Functions: past paper questions & worked solutions

The logarithm is base 10. Thus $10^y=2x+1$ and $2x=10^y-1$.

Differentiate the result of (a) with respect to y:

Take the reciprocal and use $10^y=2x+1$:

The original logarithm requires $x\gt -1/2$.

Expand $R\sin(2x-\alpha)=R\cos\alpha\sin2x-R\sin\alpha\cos2x$. Comparison gives $R\cos\alpha=\sqrt3$ and $R\sin\alpha=3$. Thus $R=\sqrt{3+9}=2\sqrt3$ and $\tan\alpha=\sqrt3$. As $0\lt \alpha\lt \pi/2$, $\alpha=\pi/3$ and

The denominator is always positive, between $2\sqrt3$ and $6\sqrt3$. The fraction is smallest when the denominator is largest:

The maximum denominator occurs when $\sin(6x-\pi/3)=1$:

The smallest positive value is $x=5\pi/36$, since $n=-1$ gives a negative value.

Since $x\ge0$, $2x^2\ge0$. The minimum is $f(0)=-5$ and there is no upper bound.

Reflect the given curve in $y=x$. Solving $x=2y^2-5$ with $y\ge0$ gives $y=\sqrt{(x+5)/2}$, $x\ge-5$. The inverse starts at $(-5,0)$, passes through $(0,\sqrt{5/2})$, and is increasing and concave down. The solid inverse curve below is paired with the dashed original curve.

Here $f$ is strictly increasing on $[0,\infty)$, so an intersection of $f$ and its inverse must be on $y=x$: if $y=f(x)\gt x$, then $f(y)\gt f(x)=y$, contradicting $f(y)=x$; the case $y\lt x$ is similar. Hence

Reject the negative root because the original curve has $x\ge0$.

For $x\ge0$, $f(x)=9-x^2$ has maximum $f(0)=9$ and decreases without a lower bound.

Apply $g$ before $f$:

The intermediate value $3/4$ is in the nonnegative domain of $f$.

Let $y=3/(2x+1)$. Then $y(2x+1)=3$, so $2xy=3-y$ and $x=(3-y)/(2y)$. Exchange the variables. Since $g$ decreases from 3 towards 0 on $[0,\infty)$, its range is $(0,3]$, which is the inverse domain.

Expand $R\cos(x+\alpha)=R\cos\alpha\cos x-R\sin\alpha\sin x$. Comparing coefficients gives

Hence

Thus $f(x)=25\cos(x+\alpha)$, using the unrounded $\alpha$ in subsequent calculations.

Since $-25\leq f(2x)\leq25$, the positive denominator $90-3f(2x)$ lies between $15$ and $165$. The smallest reciprocal occurs at the largest denominator:

The minimum requires $f(2x)=-25$, so

Since $0\lt \alpha\lt \pi/2$, the first positive solution uses $n=0$:

Set $u=x^2+k\gt 0$. Applying the quotient and chain rules,

Comparing with the required form identifies both constants.

The denominator is positive, so stationary points satisfy

There are three distinct real stationary points precisely when $e-k\gt 0$. In that case the factor $1-\ln(x^2+k)$ is positive between the outer roots and negative outside them; $2x$ changes sign at zero. Thus the derivative changes sign at all three points, so all are turning points. Together with $k\gt 0$, the three coordinates below are real and distinct; the restriction is $0\lt k\lt e$.

At $k=e$, the two outer stationary points merge with $x=0$, so there are no longer three distinct turning points. Values of $k$ can approach $e$ from below, but cannot equal it under this condition.

For $x\ge0$, $0\lt 4/(3x+2)\le2$. Thus $f(0)=3$ is attained, while $f(x)$ increases towards 5 without reaching it.

(i) Put $y=5-4/(3x+2)$. Rearranging gives $(5-y)(3x+2)=4$, hence

Exchange variables. (ii) The inverse domain is the range of $f$ found above.

The original function has an absolute value and angle $x/3+\pi/6$:

Then $fg(-\pi)=f(2)=5-4/(3\cdot2+2)=5-1/2=9/2$.

By the product rule,

Since the exponential is positive, stationary points have $x=2$ or $x=4/3$. The sign of $f'$ changes from positive to negative at $4/3$, so this is the maximum A.

The curve is nonnegative and has just one zero, $x=2$; hence $k=0$ is excluded. It tends to 0 from above as $x\to-\infty$, rises to $4e^4/9$, falls to 0, then rises without bound. For $0\lt k\lt 4e^4/9$ there are three intersections. At the maximum level there are two distinct intersections, including A. For a larger positive k there is only one.

