Modulus: past paper questions & worked solutions

For $y=f(2x)$ the horizontal coordinate is divided by 2 while the vertical coordinate is unchanged: $(-4,-3)\mapsto(-4/2,-3)$.

The inside shift moves the point right by 1; the outside factor multiplies its ordinate by 3. Thus $(-4,-3)\mapsto(-4+1,3(-3))$.

Output modulus retains the input and replaces the output by its absolute value: $(-4,-3)\mapsto(-4,|-3|)$.

The modulus is smallest when $x-5=0$. At $x=5$, $f(5)=10$.

For $x\lt 5$, $|x-5|=5-x$, so

This is consistent with $x\lt 5$. For $x\ge5$, $|x-5|=x-5$, so $2x\gt 6x$, which requires $x\lt 0$ and is inconsistent with this branch. The equality point $x=5/2$ is excluded because the inequality is strict.

The transformation $y=3f(x-2)$ moves a point $(x,y)$ to $(x+2,3y)$. Thus $P=(5,10)$ moves to $(5+2,3\cdot10)$.

\paragraph{Solution} Put $u=x^2+(3k-2)x$, so $u'=2x+3k-2$. By the product and chain rules,

As $2e^u\gt 0$, stationary points satisfy

For two distinct real roots, the discriminant must be positive:

Consequently

For these values the quadratic changes sign at both simple roots, so both stationary points are genuine turning points. Equality gives a repeated root with no derivative sign change and is excluded.

Set $u=|x|\ge0$. Then

Reject $u=-3$ since $u\ge0$. Therefore $|x|=8$.

For $x\le0$, the right side $\frac52x\le0$ and the left side is nonnegative, so every such $x$ satisfies the inequality. For $0\lt x\lt 5$, $f(x)\lt 0$ and

For $x\ge5$, $f(x)\ge0$ and

Combining the cases retains equality at both boundary points.

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

For $x\le a/3$, $y=a-3x$; for $x\ge a/3$, $y=3x-a$. These two straight branches form a V with minimum $(a/3,0)$ and $y$-intercept $(0,a)$.

Subtracting $b$ translates the previous graph vertically down by $b$. The minimum becomes $(a/3,-b)$ and the $y$-intercept becomes $(0,a-b)$, which is above the $x$-axis because $a\gt b\gt 0$.

For $x\le a/3$, the equation becomes $a-3x-b=5x$, giving $x=(a-b)/8$. This is positive and less than $a/8\lt a/3$, so it satisfies this branch condition. For $x\ge a/3$, the equation becomes $3x-a-b=5x$, giving $x=-(a+b)/2\lt 0$. This contradicts $x\ge a/3\gt 0$ and must be rejected. Therefore only one solution remains.

For $y=|3x-5a|-2a$ with $a\gt 0$: (i) At $x=0$, $y=5a-2a=3a$, so $P=(0,3a)$. (ii) At $y=0$, $|3x-5a|=2a$ gives $3x-5a=\pm2a$. Thus $x=a$ or $7a/3$; since $a\lt 7a/3$, $Q=(a,0)$ and $R=(7a/3,0)$. (iii) The modulus vanishes at $x=5a/3$, giving the minimum $S=(5a/3,-2a)$.

Split $|3x-5a|-2a=|x-2a|$ at $x=5a/3$ and $x=2a$; their order uses $a\gt 0$. For $x\lt 5a/3$: $3a-3x=2a-x$, so $x=a/2$, which is in this region. For $5a/3\le x\lt 2a$: $3x-7a=2a-x$, so $x=9a/4$; this is outside the assumed region and is rejected. For $x\ge2a$: $3x-7a=x-2a$, so $x=5a/2$, which is valid. These three regions cover the whole real line, so there are exactly two solutions.

(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).

(i) $y=k-2|x|$ is an inverted V with vertex $(0,k)$ and roots $x=\pm k/2$. (ii) $y=|2x-k/3|$ is an upright V with vertex $(k/6,0)$ and y-intercept $(0,k/3)$. Because $k\gt 0$, both sketches have the ordering shown. Axes below use coordinates normalized by k, while labels give the original coordinates.

\qquad

Consider the breakpoints $0$ and $k/6$. For $x\lt 0$, the equation is $k/3-2x=k+2x$, giving $x=-k/6$, valid in this branch. For $0\le x\lt k/6$, it is $k/3-2x=k-2x$, impossible because $k\gt 0$. For $x\ge k/6$, it is $2x-k/3=k-2x$, giving $x=k/3$, valid in this branch. Both have nonnegative right side and satisfy the original equation.

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 $21a=3$.

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.

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.

At the vertex the inner expression vanishes: $3x+a=0$, giving $x=-a/3$ and $f(x)=a$.

$g(x)=|x+5a|$ is a V translated left by $5a$, with vertex $(-5a,0)$ and y-intercept $(0,5a)$. Since $a\gt 0$, the vertex lies strictly left of the y-axis. Coordinates in the drawing are normalized by a; labels give the actual intercepts.

Solve $|x+5a|=|3x+a|+a$ using breakpoints $-5a\lt -a/3$. For $x\lt -5a$: $-x-5a=-3x$, giving $x=5a/2$, outside this branch, so reject it. For $-5a\le x\lt -a/3$: $x+5a=-3x$, giving $x=-5a/4$ and $y=15a/4$. For $x\ge-a/3$: $x+5a=3x+2a$, giving $x=3a/2$ and $y=13a/2$. The last two values belong to their respective intervals and satisfy both original functions.

At the vertex, the expression inside the modulus is zero:

Hence \(\boxed{P=\left(\frac52,3\right)}\).

Write the function in its two branches:

On the left branch,

This lies in \(0\leq x\leq\frac52\), so it is valid. On the right branch,

This satisfies \(x\geq\frac52\), so it is also valid. Thus

Each straight branch can have at most one intersection with \(y=kx+2\). Two distinct roots therefore require one intersection strictly to each side of the vertex.

For the left branch,

The condition \(0\leq x_L\lt \frac52\) first requires \(k+4\gt 0\), then gives

For the right branch,

The condition \(x_R\gt \frac52\) requires \(k\lt 4\), and then

At \(k=\frac25\), both intersections coincide at the vertex, giving only one root. At \(k=4\), the line is parallel to the right branch and has no intersection with that branch, again leaving only one root. Hence

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.

At the vertex $2x+7=0$, hence $x=-7/2$ and $f(x)=-10$.

For $x\ge-7/2$, $f(x)=x-13/2$. The inequality becomes $x-13/2\ge x/3+1$, giving $x\ge45/4$, which satisfies this branch. For $x\lt -7/2$, $f(x)=-x-27/2$. Then $-x-27/2\ge x/3+1$, giving $x\le-87/8$, which satisfies that branch. Join the two permitted outer intervals, including the equality endpoints.

The zeros of f satisfy $|2x+7|=20$, giving $x=-27/2$ and $x=13/2$. Reflect the part below the x-axis upward, leaving the positive branches unchanged. This creates minima at $(-27/2,0)$ and $(13/2,0)$ and a local maximum at $(-7/2,10)$.

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 $b=1/3$.

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 $a\gt b\gt 0$.

(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 $Q=(3/2,15/2)$.