

\(9(9x-5)-5=112.\)

Thus

\(81x-50=112.\)

Hence

\(81x=162,\)

and so

\(x=2.\)

Answer: (a)(i) \(fg\), (ii) \(g^{-1}\), (iii) \(f^{-1}\), (iv) \(g^2\). (b) It includes \(x=0\); \(a=12\), \(b=-8\).

For part (a), remember that \(fg\) means \(f(g(x))\).

Since \(g(x)=x+5\),

\(f(g(x))=\sqrt{x+5},\)

so \(\sqrt{x+5}\) is \(fg\).

The inverse of \(g(x)=x+5\) is

\(g^{-1}(x)=x-5.\)

The inverse of \(f(x)=\sqrt{x}\) is

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

Finally,

\(g^2(x)=g(g(x))=g(x+5)=x+10.\)

For part (b), the interval \(-2\leq x\leq2\) includes \(x=0\). Since \(h(x)\) contains \(\frac{b}{x^2}\), \(h\) is undefined when \(x=0\).

Now

\(h(1)=a+b=4.\)

Differentiate:

\(h'(x)=\frac{d}{dx}\left(a+bx^{-2}\right)=-2bx^{-3}.\)

Using \(h'(1)=16\),

\(-2b=16,\)

so

\(b=-8.\)

Then

\(a+b=4\)

gives

\(a-8=4,\)

so

\(a=12.\)

Therefore

\(\boxed{a=12,\quad b=-8}.\)

Answer: (i) \(gf(x)=\dfrac{8x-5}{12x-10}\). (ii) \(g^{-1}(x)=\dfrac{x+1}{3x-2}\). (iii) \(x=2\).

(i) Since \(f(x)=4x-3\),

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

Substitute \(4x-3\) into \(g\):

\(g(4x-3) = \frac{2(4x-3)+1}{3(4x-3)-1}.\)

Simplify:

\(gf(x) = \frac{8x-6+1}{12x-9-1} = \frac{8x-5}{12x-10}.\)

(ii) Let

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

Rearrange to make \(x\) the subject:

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

So

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

Collect the terms containing \(x\):

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

Factorise:

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

Therefore

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

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

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

(iii) First find \(fg(x)=f(g(x))\):

\(f(g(x)) = 4\left(\frac{2x+1}{3x-1}\right)-3.\)

Use a common denominator:

\(f(g(x)) = \frac{8x+4-3(3x-1)}{3x-1} = \frac{8x+4-9x+3}{3x-1} = \frac{7-x}{3x-1}.\)

Now solve \(fg(x)=x-1\):

\(\frac{7-x}{3x-1}=x-1.\)

Multiply by \(3x-1\):

\(7-x=(x-1)(3x-1).\)

Expand and rearrange:

\(7-x=3x^2-4x+1,\)

so

\(3x^2-3x-6=0.\)

Divide by \(3\):

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

Factorise:

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

Since the functions are defined for \(x\gt 1\), \(x=-1\) is not valid. Hence

\(\boxed{x=2}.\)

Answer: (i) \(g'(x)=-\dfrac{6}{(2x+1)^2}\), so it is always negative. (ii) \(g(x)\gt 0\). (iii) \(hg(x)=\dfrac{3k}{2x+1}+3\). (iv) \(k=\dfrac23\). (v) Domain: \(x\gt -\dfrac12\).

(i) Write \(g(x)\) as

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

Differentiate:

\(g'(x)=3(-1)(2x+1)^{-2}\cdot2 =-6(2x+1)^{-2} =-\frac{6}{(2x+1)^2}.\)

Since \((2x+1)^2\gt 0\) for every \(x\) in the domain, the denominator is always positive, and the numerator is negative. Therefore \(g'(x)\) is always negative.

(ii) Because \(x\gt -\frac12\), we have \(2x+1\gt 0\), so

\(g(x)=\frac{3}{2x+1}\gt 0.\)

Hence the range of \(g\) is

\(g(x)\gt 0.\)

(iii) Since \(h(x)=kx+3\),

\(hg(x)=h(g(x))=k\left(\frac{3}{2x+1}\right)+3=\frac{3k}{2x+1}+3.\)

(iv) Substitute \(x=0\):

\(hg(0)=\frac{3k}{2(0)+1}+3=3k+3.\)

Given that \(hg(0)=5\),

\(3k+3=5.\)

So

\(3k=2 \quad\Rightarrow\quad k=\frac23.\)

(v) The composite \(hg\) is defined wherever \(g\) is defined, so its domain is

\(x\gt -\frac12.\)