ESAT Mathematics 2 Practice Questions by Topic

Correct Answer: A. 0.000152

Fastest Approach (🚀):
Total the indices in a single pass, $\tfrac{5}{2} - \tfrac{3}{4} - \tfrac{3}{2} = \tfrac{1}{4}$, so the condition reads $\sqrt[4]{x} = 3$ and $x = 81$ at sight. The value wanted is the reciprocal of $81^{2}$, so it has to be a small positive fraction, which rules out both whole number options before any arithmetic is done.

Step-by-Step Breakdown:

1. Collapse the left side to a single power of $x$

Powers of one base multiply by ADDING indices, so the numerator combines first:
$$x^{5/2}\cdot x^{-3/4} = x^{7/4}$$
Dividing by the denominator subtracts its index:
$$\frac{x^{7/4}}{x^{3/2}} = x^{1/4}$$
Run in a single pass the indices total $\tfrac{5}{2} - \tfrac{3}{4} - \tfrac{3}{2} = \tfrac{1}{4}$, so the condition in the stem is simply $x^{1/4} = 3$.

2. Solve for $x$

$x^{1/4}$ is the fourth root of $x$, and the stem fixes $x > 0$, so raising both sides to the power $4$ is safe and leaves no second solution to check:
$$x = 3^{4} = 81$$

3. Evaluate the requested power

A negative index means a reciprocal, not a negative value:
$$x^{-2} = \frac{1}{x^{2}} = \frac{1}{81^{2}} = \frac{1}{6561} = 0.000152$$
to 3 significant figures, which is the form the stem asks for.

Sanity check

Put $x = 81$ back into the left side. Its indices are the same chain as before, $\tfrac{5}{2} - \tfrac{3}{4} - \tfrac{3}{2} = \tfrac{1}{4}$, so the expression is $81^{1/4}$, and the fourth root of $81$ is $3$, as the stem requires. Since $x = 81 > 1$, every negative power of $x$ has to be a small positive fraction, so the two whole number options were never candidates.

The key is $0.000152$.

Why the Other Options Are Wrong (❌):

• B. 0.00617 · Index treated as a multiplier
  After reaching $x = 81$ the index was read as a multiplier rather than as a power, giving $\dfrac{1}{2 \times 81} = \dfrac{1}{162}$. An index of $-2$ means one over $x$ raised to the power $2$, so the denominator is $81^{2} = 6561$, and not $2 \times 81 = 162$.
• C. 0.111 · Root not undone
  The condition $x^{1/4} = 3$ was read as $x = 3$, with the fourth root left undone, and then $3^{-2} = \dfrac{1}{9}$ was evaluated. Undoing the fourth root comes first: raising both sides to the power $4$ gives $x = 3^{4} = 81$, which is not $3$.
• D. 81.0 · Intermediate value reported
  The index chain was carried out correctly as far as $x = 81$, and that value was reported. The stem asks for $x^{-2}$ and not for $x$, so $81$ is the intermediate result rather than the answer, and a negative power of a number larger than 1 is always smaller than 1.
• E. 6561 · Negative index ignored
  The minus sign in the index was dropped, so $x^{2} = 81^{2} = 6561$ was evaluated instead of $x^{-2}$. A negative index inverts the power rather than negating it, so the value wanted is the reciprocal of $6561$.

Common Mistake (⚠️):
Adding the denominator's index instead of subtracting it, so that $\tfrac{5}{2} - \tfrac{3}{4} - \tfrac{3}{2} = \tfrac{1}{4}$ is run with the wrong sign on $3/2$. The condition then reads $x^{13/4} = 3$, whose index is not a unit fraction at all, and that awkwardness is itself the signal that the division step was mishandled.

Takeaway (📌):
Collapse a tower of powers of one base into a single index before doing anything else. Once the left side is $x^{1/4}$ the equation is one root to undo, $x = 3^{4} = 81$, and the quantity actually asked for is a single substitution away.

