ESAT Advanced Mathematics Mock Module 2 Worked Solutions

Correct Answer: D. $42$

Fastest Approach (🚀):
Odd terms vanish.
$\displaystyle\int_{-3}^{3}7\,dx = 7\times 6 = 42$.

Matches Option D.

Step-by-Step Breakdown:

1. Classify each term

$x^{5}$ is odd, $-2x$ is odd, and the constant $7$ is even.

2. Kill the odd terms

Over limits symmetric about zero every odd term contributes nothing:
$\int_{-3}^{3}x^{5}dx = 0, \qquad \int_{-3}^{3}(-2x)\,dx = 0$

3. Integrate what is left

$\int_{-3}^{3}7\,dx = 7\left[x\right]_{-3}^{3} = 7\times 6 = 42$

Geometrically that is a rectangle of height $7$ and width $6$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $63$ — Limits Error
  Using a width of $9$ rather than $6$.
• B. $21$ — Half-interval Error
  Integrating the constant over $[0,3]$ only, without doubling.
• C. $0$ — Over-generalisation
  Assuming the whole integrand is odd.
• E. $84$ — Double Counting
  Doubling an already full-interval result.

Common Mistake (⚠️):
Concluding the whole integral is zero. Only the odd terms vanish — a constant is even and always survives.

Takeaway (📌):
Split the integrand by parity before integrating. What is left is usually a constant times the width.

Correct Answer: A. $n^{2}+2n-1$

Fastest Approach (🚀):
First differences $5, 7, 9, 11$; second differences $2 \implies a = 1$.
$u_n-n^{2}: 1, 3, 5, 7, 9 = 2n-1$.
$u_n = n^{2}+2n-1$.

Matches Option A.

Step-by-Step Breakdown:

1. Take differences

First differences: $5,\ 7,\ 9,\ 11$. Second differences: $2,\ 2,\ 2$.

2. Fix the leading coefficient

The second difference is $2a$, so $a = 1$.

3. Subtract the quadratic part

$u_{n}-n^{2}$: $1, 3, 5, 7, 9$ — the arithmetic sequence $2n-1$.

$u_{n} = n^{2}+2n-1$

Check $n=4$: $16+8-1 = 23$. Testing only $n=1$ would not have separated this from two of the distractors.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $n^{2}+4n-3$ — Insufficient Checking
  Fits $n=1$ but gives $9$ instead of $7$ at $n=2$.
• C. $2n^{2}-1$ — Coefficient Error
  Taking the second difference as a rather than 2a.
• D. $n^{2}+n$ — Insufficient Checking
  Fits $n=1$ only; the constant term was dropped.
• E. $2n^{2}+n-1$ — Coefficient Error
  Doubling the leading coefficient and patching the linear term.

Common Mistake (⚠️):
Using the second difference $2$ as the leading coefficient, or accepting a formula that only matches the first term.

Takeaway (📌):
Second difference $= 2a$, then subtract $an^{2}$ and the remainder is linear. Always test on the third or fourth term.

Correct Answer: C. $x = 3$

Fastest Approach (🚀):
$x(x+6) = 3^{3} = 27 \implies (x+9)(x-3) = 0$.
$\log_{3}x$ needs $x>0$, so reject $x = -9$.
$x = 3$.

Matches Option C.

Step-by-Step Breakdown:

1. Combine and remove the logarithm

$\log_{3}\left[x(x+6)\right] = 3 \implies x(x+6) = 27$

2. Solve the quadratic

$x^{2}+6x-27 = 0 \implies (x+9)(x-3) = 0$

3. Apply the domain

$\log_{3}x$ requires $x>0$, which rejects $x = -9$.

Check $x=3$: $\log_{3}3+\log_{3}9 = 1+2 = 3$. Correct.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $x = 3$ and $x = -9$ — Domain Neglect
  Giving both roots without applying the domain.
• B. $x = -9$ — Domain Neglect
  Choosing the phantom root.
• D. $x = 27$ — Omitted Term
  Setting $x = 3^{3}$ and ignoring the second logarithm.
• E. $x = 9$ — Solving Error
  Mis-factorising the quadratic.

Common Mistake (⚠️):
Quoting both roots. The phantom root satisfies the quadratic but not the original equation, because a logarithm of a negative number is undefined.

Takeaway (📌):
Combining logs is not reversible. Write the domain down first and the phantom root is discarded on sight.

Correct Answer: E. $\dfrac{2}{\sin\theta}$

Fastest Approach (🚀):
Numerator $= \sin^{2}\theta+1+2\cos\theta+\cos^{2}\theta = 2+2\cos\theta$.
$\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \dfrac{2}{\sin\theta}$.

Matches Option E.

Step-by-Step Breakdown:

1. Combine over a common denominator

$\dfrac{\sin^{2}\theta+(1+\cos\theta)^{2}}{\sin\theta\,(1+\cos\theta)}$

2. Expand and apply the identity

$\sin^{2}\theta+1+2\cos\theta+\cos^{2}\theta = 1+1+2\cos\theta = 2(1+\cos\theta)$

3. Cancel

$\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \dfrac{2}{\sin\theta}$

Check at $\theta = 90^{\circ}$: the original gives $1+1 = 2$, and $\dfrac{2}{\sin 90^{\circ}} = 2$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{2}{\cos\theta}$ — Cancellation Error
  Cancelling to $\cos\theta$ in the denominator.
• B. $2\tan\theta$ — Cancellation Error
  Cancelling $(1+\cos\theta)$ against $\sin\theta$ incorrectly.
• C. $\dfrac{1}{\sin\theta}$ — Omitted Factor
  Losing the factor of $2$.
• D. $2\cos\theta$ — Inversion Error
  Inverting the final fraction.

