ESAT Advanced Mathematics Mock Module 1 Worked Solutions

Correct Answer: C. $5$

Fastest Approach (🚀):
$y' = 3(x-1)(x-3) = 0 \implies x = 1, 3$.
Positive cubic: left stationary point is the maximum, so take $x=1$.
$y(1) = 1-6+9+1 = 5$.

Matches Option C.

Step-by-Step Breakdown:

1. Differentiate

$\dfrac{dy}{dx} = 3x^{2}-12x+9$

2. Solve for the stationary points

$3(x^{2}-4x+3) = 3(x-1)(x-3) = 0 \implies x = 1 \text{ or } x = 3$

3. Classify them

$\dfrac{d^{2}y}{dx^{2}} = 6x-12$.
At $x=1$: $6-12 = -6 < 0$, so this is a local maximum.
At $x=3$: $18-12 = +6 > 0$, so this is a local minimum.

(For a cubic with a positive leading coefficient, the left stationary point is always the maximum — the second derivative merely confirms it.)

4. Evaluate y at the maximum

$y(1) = 1-6+9+1 = 5$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $1$ — Wrong Stationary Point
  Evaluating at $x=3$, which gives the local minimum $y = 1$.
• B. $3$ — Value Error
  Averaging the stationary $x$-values or evaluating at $x=2$, the point of inflection region.
• D. $-3$ — Arithmetic Error
  Sign slip in $y(1)$, e.g. $1-6-9+1$.
• E. $2$ — Misread Question
  Giving a stationary $x$-value rather than the $y$-coordinate.

Common Mistake (⚠️):
Reporting the $x$-coordinate ($x=1$) instead of the $y$-coordinate, or evaluating at $x=3$ and giving the minimum value $y=1$.

Takeaway (📌):
For a cubic with a positive $x^3$ coefficient the maximum always comes first. Use the second derivative only when you genuinely need it.

Correct Answer: B. $60$

Fastest Approach (🚀):
Power of $x$: $2(6-k)-k = 12-3k = 0 \implies k = 4$.
Coefficient: $\binom{6}{4}\times 2^{6-4}\times(-1)^{4} = 15\times 4\times 1 = 60$.

Matches Option B.

Step-by-Step Breakdown:

1. Write the general term

For $(a+b)^n$ the general term is $\binom{n}{k}a^{\,n-k}b^{\,k}$. Here $a = 2x^2$, $b = -x^{-1}$, $n = 6$:
$T_k = \binom{6}{k}\left(2x^{2}\right)^{6-k}\left(-x^{-1}\right)^{k}$

2. Collect the index of x

$T_k = \binom{6}{k}\,2^{\,6-k}\,(-1)^{k}\,x^{\,2(6-k)}\,x^{-k} = \binom{6}{k}2^{\,6-k}(-1)^{k}x^{\,12-3k}$

3. Impose independence from x

$12-3k = 0 \implies k = 4$

4. Evaluate that single term

$\binom{6}{4} = 15,\qquad 2^{\,6-4} = 4,\qquad (-1)^{4} = +1$
$T_4 = 15\times 4\times 1 = 60$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $15$ — Omitted Factor
  Giving $\binom{6}{4} = 15$ and forgetting the $2^{6-k}$ factor.
• C. $240$ — Index Error
  Using $k=3$ (from $12-3k$ mis-solved) or applying $2^{6-k}$ with $k=2$.
• D. $-60$ — Sign Error
  Applying the sign as $(-1)^{k}$ with $k$ odd — but $k=4$ is even, so the term is positive.
• E. $160$ — Formula Misuse
  Mixing the binomial coefficient with the power, e.g. $\binom{6}{2}\times\cdots$ with a mis-set index.

Common Mistake (⚠️):
Expanding all seven terms, or forgetting that $(-x^{-1})^k$ contributes both a sign $(-1)^k$ and an index $-k$. With $k=4$ the sign is positive, so a negative answer signals a slip.

Takeaway (📌):
Set the index of $x$ to the value you want and solve for $k$. Constant term, coefficient of $x^5$, middle term — all the same one-line method.

Correct Answer: B. $0$

Fastest Approach (🚀):
$y = (x-3)^{2}+2 \implies$ minimum value $2$ at $x=3$.
Reflection in the $x$-axis: minimum $2 \to$ maximum $-2$.
Translate $+2$: maximum $= -2+2 = 0$.

Matches Option B.

Step-by-Step Breakdown:

1. Complete the square

$y = x^{2}-6x+11 = (x-3)^{2}-9+11 = (x-3)^{2}+2$

The vertex is $(3,2)$ and, since the parabola opens upwards, $y_{\min} = 2$.

2. Reflect in the x-axis

Reflection in the $x$-axis maps $y \mapsto -y$, so the curve becomes
$y = -\left[(x-3)^{2}+2\right] = -(x-3)^{2}-2$

The vertex moves to $(3,-2)$, and the parabola now opens downwards, so $-2$ is a maximum.

3. Translate 2 units up

$y \mapsto y+2$ gives
$y = -(x-3)^{2}-2+2 = -(x-3)^{2}$

Vertex: $(3,0)$.

4. Read off the greatest value

The curve opens downwards with vertex $(3,0)$, so the greatest value of $y$ is $0$ (attained at $x=3$).

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $-2$ — Incomplete Answer
  Stopping after the reflection and forgetting the translation.
• C. $2$ — Omitted Transformation
  Giving the original minimum value, with no transformation applied.
• D. $4$ — Order Error
  Adding 2 to the original minimum instead of to the reflected one.
• E. $-4$ — Sign Error
  Reflecting and then translating downwards, or doubling the reflected vertex.

Common Mistake (⚠️):
Applying the transformations in the wrong order, or forgetting that reflecting a minimum in the $x$-axis converts it into a maximum. The word 'greatest' only makes sense after the reflection.

Takeaway (📌):
Track the vertex, not the equation. Reflection negates its $y$-coordinate and flips min into max; translation shifts it.

Correct Answer: E. $3$

Fastest Approach (🚀):
$L = \sqrt{S(0,0)} = \sqrt{0+0-0-0+9} = 3$.

Matches Option E.

Step-by-Step Breakdown:

1. The geometry

The tangent from an external point $P$ touches the circle at $T$, and $\angle PTC = 90^{\circ}$ where $C$ is the centre. By Pythagoras:
$L^{2} = d^{2}-r^{2}, \qquad d = PC$

2. Centre and radius

$x^{2}-4x+y^{2}-6y+9 = 0$
$(x-2)^{2}-4+(y-3)^{2}-9+9 = 0$
$(x-2)^{2}+(y-3)^{2} = 4$

So $C(2,3)$ and $r = 2$.

