Star problems Advanced ← Back to chapter

Special Functions: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 Distances hidden under square roots

Difficulty: 1 of 3 stars, Stretch

Find all real solutions of $\sqrt{(x-2)^2}+\sqrt{(x+3)^2}=9$ . Then find all solutions when the right side is changed to $5$ . Explain why one equation has finitely many solutions while the other has an interval of solutions.
1. Hint 1
  For real $u$ , the principal square root of $u^2$ is $|u|$ , not necessarily $u$ .
2. Hint 2
  Interpret the two terms as distances from $x$ to the points $2$ and $-3$ on a number line.
Answer

For right side $9$ : $x=-5,4$ . For right side $5$ : every $x\in[-3,2]$ .

Full solution

The principal square roots are nonnegative, so the left side is $|x-2|+|x+3|$ .
It is the sum of the distances from $x$ to $2$ and to $-3$ , which are five units apart.

If $-3\le x\le2$ , those distances add to
$(2-x)+(x+3)=5.$
If $x>2$ , their sum is
$(x-2)+(x+3)=2x+1=5+2(x-2).$
If $x<-3$ , their sum is
$(2-x)+(-x-3)=-2x-1=5+2(-3-x).$

To obtain $9$ , $x$ must lie outside the interval and be two units from its nearer endpoint.
This gives $x=4$ or $x=-5$ ; both check directly.
To obtain $5$ , every point inside the interval works, while an outside point adds a strictly positive extra distance.
Hence exactly the closed interval $[-3,2]$ is the second solution set.
The interval arises because moving within it increases one distance by exactly as much as it decreases the other.

Answer

For right side $9$ : $x=-5,4$ . For right side $5$ : every $x\in[-3,2]$ .

Key idea

Square roots of squares encode distances; a geometric interpretation can reveal entire intervals of solutions.
Problem 1: Mark complete
Problem 2 A floor and a ceiling disagree

Difficulty: 1 of 3 stars, Stretch

Find all real numbers $x$ satisfying $\lfloor x\rfloor+\lceil2x\rceil=7$ . Here $\lfloor t\rfloor$ is the greatest integer at most $t$ , and $\lceil t\rceil$ is the least integer at least $t$ . State exactly which endpoints belong to the solution set.
1. Hint 1
  Write $x=n+t$ , where $n$ is an integer and $0\le t<1$ .
2. Hint 2
  The equation becomes $3n+\lceil2t\rceil=7$ . The ceiling can be $0$ , $1$ , or $2$ , with different endpoint rules.
Answer

$2<x\le5/2$ .

Full solution

Write $x=n+t$ , where $n=\lfloor x\rfloor$ is an integer and $0\le t<1$ .
Because $2n$ is integral,
$\lceil2x\rceil=2n+\lceil2t\rceil.$
Thus the equation is
$3n+\lceil2t\rceil=7.$

There are exactly three cases.
If $t=0$ , then $\lceil2t\rceil=0$ , which would require $3n=7$ , impossible for integer $n$ .
If $0<t\le1/2$ , then $\lceil2t\rceil=1$ , giving $n=2$ .
If $1/2<t<1$ , then $\lceil2t\rceil=2$ , which would require $3n=5$ , also impossible.

Therefore $x=2+t$ with $0<t\le1/2$ , or $2<x\le5/2$ .
The lower endpoint fails because at $x=2$ the left side is $2+4=6$ .
The upper endpoint works because at $x=5/2$ it is $2+5=7$ .
The three fractional-part cases cover negative $x$ as well, so no additional intervals have been overlooked.

Answer

$2<x\le5/2$ .