Common Mistake (⚠️):
Adding numerators and denominators separately, or cancelling $(1+\cos\theta)$ before it appears as a factor of the whole numerator.

Takeaway (📌):
When two fractions are reciprocals, combining them nearly always surfaces $\sin^{2}+\cos^{2}$. Expand, identity, cancel — in that order.

Correct Answer: C. $1$

Fastest Approach (🚀):
$f(2) = 2(8)-3(4)+2-5 = 16-12+2-5 = 1$.

Matches Option C.

Step-by-Step Breakdown:

1. State the remainder theorem

If $f(x)$ is divided by $(x-a)$, the remainder is $f(a)$.

Here the divisor is $(x-2)$, so $a = 2$. Note the sign: $(x-2)$ means substituting $+2$.

2. Substitute

$f(2) = 2(2)^{3}-3(2)^{2}+(2)-5$
$= 2(8)-3(4)+2-5$
$= 16-12+2-5$

3. Evaluate

$= 1$

4. Interpretation

The remainder is not zero, so $(x-2)$ is not a factor of this cubic — which is exactly what the factor theorem is: the special case where $f(a) = 0$.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $-5$ — Substitution Error
  Giving the constant term, that is $f(0)$.
• B. $5$ — Sign Error
  Sign error on the constant term when evaluating.
• D. $3$ — Arithmetic Error
  Arithmetic slip in $16-12+2-5$.
• E. $-1$ — Sign Error
  Substituting $x=-2$ and then simplifying incorrectly.

Common Mistake (⚠️):
Substituting $x = -2$ because of the minus sign in $(x-2)$. The divisor is zero at $x = +2$, and that is the value to use.

Takeaway (📌):
Remainder on division by $(x-a)$ is $f(a)$. The factor theorem is the same statement with remainder zero.

Correct Answer: B. $(x-4)^{2}+(y-6)^{2} = 25$

Fastest Approach (🚀):
Centre $= \left(\tfrac{1+7}{2},\tfrac{2+10}{2}\right) = (4,6)$.
Diameter $= \sqrt{6^{2}+8^{2}} = 10 \implies r = 5$.
$(x-4)^{2}+(y-6)^{2} = 25$.

Matches Option B.

Step-by-Step Breakdown:

1. Centre is the midpoint

$C = \left(\dfrac{1+7}{2},\ \dfrac{2+10}{2}\right) = (4,6)$

2. Radius is half the diameter

$\Delta x = 6,\ \Delta y = 8$, a $(6,8,10)$ triple:
$\text{diameter} = 10 \implies r = 5$

3. Write the equation

$(x-h)^{2}+(y-k)^{2} = r^{2}$ with $h=4$, $k=6$, $r=5$:
$(x-4)^{2}+(y-6)^{2} = 25$

4. Check an endpoint

At $(1,2)$: $(1-4)^{2}+(2-6)^{2} = 9+16 = 25$. The point lies on the circle. Correct.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $(x-4)^{2}+(y-6)^{2} = 100$ — Radius Error
  Using the diameter 10 as the radius, giving 100 on the right.
• C. $(x-1)^{2}+(y-2)^{2} = 25$ — Centre Error
  Using one endpoint as the centre.
• D. $(x+4)^{2}+(y+6)^{2} = 25$ — Sign Error
  Sign error in the bracketed centre coordinates.
• E. $(x-4)^{2}+(y-6)^{2} = 10$ — Radius Error
  Using $r = \sqrt{10}$ or writing the diameter rather than $r^{2}$.

Common Mistake (⚠️):
Using the diameter as the radius and writing $r^{2} = 100$, or forgetting that the equation contains $r^{2}$ rather than $r$.

Takeaway (📌):
Given a diameter: midpoint for the centre, half the distance for the radius, and remember to square it in the equation.

Correct Answer: A. $-2 < x < 3$

Fastest Approach (🚀):
$-5 < 2x-1 < 5$.
Add 1: $-4 < 2x < 6$.
Divide by 2: $-2 < x < 3$.

Matches Option A.

Step-by-Step Breakdown:

1. Remove the modulus

$|A| < k$ (with $k>0$) means $A$ lies within $k$ of zero:
$-5 < 2x-1 < 5$

2. Add 1 throughout

$-5+1 < 2x < 5+1 \implies -4 < 2x < 6$

Every operation applies to all three parts of the inequality.

3. Divide by 2

$-2 < x < 3$

Dividing by a positive number leaves the directions unchanged.

4. Check the endpoints

At $x=3$: $|6-1| = 5$, not less than 5 — correctly excluded. At $x=0$: $|-1| = 1 < 5$ — correctly included.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $x < 3$ — Incomplete Answer
  Solving only the upper branch.
• C. $-3 < x < 2$ — Sign Error
  Sign error when adding 1, giving the interval reflected.
• D. $x < -2$ or $x > 3$ — Inequality Reversal
  Using the rule for $|A| > k$ instead.
• E. $0 < x < 3$ — Conceptual Error
  Assuming $x$ must be positive because of the modulus.

Common Mistake (⚠️):
Writing $|A| < k$ as two separate 'or' branches. That is the rule for $|A| > k$; a less-than modulus always gives a single interval.

Takeaway (📌):
$|A| < k$ gives one interval ($-k<A<k$); $|A| > k$ gives two branches. The direction of the inequality decides the shape of the answer.