Differentiate x with respect to y by the product rule:

Where $dx/dy\ne0$, take the reciprocal:

For $y\ne0$, the original relation gives $e^{2y}=x/y$, hence the required form $dy/dx=y/[x(1+2y)]$. At the origin, that last quotient is $0/0$; the preceding uncancelled derivative gives $dy/dx=1$. At $y=-1/2$ there is no finite $dy/dx$.

Consider $x=h(y)=ye^{2y}$. Its derivative $h'(y)=e^{2y}(1+2y)$ vanishes at $y=-1/2$. It is negative before this y value and positive after it, so the minimum x value is $h(-1/2)=-1/(2e)$. As $y\to-\infty$, $h(y)\to0$ from below; also $h(0)=0$ and h then increases without bound. Thus a level $x=k$ meets both negative-y branches precisely between the minimum and zero. At the minimum the two intersections coincide; at zero there is just the finite point $y=0$.

The principal arcsine curve increases from $(-2,-\pi/2)$ through $(0,0)$ to $(2,\pi/2)$. It is concave down on the negative-x half, concave up on the positive-x half, and vertical at both ends.

Differentiate the supplied inverse relation: $dx/dy=2\cos y$. For $-\pi/2\lt y\lt \pi/2$, cosine is positive, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2/4}$. Taking the reciprocal gives

Thus $A=4$; the endpoints have vertical tangents, not finite derivatives.

At $y=\pi/4$, $x=2\sin(\pi/4)=\sqrt2$. The gradient is $1/\sqrt{4-2}=1/\sqrt2$. Hence $y-\pi/4=(x-\sqrt2)/\sqrt2$, so $y=x/\sqrt2+\pi/4-1$.

Write $f(x)=(2x+3)(4x-1)^{-1/2}$. The product and chain rules give

The denominator of $f'(x)$ is positive throughout the domain, so the derivative is negative for $1/4\lt x\lt 2$ and positive for $x\gt 2$. Consequently $x=2$ is the global minimum, with $f(2)=7/\sqrt7=\sqrt7$. Also $f(x)\to\infty$ as $x\to1/4^+$ and as $x\to\infty$. By continuity all values at least $\sqrt7$ are attained, including the minimum.

Expanding the required form gives

Comparing coefficients with \(5\cos\theta-4\sin\theta\),

Squaring and adding, using \(R\gt 0\), gives

Also, since \(0\lt \alpha\lt \frac\pi2\),

Thus the exact form is

A stretch parallel to the \(y\)-axis with scale factor \(\boxed{\sqrt{41}}\), taking \(y=\cos\theta\) to \(y=\sqrt{41}\cos\theta\).

A translation by the vector

This moves the curve \(\alpha\) units to the left, giving \(y=\sqrt{41}\cos(\theta+\alpha)\).

Since \(\theta\) can be any real number,

Hence \(4\leq4+(f(\theta))^2\leq45\). Taking reciprocals of these positive bounds and multiplying by \(90\) gives

Both endpoints are attained: the minimum occurs when \(f(\theta)=\pm\sqrt{41}\), and the maximum when \(f(\theta)=0\).

Substitute $f(x)$ into $f$ and multiply numerator and denominator by $x-4$:

The composition requires $x\ne4$ and $f(x)\ne4$, hence also $x\ne16$.

To find $f^{-1}(12)$, solve $f(x)=12$:

Thus $f^{-1}(12)=\frac{32}{5}$.

Put $u=\ln a$. Then $fg(a)=g(a)$ means $f(u)=u$, where $u\ne4$:

Therefore $a=e^8$ or $a=e^{-2}$. Both are positive and neither is $e^4$, so both satisfy the domain restrictions.

First apply $f$, then $g$:

For $x\geq0$,

Thus $f(0)=0$ and $f(x)\lt 2/3$. Also $f'(x)=2/(3x+1)^2\gt 0$, so the range is

Therefore

On the common domain $0\leq x\lt 2/3$, both denominators below are nonzero:

Both candidates lie in the common domain, giving

Let $y=(4x+3)/(x-2)$. Then $yx-2y=4x+3$, so

Swap the input and output names. Since $f(x)=4+11/(x-2)$ can never equal 4, the inverse domain excludes 4.

