ESAT Practice Set 1A Advanced Mathematics Worked Solutions

Correct Answer: A. 144

Fastest Approach (🚀):
$p$ never has to be found. A vanishing integral with one sign change inside it forces the two regions to be equal in size, so the answer is twice the area between the roots, $2\times2\times\dfrac{6^{3}}{6}=144$.

Step-by-Step Breakdown:

1. Where the curve lies

The constant $2$ in front of the brackets is positive, so on $24<x<30$ the brackets $\left(x-24\right)$ and $\left(x-30\right)$ have opposite signs and the curve runs below the $x$ axis, while for $x>30$ both are positive and it runs above. An interval that starts at $x=24$ and reaches past $x=30$ therefore holds one region under the axis and one over it.

2. Find $p$

Put $u=x-24$, which slides the left root to the origin and leaves the gap between the roots at $6$. The curve becomes $y=2\,u\left(u-6\right)$, and with $q=p-24$,

which vanishes for $q>0$ only at $q=\dfrac{3\times6}{2}=9$. So $p=24+9=33$, and that is past $30$, so the sign change really does sit inside the interval.

3. The two pieces are the same size

The integral over the whole of $24\le x\le33$ is zero and the only crossing inside it is at $x=30$, so the negative contribution of $24\le x\le30$ and the positive contribution of $30\le x\le33$ cancel exactly. The first of them is the area a parabola cuts off between its roots:

so that region measures $72$, and so does the one beyond $x=30$.

4. Add the two sizes

Sanity check: the signed pieces are $-72$ and $+72$, which add to the zero integral the question specified.

The key is $144$.

Why the Other Options Are Wrong (❌):

• B. 0 · The vanishing integral quoted as the area
  Reports the definite integral itself. It really is zero, but that zero records $72$ of area below the axis cancelling $72$ above it, not the absence of anything to measure: over $24\le x\le33$ the curve encloses two pieces, each of size $72$.
• C. 288 · The symmetry doubling applied on top of the full sum
  Adds the two pieces correctly, $72+72=144$, then notices they are equal in size and doubles as well, reaching $2\times144=288$. The doubling is a substitute for the addition, not a step that follows it.
• D. 33 · The upper limit $p$ quoted instead of the area
  Stops as soon as $q^{2}\left(\dfrac{q}{3}-\dfrac{6}{2}\right)=0$ gives $q=9$, and reports $p=24+9=33$. That is a position on the $x$ axis, the right hand end of the region, not the size of the region.
• E. 72 · The constant in front of the brackets dropped
  Integrates $(x-24)(x-30)$ on its own, forgetting that the curve is $2$ times that product. The two pieces then come out as $\dfrac{6^{3}}{6}$ each and the total as $\dfrac{6^{3}}{3}=72$. Every ordinate of the curve carries the constant $2$, and so does every area measured under it.

Common Mistake (⚠️):
Reading "the definite integral is zero" as "there is no region to measure". The integral vanishes because a region of area $72$ below the axis is cancelled by a region of area $72$ above it, and the quantity asked for is the sum of those two rather than their difference.

Takeaway (📌):
The Cheat Code: Solve $y=0$ before integrating anything. A root strictly inside the limits splits the job in two, and the two answers are added with their signs discarded. When the integral across the whole interval is zero and only one root lies inside it, those two answers are equal, so one of them doubled is the entire calculation.

Correct Answer: B. 14

Fastest Approach (🚀):
Never solve the cubic. Only the product $(k + 8)(k - 8)$ matters, so differentiate once, read off $8$, and count the integers of the range lying outside $-8$ to $8$ directly.

Step-by-Step Breakdown:

1. Fix the shape and the turning points

$y = 4x^{3} - 12x + k$ has a positive $x^{3}$ coefficient, so it rises, turns down, then rises again. Differentiating, $\frac{dy}{dx} = 12x^{2} - 12$, which is zero when $x^{2} = 1$, that is $x = \pm 1$. Neither turning point depends on $k$: there is a local maximum at $x = -1$ and a local minimum at $x = 1$ in every run.

2. Write the two turning heights in terms of $k$

$y(-1) = -4 + 12 + k = k + 8$ and $y(1) = 4 - 12 + k = k - 8$.