Correct Answer: B. $5$

Fastest Approach (🚀):
$y = 2x^{2}-3x+c$.
$(1,2)$: $2-3+c = 2 \implies c = 3$.
$y(2) = 8-6+3 = 5$.

Matches Option B.

Step-by-Step Breakdown:

1. Integrate

$y = \int(4x-3)\,dx = 2x^{2}-3x+c$

2. Use the given point

$2 = 2(1)-3(1)+c = -1+c \implies c = 3$

3. Evaluate the curve

$y = 2x^{2}-3x+3$, so
$y(2) = 8-6+3 = 5$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $2$ — Missing Constant
  Taking $c = 0$.
• C. $8$ — Incomplete Answer
  Using $2x^{2}$ alone at $x=2$.
• D. $3$ — Misread Question
  Giving the constant itself.
• E. $11$ — Conceptual Error
  Substituting $x=2$ into the derivative and adding $c$.

Common Mistake (⚠️):
Assuming $c = 0$ and answering $2$. Without the constant the curve does not pass through the given point at all.

Takeaway (📌):
Every indefinite integration carries $+c$, and a supplied point is always there to pin it down.

Correct Answer: D. $60$

Fastest Approach (🚀):
$\binom{6}{2} = 15$ and $(-2)^{2} = 4$.
$15\times 4 = 60$.

Matches Option D.

Step-by-Step Breakdown:

1. Write the general term

With $u = -2x$ and $n = 6$:
$T_{k} = \binom{6}{k}(-2x)^{k} = \binom{6}{k}(-2)^{k}x^{k}$

2. Select k = 2

$\binom{6}{2} = 15$ and $(-2)^{2} = +4$ — the square makes it positive.

3. Multiply

$15\times 4 = 60$

For contrast the $x^{3}$ coefficient is $\binom{6}{3}(-2)^{3} = -160$, negative because the index is odd.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $-60$ — Sign Error
  Applying the minus sign without squaring it.
• B. $15$ — Omitted Factor
  Giving $\binom{6}{2}$ alone.
• C. $-30$ — Power Ignored
  Using $(-2)$ rather than $(-2)^{2}$.
• E. $30$ — Power Ignored
  Multiplying by $2$ instead of $4$.

Common Mistake (⚠️):
Carrying the minus sign through regardless of the index and answering $-60$, or forgetting to raise the $2$ to the power at all.

Takeaway (📌):
$(-a)^{k}$ is positive for even $k$ and negative for odd $k$. Bracket the whole term and the sign looks after itself.

Correct Answer: B. $8$

Fastest Approach (🚀):
$u = \log_{2}x \implies u^{2}-3u+2 = (u-1)(u-2) = 0$.
$x = 2^{1} = 2$ and $x = 2^{2} = 4$.
Product $= 8$.

Matches Option B.

Step-by-Step Breakdown:

1. Substitute

Let $u = \log_{2}x$. The equation becomes an ordinary quadratic:
$u^{2}-3u+2 = 0$

2. Factorise

$(u-1)(u-2) = 0 \implies u = 1 \text{ or } u = 2$

3. Convert back

$\log_{2}x = 1 \implies x = 2$
$\log_{2}x = 2 \implies x = 4$

Both are positive, so both are valid — unlike a sum-of-logs equation, no root is lost to the domain here.

4. Answer the question asked

$2\times 4 = 8$

(Equivalently: the roots multiply to $2^{u_1+u_2} = 2^{3} = 8$, using the sum of the roots of the quadratic.)

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $6$ — Misread Question
  Adding the roots $2+4$ and adjusting.
• C. $3$ — Substitution Not Reversed
  Giving $u_{1}+u_{2} = 3$, the sum of the logarithms.
• D. $16$ — Solving Error
  Using $x = 4$ and $x = 4$ or squaring one root.
• E. $2$ — Substitution Not Reversed
  Giving the product of the $u$ values rather than the $x$ values.

Common Mistake (⚠️):
Giving the product of the $u$ values ($1\times 2 = 2$) rather than of the $x$ values, or stopping once $u$ is found.

Takeaway (📌):
When a logarithm appears squared, substitute. Then remember to convert every root back — the question is about $x$, not $u$.

Correct Answer: A. $15^{\circ}$

Fastest Approach (🚀):
$2x \in [0^\circ,720^\circ)$.
$\sin 2x = \tfrac12 \implies 2x = 30^\circ,150^\circ,390^\circ,510^\circ$.
$x = 15^\circ,75^\circ,195^\circ,255^\circ$. Smallest: $15^\circ$.

Matches Option A.

Step-by-Step Breakdown:

1. Transform the range first

$0^{\circ}\le x < 360^{\circ} \implies 0^{\circ}\le 2x < 720^{\circ}$

Doubling the range before solving is what stops solutions going missing.

2. Solve for the whole angle

$\sin\theta = \tfrac12$ has principal value $\theta = 30^{\circ}$, and sine is positive in the first and second quadrants:
$\theta = 30^{\circ},\ 150^{\circ}$

Adding $360^{\circ}$ to each keeps them inside $720^{\circ}$:
$\theta = 30^{\circ},\ 150^{\circ},\ 390^{\circ},\ 510^{\circ}$

3. Divide by 2

$x = 15^{\circ},\ 75^{\circ},\ 195^{\circ},\ 255^{\circ}$

All four lie in the original range — a multiple angle of 2 always yields twice as many solutions.