Correct Answer: B. 0.136

Fastest Approach (🚀):
The quotient has the smaller surd expression on top and the larger underneath, so its value must be positive and below $1$. That discards the negative option and both options above $1$ without any algebra, leaving two. To choose between those two, estimate: $\sqrt{19}\approx4.36$ and $\sqrt{11}\approx3.32$ make the quotient about $\dfrac{1.04}{7.68}$, a little under $0.14$, so the key is $0.136$.

Step-by-Step Breakdown:

1. Multiply above and below by the conjugate of the denominator

The denominator is $\sqrt{19}+\sqrt{11}$, whose conjugate is $\sqrt{19}-\sqrt{11}$. Multiplying numerator and denominator by the same quantity leaves the value unchanged:
$$\dfrac{\sqrt{19}-\sqrt{11}}{\sqrt{19}+\sqrt{11}} = \frac{(\sqrt{19}-\sqrt{11})^{2}}{(\sqrt{19})^{2}-(\sqrt{11})^{2}} = \frac{(\sqrt{19}-\sqrt{11})^{2}}{19-11}$$
The denominator is now the rational number $8$.

2. Expand the numerator, then divide every term of it

$$(\sqrt{19}-\sqrt{11})^{2} = 19 - 2\sqrt{19}\sqrt{11} + 11 = 30 - 2\sqrt{209}$$
so the quotient is $\dfrac{30-2\sqrt{209}}{8}$. Both $19$ and $11$ are odd, so $30$ and $8$ are both even and the factor $2$ comes out of the whole numerator and out of the denominator together:
$$\frac{30-2\sqrt{209}}{8} = \frac{2\left(15-\sqrt{209}\right)}{2 \times 4} = \frac{15-\sqrt{209}}{4}$$
The factor cancels out of the whole bracket, not out of one term of it.

3. Put a number to the exact form

$\sqrt{209}$ is close to $14.5$, and it is smaller than $15$, since $15$ is the arithmetic mean of $19$ and $11$ while $\sqrt{209}$ is their geometric mean. The numerator is therefore positive, and the fraction $\dfrac{15-\sqrt{209}}{4}$ equals $\dfrac{4}{15+\sqrt{209}}$, because $(15-\sqrt{209})(15+\sqrt{209})=225-209=16$. That is about $\dfrac{4}{29.5}$, which is $0.136$ to 3 significant figures.

Sanity check

Take $\sqrt{19} \approx 4.36$ and $\sqrt{11} \approx 3.32$. The original quotient is then roughly $4.36 - 3.32$ over $4.36 + 3.32$, which is close to $0.136$. The value also has to be positive and below $1$, because $\sqrt{19}-\sqrt{11}$ is the smaller of two positive quantities and $\sqrt{19}+\sqrt{11}$ is the larger, and it is.

The key is $0.136$.

Why the Other Options Are Wrong (❌):

• A. 7.36 · Cross-term sign slip
  The cross term was expanded with the wrong sign, so $(\sqrt{19}-\sqrt{11})^{2}$ was written as $19 + 2\sqrt{209} + 11 = 30 + 2\sqrt{209}$. Dividing by the correct denominator $8$ then gives about $7.36$, which is greater than $1$, although the quotient in the stem has the smaller of the two positive quantities on top and so cannot exceed $1$. Squaring a difference subtracts the cross term.
• C. 3.89 · Partial cancellation
  From the correct $\dfrac{30-2\sqrt{209}}{8}$ the factor $2$ was cancelled against $8$ but applied to the surd term only, leaving $\dfrac{30-\sqrt{209}}{4}$, which is about $3.89$. The whole numerator is being divided, so $30$ has to be halved to $15$ as well.
• D. 0.0362 · Denominator added instead of differenced
  The denominator was multiplied out as $\sqrt{19}\times\sqrt{19} + \sqrt{11}\times\sqrt{11} = 19+11 = 30$ rather than as the difference of two squares $19-11 = 8$. That gives $\dfrac{30-2\sqrt{209}}{30} \approx 0.0362$, which is smaller than the true value of about $0.136$. A conjugate pair multiplies to a difference of squares, never to a sum.
• E. -2.61 · Second square subtracted in the expansion
  The second square was subtracted rather than added while expanding the numerator: $19 - 2\sqrt{209} - 11 = 8 - 2\sqrt{209}$. Dividing by $8$ gives about $-2.61$, a negative value, although $\sqrt{19}-\sqrt{11}$ and $\sqrt{19}+\sqrt{11}$ are both positive and so their quotient cannot be negative. Both squares are added when a bracket is squared; only the cross term changes sign.