3. Impose exactly one intersection

The curve meets the axis once only when both turning heights are strictly on the same side of it, that is $(k + 8)(k - 8) > 0$, so $k > 8$ or $k < -8$. At $k = 8$ the cubic factorises as $4(x - 1)^{2}(x + 2)$, which touches the axis at $x = 1$ and cuts it at $x = -2$: two distinct points, not one. $k = -8$ is the mirror image, $4(x + 1)^{2}(x - 2)$.

4. Count the integers at each end of the range

Above the window, $k$ runs from $9$ to $14$, which is $14 - 8 = 6$ values. Below it, $k$ runs from $-16$ to $-9$, which is $16 - 8 = 8$ values. Together that is $6 + 8 = 14$.

Sanity check: at $k = 14$ the two turning heights are $22$ and $6$, both above the axis, so that curve crosses only once, away to the left of $x = -1$. The run at $k = 8$ was excluded for the right reason: there the local minimum sits exactly on the axis.

The key is $14$.

Why the Other Options Are Wrong (❌):

• A. 16 · Boundary Error
  Used $|k| \ge 8$ where the condition is $|k| > 8$, so $k = 8$ and $k = -8$ join the $14$ values that genuinely qualify: $14 + 2 = 16$. At each of those two the cubic has a repeated root and one further crossing, so the curve meets the axis at two distinct points.
• C. 15 · Reversed Condition
  Answered the opposite condition and counted the values giving three distinct crossings. Those are the integers with $|k| < 8$, from $-7$ to $7$, and there are $15$ of them.
• D. 17 · Reversed Condition with Boundary Error
  The same reversal, with the two repeated-root values kept in: $-8$ to $8$ inclusive is $17$ integers. The $15$ strictly inside give three distinct points and the two ends give two, so none of them is a single-point run.
• E. 6 · One Branch Only
  Solved $k > 8$ and stopped there, counting $k = 9$ up to $k = 14$, which is $6$ values. The condition $|k| > 8$ has a second branch, $k < -8$, and the $8$ integers from $-16$ to $-9$ satisfy it too.

Common Mistake (⚠️):
Counting $k = 8$ and $k = -8$ among the single-point runs. At $k = 8$ the curve is $4(x - 1)^{2}(x + 2)$, which touches the axis at $x = 1$ and cuts it at $x = -2$, so two distinct points are reported rather than one, and $k = -8$ behaves the same way.

Takeaway (📌):
The Cheat Code: In a family $y = 4x^{3} - 12x + k$ only the constant moves, so the turning points stay at $x = \pm 1$ for every $k$. Write the two turning heights as $k + 8$ and $k - 8$, and the whole root count follows from the sign of their product, with no cubic ever solved.

Correct Answer: A. 10

Fastest Approach (🚀):
The whole question is one multiplication, $5\times(2)$. Read past the $\dfrac{x}{2}$ without working out the new $x$-coordinate at all, because nothing inside the bracket can change a height.

Step-by-Step Breakdown:

1. Decide which factor can change a height

The new curve is $y=5f\!\left(\dfrac{x}{2}\right)$, and it makes two independent changes. Replacing $x$ by $\dfrac{x}{2}$ acts on the input: it stretches the graph horizontally away from the $y$ axis by a factor of $2$, so the stationary point moves from $x=-6$ to $x=2\times(-6)=-12$ and its height is untouched. Multiplying the whole function by $5$ acts on the output, so it changes the height and nothing else.

2. Scale the height by the outside factor

Only the outside factor can reach the $y$-coordinate, so

$y=5\times(2)=10$

The stationary point of the new curve is $(-12,10)$, and the question asks for its $y$-coordinate. The multiplier is positive, so the curve is not turned over and the stationary point stays the same kind.

Check: nothing from inside the bracket was needed in that line of working, and that is the point of the question. The stretch inside the bracket settles where the stationary point sits, not how high it is.

The key is $10$.