3. Apply the formula

$d^{2} = 2^{2}+3^{2} = 13$
$L^{2} = 13-4 = 9 \implies L = 3$

4. The shortcut

Writing the circle as $S \equiv x^{2}+y^{2}+2gx+2fy+c = 0$, substituting the external point gives $d^{2}-r^{2}$ directly. Here $S(0,0) = 9$, so $L = \sqrt{9} = 3$ with no completing the square at all.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $2$ — Misread Question
  Giving the radius rather than the tangent length.
• B. $\sqrt{5}$ — Arithmetic Error
  Computing $d^2-r^2$ as $9-4$ instead of $13-4$.
• C. $\sqrt{13}$ — Misread Question
  Giving the distance from the origin to the centre.
• D. $4$ — Formula Misuse
  Using $d-r = \sqrt{13}-2$ rounded, or $L^{2}=16$ from a sign slip in the constant term.

Common Mistake (⚠️):
Giving $d = \sqrt{13}$ (the distance to the centre) or $r=2$ instead of the tangent length. The tangent is always the shortest of the three.

Takeaway (📌):
For a circle in the form $S=0$, the tangent length from any external point is $\sqrt{S(\text{point})}$. A positive value also confirms the point is outside the circle.

Correct Answer: A. $x=4$

Fastest Approach (🚀):
$x(x-2) = 2^{3} = 8 \implies x^2-2x-8 = 0 \implies x = 4$ or $x = -2$.
$\log_2 x$ requires $x>0$ and $\log_2(x-2)$ requires $x>2$, so $x=-2$ is rejected.
$x = 4$.

Matches Option A.

Step-by-Step Breakdown:

1. Combine the logarithms

$\log_{2}A+\log_{2}B = \log_{2}(AB)$, so
$\log_{2}\left[x(x-2)\right] = 3$

2. Remove the logarithm

$x(x-2) = 2^{3} = 8$

3. Solve the quadratic

$x^{2}-2x-8 = 0 \implies (x-4)(x+2) = 0 \implies x = 4 \text{ or } x = -2$

4. Apply the domain

The original equation needs $x>0$ (from $\log_2 x$) and $x>2$ (from $\log_2(x-2)$), so the domain is $x>2$. That rejects $x=-2$.

Check $x=4$: $\log_{2}4+\log_{2}2 = 2+1 = 3$. Correct.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $x=8$ — Omitted Term
  Setting $x = 2^{3}$ directly, ignoring the second logarithm.
• C. $x=2$ — Domain Error
  Solving $\log_2 x = 3-\log_2(x-2)$ incorrectly and landing on the excluded boundary $x=2$.
• D. $x=-2$ — Domain Error
  Choosing the extraneous root of the quadratic.
• E. $x=4$ and $x=-2$ — Domain Error
  Giving both roots without applying the domain restriction — the intended trap.

Common Mistake (⚠️):
Reporting both roots of the quadratic. Combining logs widens the domain, so every candidate root must be tested in the original equation.

Takeaway (📌):
Combining logarithms is not reversible. Write the domain restriction down first, then solve — it is the fastest way to discard the trap root.

Correct Answer: D. $\dfrac{1}{11}$

Fastest Approach (🚀):
$R = \sqrt{3^2+4^2} = 5$, so $3\sin\theta+4\cos\theta \in [-5,5]$.
Denominator $\in [1,11]$, and it never reaches zero, so the fraction is continuous and positive.
Least value $= \dfrac{1}{11}$.

Matches Option D.

Step-by-Step Breakdown:

1. Write the trigonometric part in harmonic form

$a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)$ with $R = \sqrt{a^{2}+b^{2}}$.
$R = \sqrt{3^{2}+4^{2}} = 5$

2. State the range of the denominator

$\sin(\theta+\alpha)\in[-1,1] \implies 3\sin\theta+4\cos\theta\in[-5,5]$
$6+3\sin\theta+4\cos\theta \in [1,\,11]$

3. Confirm the fraction is well behaved

The denominator's minimum is $1 > 0$, so the expression is defined and positive for every $\theta$ — no asymptote to worry about.

4. Minimise the reciprocal

A positive fraction with fixed numerator is least when its denominator is greatest:
$\left(\dfrac{1}{6+3\sin\theta+4\cos\theta}\right)_{\min} = \dfrac{1}{11}$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{6}$ — Conceptual Error
  Setting the trigonometric part to zero rather than to its extreme value.
• B. $\dfrac{1}{5}$ — Omitted Term
  Using $R=5$ as the denominator instead of $6+5$.
• C. $1$ — Optimisation Inversion
  Giving the greatest value $\tfrac11 = 1$ (denominator minimised) rather than the least.
• E. $11$ — Misread Question
  Giving the maximum of the denominator rather than the value of the fraction.

Common Mistake (⚠️):
Minimising the denominator instead of maximising it — that gives the greatest value $1$, not the least. Also common: taking $R = 3+4 = 7$.

Takeaway (📌):
$R=\sqrt{a^2+b^2}$ converts any $a\sin\theta+b\cos\theta$ into a single sine wave, and every max/min question about it becomes a one-line range statement.

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

Fastest Approach (🚀):
$f(1)=0: 2+a-5+b = 0 \implies a+b = 3$.
$f(-2)=0: -16+4a+10+b = 0 \implies 4a+b = 6$.
Subtracting: $3a = 3 \implies a = 1,\ b = 2$.
Product of roots $= -\dfrac{b}{2} = -1$, so the third root is $\dfrac{-1}{(1)(-2)} = \dfrac12$.

Matches Option A.

Step-by-Step Breakdown:

1. Apply the factor theorem twice

If $(x-1)$ and $(x+2)$ are factors then $f(1) = 0$ and $f(-2) = 0$.
$f(1) = 2+a-5+b = 0 \implies a+b = 3$
$f(-2) = -16+4a+10+b = 0 \implies 4a+b = 6$

2. Solve for a and b

Subtracting the first from the second:
$3a = 3 \implies a = 1, \qquad b = 2$

So $f(x) = 2x^{3}+x^{2}-5x+2$.

3. Use the product of the roots

For $Ax^{3}+Bx^{2}+Cx+D$, the product of the roots is $-\dfrac{D}{A}$:
$\alpha\beta\gamma = -\dfrac{2}{2} = -1$

4. Solve for the third root

Two roots are $1$ and $-2$, so
$(1)(-2)\gamma = -1 \implies \gamma = \dfrac{1}{2}$

Check: $2\left(\tfrac18\right)+\tfrac14-\tfrac52+2 = \tfrac14+\tfrac14-\tfrac52+2 = 0$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $-\dfrac{1}{2}$ — Sign Error
  Sign error in the product of the roots, using $+D/A$ instead of $-D/A$.
• C. $2$ — Formula Misuse
  Using the sum of the roots $-B/A = -\tfrac12$ and mis-solving for $\gamma$.
• D. $-2$ — Misread Question
  Repeating one of the given roots rather than finding the third.
• E. $1$ — Misread Question
  Repeating the known root 1.