Key idea

Separate a real input into its integer and fractional parts, and preserve endpoint inequalities through each floor or ceiling.
Problem 2: Mark complete
Problem 3 A canceled factor leaves a missing output

Difficulty: 1 of 3 stars, Stretch

The function $H(x)=(x^2-5x+6)/(x^2-4)$ has its original real domain. Find all $x$ such that $H(x)=1/2$ . Then determine the complete range of $H$ , explaining any effect of the canceled factor.
1. Hint 1
  Factor numerator and denominator, but write the original excluded inputs before canceling.
2. Hint 2
  For a proposed output $y$ , solve $y=(x-3)/(x+2)$ for $x$ , then check whether that input is allowed.
Answer

$H(x)=1/2$ only at $x=8$ . The range is $\mathbb R\setminus\{1,-1/4\}$ .

Full solution

The original denominator factors as $(x-2)(x+2)$ , so $x\ne2,-2$ .
For those allowed inputs, cancellation gives
$H(x)=(x-3)/(x+2).$
Setting this equal to $1/2$ yields $2x-6=x+2$ , hence $x=8$ , which is allowed and works.

For the range, let
$y=(x-3)/(x+2).$
If $y=1$ , this would require $x-3=x+2$ , impossible.
If $y\ne1$ , the unique candidate input is
$x=(3+2y)/(1-y).$
It never equals $-2$ , because that equality would give $3=-2$ .
It equals $2$ exactly when $3+2y=2-2y$ , or $y=-1/4$ .

Thus outputs $1$ and $-1/4$ are impossible.
Conversely, for every other real $y$ , the displayed input is real, avoids both excluded values, and gives $H(x)=y$ .
This proves the complete range.
The missing value $-1/4$ is the output that the simplified formula would assign at the removed input $x=2$ .

Answer

$H(x)=1/2$ only at $x=8$ . The range is $\mathbb R\setminus\{1,-1/4\}$ .

Key idea

Cancellation preserves values only on the original domain; reconstructing inputs detects missing outputs.
Problem 3: Mark complete
Problem 4 When a weight moves the best location

Difficulty: 2 of 3 stars, Challenge

For real $x$ and a positive real parameter $w$ , let $D_w(x)=w|x+2|+2|x-1|+|x-7|$ . For every $w>0$ , find the minimum value of $D_w$ and every input attaining it.

Identify the weight at which the set of best locations changes, and explain why an entire interval is optimal at that weight.

Builds on Absolute Value Equations and Graphs, Piecewise Functions
1. Hint 1
  The breakpoints remain $-2$ , $1$ , and $7$ , regardless of $w$ .
2. Hint 2
  Find the coefficient of $x$ in each linear branch. On $[-2,1]$ , its sign depends on $w$ ; outside that interval the direction of change does not.
Answer

If $0<w<3$ , the minimum is $3w+6$ , attained only at $x=1$ . If $w=3$ , the minimum is $15$ , attained for every $x\in[-2,1]$ . If $w>3$ , the minimum is $15$ , attained only at $x=-2$ .

Full solution

The expression is linear on the intervals cut by $-2,1,7$ .
On the far-left interval its coefficient of $x$ is $-(w+3)<0$ , so the value decreases as $x$ moves toward $-2$ .
On $[1,7]$ , the coefficient is $w+1>0$ , and on the far-right interval it is $w+3>0$ .
Thus the value strictly increases after $1$ .
Every minimum must lie in $[-2,1]$ .

For $-2\le x\le1$ , expanding the absolute values gives
$D_w(x)=w(x+2)+2(1-x)+(7-x)=(w-3)x+2w+9.$
If $0<w<3$ , this decreases strictly, so its unique minimum is at $x=1$ , with value $3w+6$ .
If $w>3$ , it increases strictly, so its unique minimum is at $x=-2$ , with value $15$ .

At the threshold $w=3$ , the coefficient of $x$ vanishes and the whole interval has value $15$ .
The outer intervals are strictly worse, so this is the full set of minimizers.
Moving within $[-2,1]$ increases the weighted distance from $-2$ at rate $w$ while decreasing the other two weighted distances at combined rate $3$ .
They cancel exactly at the threshold.

Answer

If $0<w<3$ , the minimum is $3w+6$ , attained only at $x=1$ . If $w=3$ , the minimum is $15$ , attained for every $x\in[-2,1]$ . If $w>3$ , the minimum is $15$ , attained only at $x=-2$ .

