ESAT Advanced Mathematics Mock Module 4 Worked Solutions

Correct Answer: B. $3$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = \dfrac32 x^{\frac12} = \dfrac32\sqrt{x}$.
At $x=4$: $\dfrac32\times 2 = 3$.

Matches Option B.

Step-by-Step Breakdown:

1. Apply the power rule

$\dfrac{d}{dx}x^{n} = nx^{\,n-1}$ with $n = \dfrac32$:
$\dfrac{dy}{dx} = \dfrac32 x^{\frac32-1} = \dfrac32 x^{\frac12}$

2. Rewrite the fractional power

$x^{\frac12} = \sqrt{x}$, so
$\dfrac{dy}{dx} = \dfrac{3\sqrt{x}}{2}$

3. Substitute

At $x=4$: $\sqrt{4} = 2$, so
$\dfrac{dy}{dx} = \dfrac{3\times 2}{2} = 3$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{3}{4}$ — Arithmetic Error
  Dividing by 4 rather than multiplying by $\sqrt4$.
• C. $8$ — Misread Question
  Giving $4^{3/2}$, the value of $y$ rather than the gradient.
• D. $6$ — Index Error
  Using $\tfrac32\times 4$ without taking the square root.
• E. $\dfrac{3}{2}$ — Incomplete Answer
  Giving the coefficient only, without substituting.

Common Mistake (⚠️):
Subtracting one incorrectly, for example $\tfrac32-1 = \tfrac12$ written as $\tfrac13$, or evaluating $4^{1/2}$ as $\tfrac12\times 4 = 2$ by luck rather than by rule.

Takeaway (📌):
The power rule needs no modification for fractions. Convert $x^{1/2}$ to $\sqrt x$ before substituting to keep the arithmetic exact.

Correct Answer: A. $2$

Fastest Approach (🚀):
$\tan 60^{\circ} = \sqrt3$. Period $180^{\circ} \implies x = 60^{\circ}, 240^{\circ}$.
Two solutions.

Matches Option A.

Step-by-Step Breakdown:

1. Find the principal value

$\tan 60^{\circ} = \sqrt{3}$, so $x = 60^{\circ}$ is one solution.

2. Use the period

Unlike sine and cosine, tangent repeats every $180^{\circ}$:
$x = 60^{\circ}+180^{\circ} = 240^{\circ}$

3. Check the range

Adding another $180^{\circ}$ gives $420^{\circ}$, outside $[0^{\circ},360^{\circ})$. So there are exactly two solutions.

4. Sanity check with quadrants

Tangent is positive in the first and third quadrants, and $60^{\circ}$ and $240^{\circ}$ are exactly those. Consistent.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $1$ — Incomplete Answer
  Giving only the principal value.
• C. $3$ — Quadrant Error
  Adding $120^{\circ}$ as a third solution using the sine rule.
• D. $4$ — Period Error
  Treating tangent as having period $90^{\circ}$.
• E. $0$ — Range Confusion
  Assuming $\sqrt3$ is out of range, as it would be for sine.

Common Mistake (⚠️):
Applying the sine rule of $180^{\circ}-x$ and offering $120^{\circ}$, where tangent is negative, not positive.

Takeaway (📌):
$\tan$ has period $180^{\circ}$; $\sin$ and $\cos$ have period $360^{\circ}$. So $\tan x = k$ always has two solutions per revolution.

Correct Answer: D. $-6 < k < 6$

Fastest Approach (🚀):
$b^{2}-4ac = k^{2}-36 < 0 \implies k^{2} < 36 \implies -6 < k < 6$.

Matches Option D.

Step-by-Step Breakdown:

1. Write the discriminant

$a = 1,\ b = k,\ c = 9$:
$\Delta = b^{2}-4ac = k^{2}-36$

2. Impose no real roots

$\Delta < 0 \implies k^{2}-36 < 0 \implies k^{2} < 36$

3. Solve the quadratic inequality

$k^{2} < 36$ means $|k| < 6$, that is
$-6 < k < 6$

Note this is a 'less than' quadratic inequality, so the solution is the interval between the roots.

4. Check a value

$k = 0$ gives $x^{2}+9 = 0$, which indeed has no real roots. And $k = 6$ gives $(x+3)^{2} = 0$, a repeated root — correctly excluded.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $k > 6$ — Incomplete Answer
  Taking only one branch, and the wrong inequality direction.
• B. $k < -6$ or $k > 6$ — Inequality Reversal
  Giving the condition for two distinct real roots.
• C. $k = \pm 6$ — Boundary Error
  Giving the repeated-root boundary.
• E. $k < 6$ — Incomplete Answer
  Forgetting the negative branch of $|k|<6$.

Common Mistake (⚠️):
Solving $k^{2} > 36$ and giving the outside region, which is the condition for two distinct real roots.

Takeaway (📌):
$\Delta<0$ no real roots, $\Delta=0$ repeated root, $\Delta>0$ two distinct roots. A $k^{2}<c$ inequality always gives an interval.

Correct Answer: C. $|x| < \dfrac{1}{3}$

Fastest Approach (🚀):
$|3x| < 1 \implies |x| < \dfrac13$.

Matches Option C.

Step-by-Step Breakdown:

1. State the condition

$(1+u)^{n}$ has a convergent binomial expansion for non-integer $n$ precisely when
$|u| < 1$

2. Identify u

Here $u = 3x$, not $x$.