Common Mistake (⚠️):
Dividing the cubic by $(x-1)(x+2)$ by long division. It works, but it is three times the writing — and the sign of the product of the roots is where most errors appear.

Takeaway (📌):
Sum of roots $=-B/A$, product $=-D/A$ for a cubic. With two roots known, the third follows from either relation in one line.

Correct Answer: D. $16$

Fastest Approach (🚀):
Odd terms $x^3$ and $5x$ vanish by symmetry.
$\int_{-2}^{2}3x^{2}\,dx = 2\int_{0}^{2}3x^{2}\,dx = 2\left[x^{3}\right]_{0}^{2} = 2(8) = 16$.

Matches Option D.

Step-by-Step Breakdown:

1. Classify each term

$f(-x) = -f(x)$ defines an odd function; $f(-x) = f(x)$ defines an even function.
$x^{3}$ — odd. $\quad 5x$ — odd. $\quad 3x^{2}$ — even.

2. Apply the symmetry rules

Over an interval symmetric about the origin:
$\int_{-a}^{a}\text{odd}\,dx = 0, \qquad \int_{-a}^{a}\text{even}\,dx = 2\int_{0}^{a}\text{even}\,dx$

So the $x^{3}$ and $5x$ terms contribute nothing.

3. Integrate the surviving term

$\int_{-2}^{2}3x^{2}\,dx = 2\int_{0}^{2}3x^{2}\,dx = 2\left[x^{3}\right]_{0}^{2} = 2(8-0) = 16$

4. Confirm the long way

$\left[\tfrac{x^{4}}{4}+x^{3}+\tfrac{5x^{2}}{2}\right]_{-2}^{2} = (4+8+10)-(4-8+10) = 22-6 = 16$. The symmetry route agrees.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $0$ — Over-generalisation
  Assuming the whole integrand is odd and cancelling everything.
• B. $8$ — Omitted Factor
  Computing $\int_{0}^{2}3x^{2}dx = 8$ and forgetting to double it.
• C. $24$ — Extra Term
  Including the even part correctly but also adding a spurious contribution from $5x$.
• E. $32$ — Over-application
  Doubling every term rather than only the even one.

Common Mistake (⚠️):
Doubling every term, including the odd ones, or evaluating $\left[x^3\right]_{-2}^{2}$ as $8$ rather than $16$ after correctly reducing to the even part.

Takeaway (📌):
Check the limits before integrating. Symmetric limits plus odd terms is free cancellation — often the whole question.

Correct Answer: E. $2$

Fastest Approach (🚀):
$a = 3\left(\tfrac25\right) = \tfrac65,\quad r = \tfrac25$.
$S_\infty = \dfrac{a}{1-r} = \dfrac{6/5}{3/5} = 2$.

Matches Option E.

Step-by-Step Breakdown:

1. Write out the first few terms

$n=1: 3\left(\tfrac25\right) = \tfrac65, \qquad n=2: 3\left(\tfrac25\right)^{2} = \tfrac{12}{25}, \qquad \ldots$

2. Identify a and r

$a = \dfrac{6}{5}$ (the $n=1$ term, not $3$), and each term is $\tfrac25$ of the previous one, so $r = \dfrac{2}{5}$.

3. Check convergence

$|r| = \tfrac25 < 1$, so the sum to infinity exists.

4. Apply the formula

$S_{\infty} = \dfrac{a}{1-r} = \dfrac{6/5}{1-\tfrac25} = \dfrac{6/5}{3/5} = \dfrac{6}{5}\times\dfrac{5}{3} = 2$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{15}{2}$ — First Term Error
  Using $a=3$ and $r=\tfrac25$ with an arithmetic slip in $\dfrac{3}{3/5}$.
• B. $\dfrac{6}{5}$ — Incomplete Answer
  Giving the first term $a=\tfrac65$ rather than the sum.
• C. $3$ — Misread Question
  Quoting the coefficient 3 as the answer.
• D. $5$ — First Term Error
  Taking $a=3$ (the $n=0$ term): $\dfrac{3}{3/5}=5$. The summation starts at $n=1$.

Common Mistake (⚠️):
Taking $a=3$ because 3 is the visible coefficient. That is the $n=0$ term, which the summation does not include, and it gives $5$ instead of $2$.

Takeaway (📌):
Always substitute the lower limit to get $a$. The coefficient in front of the bracket is rarely the first term.

Correct Answer: C. $x < 2$ or $x > 3$

Fastest Approach (🚀):
$x^{2}-5x+6 > 0 \implies (x-2)(x-3) > 0$.
Positive parabola, so it is above zero outside the roots.
$x<2$ or $x>3$.

Matches Option C.

Step-by-Step Breakdown:

1. Rearrange to zero

$x^{2} > 5x-6 \implies x^{2}-5x+6 > 0$

Never divide an inequality by an expression in $x$ — its sign is unknown.

2. Factorise to find the critical values

$(x-2)(x-3) > 0 \implies$ critical values $x = 2$ and $x = 3$

3. Use the shape of the parabola

The coefficient of $x^{2}$ is positive, so the curve is a U shape crossing the axis at 2 and 3. It lies above the axis outside those roots and below between them.

Strict inequality, so the roots themselves are excluded.

4. State the solution

$x < 2 \quad\text{or}\quad x > 3$

Check $x=0$: $0 > -6$ is true, and $0$ is indeed in the solution set.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $2 < x < 3$ — Inequality Reversal
  Reversing the inequality — this is the region where the quadratic is negative.
• B. $x > 3$ — Incomplete Answer
  Giving only one of the two branches.
• D. $x < -3$ or $x > -2$ — Sign Error
  Sign error in factorising, using roots $-2$ and $-3$.
• E. All real values of $x$ — Conceptual Error
  Assuming a positive quadratic is always positive, ignoring the roots.

Common Mistake (⚠️):
Giving $2<x<3$, which is where the expression is negative — the answer to $x^{2}<5x-6$. Sketching the U shape prevents the reversal.

Takeaway (📌):
Positive quadratic $> 0$: outside the roots. Positive quadratic $< 0$: between the roots. A three-second sketch settles it every time.

Correct Answer: D. $\dfrac{15}{4}$

Fastest Approach (🚀):
Set equal: $x^{2}+3x+4 = 2x+k \implies x^{2}+x+(4-k) = 0$.
Tangent $\implies b^{2}-4ac = 0$: $1-4(4-k) = 0$.
$1-16+4k = 0 \implies k = \dfrac{15}{4}$.

Matches Option D.