Substitute $f(x)$ into both occurrences of the argument:

Thus $a=19$, $b=6$, $c=2$, $d=7$. The composite retains $x\ne2$ from the inner function and also requires $f(x)\ne2$, that is $x\ne-7/2$.

For $y=2f(3x)+8$, the point mapping is $(x,y)\mapsto(x/3,2y+8)$. Hence $(3,15)\mapsto(3/3,2(15)+8)=(1,38)$.

Since $x\gt 0$, $6\sqrt{x}$ is positive. Therefore $f(x)=0$ requires $\ln(4x)=0$, so $4x=1$ and $x=1/4$.

Apply the product and chain rules:

At $Q$, set $f'(x)=0$. Since $3/\sqrt{x}\gt 0$,

Then $\sqrt{x}=1/(2e)$ and $\ln(4x)=-2$, giving $f(x)=6(1/(2e))(-2)=-6/e$.

The factor $3/\sqrt{x}$ is positive and $\ln(4x)+2$ increases through zero at $Q$, so $f'$ changes from negative to positive: $Q$ is the global minimum. As $x\to\infty$, $f(x)\to\infty$, hence the range of $f$ is $[-6/e,\infty)$. Multiplication by $-2$ reverses the inequality and includes the maximum $12/e$.

First apply $f$: $f(2)=6-21/(2\cdot2+3)=3$. Then apply $g$: $gf(2)=g(3)=3^2+5$.

Let $y=6-21/(2x+3)$. Then

Interchange $x,y$. On the original domain $x\ge0$, $f$ increases from $f(0)=-1$ towards 6, never attaining 6. Thus the inverse domain is $-1\le x\lt 6$.

Since $x^2+5\gt 0$, take $x^2+5=11$, not $-11$. Therefore $x^2=6$.

Apply $f$ twice, in order:

Let $y=(x+3)/(x-4)$. Then

Interchanging $x$ and $y$ gives the inverse. The value 1 is never an output of $f$, since $(x+3)/(x-4)=1$ would require $3=-4$.

Since $g$ is defined for positive inputs, $gf(a)=7$ requires $f(a)\gt 0$.

The negative square root is not in the domain of $g$. Hence

This value is not 4 and gives $f(a)=\sqrt2\gt 0$, as required.

(i) At $x=0$, $f(0)=|-9|-2=7$, so the y coordinate of A is 7. (ii) The minimum occurs when the modulus vanishes: $kx-9=0$, hence $x_B=9/k$.

$|kx-9|-2\lt 0$ means $|kx-9|\lt 2$, or $-2\lt kx-9\lt 2$. Add 9 and divide by the positive k, preserving both strict inequality signs.

For $x\le9/k$, the graph is $y=7-kx$; for $x\ge9/k$, it is $y=kx-11$. The intersection of the right branch with $3-2x$ has $x=14/(k+2)$. For it to lie strictly to the right of the vertex,

For these k, the left-branch intersection has $x=4/(k-2)\lt 9/k$, so both intersections exist and differ. At $k=18/5$ the two points coincide at the vertex. Each straight branch can supply at most one intersection; when the right branch is inadmissible, two distinct points cannot occur (k=2 gives parallel, unequal left lines).

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Therefore $R\cos\alpha=12$, $R\sin\alpha=5$. Thus $R=\sqrt{12^2+5^2}=13$ and $\alpha=\arctan(5/12)=0.3947911197\ldots$ radians, in the required first quadrant.

Write $g(\theta)=10+13\sin(2\theta-\pi/6-\alpha)$. (i) The minimum is $10-13=-3$. (ii) For sine to equal $-1$, $2\theta-\pi/6-\alpha=-\pi/2+2\pi n$. Hence $\theta=-\pi/6+\alpha/2+\pi n$. For $n=0$ this is negative, so the smallest permitted value uses $n=1$:

Use the unrounded phase angle for the calculation.

As $\beta$ varies over the reals, $12\sin\beta-5\cos\beta$ covers $[-13,13]$. Its square therefore covers $[0,169]$, not $[169,169]$. Subtracting from 10 reverses the bounds, giving $[-159,10]$; both endpoints are attained.

Factor $2x^2-3x-5=(2x-5)(x+1)$. Then

so

On $x\gt 4$ the cancelled factor is nonzero. One integer choice is $a=4,b=1,c=2,d=-5$.

Let $y=(4x+1)/(2x-5)$. Rearranging gives $2xy-5y=4x+1$, hence $x(2y-4)=5y+1$ and $x=(5y+1)/(2y-4)$. Swap variable names to obtain the inverse rule; its restricted domain is determined in (c).