4. Answer the question

The smallest positive solution is $15^{\circ}$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $30^{\circ}$ — Multiple Angle Ignored
  Solving $\sin x = \tfrac12$ and ignoring the double angle.
• C. $75^{\circ}$ — Misread Question
  Giving the second solution rather than the smallest.
• D. $150^{\circ}$ — Variable Confusion
  Giving $150^\circ$, a solution for $2x$ rather than for $x$.
• E. $195^{\circ}$ — Misread Question
  Giving the third solution in the list.

Common Mistake (⚠️):
Solving $\sin x = \tfrac12$ and answering $30^\circ$, ignoring the double angle entirely.

Takeaway (📌):
For $\sin kx$, widen the range by the factor $k$, solve for $kx$, then divide. Expect $k$ times as many solutions.

Correct Answer: B. $7680$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = 5\left(3x^{2}+1\right)^{4}\times 6x = 30x\left(3x^{2}+1\right)^{4}$.
At $x=1$: $30\times 4^{4} = 7680$.

Matches Option B.

Step-by-Step Breakdown:

1. Apply the chain rule

With $u = 3x^{2}+1$ and $y = u^{5}$:
$\dfrac{dy}{du} = 5u^{4}, \qquad \dfrac{du}{dx} = 6x$

2. Multiply

$\dfrac{dy}{dx} = 30x\left(3x^{2}+1\right)^{4}$

3. Substitute

$3(1)^{2}+1 = 4$, so $\dfrac{dy}{dx} = 30\times 4^{4} = 30\times 256 = 7680$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $1280$ — Chain Rule Omitted
  Forgetting the inner derivative: $5\times 4^{4}$.
• C. $1024$ — Index Error
  Using $4^{5}$ or mis-tracking the power.
• D. $5120$ — Coefficient Error
  Using $20x$ instead of $30x$.
• E. $320$ — Index Error
  Reducing the bracket's power too far.

Common Mistake (⚠️):
Omitting the inner derivative $6x$ and answering $1280$, or expanding the bracket to the fifth power first.

Takeaway (📌):
Power down, bracket untouched, times the inner derivative. Never expand a bracket raised to a power.

Correct Answer: D. $7$

Fastest Approach (🚀):
$S_{n} = \dfrac{3(2^{n}-1)}{2-1} = 3\left(2^{n}-1\right)$.
$3\left(2^{n}-1\right) = 381 \implies 2^{n}-1 = 127 \implies 2^{n} = 128$.
$n = 7$.

Matches Option D.

Step-by-Step Breakdown:

1. Write the sum formula

For $r \ne 1$:
$S_{n} = \dfrac{a\left(r^{n}-1\right)}{r-1}$

With $a=3$, $r=2$ the denominator is 1:
$S_{n} = 3\left(2^{n}-1\right)$

2. Set equal to the target

$3\left(2^{n}-1\right) = 381 \implies 2^{n}-1 = 127$

3. Match the power

$2^{n} = 128 = 2^{7} \implies n = 7$

4. Check

$3+6+12+24+48+96+192 = 381$. Seven terms exactly.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $5$ — Arithmetic Error
  Solving $2^{n} = 32$ after dividing by the wrong factor.
• B. $6$ — Off-by-one Error
  Matching $2^{n} = 64$ — one power short.
• C. $8$ — Off-by-one Error
  Adding one to the correct index, as if the exponent were $n-1$.
• E. $9$ — Counting Error
  Listing terms and miscounting past the target.

Common Mistake (⚠️):
Forgetting the $-1$ inside the bracket and solving $2^{n} = 127$, which has no integer solution — a clear signal that a term has been dropped.

Takeaway (📌):
$S_n = \dfrac{a(r^{n}-1)}{r-1}$ for $r>1$. Isolate the power completely before matching indices.

Correct Answer: E. $\dfrac{1}{16}$

Fastest Approach (🚀):
$\dfrac{20}{5} = 4$ half-lives.
$\left(\dfrac12\right)^{4} = \dfrac{1}{16}$.

Matches Option E.

Step-by-Step Breakdown:

1. Write the decay model

$N = N_{0}\left(\dfrac12\right)^{t/T}$ where $T$ is the half-life.

2. Count the half-lives

$\dfrac{t}{T} = \dfrac{20}{5} = 4$

3. Apply the halving

$\dfrac{N}{N_{0}} = \left(\dfrac12\right)^{4} = \dfrac{1}{16}$

4. Check by stepping through

After 5 years $\tfrac12$, after 10 years $\tfrac14$, after 15 years $\tfrac18$, after 20 years $\tfrac{1}{16}$. Four steps, as counted.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{4}$ — Counting Error
  Counting two half-lives, or dividing $20$ by $5$ and using it as a denominator.
• B. $\dfrac{1}{20}$ — Model Error
  Treating the decay as linear over 20 years.
• C. $\dfrac{1}{8}$ — Off-by-one Error
  Counting three half-lives instead of four.
• D. $\dfrac{1}{32}$ — Off-by-one Error
  Counting five half-lives.

Common Mistake (⚠️):
Treating the decay as linear — assuming four half-lives removes everything, or dividing 20 by 5 and answering $\tfrac14$.

Takeaway (📌):
Half-life questions are 'how many halvings'. Exponential decay never reaches zero, however many half-lives pass.

Correct Answer: E. $\dfrac{35\sqrt{3}}{4}$

Fastest Approach (🚀):
$\tfrac12\times 5\times 7 = \tfrac{35}{2}$.
$\times \sin 60^\circ = \times\dfrac{\sqrt3}{2} \implies \dfrac{35\sqrt3}{4}$.

Matches Option E.

