ESAT Practice Set 1B Advanced Mathematics Worked Solutions

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 (📌):
The Cheat Code: 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 $2\pi$, and double it. The number inside the bracket multiplies the crossings; the number outside it only decides whether there are any.

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 (📌):
The Cheat Code: 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 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 (📌):
The Cheat Code: 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: 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. Of those two the smaller comes from dividing by $30$ instead of by $8$, so the larger survivor is the key.

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 to 3 significant figures the fraction $\dfrac{15-\sqrt{209}}{4}$ is $0.136$.

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 (⚠️):
Dividing only the surd term by the rationalised denominator. 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 roughly $3.89$, which is above $1$ and therefore impossible here.

Takeaway (📌):
The Cheat Code: 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. 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 $x^{p} = 3$ for an index $p$ that is usually not a unit fraction at all, and that awkwardness is itself the signal that the division step was mishandled.

Takeaway (📌):
The Cheat Code: 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.