Rewrite $f(x)=2+11/(2x-5)$. For $x\gt 4$, the positive denominator is greater than 3 and grows without bound. Thus $0\lt 11/(2x-5)\lt 11/3$, so $2\lt f(x)\lt 17/3$. Neither bound is attained: $x=4$ is excluded and $f(x)=2$ is impossible. The range of f is the domain of its inverse.

Apply f twice, in order: $f(6)=21-2|2-6|=13$. Then $ff(6)=f(13)=21-2|2-13|=21-22=-1$. The intermediate input 13 lies in the domain $x\ge0$.

For $0\le x\le2$, $f(x)=17+2x$, so $17+2x=5x$ gives $x=17/3$, outside that branch; reject it. For $x\ge2$, $f(x)=25-2x$. Then $25-2x=5x$, giving $x=25/7$. This belongs to the second branch and satisfies the original modulus equation.

On the left branch $0\le x\le2$, f rises from $f(0)=17$ to $f(2)=21$. The right branch decreases from 21 without lower bound. A horizontal line therefore meets both branches at distinct points exactly when $17\le k\lt 21$. At $k=17$, one root is the permitted endpoint $x=0$; at $k=21$, the two intersections merge at the vertex, so 21 is excluded.

The original vertex is $(2,21)$. Under $y=af(x-b)$ it maps to $(2+b,21a)$. Comparing with $(6,3)$ gives $2+b=6$ and $21a=3$.

First evaluate the inner function:

Then apply $f$ to this result:

Put $y=f(x)$ and rearrange to make $x$ the subject:

For $x\gt 0$, the denominator $x+1\gt 1$, so $0\lt f(x)\lt 12$. Conversely, every $y$ with $0\lt y\lt 12$ gives $x=12/y-1\gt 0$. Thus the range of $f$, and hence the domain of its inverse, is $(0,12)$. Interchanging $x$ and $y$ gives

Both functions must be defined, so $0\lt x\lt 12$. Substitute the formula for the inverse and multiply by the nonzero quantity $x(x+1)$:

The algebraic candidates are $x=-4$ and $x=3$. Reject $x=-4$ because it lies outside $0\lt x\lt 12$. For $x=3$, $f^{-1}(3)=12/3-1=3=f(3)$, so the only real solution is

Since $e^{2x}\gt 0$ for every real $x$, the range is $\boxed{(3,\infty)}$.

Put $y=\sqrt{x+5}$. The original domain is $x\gt -5$, so $y\gt 0$. Squaring gives $y^2=x+5$, hence $x=y^2-5$. Interchanging the input and output,

Evaluate the inside function first:

Because $a\gt 0$, $g(a^2-5)=\sqrt{a^2}=a$, while $g(59)=8$. Thus

Here $p=\sqrt5/2$ is irrational and greater than 1, so the answer satisfies $a\gt 0$.

Since $x^2+3\ge3$, with equality at $x=0$, and the logarithm is increasing, the range of $f$ is $[\ln3,\infty)$.

Let $y=(3+5x)/(x+2)$. Then

Moreover, $g(x)=5-7/(x+2)$ with $x\gt -2$ has range $(-\infty,5)$. Thus

Therefore

Taking natural logarithms gives $a=\frac12\ln5$. The input $e^{2a}=5$ satisfies the domain of $g$.

The logarithm requires $x+1\gt 0$, so the domain is $x\gt -1$. As $x\to-1^+$, $\ln(x+1)\to-\infty$ and the ordinate tends to infinity.

(i) At the $y$-axis, $x=0$, giving $y=|2-4\ln1|=2$.

(ii) At the $x$-axis, $|2-4\ln(x+1)|=0$, so

Thus $A=(0,2)$ and $B=(e^{1/2}-1,0)$.

On $x\gt -1$, split the absolute-value inequality into two alternatives:

Dividing by $-4$ reverses the inequality in each branch:

Since the exponential is increasing, this becomes $x+1\lt e^{-1/4}$ or $x+1\gt e^{5/4}$. Intersect with the logarithmic domain; neither boundary is included because the source inequality is strict.

At $x=1/4$, $\sin^2 4y=1/4$. Since $0\le4y\le\pi/2$,

Only this branch lies in the given range of $y$.