3. Solve

$|3x| < 1 \implies |x| < \dfrac{1}{3}$

4. Why it matters

For an integer $n$ the expansion terminates and no restriction is needed. For $n = \tfrac12$ the series is infinite, and outside $|x|<\tfrac13$ it simply does not converge.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $|x| < 1$ — Substitution Error
  Applying the condition to $x$ rather than to $3x$.
• B. $|x| < 3$ — Inversion Error
  Inverting the coefficient.
• D. $x > 0$ — Conceptual Error
  Confusing convergence with the domain of the square root.
• E. All real $x$ — Conceptual Error
  Treating the expansion as terminating, as for integer $n$.

Common Mistake (⚠️):
Applying the condition to $x$ instead of $3x$ and answering $|x|<1$, or inverting the coefficient to get $|x|<3$.

Takeaway (📌):
The validity condition is on the whole bracket term. For $(1+ax)^{n}$, it is $|x| < \dfrac{1}{|a|}$.

Correct Answer: C. $2\pi\ \text{cm}$

Fastest Approach (🚀):
$s = r\theta = 12\times\dfrac{\pi}{6} = 2\pi\ \text{cm}$.

Matches Option C.

Step-by-Step Breakdown:

1. Recognise the units

The angle is given as $\dfrac{\pi}{6}$, a multiple of $\pi$ — so it is in radians, not degrees.

2. Apply the radian formula

$s = r\theta = 12\times\dfrac{\pi}{6} = 2\pi\ \text{cm}$

3. Why the formula is so short

A radian is defined as the angle subtending an arc equal to the radius. So $\theta$ radians subtends $\theta$ radii of arc, and $s = r\theta$ follows directly — no $\dfrac{\theta}{360}$ factor is needed.

4. The degree cross-check

$\dfrac{\pi}{6}$ radians is $30^{\circ}$, and $\dfrac{30}{360}\times 2\pi(12) = \dfrac{1}{12}\times 24\pi = 2\pi$. ✓ Same answer, three times the work.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $12\pi\ \text{cm}$ — Angle Ignored
  Using the full circumference $2\pi r$ and halving.
• B. $\dfrac{\pi}{2}\ \text{cm}$ — Formula Inverted
  Dividing by $r$ instead of multiplying.
• D. $6\pi\ \text{cm}$ — Wrong Formula
  Using $A = \tfrac12 r^{2}\theta$ and mis-simplifying, or halving the circumference.
• E. $4\pi\ \text{cm}$ — Angle Misread
  Using $\dfrac{\pi}{3}$ in place of $\dfrac{\pi}{6}$.

Common Mistake (⚠️):
Using the degree formula $\dfrac{\theta}{360}\times 2\pi r$ with $\theta = \dfrac{\pi}{6}$ substituted as though it were a number of degrees. Check the units before choosing the formula.

Takeaway (📌):
In radians: $s = r\theta$ and $A = \tfrac12 r^{2}\theta$. An angle written with $\pi$ in it is almost always radians.

Correct Answer: D. $3$

Fastest Approach (🚀):
$27^{x-1} = 3^{3(x-1)}$.
$2x = 3x-3 \implies x = 3$.

Matches Option D.

Step-by-Step Breakdown:

1. Write both sides in base 3

$27 = 3^{3}$, so
$27^{\,x-1} = \left(3^{3}\right)^{x-1} = 3^{\,3(x-1)}$

2. Equate the indices

$3^{2x} = 3^{\,3x-3} \implies 2x = 3x-3$

Note the bracket must be expanded fully: $3(x-1) = 3x-3$, not $3x-1$.

3. Solve

$3 = 3x-2x \implies x = 3$

4. Check

LHS $= 3^{6} = 729$. RHS $= 27^{2} = 729$. Correct.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $1$ — Bracket Error
  Expanding $3(x-1)$ as $3x-1$.
• B. $-3$ — Sign Error
  Sign error when collecting terms.
• C. $-1$ — Sign Error
  Solving $2x = 3x+3$.
• E. $\dfrac{3}{2}$ — Omitted Term
  Equating $2x = 3$ and ignoring the $-1$.

Common Mistake (⚠️):
Expanding $3(x-1)$ as $3x-1$, which gives $x=1$ — a wrong answer that looks plausible.

Takeaway (📌):
Common base, then equate indices, then expand brackets carefully. The bracket is where these questions are lost.

Correct Answer: E. $4$

Fastest Approach (🚀):
$u = x^{2}: (u-1)(u-4) = 0 \implies u = 1, 4$.
$x = \pm1, \pm2$ — four real solutions.

Matches Option E.

Step-by-Step Breakdown:

1. Substitute

Let $u = x^{2}$ (valid because only even powers appear):
$u^{2}-5u+4 = 0$

2. Solve the quadratic

$(u-1)(u-4) = 0 \implies u = 1 \text{ or } u = 4$

3. Convert back, counting carefully

$x^{2} = 1 \implies x = \pm 1$
$x^{2} = 4 \implies x = \pm 2$

Each positive value of $u$ gives two real values of $x$. A negative $u$ would give none, and $u=0$ exactly one.

4. Count

Four real solutions: $-2, -1, 1, 2$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $2$ — Substitution Not Reversed
  Counting the roots in $u$ rather than in $x$.
• B. $1$ — Incomplete Answer
  Taking only positive square roots.
• C. $0$ — Conceptual Error
  Assuming a quartic with a positive constant has no real roots.
• D. $3$ — Counting Error
  Miscounting, treating one root as repeated.

Common Mistake (⚠️):
Stopping at $u = 1, 4$ and answering 2. Those are values of $x^{2}$, not of $x$.

Takeaway (📌):
$u = x^{2}$ linearises a quartic with only even powers. Each positive root doubles; negative roots contribute nothing real.