Key idea

In a piecewise linear optimization problem, parameter thresholds occur when a branch changes its direction of increase.
Problem 4: Mark complete
Problem 5 Coupled radicals in sum-and-difference coordinates

Difficulty: 2 of 3 stars, Challenge

Find all ordered pairs of real numbers $(x,y)$ satisfying both equations
$\sqrt{x+y}=x-y,\qquad \sqrt{8(x-y)}=x+y.$
Use principal, nonnegative square roots. Prove completeness and check each candidate in both original equations.
1. Hint 1
  The repeated expressions $x+y$ and $x-y$ suggest new variables. What signs do the original equations force on them?
2. Hint 2
  Put $u=x+y$ and $v=x-y$ . The squared equations are $u=v^2$ and $u^2=8v$ . Keep the possibility $v=0$ before dividing by $v$ .
Answer

The only ordered pairs are $(x,y)=(0,0)$ and $(3,1)$ .

Full solution

Set $u=x+y$ and $v=x-y$ .
The first original equation requires $u\ge0$ and gives
$v=\sqrt u\ge0.$
The second also has nonnegative sides.
Thus both new variables must be nonnegative, and squaring gives $u=v^2$ and $u^2=8v$ .

Substitution yields $v^4=8v$ , or
$v(v^3-8)=0.$
One possibility is $v=0$ , which then gives $u=0$ .
Otherwise $v>0$ , so division by $v$ is valid and gives $v^3=8$ .
Its only nonnegative solution is $v=2$ , followed by $u=4$ .
There are no negative branches because the original square roots already forced $u,v\ge0$ .

The change of variables is reversible: $x=(u+v)/2$ and $y=(u-v)/2$ .
The two possibilities become $(0,0)$ and $(3,1)$ .
At $(0,0)$ both equations read $0=0$ .
At $(3,1)$ the first reads $\sqrt4=2$ , and the second reads
$\sqrt{16}=4.$
Both are valid.
Every original solution led to one of the two cases, so the list is complete; dividing by $v$ at the start would have lost the first pair.

Answer

The only ordered pairs are $(x,y)=(0,0)$ and $(3,1)$ .

Key idea

Use repeated combinations as coordinates, and retain zero cases before division in a radical system.
Problem 5: Mark complete
Problem 6 Divisibility inside square-root blocks

Difficulty: 2 of 3 stars, Challenge

How many integers $n$ with $1\le n\le400$ are divisible by $\lfloor\sqrt n\rfloor$ ? Find a structural description of all such $n$ , and then give the count for $1\le n\le M^2$ , where $M$ is any positive integer.
1. Hint 1
  Group inputs according to $k=\lfloor\sqrt n\rfloor$ . What interval of integers has the same $k$ ?
2. Hint 2
  Within $k^2\le n<(k+1)^2$ , list the multiples of $k$ . The final block at $M^2$ is incomplete.
Answer

There are $58$ for $1\le n\le400$ . In general the count is $3M-2$ ; full blocks contribute $k^2,k(k+1),k(k+2)$ .

Full solution

For any positive integer $n$ , set
$k=\lfloor\sqrt n\rfloor\ge1.$
Its block is
$k^2\le n<(k+1)^2=k^2+2k+1.$
If $n$ is a multiple of $k$ , write $n=kq$ and divide these inequalities by $k>0$ :
$k\le q<k+2+1/k.$

The integer choices are exactly
$q=k,k+1,k+2.$
This remains true at $k=1$ , where the upper bound is the strict inequality $q<4$ .
Hence every full block contributes exactly $k^2,k(k+1),k(k+2)$ , and these numbers all lie in their designated block.

Up to $M^2$ , the blocks $k=1,2,\ldots,M-1$ are complete and contribute $3(M-1)$ numbers.
The last block, $k=M$ , contributes only $M^2$ itself.
Distinct blocks do not overlap, so the total is
$3(M-1)+1=3M-2.$
This formula also gives $1$ when $M=1$ , correctly counting just $n=1$ .
Taking $M=20$ gives $58$ .