Step-by-Step Breakdown:

1. Choose the right formula

Two sides and the included angle:
$\text{Area} = \dfrac12 ab\sin C$

The formula requires the angle to lie between the two sides — it does here.

2. Substitute

$\text{Area} = \dfrac12(5)(7)\sin 60^{\circ}$

3. Use the exact value

$\sin 60^{\circ} = \dfrac{\sqrt3}{2}$, so
$\text{Area} = \dfrac{35}{2}\times\dfrac{\sqrt3}{2} = \dfrac{35\sqrt3}{4}$

4. Sanity check the size

$\dfrac{35\sqrt3}{4}\approx 15.2$, comfortably less than the $\tfrac12(5)(7) = 17.5$ that a right angle would give. Sensible, since $\sin 60^\circ < 1$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{35}{2}$ — Value Error
  Using $\sin 60^\circ = 1$, that is treating the angle as a right angle.
• B. $\dfrac{35\sqrt{3}}{2}$ — Omitted Factor
  Dropping the $\tfrac12$ from the area formula.
• C. $\dfrac{35}{4}$ — Value Error
  Using $\sin 60^\circ = \tfrac12$.
• D. $35\sqrt{3}$ — Omitted Factor
  Dropping both halves.

Common Mistake (⚠️):
Forgetting the factor of $\tfrac12$ in $\sin 60^\circ = \tfrac{\sqrt3}{2}$ and answering $\tfrac{35\sqrt3}{2}$, or using $\tfrac12 \times 5\times 7$ as though the angle were $90^\circ$.

Takeaway (📌):
$\tfrac12 ab\sin C$ needs the included angle. Combined with exact values for $30/45/60$, these questions never need a calculator.

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

Fastest Approach (🚀):
Intersections at $x = \pm 2$.
$2\left[4x-\tfrac{x^{3}}{3}\right]_{0}^{2} = 2\left(8-\tfrac{8}{3}\right) = \tfrac{32}{3}$.

Matches Option E.

Step-by-Step Breakdown:

1. Find the limits

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

2. Decide which is on top

At $x=0$ the line gives $4$ and the curve gives $0$, so the line is above.

3. Exploit the symmetry and evaluate

$A = 2\int_{0}^{2}\left(4-x^{2}\right)dx = 2\left[4x-\dfrac{x^{3}}{3}\right]_{0}^{2} = 2\left(8-\dfrac83\right) = \dfrac{32}{3}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{16}{3}$ — Half-interval Error
  Integrating over half the region and not doubling.
• B. $16$ — Setup Error
  Using the rectangle $4\times 4$ without subtracting the parabola.
• C. $8$ — Estimation Error
  Halving the rectangle area as an estimate.
• D. $\dfrac{8}{3}$ — Wrong Integrand
  Integrating $x^{2}$ alone over $[0,2]$.

Common Mistake (⚠️):
Integrating only from $0$ to $2$ and forgetting to double, or integrating $x^{2}-4$ and reporting a negative area.

Takeaway (📌):
Symmetric regions: integrate the half and double. It halves the arithmetic and removes the negative limit entirely.

Correct Answer: C. $6$

Fastest Approach (🚀):
$S_{20} = \dfrac{20}{2}\left[2(5)+19d\right] = 10\left[10+19d\right]$.
$10+19d = 124 \implies 19d = 114 \implies d = 6$.

Matches Option C.

Step-by-Step Breakdown:

1. Write the sum formula

$S_{n} = \dfrac{n}{2}\left[2a+(n-1)d\right]$

2. Substitute the known values

$a = 5$, $n = 20$, $S_{20} = 1240$:
$1240 = \dfrac{20}{2}\left[10+19d\right] = 10\left[10+19d\right]$

Note $n-1 = 19$, not 20 — the first term contributes no common difference.

3. Solve the linear equation

$10+19d = 124 \implies 19d = 114 \implies d = 6$

4. Check via the last term

$u_{20} = 5+19(6) = 119$, and $S_{20} = \dfrac{20}{2}(5+119) = 10(124) = 1240$. Confirmed.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $5$ — Substitution Error
  Guessing from the first term, or using $20d$ and rounding.
• B. $4$ — Formula Misuse
  Using $S_{20} = 20\left[10+19d\right]$ — forgetting the factor of a half.
• D. $12$ — Solving Error
  Solving for $2d$ rather than $d$.
• E. $3$ — Off-by-one Error
  Using $n-1 = 20$ and rounding the result.

Common Mistake (⚠️):
Using $(n)d = 20d$ instead of $(n-1)d = 19d$, which gives a non-integer common difference — usually the first sign something is wrong.

Takeaway (📌):
$n-1$ steps separate $n$ terms. If the common difference comes out ugly, check that index before checking the arithmetic.

Correct Answer: B. $7$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = 2x+3$, so at $x=2$ the gradient is $7$.
First principles confirms it: the difference quotient simplifies to $7+h$.

Matches Option B.

Step-by-Step Breakdown:

1. Write the difference quotient

$\dfrac{f(2+h)-f(2)}{h}$ with $f(x) = x^{2}+3x$.

$f(2) = 4+6 = 10$
$f(2+h) = (2+h)^{2}+3(2+h) = 4+4h+h^{2}+6+3h = 10+7h+h^{2}$

2. Subtract and cancel

$f(2+h)-f(2) = 7h+h^{2} = h(7+h)$

The constant terms always cancel — if they do not, the expansion is wrong.