Correct Answer: E. $\dfrac{32}{3}$

Fastest Approach (🚀):
Roots at $x = 0, 4$.
$\int_{0}^{4}\left(4x-x^{2}\right)dx = \left[2x^{2}-\tfrac{x^{3}}{3}\right]_{0}^{4} = 32-\tfrac{64}{3} = \tfrac{32}{3}$.

Matches Option E.

Step-by-Step Breakdown:

1. Read the roots

$y = x(4-x) = 0$ at $x = 0$ and $x = 4$. These are the limits.

2. Expand and integrate

$y = 4x-x^{2}$
$\int_{0}^{4}\left(4x-x^{2}\right)dx = \left[2x^{2}-\dfrac{x^{3}}{3}\right]_{0}^{4}$

3. Evaluate

At $x=4$: $2(16)-\dfrac{64}{3} = 32-\dfrac{64}{3} = \dfrac{96-64}{3} = \dfrac{32}{3}$
At $x=0$: $0$

4. Check the sign and size

The coefficient of $x^{2}$ is negative, so the parabola opens downwards and the region lies above the axis — a positive integral, as obtained. The bounding rectangle is $4\times 4 = 16$, and $\tfrac{32}{3}\approx 10.7$ is a sensible fraction of it.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $16$ — Estimation Error
  Using the bounding rectangle $4\times 4$.
• B. $\dfrac{16}{3}$ — Limits Error
  Integrating over $[0,2]$ — half the region.
• C. $\dfrac{64}{3}$ — Incomplete Evaluation
  Giving $\tfrac{64}{3}$ — the second term alone.
• D. $8$ — Model Error
  Using a triangle $\tfrac12\times 4\times 4$.

Common Mistake (⚠️):
Forgetting to expand before integrating, or integrating $x(4-x)$ as though the bracket were a constant.

Takeaway (📌):
Factorised form gives the limits for free. Expand, integrate, then sanity-check against the bounding rectangle.

Correct Answer: E. $\dfrac{3}{4}$

Fastest Approach (🚀):
$\dfrac{8}{\sin 30^{\circ}} = \dfrac{12}{\sin B} \implies \sin B = \dfrac{12\times 0.5}{8} = \dfrac34$.

Matches Option E.

Step-by-Step Breakdown:

1. Set up the sine rule

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$

Each side pairs with the angle opposite it — that pairing is the whole rule.

2. Substitute

$\dfrac{8}{\sin 30^{\circ}} = \dfrac{12}{\sin B}$

$\sin 30^{\circ} = \dfrac12$, so the left side is $\dfrac{8}{0.5} = 16$.

3. Solve

$\sin B = \dfrac{12}{16} = \dfrac{3}{4}$

4. Check it is possible

$\tfrac34 \le 1$, so such a triangle exists. Note that $B$ could be either $\arcsin\tfrac34$ or its supplement — the ambiguous case — but $\sin B$ itself is unambiguous.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{3}$ — Formula Inverted
  Using $\dfrac{a\sin A}{b}$ with a further slip.
• B. $\dfrac{2}{3}$ — Omitted Factor
  Using $\dfrac{8}{12}$ without the sine factor.
• C. $\dfrac{1}{2}$ — Misread Question
  Repeating $\sin A$.
• D. $\dfrac{4}{3}$ — Formula Inverted
  Inverting the ratio — impossible, since $\sin B \le 1$.

Common Mistake (⚠️):
Inverting the fraction to get $\tfrac43$, which is impossible since $\sin B$ can never exceed 1 — a built-in check.

Takeaway (📌):
Sine rule: side over sine of the opposite angle, equal across the triangle. Any $\sin$ above 1 signals an inverted fraction.

Correct Answer: A. $(3,-1)$

Fastest Approach (🚀):
$x: 1+2 = 3$.
$y: -4+3 = -1$.
$(3,-1)$.

Matches Option A.

Step-by-Step Breakdown:

1. Handle the inside change

$f(x-2)$ means the input must be 2 larger to produce the same output, so the graph moves right by 2 — the opposite of the sign written.

$x: 1 \to 3$

2. Handle the outside change

$+3$ is added to the output, so the graph moves up by 3, exactly as written.

$y: -4 \to -1$

3. Combine

The minimum moves to $(3,-1)$.

4. Note what does not change

A translation does not alter the shape, so the point remains a minimum — only its position changes.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $(-1,-1)$ — Direction Error
  Moving left instead of right.
• C. $(3,-7)$ — Sign Error
  Subtracting 3 from $y$ instead of adding.
• D. $(-1,-7)$ — Direction Error
  Both direction errors together.
• E. $(1,-1)$ — Incomplete Transformation
  Applying only the vertical shift.

Common Mistake (⚠️):
Moving left instead of right for $f(x-2)$. Inside-the-bracket changes always act in the opposite direction to their sign.

Takeaway (📌):
$f(x-a)+b$ translates by $\begin{pmatrix} a \\ b \end{pmatrix}$. Inside: opposite sign, affects $x$. Outside: as written, affects $y$.

Correct Answer: C. $\dfrac{9}{2}$

Fastest Approach (🚀):
$1-\left(-\tfrac13\right) = \tfrac43$.
$\dfrac{6}{4/3} = \dfrac{18}{4} = \dfrac92$.

Matches Option C.

Step-by-Step Breakdown:

1. Check convergence

$|r| = \tfrac13 < 1$, so the sum to infinity exists.