Answer

There are $58$ for $1\le n\le400$ . In general the count is $3M-2$ ; full blocks contribute $k^2,k(k+1),k(k+2)$ .

Key idea

A floor of a square root is constant between consecutive squares; count within those blocks before summing.
Problem 6: Mark complete
Problem 7 Returning after two folds

Difficulty: 2 of 3 stars, Challenge

On $[0,1]$ , define $T(x)=2x$ for $0\le x\le1/2$ , and $T(x)=2-2x$ for $1/2<x\le1$ .

Find all $x\in[0,1]$ for which $T(T(x))=x$ . Identify which of these return after one application and which form a genuine two-step cycle.

Text description of this figure

Axes labeled x and T of x, with a light square grid in steps of one half over the unit square. The graph of T is a tent made of two straight segments: it rises from the origin to a peak of height 1 at x equals one half, then falls back to height 0 at x equals 1. Dots mark the two ends and the peak. The horizontal axis is marked at 0, one half and 1, and the vertical axis at 1.

Builds on Piecewise Functions
1. Hint 1
  The formula used on the second application depends on whether $T(x)$ is at most $1/2$ .
2. Hint 2
  The needed breakpoints are $0,1/4,1/2,3/4,1$ . Compute $T(T(x))$ on each resulting interval and check the interval for every candidate.
Answer

The solutions are $0,2/5,2/3,4/5$ . The fixed points are $0,2/3$ ; the genuine cycle is $2/5\mapsto4/5\mapsto2/5$ .

Full solution

The output of $T$ stays in $[0,1]$ , so the second application is always defined.
On $[0,1/4]$ , the first output is $2x\le1/2$ , giving
$T(T(x))=4x.$
On $[1/4,1/2]$ , it is $2x\ge1/2$ , giving
$T(T(x))=2-4x.$
Both formulas agree at $1/4$ .

On $[1/2,3/4]$ , the first output is $2-2x\ge1/2$ , so the second output is
$2-2(2-2x)=4x-2.$
On $[3/4,1]$ , the first output is at most $1/2$ , so the second output is
$2(2-2x)=4-4x.$
Adjacent formulas also agree at $1/2$ and $3/4$ .

Equating each formula to $x$ gives, respectively,
$x=0,2/5,2/3,4/5.$
Every candidate belongs to the interval used to obtain it, so all four work and there are no others.
Directly, $T(0)=0$ and $T(2/3)=2/3$ , while $T(2/5)=4/5$ and
$T(4/5)=2/5.$
Thus two are fixed points, and the other two exchange places in a genuine two-step cycle.

Answer

The solutions are $0,2/5,2/3,4/5$ . The fixed points are $0,2/3$ ; the genuine cycle is $2/5\mapsto4/5\mapsto2/5$ .

Key idea

When composing a piecewise function, split at preimages of the original breakpoints as well as at the breakpoints themselves.
Problem 7: Mark complete
Problem 8 A radical equation with a moving target

Difficulty: 3 of 3 stars, Deep challenge

For every real $k$ , find all real solutions of $\sqrt{x+4}+\sqrt{9-x}=k$ . Classify when there are zero, one, or two distinct solutions, and give exact formulas for the solutions when they exist.
1. Hint 1
  The domain is $-4\le x\le9$ , and the two radicands have constant sum $13$ . First bound the possible value of their square-root sum.
2. Hint 2
  After one squaring, isolate the nonnegative product of the radicals. The identity $(2x-5)^2=169-4(x+4)(9-x)$ can simplify the second step.
Answer

No solutions for $k<\sqrt{13}$ or $k>\sqrt{26}$ ; one solution $x=5/2$ at $k=\sqrt{26}$ ; two solutions for $\sqrt{13}\le k<\sqrt{26}$ . The formula is $x=(5\pm k\sqrt{26-k^2})/2$ .