Common Mistake (⚠️):
Cancelling the factor $2$ from the surd term of the numerator but not from the $30$. Once the quotient reads $\dfrac{30-2\sqrt{209}}{8}$, the whole numerator is being divided, so $30$ has to be reduced to $15$ at the same time as the $2$ in front of the surd is taken away. Cancelling into one term only leaves a value of roughly $3.89$, which is above $1$ and therefore impossible here.

Takeaway (📌):
The conjugate is the whole method. It turns $\sqrt{19}+\sqrt{11}$ into $19-11=8$ and turns the numerator into $(\sqrt{19}-\sqrt{11})^{2} = 30 - 2\sqrt{209}$. Expect the exact answer in the form $\dfrac{15-\sqrt{209}}{4}$, and divide every term of the numerator, never the surd alone.

Correct Answer: A. 2.58

Fastest Approach (🚀):
The roots themselves are not needed. Collect the substitution to $3x^2 - 6x - 2 = 0$, read $(-6)^2 - 4(3)(-2) = 60$ off the discriminant, then divide $\sqrt{60}$ by $3$. That is two lines of work with no root to evaluate and no fraction to simplify.

Step-by-Step Breakdown:

1. Substitute the line into the curve

Both equations give $y$, so at an intersection

$3x + 1 = 3x^2 - 3x - 1$

Taking every term of the line across, the $x$ term and the constant together, leaves one quadratic:

$3x^2 - 6x - 2 = 0$

Its $x$ coefficient is $-6$ and its constant is $-2$.

2. Evaluate the discriminant

With leading coefficient $3$,

$(-6)^2 - 4(3)(-2) = 60$

It is positive, which is consistent with the two crossings the question describes.

3. Take the difference of the roots

$x = \dfrac{-(-6) \pm \sqrt{60}}{2(3)}$, so the two roots differ by

$\dfrac{2\sqrt{60}}{2(3)} = \dfrac{\sqrt{60}}{3} = 2.58$ to 3 significant figures.

The centre the two roots sit either side of cancels in that subtraction and never has to be worked out.

Sanity check

The roots are near $-0.291$ and $2.29$, which are roughly $2.58$ apart. Between them $3x^2 - 6x - 2$ is negative and outside them it is positive, as a quadratic with a positive $x^2$ coefficient and two real roots must be.

The key is $2.58$.

Why the Other Options Are Wrong (❌):

• B. 1.53 · Solved the wrong equation
  The line was left out of the working and the curve's own $x$-intercepts were measured instead. $3x^2 - 3x - 1 = 0$ has discriminant $(-3)^2 - 4(3)(-1) = 21$, so its two roots are $\dfrac{\sqrt{21}}{3} = 1.53$ apart to 3 significant figures. Those are the points where the curve crosses the $x$-axis, not the points where it crosses $y = 3x + 1$.
• C. 2.31 · Constant term of the line left behind
  The $x$ term of the line was taken across but its constant was not, leaving $3x^2 - 6x - 1 = 0$, whose constant is still $-1$ where the substitution makes it $-2$. Its discriminant is $(-6)^2 - 4(3)(-1) = 48$, giving $\dfrac{\sqrt{48}}{3} = 2.31$ to 3 significant figures. The constant of the line crosses over with its $x$ term, in the same move.
• D. 1.91 · x term of the line left behind
  The constant of the line was taken across but its $x$ term was not, leaving $3x^2 - 3x - 2 = 0$, whose $x$ coefficient is still $-3$ where the substitution makes it $-6$. Its discriminant is $(-3)^2 - 4(3)(-2) = 33$, giving $\dfrac{\sqrt{33}}{3} = 1.91$ to 3 significant figures. The $x$ term of the line crosses over with its constant, in the same move.
• E. 1.15 · Sign error in the discriminant
  The $4ac$ part of the discriminant was added rather than subtracted: $(-6)^2 + 4(3)(-2) = 12$ was used in place of $(-6)^2 - 4(3)(-2) = 60$, which gives $\dfrac{\sqrt{12}}{3} = 1.15$ to 3 significant figures. The constant of $3x^2 - 6x - 2 = 0$ is $-2$, and it enters $b^2 - 4ac$ carrying its own sign.