Step-by-Step Breakdown:

1. Eliminate y

$x^{2}+3x+4 = 2x+k$

2. Collect into a quadratic

$x^{2}+3x-2x+4-k = 0 \implies x^{2}+x+(4-k) = 0$

3. Impose tangency

A tangent meets the curve exactly once, so the quadratic has a repeated root:
$b^{2}-4ac = 0 \implies 1^{2}-4(1)(4-k) = 0$

4. Solve for k

$1-16+4k = 0 \implies 4k = 15 \implies k = \dfrac{15}{4}$

Check: with $k = \tfrac{15}{4}$ the quadratic is $x^{2}+x+\tfrac14 = \left(x+\tfrac12\right)^{2}$, a perfect square — one contact point at $x = -\tfrac12$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $4$ — Misread Question
  Reading off the curve's constant term rather than solving the condition.
• B. $\dfrac{15}{2}$ — Formula Misuse
  Solving $1-2(4-k)=0$ by mis-remembering the discriminant as $b^{2}-2ac$.
• C. $\dfrac{1}{4}$ — Misread Question
  Giving the repeated root $x=-\tfrac12$ squared, or the value $\tfrac14$ from the perfect square.
• E. $-\dfrac{15}{4}$ — Sign Error
  Sign error when moving $k$ across the equals sign.

Common Mistake (⚠️):
Sign slip on the constant: writing $4+k$ instead of $4-k$ when moving $k$ across, which produces $-\tfrac{15}{4}$.

Takeaway (📌):
Tangency, one solution, and repeated root are the same statement. Discriminant zero is faster than matching gradients via differentiation.

Correct Answer: B. $\dfrac{1}{6}$

Fastest Approach (🚀):
$\log_a c = \log_a b \times \log_b c = 3\times 2 = 6$.
$\log_c a = \dfrac{1}{\log_a c} = \dfrac16$.

Matches Option B.

Step-by-Step Breakdown:

1. State the change of base rule

$\log_{a}c = \dfrac{\log_{b}c}{\log_{b}a}$

Equivalently, in chain form:
$\log_{a}b \times \log_{b}c = \log_{a}c$

2. Chain the two given values

$\log_{a}c = 3\times 2 = 6$

3. Invert

Swapping base and argument gives the reciprocal:
$\log_{c}a = \dfrac{1}{\log_{a}c} = \dfrac{1}{6}$

4. Check with concrete numbers

Take $a=2$. Then $b = 2^{3} = 8$ and $c = b^{2} = 64 = 2^{6}$. So $\log_{c}a = \log_{64}2 = \tfrac16$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $6$ — Inversion Omitted
  Giving $\log_{a}c = 6$ without inverting for $\log_{c}a$.
• C. $5$ — Additive Error
  Adding the two logarithms instead of multiplying them.
• D. $\dfrac{1}{5}$ — Additive Error
  Adding the logarithms and then inverting.
• E. $\dfrac{2}{3}$ — Formula Misuse
  Dividing the given values, $\tfrac{2}{3}$, rather than multiplying and inverting.

Common Mistake (⚠️):
Stopping at $\log_a c = 6$ and answering $6$. The question asks for the logarithm with the base and argument the other way round.

Takeaway (📌):
$\log_a b\times\log_b c = \log_a c$ and $\log_a b = \dfrac{1}{\log_b a}$. Between them, any chain of logarithms collapses in one line.

Correct Answer: C. $2$

Fastest Approach (🚀):
$2\cos^2 x+3\cos x-2 = 0 \implies \cos x = \tfrac12$ or $\cos x = -2$.
$\cos x = -2$ is impossible.
$\cos x = \tfrac12$ has one solution in each of the first and fourth quadrants: $x = \tfrac{\pi}{3},\ \tfrac{5\pi}{3}$.

Matches Option C.

Step-by-Step Breakdown:

1. Eliminate the mixed functions

Use $\sin^{2}x = 1-\cos^{2}x$:
$2\left(1-\cos^{2}x\right) = 3\cos x$

2. Rearrange into a quadratic in cos x

$2-2\cos^{2}x = 3\cos x \implies 2\cos^{2}x+3\cos x-2 = 0$

3. Factorise and filter

$(2\cos x-1)(\cos x+2) = 0 \implies \cos x = \dfrac{1}{2} \ \text{ or } \ \cos x = -2$

Since $\cos x \in[-1,1]$, the root $\cos x = -2$ is rejected without further work.

4. Count the solutions in the interval

$\cos x = \tfrac12$ is positive, so solutions lie in the first and fourth quadrants:
$x = \dfrac{\pi}{3}, \qquad x = 2\pi-\dfrac{\pi}{3} = \dfrac{5\pi}{3}$

Both lie in $0\le x\le 2\pi$, so there are $2$ solutions.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0$ — Sign Error
  Rejecting both roots, or concluding no solution exists after a sign error in the quadratic.
• B. $1$ — Incomplete Case Analysis
  Giving only the principal value $x=\tfrac{\pi}{3}$ and missing $x=\tfrac{5\pi}{3}$.
• D. $3$ — Quadrant Error
  Counting $x=\tfrac{\pi}{3},\ \pi-\tfrac{\pi}{3},\ 2\pi-\tfrac{\pi}{3}$ by applying the sine quadrant rule to a cosine.
• E. $4$ — Range Violation
  Counting solutions from the impossible root $\cos x = -2$ as well.

Common Mistake (⚠️):
Counting solutions from the rejected root $\cos x = -2$, or giving only the principal value $x=\tfrac{\pi}{3}$ and missing the fourth-quadrant partner.

Takeaway (📌):
One function, one quadratic, one sanity check on $[-1,1]$. Then count quadrants — $\cos x = k$ has exactly two solutions per full revolution unless $k=\pm1$.

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

Fastest Approach (🚀):
Intersections: $x^{2}=2x \implies x = 0, 2$.
$\int_{0}^{2}\left(2x-x^{2}\right)dx = \left[x^{2}-\tfrac{x^{3}}{3}\right]_{0}^{2} = 4-\tfrac83 = \tfrac43$.

Matches Option B.

Step-by-Step Breakdown:

1. Find the intersection points

$x^{2} = 2x \implies x(x-2) = 0 \implies x = 0 \text{ or } x = 2$

These are the limits of integration.

2. Decide which curve is on top

At $x=1$: the line gives $y=2$, the parabola gives $y=1$. The line is above throughout $0<x<2$.

3. Integrate the difference

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

4. Evaluate

$A = \left(4-\dfrac{8}{3}\right)-0 = \dfrac{12-8}{3} = \dfrac{4}{3}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{2}{3}$ — Order Error
  Integrating $x^{2}-2x$ (wrong order) and taking part of the magnitude, or mis-evaluating $\tfrac83-2$.
• C. $\dfrac{8}{3}$ — Incomplete Answer
  Evaluating $\int_{0}^{2}x^{2}dx = \tfrac83$ only — the parabola's area, not the enclosed region.
• D. $2$ — Conceptual Error
  Using the triangle under the line, $\tfrac12\times 2\times 4 = 4$, or a rounded estimate.
• E. $\dfrac{1}{3}$ — Arithmetic Error
  Arithmetic slip in $4-\tfrac83$.