2. Apply the formula

$S_{\infty} = \dfrac{a}{1-r} = \dfrac{6}{1-\left(-\tfrac13\right)}$

3. Handle the double negative

$1-\left(-\tfrac13\right) = 1+\tfrac13 = \dfrac43$

This is the step the question is built around.

4. Divide

$S_{\infty} = \dfrac{6}{\tfrac43} = 6\times\dfrac34 = \dfrac{9}{2}$

Check by listing: $6-2+\tfrac23-\tfrac29+\cdots$ oscillates and settles near $4.5$. Consistent.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $9$ — Sign Error
  Using $1-\tfrac13 = \tfrac23$ and ignoring the sign of $r$.
• B. $4$ — Substitution Error
  Rounding, or using $r = -\tfrac12$.
• D. $18$ — Operation Error
  Multiplying by 3 instead of dividing by $\tfrac43$.
• E. $-9$ — Sign Error
  Carrying the negative sign through to the sum.

Common Mistake (⚠️):
Using $1-\tfrac13 = \tfrac23$ and answering 9, losing the double negative entirely.

Takeaway (📌):
A negative ratio makes $1-r$ larger than 1, so the sum is smaller than $a\div 1$. Alternating series converge to less than their first term.

Correct Answer: B. $12\pi\ \text{cm}^{2}$

Fastest Approach (🚀):
$A = \tfrac12(12)^{2}\left(\dfrac{\pi}{6}\right) = 72\times\dfrac{\pi}{6} = 12\pi\ \text{cm}^{2}$.

Matches Option B.

Step-by-Step Breakdown:

1. Apply the radian formula

$A = \tfrac12 r^{2}\theta = \tfrac12\times 144\times\dfrac{\pi}{6} = 72\times\dfrac{\pi}{6} = 12\pi\ \text{cm}^{2}$

2. Square the radius, not the diameter

$r^{2} = 144$, not $12$. The most common slip here is carrying $r$ rather than $r^{2}$, which turns $12\pi$ into $\pi$.

3. The relationship worth remembering

$A = \tfrac12 r^{2}\theta = \tfrac12 r(r\theta) = \tfrac12 rs$

So the sector area is half the radius times the arc length — the sector behaves exactly like a triangle with base $s$ and height $r$. With $s = 2\pi$ from the companion question, $\tfrac12(12)(2\pi) = 12\pi$. ✓

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $24\pi\ \text{cm}^{2}$ — Factor Dropped
  Omitting the factor of $\tfrac12$.
• C. $6\pi\ \text{cm}^{2}$ — Factor Error
  Halving twice, or using $\tfrac14 r^{2}\theta$.
• D. $2\pi\ \text{cm}^{2}$ — Wrong Quantity
  Giving the arc length rather than the area.
• E. $144\pi\ \text{cm}^{2}$ — Angle Ignored
  Using $\pi r^{2}$ for the whole circle and ignoring the angle.

Common Mistake (⚠️):
Dropping the factor of $\tfrac12$ and giving $24\pi$, or forgetting to square the radius.

Takeaway (📌):
$A = \tfrac12 r^{2}\theta$ in radians, and equivalently $\tfrac12 rs$. If you already have the arc length, the second form is faster.

Correct Answer: B. $1$

Fastest Approach (🚀):
Product $= \dfrac{10}{1} = 10$.
$\log_{10}10 = 1$.

Matches Option B.

Step-by-Step Breakdown:

1. Combine the logarithms

$\sum\log_{10}a_n = \log_{10}\left(\prod a_n\right)$, so
$= \log_{10}\left(\dfrac{2}{1}\times\dfrac{3}{2}\times\dfrac{4}{3}\times\cdots\times\dfrac{10}{9}\right)$

2. Telescope the product

Every numerator cancels the next denominator, leaving only the first denominator and the last numerator:
$= \log_{10}\dfrac{10}{1} = \log_{10}10$

3. Evaluate

$\log_{10}10 = 1$

The same cancellation appears additively in partial-fraction telescoping; here it is multiplicative.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $0$ — Conceptual Error
  Assuming the terms cancel to zero.
• C. $9$ — Misread Question
  Counting the nine terms rather than evaluating them.
• D. $\log_{10}9$ — Off-by-one Error
  Telescoping to $\tfrac{9}{1}$ by an off-by-one in the last numerator.
• E. $10$ — Incomplete Answer
  Giving the product rather than its logarithm.

Common Mistake (⚠️):
Adding the fractions rather than multiplying them — a sum of logs is the log of a product, never of a sum.

Takeaway (📌):
Telescoping is not only for partial fractions. Logs turn a telescoping product into a telescoping sum.

Correct Answer: D. $-16$

Fastest Approach (🚀):
$y' = 3x^{2}-12 = 0 \implies x = \pm 2$.
Positive cubic $\implies$ minimum at the right root, $x = 2$.
$y(2) = 8-24 = -16$.

Matches Option D.

Step-by-Step Breakdown:

1. Differentiate and solve

$\dfrac{dy}{dx} = 3x^{2}-12 = 3\left(x^{2}-4\right) = 0 \implies x = \pm 2$

2. Classify

$\dfrac{d^{2}y}{dx^{2}} = 6x$.
At $x=2$: $+12 > 0$, a minimum.
At $x=-2$: $-12 < 0$, a maximum.

3. Evaluate at the minimum

$y(2) = 2^{3}-12(2) = 8-24 = -16$

4. Note the scope

This is a local minimum: a cubic is unbounded below as $x\to -\infty$. The question asks for the minimum value of the turning point, which is $-16$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $-8$ — Incomplete Evaluation
  Evaluating $x^{3}$ alone at $x=2$.
• B. $2$ — Misread Question
  Giving the $x$-coordinate.
• C. $16$ — Wrong Stationary Point
  Evaluating at $x=-2$, the local maximum.
• E. $0$ — Conceptual Error
  Assuming the minimum lies at $x=0$.