Common Mistake (⚠️):
Taking the $x$ term of the line across but leaving its constant where it started, which collects the equation as $3x^2 - 6x - 1 = 0$ instead of $3x^2 - 6x - 2 = 0$. The constant is then $-1$ rather than $-2$, the discriminant reads $48$ rather than $60$, and every line after that is answering a different question. Both terms of the line have to be subtracted from the curve before the discriminant means anything.

Takeaway (📌):
Collect the line and the curve into a single quadratic, then the horizontal gap between the crossings is the square root of the discriminant divided by the size of the leading coefficient, here $\dfrac{\sqrt{60}}{3}$. Neither root is ever needed, and the axis of symmetry never has to be found.

Correct Answer: A. 33

Fastest Approach (🚀):
The fixed point can be discarded on sight. Solving $L=\frac{L + 9}{2}$ gives $L = 9$, and a first term equal to $L$ would make every term equal to $L$, which contradicts $u_{4} = 12$. That removes one option for free, and the rest is $L + 2^{3} \times (3) = 33$.

Step-by-Step Breakdown:

1. Invert the rule

Make the earlier term the subject:
$$u_{n+1}=\frac{u_{n} + 9}{2} \implies 2u_{n+1} = u_{n} + 9 \implies u_{n} = 2u_{n+1} - 9$$
This rule takes any term and returns the one before it, so it can be applied as many times as needed.

2. Find the fixed point

A term equal to $L$ would reproduce itself, so $L=\frac{L + 9}{2}$, giving $2L = L + 9$, then $(2-1)L = 9$ and $L = 9$. Writing every term as $L$ plus a difference is what turns the chain into one multiplication.

3. Walk back from $u_{4}$ to $u_{1}$

The difference between the given term and the fixed point is $u_{4} - L = 12 - (9) = 3$. Each application of the inverse rule multiplies that difference by $2$, and the number of applications is the difference of the indices, $4 - 1 = 3$:
$$u_{1} = L + 2^{3} \times (3) = 9 + (24) = 33$$

Check it forwards through the original rule: $\frac{33 + 9}{2} = 21$, which is $u_{2}$, and a further $2$ forward steps return $u_{4} = 12$ as stated.

The key is $33$.

Why the Other Options Are Wrong (❌):

• B. 9 · Fixed point mistaken for the term
  The fixed point was quoted as the answer. Solving $L=\frac{L + 9}{2}$ gives $2L = L + 9$, so $(2-1)L = 9$ and $L = 9$. That is the value the sequence approaches as it runs forwards, not its first term. If $u_{1}$ were $9$ then every term would equal $9$, and the fourth term would be $9$ rather than the stated $12$.
• C. 15 · Backward steps undercounted
  The inverse rule $u_{n} = 2u_{n+1} - 9$ was applied two times too few, stopping at $u_{3} = 15$ instead of $u_{1}$. The gap from index $4$ to index $1$ is $3$, so two further applications are still owed.
• D. 24 · Fixed point not restored
  The multiply-by-$2$ shortcut was used but the fixed point was never added back on. $2^{3} \times (3) = 24$ is the difference between $u_{1}$ and the fixed point $9$, not $u_{1}$ itself, so $L$ still has to be added to it.
• E. 21 · Index gap off by one
  Only $2$ applications of the inverse rule were made, so the walk stopped one index short at $u_{2} = 21$. Going from index $4$ to index $1$ needs $4 - 1 = 3$ applications, one more than were made, and it is the subscripts that fix that count rather than the terms written down.