Differentiate $x=(\sin4y)^2$ with respect to $y$:

On the interior of the given interval, $\sin4y=\sqrt{x}$ and $\cos4y=\sqrt{1-x}$ are positive. Taking the reciprocal,

Complete the square:

Thus $dy/dx=(q+r(x+s)^2)^{-1/2}$ has the constants below, with $0\lt x\lt 1$ for a finite derivative; the endpoints have vertical tangents.

The derivative is positive. Its minimum occurs when the square root in the denominator is largest. Since $(x-1/2)^2\ge0$, the largest radicand is 16, attained at $x=1/2$. Hence the smallest gradient is $1/\sqrt{16}$.

The modulus is smallest when $x-2=0$, giving $x=2$ and $f(2)=-10$. The two straight-line branches meet there, so this is the vertex of the V-shaped graph.

First $f(0)=3|0-2|-10=-4$. Then $ff(0)=f(-4)=3|-4-2|-10=18-10=8$. In particular $ff(0)$ means $f(f(0))$, not the square of $f(0)$.

For $x\lt 2$, the inequality becomes $3(2-x)-10\lt 5x+10$, so $-4-3x\lt 5x+10$, giving $x\gt -7/4$. This yields $-7/4\lt x\lt 2$. For $x\ge2$, it becomes $3x-16\lt 5x+10$, giving $x\gt -13$; every $x\ge2$ satisfies this. The union is $x\gt -7/4$.

Let $u=|x|\ge0$. Then $f(|x|)=0$ gives

Thus $u=16/3$ or $u=-4/3$. Reject $u=-4/3$ because $u\ge0$; $|x|=16/3$ has the two solutions below.

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Thus $R\cos\alpha=8$, $R\sin\alpha=15$, so $R=\sqrt{64+225}=17$ and $\alpha=\arctan(15/8)=1.0808390005\ldots$ radians, in the required quadrant.

Using the exact phase from (a),

The denominator lies between 7 and 75, so it is always positive. (i) The smallest fraction occurs at its largest denominator: $f_{\min}=15/75=1/5$. (ii) This occurs when $x-\alpha=\pi/2+2n\pi$. The smallest positive value is $x=\alpha+\pi/2=2.6516353273\ldots$, or 2.65 to three significant figures.

The transformation $2f(x)-5$ preserves the locations of minima and maps their y coordinate to $2(1/5)-5=-23/5$.

Reflection in the x-axis changes minima of f into maxima of $-f$. The argument $2x$ halves their x coordinates, so the first positive maximum is at

$g(5)=2(5)-7=3$, then $fg(5)=f(3)=(5-3)/(9+2)=2/11$.

Let $y=(5-x)/(3x+2)$. Then $3xy+2y=5-x$, so $x(3y+1)=5-2y$ and $x=(5-2y)/(3y+1)$. Exchange x and y. The inverse cannot have input $-1/3$, since this is not an output of f: $(5-x)/(3x+2)=-1/3$ would imply $15=-2$.

Here $a\ne0$ and $a\ne-3/2$ are required by the input and denominator. Substitute into both functions:

Thus $5a-1=(2a-1)(2a+3)=4a^2+4a-3$, giving $4a^2-a-2=0$. By the quadratic formula, $a=(1\pm\sqrt{33})/8$. Neither value is excluded.

Factor $x^2+7x+12=(x+3)(x+4)$. A common denominator gives

The cancelled factor is nonzero throughout the original domain $x\gt 0$.

Let $y=5x/(x+3)$. Then $yx+3y=5x$, so $x=3y/(5-y)$. Exchange variable names. Since $f(x)=5-15/(x+3)$ for $x\gt 0$, its range is $0\lt f(x)\lt 5$. This becomes the inverse domain.

(i) By the quotient rule, $f'(x)=[5(x+3)-5x]/(x+3)^2=15/(x+3)^2$. (ii) The numerator and denominator are positive on $x\gt 0$, so $f'(x)\gt 0$ throughout its domain. Therefore f is increasing.

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Matching coefficients gives $R\cos\alpha=1$, $R\sin\alpha=4$. Thus $R=\sqrt{1^2+4^2}=\sqrt{17}$ and $\tan\alpha=4$. Since $0\lt \alpha\lt \pi/2$, $\alpha=\arctan4=1.32581766\ldots$ radians.

The model becomes $H=24/[3+\sqrt{17}\cos(t/2-\alpha)]$. For $0\le t\le6.5$, the phase lies between $-\alpha$ and $3.25-\alpha$; throughout this interval the denominator remains positive. Its maximum $3+\sqrt{17}$ is attained at $t=2\alpha=2.651635\ldots$, inside the time interval. The height is then smallest:

To the nearest centimetre this is $3.37\ \mathrm m$ (337 cm).