Why the Other Options Are Wrong (❌):

• B. 4 · Inside factor applied to the height
  Multiplies the height by the number written under the $x$: $2\times(2)=4$. The $2$ inside the bracket changes the input, so it stretches the graph horizontally and leaves every height exactly as it was. Only the $5$ written outside the function can change a height.
• C. 2 · Height taken as unchanged
  Treats the whole transformation as horizontal and quotes the original height, $2$. The stretch inside the bracket does leave heights alone, but the $5$ written in front of $f$ acts on the output, so the height becomes $5\times(2)=10$.
• D. 7 · Multiplier read as a vertical translation
  Reads $y=5f\!\left(\dfrac{x}{2}\right)$ as $y=f\!\left(\dfrac{x}{2}\right)+5$ and shifts the height instead of scaling it: $2+5=7$. A number written in front of $f$ multiplies the output; only a number added on after $f$ moves the graph up or down.
• E. -30 · Multiplier applied to the wrong coordinate
  Applies the outside factor to the $x$-coordinate, $5\times(-6)=-30$, and quotes that as the height. The factor in front of $f$ acts on the output of the function, so it scales $y$; the $x$-coordinate is changed only by what is inside the bracket, and it moves from $-6$ to $-12$.

Common Mistake (⚠️):
Multiplying the height by the number written under the $x$. The $2$ inside the bracket acts on the input, so it moves points sideways and cannot change how high any of them is. The height is scaled by the outside factor $5$ and by nothing else, which gives $10$ rather than $4$.

Takeaway (📌):
The Cheat Code: Outside the bracket acts on $y$ and does exactly what it says; inside the bracket acts on $x$ and does the opposite of what it says. When only a height is asked for, everything inside the bracket can be ignored.

Correct Answer: B. 500

Fastest Approach (🚀):
Nothing outside the below-axis stretch matters. With $G(x)=x^{3}-75x$, the change in $G$ from $x=0$ to $x=5$ has size $250$, and doubling it gives $500$ without the total area or the full integral ever being computed. The area is bound to be the larger of the two reported values, so the gap is positive whichever way the integral came out.

Step-by-Step Breakdown:

1. Find where the curve meets the axis

$3x^{2}-75$ is $3(x-5)(x+5)$, so the curve meets the $x$ axis where $x^{2}=25$, at $x=-5$ and $x=5$. Only $x=5$ lies inside $0\leq x\leq 9$. On $0\leq x<5$ the curve runs below the axis, and on $5<x\leq 9$ it runs above it.

2. Evaluate the definite integral across the whole range

An antiderivative is $G(x)=x^{3}-75x$, with $G(0)=0$, $G(5)=-250$ and $G(9)=54$. The reported integral is $G(9)-G(0)=54$.

3. Build the total area from the two pieces

The stretch from $x=0$ to $x=5$ lies below the axis, and its area is $|G(5)-G(0)|=250$. The stretch from $x=5$ to $x=9$ lies above the axis, and its area is $|G(9)-G(5)|=304$. The total area is $250+304=554$.

4. Take the difference and check

The two reported values are $554$ and $54$. The area counts the below-axis piece positively where the integral counts it negatively, so the area is the larger of the two whatever sign the integral came out with, and the gap is

Sanity check: the below-axis piece is subtracted by the integral and added by the area, so the gap must be exactly $2\times250=500$, which it is. The above-axis piece of $304$ enters both numbers in the same way and cannot affect the gap at all.

The key is $500$.

Why the Other Options Are Wrong (❌):

• A. 608 · Wrong stretch doubled
  Doubled the stretch from $x=5$ to $x=9$, which lies above the axis, giving $2\times304=608$. An above-axis stretch is counted positively by the integral and positively by the area, so it enters both reported numbers identically and cannot open a gap between them. Only the stretch from $x=0$ to $x=5$, which lies below the axis, is treated differently by the two.
• C. 250 · Doubling omitted
  Found the below-axis area, from $x=0$ to $x=5$, correctly as $250$ and reported that as the gap. The integral subtracts that piece while the total area adds it, so it is counted once with a minus sign and once with a plus sign: the two quantities are separated by $2\times250=500$, not by $250$.
• D. 54 · Integral quoted instead of the gap
  Quoted the size of the definite integral, $54$, instead of comparing the integral with the area. The integral is one of the two numbers in the comparison, $54$ against a total area of $554$, and the question asks how far apart those two numbers are.
• E. 554 · Wrong quantity reported
  Reported the total area itself, $250+304=554$, rather than the difference between the two reported values. The area was built correctly; the question then asks for the separation between $554$ and $54$, which is $500$.