Common Mistake (⚠️):
Working with the difference from the fixed point and then forgetting to put the fixed point back. The power of $2$ scales the difference, $2^{3} \times (3) = 24$, and that is a distance rather than a term: the first term is $9 + (24) = 33$.

Takeaway (📌):
For $u_{n+1}=\frac{u_{n}+c}{k}$, solve $L=\frac{L+c}{k}$ for the fixed point first. The difference from $L$ is divided by $k$ at every forward step and multiplied by $k$ at every backward one, so $u_{1} = L + k^{m}(u_{1+m} - L)$ replaces the whole chain of substitutions. Here that reads $u_{1} = 9 + 2^{3}(12 - (9)) = 33$.

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 (📌):
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. 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 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 (📌):
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.

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 (📌):
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: B. 6

Fastest Approach (🚀):
The amplitude $9$ does no work here beyond confirming that $5$ sits inside the range, so read the $3$ inside the bracket, take $2\times3$ and move on. Where between the peak and the trough the level sits never changes the count.

Step-by-Step Breakdown:

1. Check that the level is reached at all

The instants to count are the solutions of $9\cos(3t)=5$. The curve rises to $9$ and falls to $-9$, and $5$ lies strictly between those two values, so the horizontal line at height $5$ does cut the curve, and it cuts it away from both the peak and the trough.

2. Count the oscillations inside the window

The $3$ inside the bracket compresses one complete oscillation into $\frac{2\pi}{3}$, so the window $0\leq t\leq 2\pi$ holds exactly $3$ complete oscillations of the cosine wave.

3. Two crossings in each oscillation

A horizontal line strictly between the peak and the trough is met twice in every complete oscillation, once as the curve climbs and once as it falls back. The curve takes the same value at $t=2\pi$ as at $t=0$, and that value is not $5$, so nothing is gained or lost at the ends of the window.

$2\times3=6$

Sanity check: the amplitude $9$ decides only whether the level is reached, never how often, so $18$ answers a question about a different curve.

The key is $6$.

Why the Other Options Are Wrong (❌):

• A. 3 · One crossing counted per oscillation
  Takes the level to be reached once each time the wave comes round. The window $0\leq t\leq 2\pi$ holds $3$ complete oscillations, so this gives $3$. A horizontal line strictly between the peak $9$ and the trough $-9$ is met twice in every oscillation, once as the curve climbs and once as it falls, so the count is $2\times3=6$.
• C. 12 · Crossings of the negative level included
  Counts the crossings of $-5$ alongside those of $5$, on the ground that the value has size $5$ either way: $6+6=12$. On those extra occasions the curve is at $-5$, which is not the level the question asks about.
• D. 18 · Amplitude read as the oscillation count
  Reads $9$, the number in front of the cosine, as the count of oscillations in the window, and doubles it: $2\times9=18$. That number is a vertical stretch and sets only how high the curve rises; the $3$ inside the bracket is what decides how many oscillations fit into $0\leq t\leq 2\pi$.
• E. 10 · Level read as the oscillation count
  Reads $5$, the level the curve has to reach, as the number multiplying $t$ inside the bracket, and doubles it: $2\times5=10$. The level says what is being looked for; the $3$ inside the bracket says how often the curve comes round to it.

Common Mistake (⚠️):
Counting one crossing per oscillation. It is tempting to picture the wave reaching $5$ once each time it comes round, which gives $3$, but a horizontal line strictly between the peak $9$ and the trough $-9$ is cut twice in every complete oscillation, once on the climb and once on the fall. The count is $2\times3=6$.

Takeaway (📌):
For $a\cos(bt)=c$ with $c$ strictly between $-a$ and $a$, never solve for $t$. Count the oscillations the window holds, which is $b$ of them across $0$ to $2\pi$ when $b$ is a whole number, and double it. The number inside the bracket multiplies the crossings; the number outside it only decides whether there are any.

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 (📌):
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.