ESAT Advanced Mathematics Mock Module 3 Worked Solutions

Correct Answer: E. $\dfrac{1}{3}$

Fastest Approach (🚀):
$\int\cos 3x\,dx = \dfrac13\sin 3x$.
At $x=\tfrac{\pi}{6}$: $3x = \tfrac{\pi}{2}$, $\sin\tfrac{\pi}{2} = 1$.
$\dfrac13(1)-\dfrac13(0) = \dfrac13$.

Matches Option E.

Step-by-Step Breakdown:

1. Integrate

$\int\cos(kx)\,dx = \dfrac{1}{k}\sin(kx)+c$

With $k=3$:
$\int\cos 3x\,dx = \dfrac13\sin 3x+c$

2. Convert the limits

At $x = \dfrac{\pi}{6}$: $3x = \dfrac{\pi}{2}$
At $x = 0$: $3x = 0$

3. Evaluate

$\left[\dfrac13\sin 3x\right]_{0}^{\pi/6} = \dfrac13\sin\dfrac{\pi}{2}-\dfrac13\sin 0 = \dfrac13(1)-\dfrac13(0)$

4. Simplify

$= \dfrac{1}{3}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $1$ — Omitted Factor
  Forgetting the factor $\tfrac13$ entirely.
• B. $\dfrac{1}{2}$ — Limits Error
  Using $\sin\tfrac{\pi}{6} = \tfrac12$ without scaling the angle by 3.
• C. $3$ — Chain Rule Inverted
  Multiplying by 3 instead of dividing.
• D. $-\dfrac{1}{3}$ — Sign Error
  Applying the minus sign from $\int\sin$ to the cosine integral.

Common Mistake (⚠️):
Multiplying by 3 instead of dividing — integration divides by the inner coefficient, differentiation multiplies by it.

Takeaway (📌):
$\int\cos kx = \tfrac1k\sin kx$ and $\int\sin kx = -\tfrac1k\cos kx$. Note the minus belongs to the sine integral, not the cosine one.

Correct Answer: B. $x = 3$

Fastest Approach (🚀):
$2x = 6 \implies x = 3$.

Matches Option B.

Step-by-Step Breakdown:

1. Use the given information

$f(x) = 0$ at $x = 6$.

2. Set the new function to zero

$y = f(2x)$ is zero when its input is 6:
$2x = 6 \implies x = 3$

3. Interpret the transformation

$y = f(2x)$ is a horizontal stretch of scale factor $\tfrac12$ — the graph is squashed towards the $y$-axis, and every $x$-coordinate halves.

4. Contrast with the outside case

$y = 2f(x)$ would be a vertical stretch, doubling $y$-values and leaving the roots at $x=6$ unchanged. Inside the bracket affects $x$; outside affects $y$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $x = 12$ — Transformation Inverted
  Doubling instead of halving — reading $f(2x)$ as a stretch by factor 2.
• C. $x = 6$ — Inside/Outside Confusion
  Assuming the transformation does not move the roots, as for $y = 2f(x)$.
• D. $x = -6$ — Transformation Error
  Treating $f(2x)$ as a reflection.
• E. $x = \tfrac16$ — Transformation Error
  Taking the reciprocal of the root.

Common Mistake (⚠️):
Doubling the root to $x=12$, reading the 2 as a stretch rather than a compression. Transformations inside the bracket always act inversely.

Takeaway (📌):
Inside the bracket: acts on $x$, and does the opposite. Outside: acts on $y$, and does what it says.

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

Fastest Approach (🚀):
First differences: $5,7,9,11$. Second differences: $2 \implies a = 1$.
$u_{n}-n^{2}: 2,4,6,8 \implies 2n$.
$u_{n} = n^{2}+2n$.

Matches Option A.

Step-by-Step Breakdown:

1. Take differences

Sequence: $3,\ 8,\ 15,\ 24,\ 35$
First differences: $5,\ 7,\ 9,\ 11$
Second differences: $2,\ 2,\ 2$

Constant second differences confirm a quadratic $n$th term.

2. Find the leading coefficient

For $u_{n} = an^{2}+bn+c$, the second difference is $2a$:
$2a = 2 \implies a = 1$

3. Subtract the quadratic part

$u_{n}-n^{2}$ gives
$3-1 = 2,\quad 8-4 = 4,\quad 15-9 = 6,\quad 24-16 = 8$

That remainder is the arithmetic sequence $2n$, so $b = 2$ and $c = 0$.

4. State and verify

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

Check $n=5$: $25+10 = 35$. Correct.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $n^{2}+n+1$ — Insufficient Checking
  Fits $n=1$ but not $n=2$: gives $7$ rather than $8$.
• C. $2n^{2}+1$ — Coefficient Error
  Using the second difference $2$ as the coefficient $a$.
• D. $n^{2}+4$ — Insufficient Checking
  Fits $n=1$ only; the linear term was missed entirely.
• E. $3n^{2}$ — Insufficient Checking
  Matching the first term alone by scaling $n^{2}$.

Common Mistake (⚠️):
Using the second difference itself as the leading coefficient, giving $2n^{2}$, or checking only the first term — several of the distractors also give 3 at $n=1$.

Takeaway (📌):
Second difference $= 2a$. Subtract $an^{2}$ and what remains is always linear, which is then read off directly.