For $H=10$, $3+\sqrt{17}\cos(t/2-\alpha)=24/10$, so

Let $\gamma=\arccos[-3/(5\sqrt{17})]$. In the permitted phase interval $[-\alpha,3.25-\alpha]$, only $+\gamma$ occurs: $-\gamma\lt -\alpha$, and shifts by $2\pi$ are outside it. Using the unrounded $\alpha=\arctan4$,

At the vertex, $3x-11=0$, so $P=(\frac{11}{3},-4)$.

Thus $3x-11=12$ or $3x-11=-12$, giving $x=\frac{23}{3}$ or $x=-\frac13$.

Write the two branches explicitly:

On the left branch, $mx=7-3x$ gives $x=7/(m+3)$ for $m\ne-3$. This lies on the left branch precisely when $m\lt -3$ or $m\ge-12/11$. On the right branch, $mx=3x-15$ gives $x=15/(3-m)$ for $m\ne3$. This lies on the right branch precisely when $-12/11\le m\lt 3$. At $m=-12/11$, the two branch intersections coincide at the vertex, so there is only one distinct point. For $m\lt -3$ or $m\ge3$, only the left intersection exists. At $m=-3$ there is no intersection; at $m=3$ there is one. Consequently the complete answer is

Under $y=af(x-b)$, the vertex becomes $(\frac{11}{3}+b,-4a)$. Equating this with $(5,16)$ gives

At $x=0$, $f(0)=|-10|+k$, so $\boxed{y_A=10+k}$.

At the vertex the expression inside the modulus vanishes:

Since $k\gt 0$, the two branches give

Consider the vertical difference $d(x)=f(x)-(3x+1)$:

The left branch is strictly decreasing. If $0\lt k\lt 3$, the right branch also decreases, so there cannot be two roots. If $k=3$, the right branch is the nonzero constant $-8$, again giving no second root. For $k\gt 3$ the right branch increases and both outer ends tend to $+\infty$. There are two distinct roots precisely when the vertex lies below zero:

Multiplying by $k\gt 0$ gives

Combining with $k\gt 3$ yields $\boxed{3\lt k\lt 6}$. At $k=6$ the two roots merge at the vertex, so the endpoint is excluded.

For $x\gt -7/4$, apply the product and chain rules:

The interior stationary values satisfy $x^2(2x+3)=0$, so $x=0$ or $x=-3/2$. Near $x=-3/2$, the derivative changes from negative to positive; near $x=0$ it stays positive on both sides. Thus the minimum turning point is at $x=-3/2$.

The endpoint has $f(-7/4)=0$, the global minimum is $-27/8$ and $f(x)$ tends to infinity as $x$ increases. Hence the range of $f$ is $[-27/8,\infty)$. Multiplication by $-4$ reverses the inequality and makes the upper endpoint $27/2$.

The transformation is a shift right by $3/2$, vertical stretch by 40, then shift down by 8:

Therefore $Q=(1/2,3/8)$ maps to $(1/2+3/2,\;40\cdot3/8-8)$.

To find $f^{-1}(22)$, solve $f(u)=22$: $2+5\ln u=22$, so $\ln u=4$ and $u=e^4\gt 0$.

Using the quotient rule,

For every $x\gt 1/3$, the denominator is positive, so $g'(x)\gt 0$. Thus g is strictly increasing on its entire domain.

Set $y=(6x-2)/(2x+1)$. Then

As $x\downarrow1/3$, $g(x)\to0$; as $x\to\infty$, $g(x)\to3$. Since g is strictly increasing, its range is $(0,3)$; neither limiting value is attained. Interchange the input and output symbols.

The inner function g takes every value in $(0,3)$. Since $f(u)=2+5\ln u$ is increasing for $u\gt 0$, its values on this interval run from $-\infty$ to, but not including, $2+5\ln3$.

The exponential is positive and the squared factor is nonnegative, so $f(x)\ge0$. The value zero is attained at $x=\pm\sqrt{3/2}$. The graph has its highest point at $x=0$, where $f(0)=9$; the outer maxima are lower. Thus both endpoints of the range are included.