Common Mistake (⚠️):
Evaluating at $x=-2$ and giving $+16$, which is the local maximum, or reporting the $x$-coordinate 2 as the value.

Takeaway (📌):
For a positive cubic the right-hand stationary point is the minimum. Differentiate, choose, substitute — and answer with $y$, not $x$.

Correct Answer: A. $14$

Fastest Approach (🚀):
$\cos 120^{\circ} = -\tfrac12$.
$c^{2} = 136-120\left(-\tfrac12\right) = 136+60 = 196 \implies c = 14$.

Matches Option A.

Step-by-Step Breakdown:

1. Choose the rule

Two sides and the included angle: the cosine rule.
$c^{2} = a^{2}+b^{2}-2ab\cos C$

2. Use the exact value

$\cos 120^{\circ} = -\dfrac12$ — obtuse, so negative.

3. Substitute

$c^{2} = 6^{2}+10^{2}-2(6)(10)\left(-\dfrac12\right) = 36+100+60 = 196$

The double negative turns the subtraction into an addition, which is why the third side exceeds both given sides.

4. Square root

$c = 14$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\sqrt{76}$ — Sign Error
  Using $\cos 120^{\circ} = +\tfrac12$.
• C. $16$ — Formula Misuse
  Adding the sides and adjusting.
• D. $8$ — Formula Misuse
  Using Pythagoras as though the angle were $90^{\circ}$ and mis-rooting.
• E. $2\sqrt{34}$ — Term Omitted
  Dropping the $-2ab\cos C$ term entirely, leaving $c^2 = 6^2 + 10^2 = 136$.

Common Mistake (⚠️):
Taking $\cos 120^{\circ}$ as $+\tfrac12$, giving $\sqrt{76}$ — a third side shorter than the longest given side, which an obtuse angle makes impossible.

Takeaway (📌):
An obtuse included angle always lengthens the opposite side. If your answer is shorter than both given sides, the cosine's sign is wrong.

Correct Answer: B. $(5,1)$

Fastest Approach (🚀):
$x: 3+2 = 5$. $y: -4+5 = 1$.
Minimum at $(5,1)$.

Matches Option B.

Step-by-Step Breakdown:

1. Handle the inside change

$f(x-2)$ is a translation of $+2$ in the $x$-direction — the minus inside the bracket moves the curve right, not left.

$x: 3 \to 5$

2. Handle the outside change

$+5$ outside is a translation of $+5$ in the $y$-direction, exactly as written.

$y: -4 \to 1$

3. Combine

The minimum sits at $(5,1)$. A translation does not change the nature of a stationary point, so it is still a minimum.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $(1,1)$ — Transformation Inverted
  Moving left instead of right.
• C. $(5,-9)$ — Sign Error
  Subtracting 5 from the $y$-coordinate.
• D. $(1,-9)$ — Transformation Inverted
  Both directions inverted.
• E. $(3,1)$ — Omitted Transformation
  Applying the vertical shift only.

Common Mistake (⚠️):
Moving left because of the minus sign inside the bracket, giving $(1,1)$. Inside changes always act in the opposite sense.

Takeaway (📌):
Inside the bracket: acts on $x$, opposite direction. Outside: acts on $y$, as written. Track one point rather than the whole curve.

Correct Answer: E. $\dfrac{50}{101}$

Fastest Approach (🚀):
$\dfrac12\left(1-\dfrac{1}{101}\right) = \dfrac12\times\dfrac{100}{101} = \dfrac{50}{101}$.

Matches Option E.

Step-by-Step Breakdown:

1. Split into partial fractions

$\dfrac{1}{(2r-1)(2r+1)} = \dfrac{A}{2r-1}+\dfrac{B}{2r+1}$

gives $1 = A(2r+1)+B(2r-1)$. Setting $r = \tfrac12$ gives $A = \tfrac12$; setting $r = -\tfrac12$ gives $B = -\tfrac12$.

$= \dfrac12\left(\dfrac{1}{2r-1}-\dfrac{1}{2r+1}\right)$

The factor of $\tfrac12$ is what distinguishes this from the consecutive-integer case.

2. Write out the sum

$\dfrac12\left[\left(\dfrac11-\dfrac13\right)+\left(\dfrac13-\dfrac15\right)+\cdots+\left(\dfrac{1}{99}-\dfrac{1}{101}\right)\right]$

3. Cancel and evaluate

Only the first and last fragments survive:
$\dfrac12\left(1-\dfrac{1}{101}\right) = \dfrac12\times\dfrac{100}{101} = \dfrac{50}{101}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{100}{101}$ — Omitted Factor
  Omitting the factor of $\tfrac12$.
• B. $\dfrac{1}{101}$ — Incomplete Answer
  Giving the surviving fragment $\tfrac{1}{101}$ rather than the sum.
• C. $\dfrac{50}{51}$ — Setup Error
  Using consecutive integers instead of consecutive odd numbers.
• D. $\dfrac{25}{101}$ — Double Correction
  Halving twice.

Common Mistake (⚠️):
Forgetting the factor of $\tfrac12$ from the partial fractions, which doubles the answer to $\tfrac{100}{101}$.

Takeaway (📌):
When the two factors differ by $d$, the partial fractions carry a factor of $\tfrac1d$. Consecutive integers hide it because $d=1$.