Correct Answer: C. $3$

Fastest Approach (🚀):
Root of the divisor: $x = \tfrac12$.
$f\!\left(\tfrac12\right) = 2\!\left(\tfrac18\right)-5\!\left(\tfrac14\right)+4 = \tfrac14-\tfrac54+4 = 3$.

Matches Option C.

Step-by-Step Breakdown:

1. Find the root of the divisor

$2x-1 = 0 \implies x = \dfrac12$

The theorem is about the root, never about the coefficients.

2. Substitute

$f\!\left(\tfrac12\right) = 2\!\left(\tfrac18\right)-5\!\left(\tfrac14\right)+4$

3. Evaluate

$= \dfrac14-\dfrac54+4 = -1+4 = 3$

Since the remainder is not zero, $(2x-1)$ is not a factor.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $1$ — Root Error
  Substituting $x=1$ from the constant in the divisor.
• B. $4$ — Substitution Error
  Giving $f(0)$, the constant term.
• D. $\dfrac{1}{2}$ — Misread Question
  Giving the root of the divisor rather than the remainder.
• E. $-3$ — Sign Error
  Sign slip when combining the fractions.

Common Mistake (⚠️):
Substituting $x = 2$ or $x = 1$ by reading the coefficients rather than solving $2x-1 = 0$.

Takeaway (📌):
Divide by $(ax-b)$ and the remainder is $f\!\left(\tfrac{b}{a}\right)$. Solve the divisor for zero every time.

Correct Answer: E. $1$

Fastest Approach (🚀):
$\log_{2}\left(8\times\tfrac14\right) = \log_{2}2 = 1$.

Matches Option E.

Step-by-Step Breakdown:

1. Combine using the product law

$\log_{b}A+\log_{b}B = \log_{b}(AB)$

$\log_{2}8+\log_{2}\dfrac14 = \log_{2}\left(8\times\dfrac14\right) = \log_{2}2$

2. Evaluate

$\log_{2}2 = 1$

3. The term-by-term route

$\log_{2}8 = 3$ because $2^{3} = 8$.
$\log_{2}\dfrac14 = -2$ because $2^{-2} = \dfrac14$.
$3+(-2) = 1$. Same answer — and note that a logarithm of a number below 1 is negative.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $5$ — Sign Error
  Taking $\log_{2}\tfrac14 = +2$ instead of $-2$.
• B. $-1$ — Operation Error
  Subtracting the logarithms: $\log_{2}\left(8\div\tfrac14\right)$ mishandled, or $2-3$.
• C. $\dfrac{3}{4}$ — Formula Misuse
  Dividing the arguments as a fraction rather than using log laws.
• D. $2$ — Substitution Error
  Evaluating $\log_{2}4$ instead.

Common Mistake (⚠️):
Treating $\log_2\tfrac14$ as positive $2$, giving $5$. Arguments between 0 and 1 always produce negative logarithms.

Takeaway (📌):
Sum of logs is the log of the product. Arguments below 1 give negative values — check the sign before adding.

Correct Answer: D. $-5 < c < 5$

Fastest Approach (🚀):
Substitute: $x^{2}+(2x+c)^{2} = 5 \implies 5x^{2}+4cx+c^{2}-5 = 0$.
$b^{2}-4ac = 16c^{2}-20\left(c^{2}-5\right) = 100-4c^{2} > 0$.
$c^{2} < 25 \implies -5 < c < 5$.

Matches Option D.

Step-by-Step Breakdown:

1. Substitute the line into the circle

$x^{2}+(2x+c)^{2} = 5$
$x^{2}+4x^{2}+4cx+c^{2} = 5$
$5x^{2}+4cx+\left(c^{2}-5\right) = 0$

2. Form the discriminant

$\Delta = b^{2}-4ac = (4c)^{2}-4(5)\left(c^{2}-5\right) = 16c^{2}-20c^{2}+100 = 100-4c^{2}$

3. Impose two distinct roots

Two distinct intersection points means two distinct real roots:
$100-4c^{2} > 0 \implies c^{2} < 25 \implies -5 < c < 5$

4. Interpret the boundary

$c = \pm 5$ gives $\Delta = 0$ — the tangent cases. Outside that interval the line misses the circle entirely.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $c < 5$ — Incomplete Answer
  Solving only the upper branch of the inequality.
• B. $c > -5$ — Incomplete Answer
  Solving only the lower branch.
• C. $c = \pm 5$ — Boundary Error
  Giving the tangent condition rather than the two-point condition.
• E. $c < -5$ or $c > 5$ — Inequality Reversal
  Reversing the inequality after $c^{2}<25$.

Common Mistake (⚠️):
Solving $\Delta = 0$ and giving $c = \pm 5$. That is the tangent condition; two distinct points requires the strict inequality.

Takeaway (📌):
$\Delta > 0$ two points, $\Delta = 0$ tangent, $\Delta < 0$ no intersection. Read which the question wants before solving.

Correct Answer: E. $\sin\theta$

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

Matches Option E.