For $f(x)=e^{-x^2}(2x^2-3)^2$, product and chain rules give

Factoring yields

The stationary abscissae are $0$, $\pm\sqrt{3/2}$ and $\pm\sqrt{7/2}$. At the outer local maxima,

The central maximum is 9 and the two minima are zero. Also $f(x)\to0$ as $x\to\pm\infty$. A positive horizontal line above both outer maxima but below 9 meets only the two sides of the central peak. At $k=16e^{-7/2}$ it also touches both outer peaks, giving four distinct intersections; at $k=9$ there is only one. Both bounds must be excluded.

Use product and chain rules:

With a common denominator,

This derivative formula applies for $x\gt -9/4$; the endpoint derivative is not finite.

The denominator is positive in the derivative domain. Thus $5x^2+9x-2=(5x-1)(x+2)=0$.

The derivative changes from positive to negative at $x=-2$, so this is the local maximum P. Substitute into f: $f(-2)=5(4-2)\sqrt{-8+9}=10$.

On $[-9/4,0]$, f rises from $f(-9/4)=0$ to $f(-2)=10$, then decreases to $f(0)=-30$. The other stationary point $1/5$ is outside this restricted interval. Thus the minimum and maximum of f on it are $-30$ and 10. Applying the positive vertical transformation $g=2f+4$ gives endpoints $2(-30)+4=-56$ and $2(10)+4=24$, both attained.

The modulus vanishes at $3x-13=0$, giving $x=13/3$ and $y=5$.

(i) The modulus is nonnegative and attains zero, so the range is $[5,\infty)$. (ii) $f(4)=|12-13|+5=6$, then $ff(4)=f(6)=|18-13|+5=10$.

For $x\lt 13/3$, $|3x-13|+5=18-3x$. Thus $16-2x\gt 18-3x$ gives $x\gt 2$, so this branch contributes $2\lt x\lt 13/3$. For $x\ge13/3$, the right side is $3x-8$. Then $16-2x\gt 3x-8$ gives $x\lt 24/5$, so this branch contributes $13/3\le x\lt 24/5$. Combine the adjacent intervals.

The new vertex is at $(13/3-b,5a)$. Therefore $5a=20$ and $13/3-b=4$, giving $a=4$ and $b=1/3$.

Substitute $g(x)=5-2x^2$ into f:

The composite requires $x\le0$ from g and $g(x)\ne5$ from f; therefore $x\lt 0$. Now $4-13/x^2=3$ gives $x^2=13$. Only the negative root belongs to the composite domain.

Let $y=(4x+6)/(x-5)$. Then $yx-5y=4x+6$, so $x(y-4)=5y+6$ and $x=(5y+6)/(y-4)$. Exchange the variable names. The original function is $4+26/(x-5)$, whose range excludes 4.

The graph of g is only the left half of the downward parabola: $y=5-2x^2$, $x\le0$. Its intercepts are $(0,5)$ and $(-\sqrt{5/2},0)$. For the inverse, $x=5-2y^2$ and $y\le0$, hence $g^{-1}(x)=-\sqrt{(5-x)/2}$, $x\le5$. Reflect the original branch in $y=x$. The inverse intercepts are $(5,0)$ and $(0,-\sqrt{5/2})$.

The absolute value allows both intervals on which the integrand is defined.

The largest value of $a-|bx-1|$ occurs when $bx-1=0$. The maximum point is therefore $\boxed{(1/b,a)}$.

At $x=0$, $9=a-1$, so $\boxed{a=10}$. At the intersection $x=3$, the first curve has height $11/11+4=5$. Thus $10-|3b-1|=5$, giving $3b-1=5$ or $-5$. These give $b=2$ or $-4/3$; only the positive integer is allowed, so $\boxed{b=2}$.

For $x\geq3$, the required branch of $C_2$ is $y=10-(2x-1)=11-2x$. Its $x$-intercept is $x=11/2$. The shaded area is the area under $C_1$ from 0 to 3 plus the triangle from 3 to $11/2$:

Hence $\boxed{A=14.4\text{ square units}}$ to 3 significant figures.

(i) Since $|2x-b|\ge0$, the maximum occurs at $2x-b=0$: $x=b/2$ and $y=a$. (ii) At $x=0$, $y=a-|-b|=a-b$ because $b\gt 0$. (iii) At an x-axis crossing, $|2x-b|=a$, hence $2x-b=\pm a$. Since $a\gt b\gt 0$, $(b-a)/2\lt 0$ and $(b+a)/2\gt 0$, matching the sketch.