$\dfrac{h(7+h)}{h} = 7+h \qquad (h \ne 0)$

3. Take the limit

$\lim_{h\to 0}(7+h) = 7$

4. Check against the standard rule

$\dfrac{dy}{dx} = 2x+3$, and at $x=2$ that is $4+3 = 7$. Agreed.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $4$ — Omitted Term
  Giving $2x$ at $x=2$ and forgetting the $+3$ from the linear term.
• C. $10$ — Misread Question
  Giving $f(2) = 10$, the $y$-value rather than the gradient.
• D. $3$ — Omitted Term
  Giving only the derivative of the $3x$ term.
• E. $2$ — Misread Question
  Giving the $x$-coordinate.

Common Mistake (⚠️):
Substituting $h=0$ before cancelling, which gives $\tfrac00$. The cancellation must happen first — that is the entire technique.

Takeaway (📌):
First principles: expand, cancel the constant terms, factor out $h$, then let $h\to 0$. Always sanity-check against the standard derivative.

Correct Answer: D. $40$

Fastest Approach (🚀):
$\binom{5}{3} = 10,\quad 2^{5-3} = 4$.
$10\times 4 = 40$.

Matches Option D.

Step-by-Step Breakdown:

1. Write the general term

$(a+b)^{n}$ has general term $\binom{n}{k}a^{\,n-k}b^{\,k}$. With $a=2$, $b=x$, $n=5$:
$T_{k} = \binom{5}{k}2^{\,5-k}x^{k}$

2. Select the term wanted

$x^{3}$ needs $k=3$.

3. Evaluate the coefficient

$\binom{5}{3} = 10$ and $2^{\,5-3} = 2^{2} = 4$, so
$\text{coefficient} = 10\times 4 = 40$

4. Interpret it combinatorially

Choosing $x$ from three of the five brackets can be done in $\binom53 = 10$ ways, and the remaining two brackets each contribute a 2. That is where the factor of 4 comes from.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $10$ — Omitted Factor
  Giving $\binom{5}{3}$ alone, without the power of 2.
• B. $80$ — Index Error
  Using $2^{3}$ instead of $2^{2}$ — the indices swapped.
• C. $20$ — Formula Misuse
  Using $\binom{5}{2}\times 2$ or another mismatched pairing.
• E. $32$ — Misread Question
  Giving $2^{5}$, the constant term of the expansion.

Common Mistake (⚠️):
Forgetting the $2^{2}$ and answering $10$, or using $2^{3}$ by attaching the power of 2 to the same index as $x$.

Takeaway (📌):
The two indices always sum to $n$. If $x$ carries $k$, the constant carries $n-k$.

Correct Answer: D. $-\dfrac{1}{8}$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = 3x^{2}-4$. At $x=2$: $12-4 = 8$.
Normal $= -\dfrac{1}{8}$.

Matches Option D.

Step-by-Step Breakdown:

1. Differentiate

$y = x^{3}-4x \implies \dfrac{dy}{dx} = 3x^{2}-4$

2. Evaluate the tangent gradient

At $x=2$:
$m_{\text{tangent}} = 3(4)-4 = 12-4 = 8$

3. Take the negative reciprocal

Tangent and normal are perpendicular, so their gradients multiply to $-1$:
$m_{\text{normal}} = -\dfrac{1}{m_{\text{tangent}}} = -\dfrac{1}{8}$

4. Note what was not needed

The point itself is $(2,0)$, but the question asks only for a gradient — no equation of a line is required.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $8$ — Misread Question
  Giving the tangent gradient rather than the normal gradient.
• B. $-8$ — Formula Misuse
  Negating the tangent gradient instead of inverting it as well.
• C. $\dfrac{1}{8}$ — Sign Error
  Taking the reciprocal but dropping the minus sign.
• E. $0$ — Misread Question
  Giving the $y$-coordinate at $x=2$, which happens to be 0.

Common Mistake (⚠️):
Giving the tangent gradient $8$, or negating it to $-8$ instead of taking the negative reciprocal.

Takeaway (📌):
Tangent gradient from the derivative; normal gradient is $-1/m$. Negating is not the same as inverting.

Correct Answer: A. $\dfrac{2}{3}$

Fastest Approach (🚀):
$1-r = \dfrac{a}{S_{\infty}} = \dfrac{9}{27} = \dfrac13$.
$r = 1-\dfrac13 = \dfrac23$.

Matches Option A.

Step-by-Step Breakdown:

1. Write the formula

$S_{\infty} = \dfrac{a}{1-r} \qquad (|r|<1)$

2. Substitute and rearrange

$27 = \dfrac{9}{1-r} \implies 1-r = \dfrac{9}{27} = \dfrac{1}{3}$

3. Solve

$r = 1-\dfrac13 = \dfrac{2}{3}$

4. Check convergence and the sum

$\left|\tfrac23\right|<1$, so the series converges. And $\dfrac{9}{1-\tfrac23} = \dfrac{9}{\tfrac13} = 27$. Confirmed.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\dfrac{1}{3}$ — Incomplete Answer
  Giving $1-r = \tfrac13$ as the answer rather than solving for $r$.
• C. $3$ — Inversion Error
  Inverting the ratio: using $\dfrac{S_\infty}{a} = 3$ as $r$.
• D. $\dfrac{1}{2}$ — Estimation Error
  Guessing a convergent-looking value without solving.
• E. $-\dfrac{2}{3}$ — Sign Error
  Sign error when rearranging $1-r$.

Common Mistake (⚠️):
Solving $1-r = 3$ by inverting the fraction the wrong way, which gives $r = -2$ — outside the convergence range and therefore impossible.