Step-by-Step Breakdown:

1. Apply the identity

$\sin^{2}\theta+\cos^{2}\theta = 1 \implies 1-\cos^{2}\theta = \sin^{2}\theta$

2. Substitute

$\dfrac{1-\cos^{2}\theta}{\sin\theta} = \dfrac{\sin^{2}\theta}{\sin\theta}$

3. Cancel

$= \sin\theta \qquad (\sin\theta \ne 0)$

4. Check numerically

At $\theta = 30^{\circ}$: $\dfrac{1-\left(\tfrac{\sqrt3}{2}\right)^{2}}{\tfrac12} = \dfrac{1-\tfrac34}{\tfrac12} = \dfrac{\tfrac14}{\tfrac12} = \tfrac12 = \sin 30^{\circ}$. Correct.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\cos\theta$ — Identity Error
  Cancelling to a cosine by misapplying the identity.
• B. $\tan\theta$ — Identity Error
  Assuming the result is $\sin/\cos$.
• C. $\dfrac{1}{\sin\theta}$ — Cancellation Error
  Cancelling the numerator to 1.
• D. $\sin^{2}\theta$ — Incomplete Simplification
  Forgetting to divide by $\sin\theta$.

Common Mistake (⚠️):
Cancelling the $1$ against the $\sin\theta$, or reading $1-\cos^{2}\theta$ as $(1-\cos\theta)^{2}$.

Takeaway (📌):
Learn the identity in all three forms: $\sin^{2}=1-\cos^{2}$, $\cos^{2}=1-\sin^{2}$, and the sum equals 1.

Correct Answer: B. $128\ \text{cm}^{3}$

Fastest Approach (🚀):
$V = x(12-2x)^{2}$, valid for $0<x<6$.
$\dfrac{dV}{dx} = 12(x-2)(x-6) = 0 \implies x = 2$ or $x = 6$.
$x=6$ gives zero volume, so $x=2$: $V = 2\times 8^{2} = 128\ \text{cm}^{3}$.

Matches Option B.

Step-by-Step Breakdown:

1. Build the volume function

Cutting $x$ from each corner leaves a base of side $12-2x$ and a height of $x$:
$V = x(12-2x)^{2}$

The physical range is $0 < x < 6$ — beyond that the base has no width.

2. Expand and differentiate

$V = x\left(144-48x+4x^{2}\right) = 144x-48x^{2}+4x^{3}$
$\dfrac{dV}{dx} = 144-96x+12x^{2} = 12\left(x^{2}-8x+12\right) = 12(x-2)(x-6)$

3. Solve and filter

$x = 2$ or $x = 6$. At $x=6$ the base has side zero and the volume is zero, so it is the minimum, not the maximum. The valid stationary point is $x = 2$.

4. Evaluate the maximum

$V = 2\left(12-4\right)^{2} = 2\times 64 = 128\ \text{cm}^{3}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $108\ \text{cm}^{3}$ — Solving Error
  Using $x = 1$ or a rounded value rather than the true stationary point.
• C. $144\ \text{cm}^{3}$ — Setup Error
  Using base $12-2x$ times $12$ rather than squaring.
• D. $64\ \text{cm}^{3}$ — Setup Error
  Using $x=2$ but a base of $12-2x$ unsquared, or $4\times 16$.
• E. $216\ \text{cm}^{3}$ — Domain Error
  Using $x=6$ and mis-evaluating, or taking $\tfrac{12}{2}$ cubed.

Common Mistake (⚠️):
Taking $x = 6$ without checking the physical range, or forgetting to square $(12-2x)$ and treating the base as $12-2x$ by $12$.

Takeaway (📌):
Optimisation is three steps: one-variable expression, differentiate, then filter the stationary points against the physical domain.

Correct Answer: B. $3$

Fastest Approach (🚀):
$2x+5 = 11 \implies 2x = 6 \implies x = 3$.

Matches Option B.

Step-by-Step Breakdown:

1. Read what the inverse asks

$f^{-1}(11)$ is the value of $x$ for which $f(x) = 11$.

2. Solve directly

$2x+5 = 11 \implies 2x = 6 \implies x = 3$

3. Or derive the inverse in full

$y = 2x+5 \implies x = \dfrac{y-5}{2}$, so $f^{-1}(x) = \dfrac{x-5}{2}$ and $f^{-1}(11) = 3$. Same answer, more writing.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $27$ — Inverse Confusion
  Computing $f(11)$ rather than $f^{-1}(11)$.
• C. $8$ — Coefficient Ignored
  Solving $x+5 = 11$ and ignoring the coefficient.
• D. $\dfrac{1}{27}$ — Notation Error
  Treating the inverse function as a reciprocal.
• E. $16$ — Operation Error
  Computing $11+5$.

Common Mistake (⚠️):
Computing $f(11) = 27$ instead of $f^{-1}(11)$ — evaluating the function rather than inverting it.

Takeaway (📌):
$f^{-1}(k)$ means 'solve $f(x)=k$'. Deriving the full inverse is optional when only one value is wanted.

Correct Answer: D. $20$

Fastest Approach (🚀):
Sum of the first $n$ odd numbers $= n^{2}$.
$n^{2} = 400 \implies n = 20$.

Matches Option D.

Step-by-Step Breakdown:

1. Identify the series

$1, 3, 5, 7, \ldots$ is arithmetic with $a = 1$ and $d = 2$.

2. Apply the sum formula

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

The sum of the first $n$ odd numbers is exactly $n^{2}$ — a result worth memorising.

3. Solve

$n^{2} = 400 \implies n = 20$

($n = -20$ is rejected: a count of terms must be positive.)