Common Mistake (⚠️):
Integrating the two curves separately and subtracting the results in the wrong order (giving $-\tfrac43$), or using the wrong upper limit by solving $x^2=2x$ as $x=2$ only.

Takeaway (📌):
Area between curves is a single integral of (upper − lower). Test one interior point to fix the order and the sign takes care of itself.

Correct Answer: D. $\dfrac{99}{100}$

Fastest Approach (🚀):
$\dfrac{1}{r(r+1)} = \dfrac1r-\dfrac1{r+1}$.
Sum $= 1-\dfrac{1}{100} = \dfrac{99}{100}$.

Matches Option D.

Step-by-Step Breakdown:

1. Split into partial fractions

$\dfrac{1}{r(r+1)} = \dfrac{A}{r}+\dfrac{B}{r+1} \implies 1 = A(r+1)+Br$

Setting $r=0$ gives $A=1$; setting $r=-1$ gives $B=-1$.
$\dfrac{1}{r(r+1)} = \dfrac{1}{r}-\dfrac{1}{r+1}$

2. Write out the sum

$\left(\dfrac11-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\cdots+\left(\dfrac{1}{99}-\dfrac{1}{100}\right)$

3. Cancel

Every negative fragment is destroyed by the positive fragment of the next bracket. Only the first and last survive:
$S = 1-\dfrac{1}{100}$

4. Evaluate

$S = \dfrac{99}{100}$

In general $\displaystyle\sum_{r=1}^{n}\dfrac{1}{r(r+1)} = \dfrac{n}{n+1}$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{100}$ — Incomplete Answer
  Giving the surviving negative fragment $\tfrac1{100}$ instead of $1-\tfrac1{100}$.
• B. $\dfrac{49}{50}$ — Arithmetic Error
  Cancelling incorrectly and halving, e.g. taking $n/(2n+2)$.
• C. $\dfrac{98}{99}$ — Off-by-one Error
  Off-by-one: using $n=98$ or stopping the cancellation one bracket early.
• E. $\dfrac{100}{101}$ — Off-by-one Error
  Off-by-one in the other direction: summing to $r=100$.

Common Mistake (⚠️):
Off-by-one on the final fragment: with $r$ running to $99$, the last negative term is $\tfrac1{100}$, not $\tfrac1{99}$ or $\tfrac1{101}$.

Takeaway (📌):
Consecutive factors in a denominator almost always telescope. Write the first two and last two brackets explicitly to fix the surviving terms.

Correct Answer: A. $3$

Fastest Approach (🚀):
$y' = 3x^{2}-3 = 0 \implies x = \pm 1$.
$y(-1) = -1+3+1 = 3$ (local maximum, above the axis).
$y(1) = 1-3+1 = -1$ (local minimum, below the axis).
Maximum positive and minimum negative $\implies 3$ crossings.

Matches Option A.

Step-by-Step Breakdown:

1. Locate the stationary points

$\dfrac{dy}{dx} = 3x^{2}-3 = 3(x^{2}-1) = 0 \implies x = -1 \text{ or } x = 1$

2. Evaluate the stationary values

$y(-1) = (-1)^{3}-3(-1)+1 = -1+3+1 = 3$
$y(1) = 1-3+1 = -1$

Since the cubic has a positive leading coefficient, $x=-1$ is the local maximum and $x=1$ the local minimum.

3. Apply the sign test

The local maximum is above the axis and the local minimum is below it. The curve therefore comes up from $-\infty$, crosses, turns down through the axis, turns again, and crosses a third time.

4. Conclude

Three crossings — the cubic has three distinct real roots (approximately $-1.88$, $0.35$ and $1.53$).

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $1$ — Conceptual Error
  Assuming one root because the cubic does not factorise with integer roots.
• C. $2$ — Counting Error
  Counting only the visible crossings near the origin, or assuming a repeated root.
• D. $0$ — Conceptual Error
  Concluding no real roots, as for a quadratic with negative discriminant.
• E. $6$ — Conceptual Error
  Confusing the number of crossings with the degree or the stationary point count doubled.

Common Mistake (⚠️):
Trying to factorise $x^{3}-3x+1$. It has no rational roots, so the factor theorem gives nothing and the time is wasted. The stationary-value test needs no roots at all.

Takeaway (📌):
For root counting, differentiate rather than factorise. Opposite-signed stationary values mean three roots; same-signed means one.

Correct Answer: E. $7\sqrt{2}$

Fastest Approach (🚀):
$d = \dfrac{|0-0+1|}{\sqrt{1^{2}+(-1)^{2}}} = \dfrac{1}{\sqrt2}$.
Half-chord $= \sqrt{r^{2}-d^{2}} = \sqrt{25-\tfrac12} = \sqrt{\tfrac{49}{2}} = \tfrac{7}{\sqrt{2}}$.
Chord $= \dfrac{14}{\sqrt2} = 7\sqrt{2}$.

Matches Option E.

Step-by-Step Breakdown:

1. Identify the centre and radius

$x^{2}+y^{2} = 25$ has centre $O(0,0)$ and radius $r = 5$.

2. Perpendicular distance from the centre to the line

Write the line as $x-y+1 = 0$ and use the standard formula:
$d = \dfrac{|ax_{0}+by_{0}+c|}{\sqrt{a^{2}+b^{2}}} = \dfrac{|0-0+1|}{\sqrt{1+1}} = \dfrac{1}{\sqrt{2}}$

Since $d < r$, the line genuinely cuts the circle.

3. Half the chord by Pythagoras

The perpendicular from the centre bisects the chord, so
$\left(\dfrac{L}{2}\right)^{2} = r^{2}-d^{2} = 25-\dfrac12 = \dfrac{49}{2}$
$\dfrac{L}{2} = \sqrt{\dfrac{49}{2}} = \dfrac{7}{\sqrt2}$

4. Double it

$L = \dfrac{14}{\sqrt{2}} = \dfrac{14\sqrt2}{2} = 7\sqrt{2}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $7$ — Simplification Error
  Dropping the $\sqrt2$ when rationalising $\tfrac{14}{\sqrt2}$.
• B. $2\sqrt{24}$ — Distance Error
  Using $d = 1$ instead of $\tfrac{1}{\sqrt2}$: $2\sqrt{25-1}$.
• C. $\dfrac{\sqrt{98}}{2}$ — Incomplete Answer
  Giving the half-chord rather than the full chord.
• D. $10$ — Conceptual Error
  Giving the diameter, as if the line passed through the centre.