Takeaway (📌):
$1-r = \dfrac{a}{S_\infty}$ is the rearrangement worth remembering. Any answer with $|r|\ge 1$ is automatically wrong for a convergent series.

Correct Answer: C. $3e$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = e^{x}\left(x^{2}+2x\right)$.
At $x=1$: $e(1+2) = 3e$.

Matches Option C.

Step-by-Step Breakdown:

1. Apply the product rule

With $u = x^{2}$ and $v = e^{x}$:
$\dfrac{dy}{dx} = u'v+uv' = 2x\,e^{x}+x^{2}e^{x}$

Recall $\dfrac{d}{dx}e^{x} = e^{x}$ — the exponential is its own derivative.

2. Factor out the exponential

$\dfrac{dy}{dx} = e^{x}\left(2x+x^{2}\right)$

Factoring first keeps the substitution to one line and makes stationary points obvious, since $e^{x} \ne 0$ ever.

3. Substitute x = 1

$e^{1}\left(2+1\right) = 3e$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $e$ — Omitted Term
  Keeping only the $x^{2}e^{x}$ term.
• B. $2e$ — Omitted Term
  Keeping only the $2xe^{x}$ term — the product rule half-applied.
• D. $e^{2}$ — Formula Misuse
  Multiplying the derivatives: $2x\cdot e^{x}$ evaluated as $e^{2}$.
• E. $0$ — Conceptual Error
  Assuming a stationary point at $x=1$.

Common Mistake (⚠️):
Differentiating the two factors separately and multiplying: $2x\times e^{x} = 2xe^{x}$ only, which misses the second product-rule term.

Takeaway (📌):
$\dfrac{d}{dx}\left[f(x)e^{x}\right] = e^{x}\left[f(x)+f'(x)\right]$. Factoring out $e^{x}$ makes every follow-up part easier.

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

Fastest Approach (🚀):
$4-x^{2} = x+2 \implies x^{2}+x-2 = 0 \implies x = -2, 1$.
$\int_{-2}^{1}\left(2-x-x^{2}\right)dx = \left[2x-\tfrac{x^{2}}{2}-\tfrac{x^{3}}{3}\right]_{-2}^{1} = \tfrac76-\left(-\tfrac{10}{3}\right) = \tfrac92$.

Matches Option B.

Step-by-Step Breakdown:

1. Find the intersections

$4-x^{2} = x+2 \implies 0 = x^{2}+x-2 = (x+2)(x-1)$
$x = -2 \quad\text{and}\quad x = 1$

2. Decide which is on top

At $x=0$: the curve gives $4$, the line gives $2$. The parabola is above throughout.

3. Integrate the difference

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

$= \left[2x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{-2}^{1}$

4. Evaluate carefully at both limits

At $x=1$: $2-\dfrac12-\dfrac13 = \dfrac{7}{6}$
At $x=-2$: $-4-2+\dfrac83 = -\dfrac{10}{3}$

$A = \dfrac76-\left(-\dfrac{10}{3}\right) = \dfrac76+\dfrac{20}{6} = \dfrac{27}{6} = \dfrac{9}{2}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{7}{2}$ — Sign Error
  Sign slip at the lower limit, losing a term.
• C. $\dfrac{27}{2}$ — Arithmetic Error
  Using $\left[\ldots\right]$ evaluated as $\tfrac{27}{2}$ by omitting the division by 3.
• D. $9$ — Setup Error
  Doubling the correct area, or integrating only the curve.
• E. $\dfrac{5}{2}$ — Limits Error
  Using limits $0$ and $1$ instead of $-2$ and $1$.

Common Mistake (⚠️):
Sign errors when substituting the negative limit, particularly $-\dfrac{(-2)^{3}}{3} = +\dfrac83$. Two negatives are involved and both must survive.

Takeaway (📌):
Upper minus lower, integrated between the intersections. Evaluate each limit separately and write both values down before subtracting.

Correct Answer: D. $6x\cos\left(3x^{2}\right)$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = \cos\left(3x^{2}\right)\times 6x = 6x\cos\left(3x^{2}\right)$.

Matches Option D.

Step-by-Step Breakdown:

1. Identify the layers

Let $u = 3x^{2}$, so $y = \sin u$.

2. Differentiate each layer

$\dfrac{dy}{du} = \cos u, \qquad \dfrac{du}{dx} = 6x$

3. Multiply

$\dfrac{dy}{dx} = \dfrac{dy}{du}\times\dfrac{du}{dx} = \cos\left(3x^{2}\right)\times 6x = 6x\cos\left(3x^{2}\right)$

The inside of the cosine is unchanged — only the multiplier is new.

4. Check the special case

If the inside were simply $x$, the multiplier would be 1 and the derivative would reduce to $\cos x$, as it must.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\cos\left(3x^{2}\right)$ — Chain Rule Omitted
  Forgetting the inner derivative entirely.
• B. $6x\sin\left(3x^{2}\right)$ — Derivative Error
  Keeping $\sin$ instead of differentiating it to $\cos$.
• C. $3x^{2}\cos\left(3x^{2}\right)$ — Chain Rule Misuse
  Multiplying by the inside function rather than its derivative.
• E. $\cos(6x)$ — Bracket Error
  Differentiating inside the bracket, changing $3x^{2}$ to $6x$ within the cosine.

Common Mistake (⚠️):
Forgetting the inner derivative and answering $\cos\left(3x^{2}\right)$, or differentiating the inside of the bracket as well as outside it.

Takeaway (📌):
Chain rule: derivative of the outside (inside unchanged) times derivative of the inside. The bracket contents never change.