4. Check

The 20th odd number is $2(20)-1 = 39$, and $\dfrac{20}{2}(1+39) = 10\times 40 = 400$. Confirmed.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $40$ — Term vs Count
  Giving the 20th odd number region, or using $2n$ instead of $n$.
• B. $100$ — Formula Misuse
  Using $S_n = 4n$ or dividing 400 by 4.
• C. $25$ — Arithmetic Error
  Solving $n^{2} = 625$ or mis-rooting 400.
• E. $200$ — Operation Error
  Halving 400 rather than taking its square root.

Common Mistake (⚠️):
Giving 400 as the last term rather than the total, or answering 40 by confusing the final term of the series with the number of terms.

Takeaway (📌):
$1+3+5+\cdots+(2n-1) = n^{2}$. Recognising it turns this from a formula exercise into a square root.

Correct Answer: A. $5$

Fastest Approach (🚀):
$f(3) = 27-36+3k-6 = 3k-15 = 0$.
$k = 5$.

Matches Option A.

Step-by-Step Breakdown:

1. Turn the factor into a root

$(x-3)$ is a factor $\implies f(3) = 0$

2. Substitute

$f(3) = 27-4(9)+3k-6 = 27-36+3k-6$

3. Solve

$3k-15 = 0 \implies k = 5$

Check: $x^{3}-4x^{2}+5x-6$ at $x=3$ gives $27-36+15-6 = 0$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $-5$ — Sign Error
  Substituting $x=-3$ and solving.
• C. $3$ — Misread Question
  Giving the root rather than $k$.
• D. $6$ — Misread Question
  Giving the constant term of the cubic.
• E. $15$ — Incomplete Answer
  Stopping at $3k = 15$.

Common Mistake (⚠️):
Substituting $x = -3$, which is the root of $(x+3)$ rather than of $(x-3)$.

Takeaway (📌):
Factor theorem: $(x-a)$ is a factor exactly when $f(a) = 0$. One substitution, one linear equation.

Correct Answer: B. $27$

Fastest Approach (🚀):
$x = 9^{3/2} = 3^{3} = 27$.

Matches Option B.

Step-by-Step Breakdown:

1. Convert to exponential form

$\log_{9}x = \dfrac32 \iff x = 9^{\frac32}$

2. Split the fractional index

$9^{\frac32} = \left(9^{\frac12}\right)^{3} = \left(\sqrt{9}\right)^{3} = 3^{3}$

Root first, then power — otherwise you would need $\sqrt{729}$.

3. Evaluate

$x = 27$

4. Check

$\log_{9}27$: is $9^{3/2} = 27$? $9^{1} = 9$, $9^{2} = 81$, so a value between 1 and 2 is right, and $\tfrac32$ fits.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $3$ — Index Error
  Evaluating $9^{1/2} = 3$ and stopping, instead of cubing that result.
• C. $\dfrac{3}{2}$ — Misread Question
  Repeating the given logarithm value.
• D. $13.5$ — Operation Error
  Computing $9\times 1.5$ instead of $9^{1.5}$.
• E. $81$ — Index Error
  Using $9^{2}$ — rounding the index up.

Common Mistake (⚠️):
Multiplying the base by the logarithm ($9\times\tfrac32 = 13.5$) instead of raising the base to that power.

Takeaway (📌):
$\log_b x = k \iff x = b^{k}$. Convert first, then handle the fractional index as root-then-power.

Correct Answer: C. $-\dfrac{3}{4}$

Fastest Approach (🚀):
$m_{\text{radius}} = \dfrac{4-0}{3-0} = \dfrac43$.
$m_{\text{tangent}} = -\dfrac34$.

Matches Option C.

Step-by-Step Breakdown:

1. Confirm the point is on the circle

$3^{2}+4^{2} = 9+16 = 25$. Yes — $(3,4)$ lies on the circle.

2. Find the gradient of the radius

The centre is the origin, so
$m_{\text{radius}} = \dfrac{4-0}{3-0} = \dfrac{4}{3}$

3. Use perpendicularity

The tangent at any point of a circle is perpendicular to the radius drawn to that point:
$m_{\text{tangent}} = -\dfrac{1}{m_{\text{radius}}} = -\dfrac{3}{4}$

4. Cross-check by implicit differentiation

$2x+2y\dfrac{dy}{dx} = 0 \implies \dfrac{dy}{dx} = -\dfrac{x}{y} = -\dfrac34$. Same answer, more work.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{4}{3}$ — Misread Question
  Giving the gradient of the radius rather than the tangent.
• B. $\dfrac{3}{4}$ — Sign Error
  Taking the reciprocal but not the negative.
• D. $-\dfrac{4}{3}$ — Formula Misuse
  Negating the radius gradient without inverting it.
• E. $0$ — Conceptual Error
  Assuming the tangent is horizontal.

Common Mistake (⚠️):
Giving the radius gradient $\tfrac43$, or negating without inverting to get $-\tfrac43$.

Takeaway (📌):
Tangent to a circle: negative reciprocal of the radius gradient. For a circle centred at the origin that is simply $-\dfrac{x}{y}$.

Correct Answer: B. $\dfrac{19}{2}$

Fastest Approach (🚀):
$h = 1$, ordinates $0, 1, 4, 9$.
$\tfrac12\left[(0+9)+2(1+4)\right] = \tfrac12(19) = \tfrac{19}{2}$.

Matches Option B.