Full solution

The domain is $[-4,9]$ .
Write $u=\sqrt{x+4}$ and $v=\sqrt{9-x}$ , so $u,v\ge0$ and
$u^2+v^2=13.$
Consequently
$(u+v)^2=13+2uv\ge13.$
Also $(u-v)^2\ge0$ implies $2uv\le13$ , so
$(u+v)^2\le26.$
Since the sum is nonnegative, necessarily
$\sqrt{13}\le k\le\sqrt{26}.$

For such a $k$ , the equation gives
$2uv=k^2-13\ge0.$
Squaring and using $u^2v^2=(x+4)(9-x)$ yields
$4(x+4)(9-x)=(k^2-13)^2.$
Therefore $(2x-5)^2=169-(k^2-13)^2=k^2(26-k^2)$ , producing the stated two candidate formulas.

We must check the squaring.
For the allowed $k$ , $k^2(26-k^2)=169-(k^2-13)^2$ lies between $0$ and $169$ .
Thus the candidates satisfy $|2x-5|\le13$ , or
$-4\le x\le9.$
Their product of radicands gives
$2uv=|k^2-13|=k^2-13.$
Hence $(u+v)^2=k^2$ and, with both sides nonnegative, $u+v=k$ .
Every candidate is valid.

The candidates coincide only when $26-k^2=0$ , giving the single solution $x=5/2$ at
$k=\sqrt{26}.$
At every smaller allowed $k$ they are distinct; in particular $k=\sqrt{13}$ gives both endpoints $-4,9$ .

Answer

No solutions for $k<\sqrt{13}$ or $k>\sqrt{26}$ ; one solution $x=5/2$ at $k=\sqrt{26}$ ; two solutions for $\sqrt{13}\le k<\sqrt{26}$ . The formula is $x=(5\pm k\sqrt{26-k^2})/2$ .

Key idea

Bound the output before squaring, then use those bounds to verify every candidate and count merged branches.
Problem 8: Mark complete
Problem 9 Which integers does a step function skip?

Difficulty: 3 of 3 stars, Deep challenge

Define $S(x)=\lfloor x\rfloor+\lfloor2x\rfloor+\lceil3x\rceil$ for every real $x$ . Determine exactly which integers occur as outputs. For every integer $N$ that occurs, give the full set of real solutions of $S(x)=N$ . Your classification must include negative inputs and all interval endpoints.

Builds on Piecewise Functions
1. Hint 1
  Write $x=n+t$ with integer $n$ and $0\le t<1$ . Then $S(x)=6n+\lfloor2t\rfloor+\lceil3t\rceil$ .
2. Hint 2
  The fractional-part formula changes at $t=0,1/3,1/2,2/3$ . Analyze those points separately when needed, then classify $N$ by its remainder upon division by $6$ .
Answer

Exactly the integers not congruent to $5$ modulo $6$ occur. Write $N=6q+r$ uniquely with $q\in\mathbb Z$ and $0\le r\le5$ . For $r=0,1,2,3,4$ , the solution sets are respectively $\{q\}$ , $(q,q+1/3]$ , $(q+1/3,q+1/2)$ , $[q+1/2,q+2/3]$ , and $(q+2/3,q+1)$ ; for $r=5$ the set is empty.

Full solution

For every real $x$ , including negative $x$ , write $x=n+t$ with $n=\lfloor x\rfloor\in\mathbb Z$ and $0\le t<1$ .
Integer shifts pass through floors and ceilings, giving $S(x)=6n+R(t)$ , where
$R(t)=\lfloor2t\rfloor+\lceil3t\rceil.$

At $t=0$ , $R=0$ .
For $0<t\le1/3$ , the floor is $0$ and the ceiling is $1$ , so $R=1$ .
For $1/3<t<1/2$ , they are $0,2$ , so $R=2$ .
For $1/2\le t\le2/3$ , they are $1,2$ , so $R=3$ .
Finally, for $2/3<t<1$ , they are $1,3$ , so $R=4$ .
These cases cover every allowed $t$ without overlaps.