Common Mistake (⚠️):
Reporting the half-chord instead of the whole chord, or solving the simultaneous equations to find both intersection points and then applying the distance formula — correct, but three times the work.

Takeaway (📌):
Chord length $= 2\sqrt{r^{2}-d^{2}}$. The perpendicular distance formula does all the work; simultaneous equations are never needed.

Correct Answer: C. $15$ hours

Fastest Approach (🚀):
$32 = 2^{5} \implies 5$ doublings.
$5\times 3 = 15$ hours.

Matches Option C.

Step-by-Step Breakdown:

1. Write the model

$N = N_{0}\,2^{\,t/3}$ where $t$ is in hours, since $t=3$ gives one doubling.

2. Set the required multiple

$\dfrac{N}{N_{0}} = 32 \implies 2^{\,t/3} = 32$

3. Match the powers

$32 = 2^{5}$, so
$\dfrac{t}{3} = 5 \implies t = 15\ \text{hours}$

4. The counting argument

Equivalently: $\times2, \times4, \times8, \times16, \times32$ is five steps of 3 hours. No logarithm is needed because 32 is a power of 2 — if the question had asked for 30 times, $t = 3\log_{2}30$ would be unavoidable.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $9$ hours — Counting Error
  Counting three doublings ($\times 8$) rather than five.
• B. $12$ hours — Off-by-one Error
  Counting four doublings ($\times 16$).
• D. $16$ hours — Conceptual Error
  Using $32/2 = 16$ as the answer in hours.
• E. $96$ hours — Model Error
  Treating the growth as linear: $32\times 3$.

Common Mistake (⚠️):
Dividing 32 by the doubling time, or treating the growth as linear and answering 96 hours (32 × 3).

Takeaway (📌):
Check whether the target multiple is a power of the growth factor before reaching for logarithms. On the ESAT it usually is.

Correct Answer: E. $\tan\theta$

Fastest Approach (🚀):
Numerator: $\sin 2\theta = 2\sin\theta\cos\theta$.
Denominator: $1+\cos2\theta = 2\cos^{2}\theta$.
Ratio $= \dfrac{2\sin\theta\cos\theta}{2\cos^{2}\theta} = \tan\theta$.

Matches Option E.

Step-by-Step Breakdown:

1. Choose the right form of cos 2θ

The three standard forms are
$\cos2\theta = \cos^{2}\theta-\sin^{2}\theta = 2\cos^{2}\theta-1 = 1-2\sin^{2}\theta$

The denominator is $1+\cos2\theta$, so the form with $-1$ in it is the one that simplifies:
$1+\cos2\theta = 1+2\cos^{2}\theta-1 = 2\cos^{2}\theta$

2. Expand the numerator

$\sin2\theta = 2\sin\theta\cos\theta$

3. Divide

$\dfrac{2\sin\theta\cos\theta}{2\cos^{2}\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$

4. Check numerically

At $\theta = 45^\circ$: $\dfrac{\sin 90^\circ}{1+\cos 90^\circ} = \dfrac{1}{1} = 1 = \tan 45^\circ$. Correct.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\tan 2\theta$ — Formula Misuse
  Treating the expression as if the identity $\tfrac{\sin A}{\cos A}$ applied with $A=2\theta$.
• B. $2\tan\theta$ — Cancellation Error
  Cancelling only one $\cos\theta$ from the denominator.
• C. $\cot\theta$ — Inversion Error
  Inverting the final ratio.
• D. $\sin\theta$ — Cancellation Error
  Cancelling $\cos\theta$ from the numerator against the $2$ in the denominator.

Common Mistake (⚠️):
Using $\cos2\theta = 1-2\sin^{2}\theta$, which turns the denominator into $2-2\sin^{2}\theta$ and leads nowhere. Picking the wrong form is the only real difficulty here.

Takeaway (📌):
Match the identity to the constant already present: $1+\cos2\theta \to 2\cos^{2}\theta$ and $1-\cos2\theta \to 2\sin^{2}\theta$.

Correct Answer: A. $-1$

Fastest Approach (🚀):
$2x+3\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$.
At $(1,1)$: $2+3+ (3+2)\dfrac{dy}{dx}=0 \implies \dfrac{dy}{dx} = -1$.

Matches Option A.

Step-by-Step Breakdown:

1. Differentiate each term with respect to x

$\dfrac{d}{dx}\left(x^{2}\right) = 2x$
$\dfrac{d}{dx}\left(3xy\right) = 3y+3x\dfrac{dy}{dx}$ (product rule)
$\dfrac{d}{dx}\left(y^{2}\right) = 2y\dfrac{dy}{dx}$ (chain rule)
$\dfrac{d}{dx}(5) = 0$

2. Assemble

$2x+3y+3x\dfrac{dy}{dx}+2y\dfrac{dy}{dx} = 0$

3. Substitute the point before rearranging

At $(1,1)$:
$2+3+3\dfrac{dy}{dx}+2\dfrac{dy}{dx} = 0 \implies 5+5\dfrac{dy}{dx} = 0$

4. Solve

$\dfrac{dy}{dx} = -1$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $1$ — Sign Error
  Sign error when collecting the terms.
• C. $-\dfrac{5}{3}$ — Omitted Term
  Omitting the $2y\dfrac{dy}{dx}$ term, leaving $5+3\dfrac{dy}{dx}=0$.
• D. $-\dfrac{2}{5}$ — Inversion Error
  Omitting the product rule on $3xy$ and using $2+3$ over $3+2$ inverted.
• E. $0$ — Conceptual Error
  Assuming the point is stationary because the curve is symmetric in $x$ and $y$.

Common Mistake (⚠️):
Treating $3xy$ as $3x$ times a constant and differentiating it to $3y$ only. The product rule is essential: both factors depend on $x$.

Takeaway (📌):
Substitute the coordinates as early as possible in an implicit differentiation question. Rearranging the general formula first is slower and invites algebraic slips.

Correct Answer: A. $59$

Fastest Approach (🚀):
$S_{10} = 300+20 = 320$.
$S_{9} = 243+18 = 261$.
$u_{10} = 320-261 = 59$.

Matches Option A.

Step-by-Step Breakdown:

1. State the relationship

The sum of the first $n$ terms differs from the sum of the first $n-1$ terms by exactly the $n$th term:
$u_{n} = S_{n}-S_{n-1}$

2. Evaluate both sums

$S_{10} = 3(100)+2(10) = 320$
$S_{9} = 3(81)+2(9) = 243+18 = 261$

3. Subtract

$u_{10} = 320-261 = 59$

4. Confirm the general term

$u_{n} = \left[3n^{2}+2n\right]-\left[3(n-1)^{2}+2(n-1)\right] = 6n-1$

At $n=10$: $60-1 = 59$. The sequence is arithmetic with $d=6$, as the quadratic $S_n$ implied.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $320$ — Misread Question
  Giving $S_{10}$, the sum rather than the term.
• C. $62$ — Off-by-one Error
  Using $S_{10}-S_{8}$ or an off-by-one in the second sum.
• D. $32$ — Conceptual Error
  Using $u_n = 3n+2$ by treating $S_n$ as the term formula.
• E. $61$ — Sign Error
  Deriving $u_n = 6n+1$ with a sign slip in the expansion.