Step-by-Step Breakdown:

1. Strip width and ordinates

$h = \dfrac{3-0}{3} = 1$, so the ordinates are at $x = 0, 1, 2, 3$:
$y = 0,\ 1,\ 4,\ 9$

2. Apply the rule

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

Only the two interior ordinates are doubled.

3. Evaluate

$\dfrac12\left[0+9+2(1+4)\right] = \dfrac12\left[9+10\right] = \dfrac{19}{2}$

4. Judge the error

The exact value is $9$, so the rule overestimates by $\tfrac12$ — as it must, because $y = x^{2}$ is convex and the trapezia sit above the curve.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $9$ — Misread Question
  Giving the exact integral rather than the trapezium estimate.
• C. $10$ — Weighting Error
  Doubling every ordinate.
• D. $\dfrac{17}{2}$ — Weighting Error
  Doubling the ends instead of the middles.
• E. $14$ — Strip Width Error
  Using $h = 3$ somewhere in the calculation.

Common Mistake (⚠️):
Doubling the end ordinates as well, or using $h = 3$ instead of the strip width $1$.

Takeaway (📌):
Ends once, middles twice, times $\tfrac h2$. For a convex curve the estimate always exceeds the true value.

Correct Answer: A. $3$

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

Matches Option A.

Step-by-Step Breakdown:

1. Set the function equal to the value

$x^{3}-1 = 26$

2. Undo in reverse order

The function cubes then subtracts $1$, so the inverse adds $1$ then takes a cube root:
$x^{3} = 27 \implies x = \sqrt[3]{27} = 3$

3. State the inverse

$f^{-1}(x) = \sqrt[3]{x+1}$, and $f^{-1}(26) = \sqrt[3]{27} = 3$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $27$ — Incomplete Answer
  Stopping at $x^{3} = 27$.
• C. $\sqrt[3]{25}$ — Order Error
  Cube-rooting before adding the 1.
• D. $9$ — Root Error
  Taking a square root of 81 or otherwise mis-rooting.
• E. $25$ — Order Error
  Computing $26-1$.

Common Mistake (⚠️):
Taking the cube root before adding the $1$, giving $\sqrt[3]{25}$ — the operations must be undone in reverse order.

Takeaway (📌):
Inverting means undoing each operation in the opposite order to the one applied. Last on, first off.

Correct Answer: A. $2842$

Fastest Approach (🚀):
Largest multiple below 200: $7\times 28 = 196$.
$7\left(1+2+\cdots+28\right) = 7\times\dfrac{28\times 29}{2} = 7\times 406 = 2842$.

Matches Option A.

Step-by-Step Breakdown:

1. Count the terms

The multiples are $7,14,\ldots$ up to the largest below 200:
$7\times 28 = 196$ and $7\times 29 = 203 > 200$

So there are 28 terms.

2. Factor out the 7

$7+14+\cdots+196 = 7\left(1+2+\cdots+28\right)$

3. Use the standard sum

$1+2+\cdots+28 = \dfrac{28\times 29}{2} = 406$

4. Multiply back

$7\times 406 = 2842$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $2800$ — Arithmetic Error
  Rounding the count or the triangular sum.
• C. $2842$ minus $7$ — Off-by-one Error
  Excluding a term unnecessarily — 196 is below 200 and belongs in the sum.
• D. $2900$ — Off-by-one Error
  Using 29 terms, including $203$.
• E. $1421$ — Arithmetic Error
  Giving half the sum, or forgetting to double after $\tfrac{n(n+1)}{2}$.

Common Mistake (⚠️):
Including $203$ by using 29 terms, or forgetting to multiply the triangular sum back by 7.

Takeaway (📌):
Sum of multiples of $k$ below $N$: count the terms, use $\tfrac{n(n+1)}{2}$, multiply by $k$. Check the largest term really is below $N$.

Correct Answer: C. $14$

Fastest Approach (🚀):
$\left(\sum\alpha\right)^{2} = \sum\alpha^{2}+2\sum\alpha\beta$.
$36 = \sum\alpha^{2}+22 \implies \sum\alpha^{2} = 14$.

Matches Option C.

Step-by-Step Breakdown:

1. Read the symmetric sums off the coefficients

For $x^{3}+px^{2}+qx+r = 0$:
$\sum\alpha = -p = 6$
$\sum\alpha\beta = q = 11$
$\alpha\beta\gamma = -r = 6$

2. Use the squares identity

$\left(\alpha+\beta+\gamma\right)^{2} = \alpha^{2}+\beta^{2}+\gamma^{2}+2\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)$

3. Rearrange and substitute

$\sum\alpha^{2} = \left(\sum\alpha\right)^{2}-2\sum\alpha\beta = 36-22 = 14$

4. Verify

The roots are $1, 2, 3$, so $1+4+9 = 14$. The identity route agrees — and needs no factorising.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $36$ — Incomplete Identity
  Giving $\left(\sum\alpha\right)^{2}$ without the correction.
• B. $22$ — Misread Question
  Giving $2\sum\alpha\beta$.
• D. $11$ — Misread Question
  Giving $\sum\alpha\beta$.
• E. $6$ — Misread Question
  Giving $\sum\alpha$ or the product of the roots.

Common Mistake (⚠️):
Giving $\left(\sum\alpha\right)^{2} = 36$ without subtracting $2\sum\alpha\beta$, or using the product of the roots in place of the pairwise sum.