Correct Answer: A. $16$

Fastest Approach (🚀):
$y = 3x^{2}-4x+c$.
$(2,5)$: $12-8+c = 5 \implies c = 1$.
$y(3) = 27-12+1 = 16$.

Matches Option A.

Step-by-Step Breakdown:

1. Integrate

$y = \int (6x-4)\,dx = 3x^{2}-4x+c$

2. Use the point to find c

Substituting $(2,5)$:
$5 = 3(4)-4(2)+c = 12-8+c = 4+c \implies c = 1$

3. State the curve

$y = 3x^{2}-4x+1$

4. Evaluate at x = 3

$y(3) = 3(9)-4(3)+1 = 27-12+1 = 16$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $15$ — Omitted Constant
  Forgetting the constant of integration, leaving $27-12$.
• C. $14$ — Sign Error
  Using $c = -1$ from a sign slip.
• D. $27$ — Incomplete Answer
  Giving $3x^{2}$ at $x=3$ only.
• E. $19$ — Conceptual Error
  Substituting $x=3$ into $\dfrac{dy}{dx}$ and adding $c$.

Common Mistake (⚠️):
Omitting the constant of integration entirely, which gives $y(3) = 15$, or substituting $x=3$ into the derivative instead of the curve.

Takeaway (📌):
Indefinite integration always produces $+c$, and a stated point is always there to find it. No point given means no numerical answer is possible.

Correct Answer: C. $2$

Fastest Approach (🚀):
$\dfrac{dy}{dx} = 1-\dfrac{1}{x^{2}} = 0 \implies x^{2} = 1 \implies x = 1$ (since $x>0$).
$y(1) = 1+1 = 2$.

Matches Option C.

Step-by-Step Breakdown:

1. Rewrite with a negative index

$y = x+x^{-1}$

2. Differentiate

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

3. Solve for stationary points

$1-\dfrac{1}{x^{2}} = 0 \implies x^{2} = 1 \implies x = \pm 1$

The restriction $x>0$ leaves $x = 1$.

4. Confirm it is a minimum and evaluate

$\dfrac{d^{2}y}{dx^{2}} = 2x^{-3} = \dfrac{2}{x^{3}}$, which at $x=1$ is $+2 > 0$ — a minimum.
$y(1) = 1+\dfrac11 = 2$

(The discarded root $x=-1$ gives $y=-2$, a maximum on the negative branch — which is why the restriction matters.)

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0$ — Conceptual Error
  Setting $y=0$ rather than $\dfrac{dy}{dx}=0$.
• B. $1$ — Misread Question
  Giving the $x$-coordinate of the minimum rather than the value of $y$.
• D. $-2$ — Domain Error
  Using $x=-1$, which is excluded by the domain.
• E. $\dfrac12$ — Arithmetic Error
  Evaluating $y$ at $x=2$ or mis-simplifying the reciprocal.

Common Mistake (⚠️):
Ignoring $x>0$ and quoting $-2$, which is the maximum of the other branch, or differentiating $\tfrac1x$ to $\tfrac{1}{x^{2}}$ without the minus sign.

Takeaway (📌):
$\dfrac{d}{dx}x^{-1} = -x^{-2}$. Convert reciprocals to negative indices, and always check the stated domain before choosing a root.

Correct Answer: E. $\dfrac{7}{6}$

Fastest Approach (🚀):
$h = \dfrac{3-1}{2} = 1$; ordinates $y_{0}=1,\ y_{1}=\tfrac12,\ y_{2}=\tfrac13$.
$\dfrac{1}{2}\left[\left(1+\tfrac13\right)+2\left(\tfrac12\right)\right] = \dfrac12\left[\tfrac43+1\right] = \dfrac12\times\dfrac73 = \dfrac76$.

Matches Option E.

Step-by-Step Breakdown:

1. Find the strip width and the ordinates

Two strips across $[1,3]$:
$h = \dfrac{3-1}{2} = 1$

Ordinates at $x = 1, 2, 3$:
$y_{0} = 1, \qquad y_{1} = \dfrac12, \qquad y_{2} = \dfrac13$

2. Apply the rule

$\int_{a}^{b}y\,dx \approx \dfrac{h}{2}\left[y_{0}+y_{n}+2\left(y_{1}+\cdots+y_{n-1}\right)\right]$

Only $y_{1}$ is interior here, so only it is doubled.

3. Substitute

$\approx \dfrac{1}{2}\left[1+\dfrac13+2\left(\dfrac12\right)\right] = \dfrac12\left[\dfrac43+1\right] = \dfrac12\times\dfrac{7}{3} = \dfrac{7}{6}$

4. Judge the estimate

$\tfrac76 \approx 1.167$ against the true value $\ln 3 \approx 1.099$. The curve is convex, so the trapezia sit above it and the rule overestimates — exactly as expected.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{5}{6}$ — Weighting Error
  Doubling the ends instead of the middle, or omitting $y_{0}$.
• B. $\dfrac{11}{6}$ — Weighting Error
  Doubling every ordinate.
• C. $\dfrac{4}{3}$ — Strip Width Error
  Using $h=2$ and halving once too few times.
• D. $1$ — Strip Count Error
  Averaging the two end ordinates only — the one-strip estimate.

Common Mistake (⚠️):
Doubling the end ordinates as well as the middle one, or using $h = 2$ (the whole interval) instead of the strip width.

Takeaway (📌):
Ends once, middles twice, all multiplied by $\tfrac h2$. For a convex curve the trapezium rule always overestimates.