Correct Answer: A. $6\pi-9\sqrt{3}$

Fastest Approach (🚀):
Sector: $\tfrac12(36)\left(\dfrac{\pi}{3}\right) = 6\pi$.
Triangle: $\tfrac12(36)\sin\dfrac{\pi}{3} = 18\times\dfrac{\sqrt3}{2} = 9\sqrt3$.
$6\pi-9\sqrt3$.

Matches Option A.

Step-by-Step Breakdown:

1. See the segment as a difference

The chord cuts the sector into a triangle (between the two radii and the chord) and the segment (between the chord and the arc):
$\text{segment} = \text{sector}-\text{triangle}$

2. The sector

$\tfrac12 r^{2}\theta = \tfrac12\times 36\times\dfrac{\pi}{3} = 6\pi$

3. The triangle

Two sides of length $r$ with the angle $\theta$ between them, so use $\tfrac12 ab\sin C$:
$\tfrac12\times 6\times 6\times\sin\dfrac{\pi}{3} = 18\times\dfrac{\sqrt3}{2} = 9\sqrt3$

4. Subtract, and sanity-check

$6\pi-9\sqrt3 \approx 18.85-15.59 = 3.26$

A small positive number, as a thin segment should be. A negative result would mean the two formulas had been used the wrong way round.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $6\pi-18\sqrt{3}$ — Factor Dropped
  Omitting the $\tfrac12$ in the triangle area.
• C. $6\pi-9$ — Exact Value Error
  Using $\sin\dfrac{\pi}{3} = \tfrac12$ instead of $\dfrac{\sqrt3}{2}$.
• D. $12\pi-9\sqrt{3}$ — Factor Dropped
  Omitting the $\tfrac12$ in the sector area.
• E. $6\pi+9\sqrt{3}$ — Sign Error
  Adding the triangle rather than subtracting it — this is the major segment plus an error.

Common Mistake (⚠️):
Using $\tfrac12 r^{2}\sin\theta$ with $r^{2}$ replaced by $r$, or forgetting the $\tfrac12$ in the triangle and getting $18\sqrt3$.

Takeaway (📌):
Segment $= \tfrac12 r^{2}(\theta-\sin\theta)$ in radians. Both terms share $\tfrac12 r^{2}$, so factorise it out and the arithmetic halves.

Correct Answer: B. $\dfrac{(2x+1)^{5}}{10}+c$

Fastest Approach (🚀):
Raise: $(2x+1)^{5}$. Divide by $5$ and by the inner coefficient $2$:
$\dfrac{(2x+1)^{5}}{10}+c$.

Matches Option B.

Step-by-Step Breakdown:

1. Use the standard result

$\int\left(ax+b\right)^{n}dx = \dfrac{\left(ax+b\right)^{n+1}}{a(n+1)}+c$

The extra $\dfrac1a$ compensates for the chain rule that differentiation would apply.

2. Substitute

$a = 2$, $b = 1$, $n = 4$:
$\int\left(2x+1\right)^{4}dx = \dfrac{\left(2x+1\right)^{5}}{2\times 5}+c = \dfrac{\left(2x+1\right)^{5}}{10}+c$

3. Check by differentiating

$\dfrac{d}{dx}\left[\dfrac{(2x+1)^{5}}{10}\right] = \dfrac{5(2x+1)^{4}\times 2}{10} = (2x+1)^{4}$. Correct.

4. Why expansion is a bad idea

Expanding $(2x+1)^{4}$ gives five terms to integrate separately — perfectly valid, and roughly ten times the work.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{(2x+1)^{5}}{5}+c$ — Chain Rule Omitted
  Forgetting to divide by the inner coefficient 2.
• C. $\dfrac{(2x+1)^{5}}{2}+c$ — Omitted Divisor
  Dividing by the coefficient only, not by the new power.
• D. $8(2x+1)^{3}+c$ — Operation Error
  Differentiating instead of integrating.
• E. $(2x+1)^{5}+c$ — Omitted Divisor
  Raising the power without dividing at all.

Common Mistake (⚠️):
Dividing only by the new power and forgetting the inner coefficient, giving $\dfrac{(2x+1)^{5}}{5}$.

Takeaway (📌):
$\int(ax+b)^{n}dx = \dfrac{(ax+b)^{n+1}}{a(n+1)}$. Two divisions: the new power and the inner coefficient.

Correct Answer: D. $6$

Fastest Approach (🚀):
$3x \in [0^{\circ},1080^{\circ})$ — three revolutions.
$\cos\theta = \tfrac12$ has $2$ solutions per revolution.
$3\times 2 = 6$.

Matches Option D.

Step-by-Step Breakdown:

1. Widen the range first

$0^{\circ}\le x \lt 360^{\circ} \implies 0^{\circ}\le 3x \lt 1080^{\circ}$

2. Count solutions for the whole angle

$\cos\theta = \tfrac12$ has principal value $60^{\circ}$, and cosine is positive in the first and fourth quadrants, so within one revolution:
$\theta = 60^{\circ},\ 300^{\circ}$

3. Repeat across three revolutions

$\theta = 60^{\circ},\ 300^{\circ},\ 420^{\circ},\ 660^{\circ},\ 780^{\circ},\ 1020^{\circ}$

4. Divide by 3

$x = 20^{\circ},\ 100^{\circ},\ 140^{\circ},\ 220^{\circ},\ 260^{\circ},\ 340^{\circ}$ — six solutions, all inside the original range.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $2$ — Multiple Angle Ignored
  Solving $\cos x = \tfrac12$ and ignoring the multiple angle.
• B. $3$ — Quadrant Error
  Counting one solution per revolution.
• C. $4$ — Range Error
  Widening the range by $2$ rather than by $3$.
• E. $8$ — Boundary Error
  Widening by $4$, or counting the endpoint twice.