Takeaway (📌):
$\sum\alpha^{2} = \left(\sum\alpha\right)^{2}-2\sum\alpha\beta$ works for any number of roots. Vieta supplies both pieces directly.

Correct Answer: A. $2$

Fastest Approach (🚀):
$u = 2^{x}: u^{2}-5u+4 = (u-1)(u-4) = 0$.
$2^{x} = 1 \implies x = 0;\quad 2^{x} = 4 \implies x = 2$.
Sum $= 2$.

Matches Option A.

Step-by-Step Breakdown:

1. Express everything in one base

$4^{x} = \left(2^{2}\right)^{x} = \left(2^{x}\right)^{2}$

2. Substitute

Let $u = 2^{x}$:
$u^{2}-5u+4 = 0$

3. Solve the quadratic

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

Both are positive, so both are attainable — $2^{x}$ can never be negative or zero, and any negative root would have to be discarded.

4. Convert back and add

$2^{x} = 1 \implies x = 0$
$2^{x} = 4 \implies x = 2$
Sum $= 0+2 = 2$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $0$ — Conceptual Error
  Assuming the roots are $\pm$ symmetric.
• C. $4$ — Substitution Not Reversed
  Giving one of the $u$ values.
• D. $5$ — Substitution Not Reversed
  Summing the $u$ values instead of the $x$ values.
• E. $1$ — Index Error
  Taking $2^{x} = 1 \implies x = 1$.

Common Mistake (⚠️):
Giving the sum of the $u$ values ($1+4 = 5$) rather than of the $x$ values, or forgetting that $2^{x}=1$ gives $x=0$, not $x=1$.

Takeaway (📌):
$a^{2x}$ alongside $a^{x}$ always signals a hidden quadratic. Substitute, solve, discard non-positive roots, then convert back.

Correct Answer: A. $\dfrac{28}{15}$

Fastest Approach (🚀):
$h = 2$, so $\tfrac h2 = 1$.
$\left(1+\tfrac15\right)+2\left(\tfrac13\right) = \tfrac65+\tfrac23 = \tfrac{18+10}{15} = \tfrac{28}{15}$.

Matches Option A.

Step-by-Step Breakdown:

1. Strip width and ordinates

$h = \dfrac{5-1}{2} = 2$, so the ordinates are at $x = 1, 3, 5$:
$y = 1,\ \dfrac13,\ \dfrac15$

2. Apply the rule

$\approx \dfrac{2}{2}\left[1+\dfrac15+2\left(\dfrac13\right)\right] = 1\left[\dfrac65+\dfrac23\right]$

3. Common denominator

$\dfrac{18}{15}+\dfrac{10}{15} = \dfrac{28}{15}$

4. Judge the error

$\tfrac{28}{15}\approx 1.87$ against the true $\ln 5 \approx 1.61$. The curve is convex, so the estimate is high — and with only two wide strips, noticeably so.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\dfrac{8}{5}$ — Weighting Error
  Dropping the doubling of the middle ordinate.
• C. $\dfrac{14}{15}$ — Strip Width Error
  Halving the correct result by using $\tfrac h2 = \tfrac12$.
• D. $2$ — Estimation Error
  Rounding to a convenient integer.
• E. $\dfrac{23}{15}$ — Arithmetic Error
  Arithmetic slip when combining the fifteenths.

Common Mistake (⚠️):
Using $h = 4$ (the whole interval) or forgetting that the middle ordinate alone is doubled.

Takeaway (📌):
Wide strips on a sharply curved function give a poor estimate. The method is right; the resolution is not.

Correct Answer: B. $\dfrac{3x+1}{x-2}$

Fastest Approach (🚀):
$y = \dfrac{2x+1}{x-3} \implies y(x-3) = 2x+1 \implies x(y-2) = 3y+1$.
$f^{-1}(x) = \dfrac{3x+1}{x-2}$.

Matches Option B.

Step-by-Step Breakdown:

1. Set y equal to the function

$y = \dfrac{2x+1}{x-3}$

2. Clear the denominator

$y(x-3) = 2x+1 \implies xy-3y = 2x+1$

3. Collect the x terms

$xy-2x = 3y+1 \implies x(y-2) = 3y+1$

Factorising out $x$ is the step that makes this solvable — it cannot be done term by term.

4. Solve and rename

$x = \dfrac{3y+1}{y-2} \implies f^{-1}(x) = \dfrac{3x+1}{x-2}$

Check: $f(4) = \dfrac{9}{1} = 9$, and $f^{-1}(9) = \dfrac{28}{7} = 4$. Correct.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{x-3}{2x+1}$ — Notation Confusion
  Taking the reciprocal of $f(x)$ rather than the inverse function.
• C. $\dfrac{2x+1}{x+3}$ — Sign Error
  Changing only the sign of the 3 in the denominator.
• D. $\dfrac{x+3}{2x-1}$ — Rearrangement Error
  Swapping numerator and denominator after a partial rearrangement.
• E. $\dfrac{3x-1}{x+2}$ — Sign Error
  Sign errors when collecting the $x$ terms.

Common Mistake (⚠️):
Taking the reciprocal of $f(x)$ and calling it the inverse. $f^{-1}$ means the inverse function, not $\dfrac{1}{f}$.

Takeaway (📌):
Inverse: swap and solve. When $x$ appears twice, collect and factorise — and verify with one numerical round trip.