Common Mistake (⚠️):
Reporting the below-axis area, $250$, as the gap. That piece is not merely missing from the integral, it is actively subtracted by it, so across the two quantities it is counted once with a minus sign and once with a plus sign. The gap is therefore twice the piece, not the piece itself.

Takeaway (📌):
The Cheat Code: The total area and the definite integral differ by exactly twice the area lying below the axis inside the limits. Find the crossings, integrate the below-axis stretch alone, and double the size of what you get: every above-axis stretch enters both quantities identically and never shows up in the difference.

Correct Answer: A. -333

Fastest Approach (🚀):
Work the two limits separately rather than expanding anything. The antiderivative is worth $-445$ at $t = 25$ and $-112$ at $t = 16$, and one subtraction finishes it. The value $-445$ sits among the options for whoever stops after the first of those.

Step-by-Step Breakdown:

1. Write every term as a power of $t$

$\dfrac{1}{\sqrt{t}} = t^{-1/2}$, so the integrand is

2. Integrate term by term

Add one to each index, then divide by the index it has become. The first index becomes $\tfrac{1}{2}$, and dividing by $\tfrac{1}{2}$ multiplies the coefficient by $2$; the second becomes $2$, which halves it:

3. Evaluate at the upper limit

$\sqrt{25} = 5$, so at $t = 25$ the antiderivative is

4. Subtract the value at the lower limit

$\sqrt{16} = 4$, so at $t = 16$ the antiderivative is $36(4) - (16)^{2} = 144 - 256 = -112$. The definite integral is the first value minus the second:

Both limits are perfect squares, so $\sqrt{t}$ is a whole number at each end and every line above stays an integer.

The key is $-333$.

Why the Other Options Are Wrong (❌):

• B. 36 · Lost a term
  Only the root term was integrated and the second term, $-2t$, never reached the antiderivative: $\left[36\sqrt{t}\right]_{16}^{25} = 180 - (144) = 36$. An integral sign distributes over a sum, so both terms are integrated and both are then evaluated at both limits.
• C. -405 · Divided by the old index
  The coefficient was divided by the index the term arrived with, $-\tfrac{1}{2}$, rather than by the index it becomes, $\tfrac{1}{2}$. That turns the antiderivative into $-36\sqrt{t} - t^{2}$ and the same two limits then give $-405$. The rule raises the index first and divides by the new one, so the sign of that term does not change.
• D. -351 · Forgot to divide by the new index
  The index was raised to $\tfrac{1}{2}$ but the coefficient was never divided by it, leaving $18\sqrt{t} - t^{2}$ where the antiderivative is $36\sqrt{t} - t^{2}$, and the limits then give $-351$. Dividing by $\tfrac{1}{2}$ multiplies a coefficient by $2$, which is the step that is easiest to lose, because division is expected to make a number smaller.
• E. -445 · Omitted the lower limit
  The antiderivative was evaluated at $t = 25$ and the value at $t = 16$ was never subtracted: $180 - 625 = -445$. The antiderivative is worth $-112$ at $t = 16$, and a definite integral is the value at the top minus the value at the bottom.

Common Mistake (⚠️):
Dividing the coefficient by the index the term arrived with, $-\tfrac{1}{2}$, instead of by the index it becomes, $\tfrac{1}{2}$. The rule adds one to the index first and divides by what it turns into, so $18t^{-1/2}$ integrates to $36t^{1/2}$ and not to $-36t^{1/2}$.

Takeaway (📌):
The Cheat Code: A root in a denominator is a negative index in disguise. Rewrite $\dfrac{1}{\sqrt{t}}$ as $t^{-1/2}$ before anything else, and remember that the new index is $\tfrac{1}{2}$: dividing by $\tfrac{1}{2}$ multiplies a coefficient by $2$, which is the one step people expect to shrink it.