Common Mistake (⚠️):
Solving $\cos x = \tfrac12$ and answering $2$, ignoring the multiple angle entirely.

Takeaway (📌):
For $\cos kx = c$ over one revolution of $x$, expect $2k$ solutions — provided $|c|<1$.

Correct Answer: A. $\dfrac{1}{x}$

Fastest Approach (🚀):
$\ln(3x) = \ln 3+\ln x \implies \dfrac{dy}{dx} = \dfrac1x$.

Matches Option A.

Step-by-Step Breakdown:

1. Split the logarithm

$\ln(3x) = \ln 3+\ln x$

$\ln 3$ is just a number.

2. Differentiate term by term

$\dfrac{d}{dx}\left(\ln 3\right) = 0$
$\dfrac{d}{dx}\left(\ln x\right) = \dfrac1x$

So $\dfrac{dy}{dx} = \dfrac1x$.

3. Confirm with the chain rule

$\dfrac{d}{dx}\ln(u) = \dfrac{u'}{u}$ with $u = 3x$:
$\dfrac{3}{3x} = \dfrac{1}{x}$

The 3 cancels — the same result by a different route.

4. The general fact

$\ln(kx)$ has derivative $\tfrac1x$ for every positive constant $k$. The curves are vertical translations of each other, so they share a gradient function.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\dfrac{3}{x}$ — Chain Rule Error
  Chain rule with the numerator kept but the denominator left as $x$.
• C. $\dfrac{1}{3x}$ — Chain Rule Error
  Dividing by $3x$ without the numerator $3$.
• D. $3\ln x$ — Log Law Error
  Applying the log law incorrectly as $3\ln x$.
• E. $\dfrac{1}{3}$ — Conceptual Error
  Differentiating the constant instead of the variable.

Common Mistake (⚠️):
Applying the chain rule but forgetting to divide by the whole of $3x$, giving $\tfrac3x$.

Takeaway (📌):
$\dfrac{d}{dx}\ln(kx) = \dfrac1x$ for any constant $k$. Splitting the log makes it obvious; the chain rule confirms it.

Correct Answer: C. $-5$

Fastest Approach (🚀):
$y = x^{2}+5x+c$.
$3 = 4+10+c \implies c = -11$.
$y(1) = 6-11 = -5$.

Matches Option C.

Step-by-Step Breakdown:

1. Integrate

$y = \int(2x+5)\,dx = x^{2}+5x+c$

2. Fix the constant with the point

$3 = 2^{2}+5(2)+c = 14+c \implies c = -11$

3. Evaluate at the new value

$y(1) = 1+5-11 = -5$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $-11$ — Misread Question
  Giving the constant itself.
• B. $6$ — Missing Constant
  Assuming $c = 0$.
• D. $1$ — Conceptual Error
  Substituting $x=1$ into the derivative and subtracting.
• E. $-3$ — Arithmetic Error
  Arithmetic slip when combining the terms.

Common Mistake (⚠️):
Taking $c = 0$ and answering $6$ — the curve would then miss the given point by $11$.

Takeaway (📌):
Integrate, substitute the point, then evaluate. Three steps, always in that order.

Correct Answer: E. $3$

Fastest Approach (🚀):
$-\tfrac23+1 = \tfrac13$, and dividing by $\tfrac13$ gives $3x^{1/3}$.
$3\sqrt[3]{8}-3\sqrt[3]{1} = 6-3 = 3$.

Matches Option E.

Step-by-Step Breakdown:

1. Add one to the index

$-\dfrac23+1 = \dfrac13$

2. Divide by the new index

Dividing by $\tfrac13$ is multiplying by $3$:
$\int x^{-2/3}dx = \dfrac{x^{1/3}}{1/3} = 3x^{1/3}+c$

3. Apply the limits

$\left[3x^{1/3}\right]_{1}^{8} = 3\sqrt[3]{8}-3\sqrt[3]{1} = 3(2)-3(1)$

4. Evaluate

$= 6-3 = 3$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $9$ — Root Error
  Cubing rather than cube-rooting the limit.
• B. $1$ — Reciprocal Error
  Using $\tfrac13 x^{1/3}$ instead of $3x^{1/3}$.
• C. $6$ — Limits Error
  Evaluating at the upper limit only.
• D. $\dfrac{3}{2}$ — Arithmetic Error
  Halving the correct result.

Common Mistake (⚠️):
Dividing by $\tfrac13$ as though it were multiplying by $\tfrac13$, which scales the answer by $9$ in the wrong direction.

Takeaway (📌):
Add one to the index, then divide by the result. Dividing by a fraction below $1$ always makes the coefficient larger.

Correct Answer: A. $\dfrac{1}{2}$

Fastest Approach (🚀):
$\int x^{-2}dx = -x^{-1}$.
$\left[-\tfrac1x\right]_{1}^{2} = -\tfrac12-(-1) = \tfrac12$.

Matches Option A.

Step-by-Step Breakdown:

1. Rewrite with a negative index

$\dfrac{1}{x^{2}} = x^{-2}$

2. Integrate

$\int x^{-2}dx = \dfrac{x^{-1}}{-1} = -\dfrac{1}{x}+c$

The rule $\dfrac{x^{n+1}}{n+1}$ works here because $n = -2 \ne -1$. Only $x^{-1}$ itself is the exception, integrating to $\ln|x|$.