Correct Answer: E. $31$

Fastest Approach (🚀):
$S_{\infty} = \dfrac{16}{1/2} = 32$.
$S_{5} = 32\left(1-\tfrac{1}{32}\right) = 31$.

Matches Option E.

Step-by-Step Breakdown:

1. Write the formula

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

2. Substitute

$a = 16$, $r = \tfrac12$, $n = 5$:
$S_{5} = \dfrac{16\left(1-\left(\tfrac12\right)^{5}\right)}{1-\tfrac12} = \dfrac{16\left(1-\tfrac{1}{32}\right)}{\tfrac12}$

3. Simplify

Dividing by $\tfrac12$ doubles:
$S_{5} = 32\left(1-\dfrac{1}{32}\right) = 32-1 = 31$

4. Check by listing

$16+8+4+2+1 = 31$. And the sum to infinity is 32, so a five-term sum of 31 is exactly the right size.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $32$ — Misread Question
  Giving the sum to infinity rather than five terms.
• B. $30$ — Off-by-one Error
  Using $n=4$ terms.
• C. $\dfrac{31}{2}$ — Formula Misuse
  Halving the result, or omitting the division by $1-r$.
• D. $\dfrac{31}{16}$ — Formula Misuse
  Forgetting to multiply by the first term: computing $\tfrac{1-r^5}{1-r}$ alone and omitting the factor $a = 16$.

Common Mistake (⚠️):
Giving the sum to infinity, 32, or using the $r>1$ form and producing a negative denominator.

Takeaway (📌):
A finite geometric sum with $r<1$ always falls just short of $S_\infty$. Computing $S_\infty$ first gives a free sanity check.

Correct Answer: B. $-4$

Fastest Approach (🚀):
$y(2) = 4-6 = -2$; $\dfrac{dy}{dx} = 2x-3 = 1$.
$c = y-mx = -2-2 = -4$.

Matches Option B.

Step-by-Step Breakdown:

1. Find the point

$y(2) = 2^{2}-3(2) = 4-6 = -2$, so the point is $(2,-2)$.

2. Find the gradient

$\dfrac{dy}{dx} = 2x-3$, which at $x=2$ is $1$.

3. Form the tangent

$y-(-2) = 1(x-2) \implies y = x-4$

4. Read the intercept

At $x=0$, $y = -4$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $-2$ — Misread Question
  Giving the y-coordinate of the point of contact.
• C. $2$ — Sign Error
  Sign error when computing $c$.
• D. $-6$ — Gradient Error
  Using the normal gradient $-1$ instead of the tangent gradient.
• E. $0$ — Assumption Error
  Assuming the tangent passes through the origin.

Common Mistake (⚠️):
Giving the $y$-coordinate of the point of contact, $-2$, instead of the intercept of the tangent line.

Takeaway (📌):
Tangent questions need exactly two things: a point and a gradient. Compute both before writing any line equation.

Correct Answer: D. $-2$

Fastest Approach (🚀):
$y(3) = 9-12+7 = 4;\quad y' = 2x-4 \implies m = 2$.
$c = y-mx = 4-6 = -2$.

Matches Option D.

Step-by-Step Breakdown:

1. Find the point of contact

$y(3) = 3^{2}-4(3)+7 = 9-12+7 = 4$

The tangent touches at $(3,4)$.

2. Find the gradient there

$\dfrac{dy}{dx} = 2x-4 \implies m = 2(3)-4 = 2$

3. Form the tangent equation

$y-4 = 2(x-3) \implies y = 2x-2$

4. Read off the intercept

At $x=0$: $y = -2$.

(Shortcut: $c = y-mx = 4-2(3) = -2$, skipping the expansion entirely.)

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $4$ — Misread Question
  Giving the $y$-coordinate of the point of contact.
• B. $7$ — Misread Question
  Giving the constant term of the original curve.
• C. $2$ — Misread Question
  Giving the gradient rather than the intercept.
• E. $-4$ — Sign Error
  Sign error in $4-6$.

Common Mistake (⚠️):
Using the $y$-coordinate $4$ as the intercept, or substituting $x=3$ into the derivative and the original in the wrong order.

Takeaway (📌):
Tangent: point from the curve, gradient from the derivative, intercept from $c = y-mx$. Three substitutions, no algebra.

Correct Answer: A. $\dfrac{16}{3}$

Fastest Approach (🚀):
$\left[\tfrac23 x^{3/2}\right]_{0}^{4} = \tfrac23\times 4^{3/2} = \tfrac23\times 8 = \tfrac{16}{3}$.

Matches Option A.

Step-by-Step Breakdown:

1. Rewrite the root

$\sqrt{x} = x^{\frac12}$

2. Integrate

$\int x^{\frac12}dx = \dfrac{x^{\frac32}}{\frac32} = \dfrac{2}{3}x^{\frac32}$

Dividing by $\tfrac32$ is multiplying by $\tfrac23$ — the step most often fumbled.

3. Evaluate the limits

$4^{\frac32} = \left(\sqrt{4}\right)^{3} = 2^{3} = 8$

Taking the root before the power keeps the numbers small.