Every integer $N$ has a unique expression $6q+r$ with integer $q$ and remainder $r\in\{0,1,2,3,4,5\}$ , even when $N$ is negative.
Since $R$ takes only values $0$ through $4$ , the remainder $5$ is impossible.
For the other remainders, the equality forces $n=q$ and the corresponding listed fractional-part interval.
Adding $q$ gives exactly the solution sets in the answer.
Each nonempty set supplies actual inputs, proving attainability as well as exclusion.

Answer

Exactly the integers not congruent to $5$ modulo $6$ occur. Write $N=6q+r$ uniquely with $q\in\mathbb Z$ and $0\le r\le5$ . For $r=0,1,2,3,4$ , the solution sets are respectively $\{q\}$ , $(q,q+1/3]$ , $(q+1/3,q+1/2)$ , $[q+1/2,q+2/3]$ , and $(q+2/3,q+1)$ ; for $r=5$ the set is empty.

Key idea

An integer-plus-fractional-part decomposition can classify the entire range and every level set of a step function.
Problem 9: Mark complete
Problem 10 When a rational function reaches every height

Difficulty: 3 of 3 stars, Deep challenge

For real $a$ , let $R_a(x)=(x^2+ax+1)/(x^2-1)$ with domain $x\ne-1,1$ . Find all parameters $a$ for which every real number is an output of $R_a$ . Prove both directions.

For each boundary parameter of your answer, determine the exact range of $R_a$ , taking account of any canceled factor.
1. Hint 1
  If every output is possible, output $0$ must be possible. Examine the numerator and its roots.
2. Hint 2
  For an arbitrary target $y$ , rearrange $R_a(x)=y$ as $(y-1)x^2-ax-(y+1)=0$ . Handle $y=1$ separately, and make sure any quadratic roots avoid $x=\pm1$ .
Answer

Every real output occurs exactly when $|a|>2$ . At both boundary values $a=2$ and $a=-2$ , the range is $\mathbb R\setminus\{0,1\}$ .

Full solution

For output $0$ , the numerator must vanish at an allowed input.
Its discriminant is $a^2-4$ .
If $|a|<2$ , it has no real root.
If $a=2$ , its only root is $x=-1$ ; if $a=-2$ , its only root is $x=1$ .
Both are forbidden.
Hence reaching every real output requires $|a|>2$ .

Now assume $|a|>2$ and choose any real target $y$ .
For $y\ne1$ , the equation $(y-1)x^2-ax-(y+1)=0$ is quadratic with discriminant
$a^2+4(y^2-1)>0.$
It has real roots.
Substituting $x=1$ into its left side gives $-a-2$ , and substituting $x=-1$ gives $a-2$ ; neither is zero.
Thus its roots are allowed and give output $y$ .
If $y=1$ , the equation is linear: $-ax-2=0$ , giving $x=-2/a$ .
This also avoids $\pm1$ because $|a|>2$ .
Every target is therefore reached.

At $a=2$ , cancellation gives $R_2(x)=(x+1)/(x-1)$ while still excluding both $\pm1$ .
Output $1$ is impossible, and output $0$ would require the removed input $-1$ .
For any $y\ne0,1$ , the input $x=(y+1)/(y-1)$ is allowed and gives that output.
At $a=-2$ , the simplified form is $(x-1)/(x+1)$ , with the same exclusions.
For $y\ne0,1$ , the input $x=(1+y)/(1-y)$ works and avoids both forbidden inputs.
Thus both boundary ranges omit exactly $0$ and $1$ .

Answer

Every real output occurs exactly when $|a|>2$ . At both boundary values $a=2$ and $a=-2$ , the range is $\mathbb R\setminus\{0,1\}$ .

Key idea

Test a necessary output first; then prove full range by solving for an arbitrary target and checking degenerate equations and excluded inputs.
Problem 10: Mark complete