3. Apply the limits

$\left[-\dfrac1x\right]_{1}^{2} = \left(-\dfrac12\right)-\left(-\dfrac11\right) = -\dfrac12+1$

4. Evaluate

$= \dfrac12$

Positive, as it must be: the integrand is positive throughout $[1,2]$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\ln 2$ — Special Case Error
  Using the $\ln$ rule, which applies only to $x^{-1}$.
• C. $-\dfrac{1}{2}$ — Limits Order
  Subtracting the limits in the wrong order.
• D. $\dfrac{3}{8}$ — Index Error
  Integrating to $\tfrac{x^{-1}}{-1}$ but evaluating $\tfrac{1}{x^{3}}$ style.
• E. $\dfrac{7}{24}$ — Index Error
  Using $\tfrac{x^{-3}}{-3}$ — index reduced instead of increased.

Common Mistake (⚠️):
Using $\ln$ because the integrand is a fraction. Only $\dfrac1x$ integrates to a logarithm; $\dfrac{1}{x^{2}}$ does not.

Takeaway (📌):
$\int x^{n} = \dfrac{x^{n+1}}{n+1}$ for every $n$ except $-1$. Check the index before reaching for $\ln$.

Correct Answer: B. $-\pi$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = \sin x+x\cos x$.
At $x=\pi$: $\sin\pi = 0,\ \cos\pi = -1 \implies 0+\pi(-1) = -\pi$.

Matches Option B.

Step-by-Step Breakdown:

1. Apply the product rule

With $u = x$ and $v = \sin x$:
$\dfrac{dy}{dx} = u'v+uv' = (1)\sin x+x\cos x$

2. Substitute the exact values

At $x = \pi$:
$\sin\pi = 0, \qquad \cos\pi = -1$

3. Evaluate

$\dfrac{dy}{dx} = 0+\pi(-1) = -\pi$

4. Interpret

The gradient is negative, so the curve is falling at $x=\pi$ — consistent with $x\sin x$ dropping below the axis just past $\pi$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $0$ — Omitted Term
  Keeping only the $\sin x$ term, which vanishes at $\pi$.
• C. $\pi$ — Value Error
  Using $\cos\pi = +1$.
• D. $1$ — Formula Misuse
  Differentiating the factors separately: $1\times\cos x$ at a convenient point.
• E. $-1$ — Incomplete Answer
  Giving $\cos\pi$ rather than the full derivative.

Common Mistake (⚠️):
Using $\cos\pi = 1$ and answering $+\pi$, or differentiating the factors separately to get $\cos x$ alone.

Takeaway (📌):
Product rule needs both terms. Then exact values at $0$, $\tfrac{\pi}{2}$ and $\pi$ usually collapse one of them to zero.

Correct Answer: C. $27\ \text{cm}^{2}$

Fastest Approach (🚀):
$\dfrac{dV}{dx} = 3x^{2} = 3(9) = 27\ \text{cm}^{2}$.

Matches Option C.

Step-by-Step Breakdown:

1. Write the volume

$V = x^{3}$

2. Differentiate with respect to x

$\dfrac{dV}{dx} = 3x^{2}$

3. Substitute

At $x = 3$:
$\dfrac{dV}{dx} = 3(3)^{2} = 3\times 9 = 27\ \text{cm}^{2}$

4. Check the units and meaning

A volume (cm³) differentiated with respect to a length (cm) gives cm² — an area, as expected. It equals half the surface area $6x^{2}$, because growing a cube pushes out only three of its six faces at once from a fixed corner.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $9\ \text{cm}^{2}$ — Omitted Factor
  Giving $x^{2}$ without the factor 3.
• B. $54\ \text{cm}^{2}$ — Formula Confusion
  Giving the surface area $6x^{2}$.
• D. $3\ \text{cm}^{2}$ — Misread Question
  Giving $x$ itself.
• E. $81\ \text{cm}^{2}$ — Index Error
  Using $3x^{3}$ or $x^{4}$ style index errors.

Common Mistake (⚠️):
Giving the volume $27\ \text{cm}^{3}$ by coincidence of value, or using the surface area $54$ instead of the derivative.

Takeaway (📌):
Differentiating volume with respect to length yields an area. Check the units — they confirm which quantity you have found.

Correct Answer: D. $6$

Fastest Approach (🚀):
Odd term vanishes.
$2\int_{0}^{1}\left(5x^{4}+2\right)dx = 2\left[x^{5}+2x\right]_{0}^{1} = 2(3) = 6$.

Matches Option D.

Step-by-Step Breakdown:

1. Classify the terms

$5x^{4}$ — even. $\quad -3x$ — odd. $\quad 2$ — even.

2. Kill the odd term

$\int_{-1}^{1}(-3x)\,dx = 0$

3. Double the even part

$\int_{-1}^{1}\left(5x^{4}+2\right)dx = 2\int_{0}^{1}\left(5x^{4}+2\right)dx = 2\left[x^{5}+2x\right]_{0}^{1}$

4. Evaluate

$= 2\left(1+2\right) = 6$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $3$ — Half-interval Error
  Integrating over the half-interval without doubling.
• B. $4$ — Incomplete Doubling
  Doubling only one of the two even terms.
• C. $2$ — Omitted Term
  Keeping only the constant term.
• E. $0$ — Over-generalisation
  Assuming the whole integrand is odd.

Common Mistake (⚠️):
Computing the half-interval and forgetting to double, giving $3$; or assuming the whole integral vanishes.

Takeaway (📌):
Odd terms give zero, even terms give twice the half-interval. Classify every term before integrating any of them.