The graph $y=|x|-1$ is a V with vertex $(0,-1)$ and intercepts $(-1,0)$ and $(1,0)$. Its left branch is $y=-x-1$ for $x\lt 0$ and its right branch is $y=x-1$ for $x\ge0$. Add this V to the supplied axes; the original inverted V remains unchanged.

The grey original curve is schematic; the three labelled points belong to the added graph.

At $x=-3$, the common ordinate is $|-3|-1=2$. Since $b\gt 0$, $|-6-b|=6+b$, so $a-(6+b)=2$, giving $a=b+8$. At $x=5$, the common ordinate is $|5|-1=4$, so $a-|10-b|=4$. Substitute $a=b+8$ to obtain $|10-b|=b+4$. If $b\le10$, then $10-b=b+4$, giving $b=3$, $a=11$. If $b\gt 10$, then $b-10=b+4$, impossible. Both accepted values satisfy $a\gt b\gt 0$.

By the quotient rule,

For every $x\gt 4$, the denominator is positive and the numerator is negative. Therefore $f'(x)\lt 0$ throughout its domain, so f is decreasing.

Set $y=(5x-3)/(x-4)$. Then $yx-4y=5x-3$, giving $x(y-5)=4y-3$ and $x=(4y-3)/(y-5)$. Exchange the variable names. Since $f(x)=5+17/(x-4)$ on $x\gt 4$, its range is $(5,\infty)$, which becomes the inverse's domain.

(i) For $x\gt 4$, $f(x)\gt 5\gt 4$, so the second application is defined.

Thus $a=22$, $b=-3$, $c=13$. (ii) Rewrite $ff(x)=22-289/(x+13)$ with the original domain $x\gt 4$. It is strictly increasing, tends to 5 as $x\to4^+$ and to 22 as $x\to\infty$. Neither limiting value is attained.

(i) The modulus is zero at $3x-22=0$, giving the vertex $(22/3,5)$. (ii) At $x=0$, $y=5-22=-17$, so the y-axis intersection is $(0,-17)$.

Set $y=0$: $|3x-22|=5$. Thus $3x-22=5$ or $-5$, giving $x=9$ or $17/3$.

The curve $C_2:y=x^2/9-9$ is an upward-opening parabola, symmetric about the y-axis. Its vertex and y-intercept are $(0,-9)$; setting y=0 gives $x^2=81$, hence x-intercepts $(-9,0)$ and $(9,0)$. The sketch below includes $C_1$ for context and labels all three axis intersections of $C_2$.

For $x\le22/3$, $C_1$ is $y=3x-17$. Equating to $C_2$ gives

Only $x=3$ belongs to this branch; $y=-8$. For $x\ge22/3$, $C_1$ is $y=27-3x$. Then

Only $x=9$ belongs to this branch; $y=0$. The rejected candidates 24 and $-36$ lie outside their assumed branch regions.

Point $P$ lies on the descending branch of $C_2$, where $5x+b\lt 0$. At $P=(-2,25)$,

hence $a=15+b$.

Point $R=(2,9)$ lies on the ascending branch, so

Combining with $a-b=15$ gives $2a=14$, hence $a=7$ and $b=-8$.

The vertex occurs at $5x-8=0$, giving $(x,y)=(8/5,7)$.

The diagram places $Q$ on the descending branch $y=7-(5x-8)=15-5x$, with $x\lt 8/5$. Equate this with the cubic:

Since $x=-2$ is the known intersection $P$, divide by $x+2$:

The candidates are $-2,3/2,3$. Reject $x=3$ because it is outside the descending branch, and $x=-2$ is $P$, not $Q$. Thus $x_Q=3/2$ and $y_Q=15-5(3/2)=15/2$, giving $Q=(3/2,15/2)$.

Factor the first denominator:

On the stated domain $x\gt 2$, both denominators and $x+4$ are nonzero. Hence

The quotient rule gives

For every $x\gt 2$, the squared denominator is positive; hence $f'(x)\lt 0$. Therefore f is strictly decreasing on its whole domain.

Set $y=3+2\ln x$. Then $\ln x=(y-3)/2$, so $x=e^{(y-3)/2}$. For $x\ge1$, the original range is $y\ge3$. Interchanging the input and output symbols gives the inverse with its domain.

The equation $g(f(a))=5$ gives $3+2\ln(f(a))=5$, so $\ln(f(a))=1$ and $f(a)=e$. Using the simplified f,

The resulting value is approximately $2.2083\gt 2$, and the inner output $e\ge1$ is admissible for g.