Common Mistake (⚠️):
Substituting $n=10$ into $S_n$ and giving $320$, which is the sum of the first ten terms, not the tenth term.

Takeaway (📌):
$u_n = S_n-S_{n-1}$ works for any sequence, arithmetic or not. A quadratic $S_n$ always means an arithmetic sequence.

Correct Answer: C. $x = \pm 3$

Fastest Approach (🚀):
$fg(x) = 3x^{2}-2$.
$3x^{2}-2 = 25 \implies 3x^{2} = 27 \implies x^{2} = 9$.
$x = \pm 3$.

Matches Option C.

Step-by-Step Breakdown:

1. Build the composite in the right order

$fg(x)$ means $f\left(g(x)\right)$ — $g$ acts first:
$fg(x) = f\left(x^{2}\right) = 3x^{2}-2$

2. Solve the equation

$3x^{2}-2 = 25 \implies 3x^{2} = 27 \implies x^{2} = 9$

3. Take both roots

$x = \pm 3$

Squaring loses no solutions here: both values are valid because $g$ is defined for all real $x$.

4. Contrast with the other order

$gf(x) = \left(3x-2\right)^{2} = 25$ would give $3x-2 = \pm5$, so $x = \tfrac73$ or $x = -1$. A completely different answer — which is exactly why the order matters.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $x = 3$ only — Incomplete Answer
  Taking only the positive square root.
• B. $x = \dfrac{7}{3}$ or $x = -1$ — Order Error
  Computing $gf(x) = (3x-2)^{2}$ instead — the order is reversed.
• D. $x = \pm\sqrt{3}$ — Arithmetic Error
  Solving $x^{2} = 3$ by dividing 27 by 9 or mishandling the coefficient.
• E. $x = \pm 9$ — Incomplete Answer
  Stopping at $x^{2}=9$ and reporting $9$ with a sign, rather than taking the root.

Common Mistake (⚠️):
Computing $gf(x)$ instead of $fg(x)$. Composition is read right to left, so the function written next to the $x$ acts first.

Takeaway (📌):
$fg(x) = f(g(x))$. Write the inner function's output explicitly before applying the outer one, and always take both square roots.

Correct Answer: B. $\dfrac{3}{4\pi}\ \text{cm}\,\text{s}^{-1}$

Fastest Approach (🚀):
$\dfrac{dV}{dr} = 4\pi r^{2} = 4\pi(4) = 16\pi$.
$\dfrac{dr}{dt} = \dfrac{dV/dt}{dV/dr} = \dfrac{12}{16\pi} = \dfrac{3}{4\pi}\ \text{cm}\,\text{s}^{-1}$.

Matches Option B.

Step-by-Step Breakdown:

1. Identify what is given and what is wanted

Given: $\dfrac{dV}{dt} = 12\ \text{cm}^{3}\text{s}^{-1}$. Wanted: $\dfrac{dr}{dt}$ at $r=2$.

2. Build the chain

$\dfrac{dr}{dt} = \dfrac{dV}{dt}\times\dfrac{dr}{dV} = \dfrac{dV/dt}{dV/dr}$

3. Differentiate the volume formula

$V = \dfrac{4}{3}\pi r^{3} \implies \dfrac{dV}{dr} = 4\pi r^{2}$

At $r=2$: $\dfrac{dV}{dr} = 4\pi(2)^{2} = 16\pi$ (this is also the surface area, as it must be).

4. Combine

$\dfrac{dr}{dt} = \dfrac{12}{16\pi} = \dfrac{3}{4\pi}\ \text{cm}\,\text{s}^{-1}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{3}{16\pi}\ \text{cm}\,\text{s}^{-1}$ — Arithmetic Error
  Using $\dfrac{dV}{dr} = 4\pi r^{2}$ but with the factor 12 mis-divided, e.g. $\tfrac{3}{16\pi}$ from dividing 12 by $64\pi$.
• C. $\dfrac{4}{3\pi}\ \text{cm}\,\text{s}^{-1}$ — Inversion Error
  Inverting the final fraction: $\tfrac{16\pi}{12}$ mis-simplified.
• D. $\dfrac{12}{\pi}\ \text{cm}\,\text{s}^{-1}$ — Omitted Factor
  Dividing by $4\pi$ only, forgetting the $r^{2}=4$ factor.
• E. $\dfrac{3\pi}{4}\ \text{cm}\,\text{s}^{-1}$ — Inversion Error
  Multiplying by $\dfrac{dr}{dV}$ upside down, leaving $\pi$ in the numerator.

Common Mistake (⚠️):
Multiplying by $\dfrac{dV}{dr}$ instead of dividing by it, which gives $192\pi$ and units of cm$^4$s$^{-1}$ — a dimensional check catches this instantly.

Takeaway (📌):
Related rates are always one chain-rule fraction: the rate you are given, divided by the derivative that connects the two variables. Check the units of the answer.

Correct Answer: E. $-\dfrac{9}{8}$

Fastest Approach (🚀):
$\dfrac{n(n-1)}{2!} = \dfrac{\left(\tfrac12\right)\left(-\tfrac12\right)}{2} = -\dfrac18$.
Multiply by $3^{2} = 9$: coefficient $= -\dfrac98$.

Matches Option E.

Step-by-Step Breakdown:

1. Write the general binomial series

For any real $n$ and $|u|<1$:
$(1+u)^{n} = 1+nu+\dfrac{n(n-1)}{2!}u^{2}+\cdots$

Here $u = 3x$ and $n = \tfrac12$.

2. Compute the x² coefficient of the series

$\dfrac{n(n-1)}{2!} = \dfrac{\tfrac12\left(\tfrac12-1\right)}{2} = \dfrac{\tfrac12\times\left(-\tfrac12\right)}{2} = \dfrac{-\tfrac14}{2} = -\dfrac18$

3. Substitute u = 3x

$-\dfrac18 u^{2} = -\dfrac18 (3x)^{2} = -\dfrac18\times 9x^{2} = -\dfrac98 x^{2}$

The factor of $3$ is squared — that is the step the question is testing.

4. State the answer

Coefficient of $x^{2}$ is $-\dfrac98$.