4. Compute

$\dfrac23(8)-\dfrac23(0) = \dfrac{16}{3}$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $8$ — Omitted Factor
  Giving $4^{3/2} = 8$ without the $\tfrac23$ factor.
• C. $\dfrac{8}{3}$ — Arithmetic Error
  Using $2^{3/2}$ or halving the correct result.
• D. $4$ — Formula Misuse
  Integrating to $\tfrac12 x^{2}$ style, treating the root as a coefficient.
• E. $\dfrac{32}{3}$ — Coefficient Error
  Using $\tfrac43 x^{3/2}$ or doubling the correct answer.

Common Mistake (⚠️):
Using $\tfrac32$ instead of $\tfrac23$ as the multiplier, or evaluating $4^{3/2}$ as $6$ by multiplying rather than taking a root and a power.

Takeaway (📌):
$\int x^{n} = \dfrac{x^{n+1}}{n+1}$ works for every $n \ne -1$, fractional included. Dividing by a fraction means multiplying by its reciprocal.

Correct Answer: C. $-4$

Fastest Approach (🚀):
$y = x^{1/2} \implies \dfrac{dy}{dx} = \tfrac12 x^{-1/2} = \dfrac{1}{2\sqrt x}$.
At $x=4$: $\dfrac14$. Normal $= -4$.

Matches Option C.

Step-by-Step Breakdown:

1. Rewrite and differentiate

$y = x^{1/2} \implies \dfrac{dy}{dx} = \dfrac12 x^{-1/2} = \dfrac{1}{2\sqrt{x}}$

2. Evaluate the tangent gradient

At $x = 4$: $\dfrac{1}{2\times 2} = \dfrac14$

3. Take the negative reciprocal

$m_{\text{normal}} = -\dfrac{1}{1/4} = -4$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{4}$ — Misread Question
  Giving the tangent gradient.
• B. $4$ — Sign Error
  Inverting without negating.
• D. $-\dfrac{1}{4}$ — Formula Misuse
  Negating without inverting.
• E. $-2$ — Misread Question
  Giving the $y$-coordinate at $x=4$.

Common Mistake (⚠️):
Stopping at the tangent gradient $\tfrac14$, or negating it to $-\tfrac14$ instead of inverting as well.

Takeaway (📌):
Rewrite roots as fractional powers before differentiating, and remember the normal inverts as well as negates.

Correct Answer: C. $1$

Fastest Approach (🚀):
$y' = 3x^{2}-6x,\quad y'' = 6x-6$.
$6x-6 = 0 \implies x = 1$.

Matches Option C.

Step-by-Step Breakdown:

1. Differentiate twice

$y = x^{3}-3x^{2}+4$
$\dfrac{dy}{dx} = 3x^{2}-6x$
$\dfrac{d^{2}y}{dx^{2}} = 6x-6$

2. Set the second derivative to zero

$6x-6 = 0 \implies x = 1$

3. Confirm the concavity changes

$y'' < 0$ for $x<1$ and $y'' > 0$ for $x>1$, so the curve genuinely changes from concave down to concave up. That sign change is what makes it an inflection rather than merely a zero of $y''$.

4. Cross-check with the stationary points

$y' = 3x(x-2) = 0$ at $x = 0$ and $x = 2$. The inflection at $x=1$ sits exactly midway — always true for a cubic.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0$ — Wrong Derivative
  Giving a stationary point from $y' = 0$.
• B. $2$ — Wrong Derivative
  Giving the other stationary point.
• D. $3$ — Substitution Error
  Reading the coefficient 3 from the original cubic.
• E. $-1$ — Sign Error
  Sign error when solving $6x-6 = 0$.

Common Mistake (⚠️):
Solving $y' = 0$ and giving a stationary point ($x=0$ or $x=2$) instead of the inflection.

Takeaway (📌):
$y'=0$ locates stationary points, $y''=0$ locates inflections. For a cubic the inflection is the midpoint of the two stationary points.

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

Fastest Approach (🚀):
Roots at $x = \pm 2$.
$\int_{-2}^{2}\left(x^{2}-4\right)dx = \left[\tfrac{x^{3}}{3}-4x\right]_{-2}^{2} = -\tfrac{16}{3}-\tfrac{16}{3} = -\tfrac{32}{3}$.
Area $= \dfrac{32}{3}$.

Matches Option D.

Step-by-Step Breakdown:

1. Find the limits

$x^{2}-4 = 0 \implies x = \pm 2$

Between these roots the curve lies below the $x$-axis.

2. Integrate

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

At $x=2$: $\dfrac83-8 = -\dfrac{16}{3}$
At $x=-2$: $-\dfrac83+8 = \dfrac{16}{3}$

3. Subtract

$-\dfrac{16}{3}-\dfrac{16}{3} = -\dfrac{32}{3}$

4. Interpret the sign

The negative value confirms the region lies below the axis. Geometric area is a positive quantity:
$\text{Area} = \dfrac{32}{3}$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $-\dfrac{32}{3}$ — Sign Interpretation
  Giving the signed integral rather than the area.
• B. $0$ — Conceptual Error
  Assuming symmetry cancels the region.
• C. $\dfrac{16}{3}$ — Limits Error
  Integrating over $[0,2]$ only — half the symmetric region.
• E. $8$ — Estimation Error
  Using a rectangle $4\times 2$ as an estimate.

Common Mistake (⚠️):
Reporting the signed value $-\tfrac{32}{3}$ as an area. Areas are never negative — the sign tells you which side of the axis the region is on.

Takeaway (📌):
Integrate to get the signed value, then take the magnitude for area. If a region crosses the axis, split it at the crossing first.