For reference the expansion begins $1+\tfrac32 x-\tfrac98 x^{2}+\cdots$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{9}{8}$ — Sign Error
  Correct magnitude but the sign from $n-1 = -\tfrac12$ dropped.
• B. $\dfrac{3}{2}$ — Misread Question
  Giving the coefficient of $x$ rather than of $x^{2}$.
• C. $-\dfrac{9}{4}$ — Factorial Error
  Dividing by $2$ instead of $2!$ at the wrong stage, doubling the magnitude.
• D. $-\dfrac{3}{8}$ — Omitted Factor
  Forgetting to square the 3 in $(3x)^{2}$.

Common Mistake (⚠️):
Forgetting to square the $3$, giving $-\tfrac38$, or dropping the negative sign that comes from $n-1 = -\tfrac12$.

Takeaway (📌):
With $(1+ax)^{n}$, every term carries $a$ to the same power as $x$. Compute the series coefficient first, then multiply by $a^{k}$.

Correct Answer: C. $-7$

Fastest Approach (🚀):
Numerator: $2(x-3)-(2x+1) = -7$ (a constant).
$\dfrac{dy}{dx} = \dfrac{-7}{(x-3)^{2}}$.
At $x=4$: $(x-3)^2 = 1$, so the gradient is $-7$.

Matches Option C.

Step-by-Step Breakdown:

1. Apply the quotient rule

With $u = 2x+1$ and $v = x-3$:
$\dfrac{dy}{dx} = \dfrac{u'v-uv'}{v^{2}} = \dfrac{2(x-3)-(2x+1)(1)}{(x-3)^{2}}$

2. Simplify the numerator first

$2(x-3)-(2x+1) = 2x-6-2x-1 = -7$

Every $x$ cancels — that is guaranteed whenever both parts are linear.

3. State the derivative

$\dfrac{dy}{dx} = \dfrac{-7}{(x-3)^{2}}$

Note it is negative everywhere: the curve is decreasing on both branches.

4. Substitute

At $x=4$: $(4-3)^{2} = 1$, so
$\dfrac{dy}{dx} = -7$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $-1$ — Formula Misuse
  Differentiating numerator and denominator separately: $\tfrac{2}{1}$ mishandled.
• B. $2$ — Omitted Term
  Differentiating only the numerator and ignoring the denominator.
• D. $7$ — Sign Error
  Reversing the order in the quotient rule, flipping the sign.
• E. $-\dfrac{1}{7}$ — Inversion Error
  Inverting the final fraction, or dividing by $-7$ instead.

Common Mistake (⚠️):
Reversing the quotient rule to $\dfrac{uv'-u'v}{v^{2}}$, which flips the sign and gives $+7$. The derivative of the numerator comes first.

Takeaway (📌):
Simplify the quotient-rule numerator before substituting. For linear over linear it is always a constant, so the sign is the only thing left to get right.

Correct Answer: D. $4$

Fastest Approach (🚀):
$\int e^{2x}dx = \tfrac12 e^{2x}$.
$e^{2\ln3} = \left(e^{\ln3}\right)^{2} = 9$.
$\tfrac12(9-1) = 4$.

Matches Option D.

Step-by-Step Breakdown:

1. Integrate

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

The $\tfrac12$ comes from reversing the chain rule on the inner function $2x$.

2. Simplify the upper limit before substituting

$e^{2\ln 3} = e^{\ln 3^{2}} = 3^{2} = 9$

Doing this first keeps the arithmetic in integers.

3. Apply the limits

$\left[\dfrac12 e^{2x}\right]_{0}^{\ln 3} = \dfrac12 e^{2\ln 3}-\dfrac12 e^{0} = \dfrac12(9)-\dfrac12(1)$

4. Evaluate

$\dfrac{9}{2}-\dfrac{1}{2} = \dfrac{8}{2} = 4$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $8$ — Omitted Factor
  Omitting the factor $\tfrac12$ from integrating $e^{2x}$.
• B. $\dfrac{9}{2}$ — Limits Error
  Dropping the lower-limit term $\tfrac12 e^{0}$.
• C. $3$ — Log Law Error
  Using $e^{2\ln3} = 2\times 3 = 6$ and mis-simplifying.
• E. $9$ — Incomplete Answer
  Giving $e^{2\ln3} = 9$ without integrating at all.

Common Mistake (⚠️):
Forgetting the $\tfrac12$ and answering $8$, or dropping the lower limit's contribution ($e^{0}=1$) and answering $\tfrac92$.

Takeaway (📌):
$e^{k\ln A} = A^{k}$ turns logarithmic limits into integers. Simplify the limit before substituting, not after.

Correct Answer: D. $\dfrac{15}{8}$

Fastest Approach (🚀):
Substitute $u = 1+x^{2}$, $du = 2x\,dx$, limits $u: 1 \to 2$.
$\tfrac12\displaystyle\int_{1}^{2}u^{3}du = \tfrac12\left[\tfrac{u^{4}}{4}\right]_{1}^{2} = \tfrac12\left(4-\tfrac14\right) = \tfrac{15}{8}$.

Matches Option D.

Step-by-Step Breakdown:

1. Recognise the structure

The inner function is $1+x^{2}$, whose derivative is $2x$. The integrand has an $x$ outside the bracket — the same up to a factor of $\tfrac12$.

2. Substitute

$u = 1+x^{2} \implies du = 2x\,dx \implies x\,dx = \tfrac12\,du$

Change the limits with the variable, so no back-substitution is needed:
$x=0 \implies u=1, \qquad x=1 \implies u=2$

3. Integrate in u

$\int_{0}^{1}x(1+x^{2})^{3}dx = \dfrac12\int_{1}^{2}u^{3}\,du = \dfrac12\left[\dfrac{u^{4}}{4}\right]_{1}^{2}$

4. Evaluate

$\dfrac12\left(\dfrac{16}{4}-\dfrac{1}{4}\right) = \dfrac12\left(\dfrac{15}{4}\right) = \dfrac{15}{8}$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{15}{4}$ — Omitted Factor
  Dropping the factor $\tfrac12$ from $x\,dx = \tfrac12du$.
• B. $\dfrac{15}{16}$ — Arithmetic Error
  Dividing by 4 twice, or applying $\tfrac12$ to an already-halved result.
• C. $4$ — Limits Error
  Using the original limits $0$ and $1$ in $u$, giving $\left[\tfrac{u^4}{8}\right]_0^1$ style errors.
• E. $\dfrac{1}{8}$ — Evaluation Error
  Integrating to $\tfrac18(1+x^2)^4$ but evaluating only at the lower limit.

Common Mistake (⚠️):
Forgetting the factor of $\tfrac12$ from $du = 2x\,dx$, which doubles the answer to $\tfrac{15}{4}$. Also common: keeping the original limits $0$ and $1$ after switching to $u$.

Takeaway (📌):
Change the limits when you change the variable. It removes the back-substitution step and the most common source of error.