ESAT Advanced Mathematics Mock Module 5 Worked Solutions

Correct Answer: A. $3$

Fastest Approach (🚀):
$\bar x = \dfrac{50}{10} = 5$.
$\sigma^{2} = 34-25 = 9 \implies \sigma = 3$.

Matches Option A.

Step-by-Step Breakdown:

1. Find the mean

$\bar x = \dfrac{\sum x}{n} = \dfrac{50}{10} = 5$

2. Apply the computational formula

$\sigma^{2} = \dfrac{\sum x^{2}}{n}-\bar x^{2} = \dfrac{340}{10}-5^{2} = 34-25 = 9$

In words: the mean of the squares minus the square of the mean.

3. Take the square root

$\sigma = \sqrt{9} = 3$

4. Why this form exists

The definition $\sigma^{2} = \dfrac{\sum(x-\bar x)^{2}}{n}$ requires every deviation. Expanding it gives the form above, which needs only two running totals — the reason calculators and spreadsheets store $\sum x$ and $\sum x^{2}$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $9$ — Incomplete Answer
  Giving the variance rather than the standard deviation.
• C. $6$ — Arithmetic Error
  Using $\sqrt{34}-\ldots$ or mis-rooting the variance.
• D. $\sqrt{34}$ — Omitted Term
  Taking the root of the mean of the squares alone.
• E. $5$ — Misread Question
  Giving the mean rather than the standard deviation.

Common Mistake (⚠️):
Reversing the two terms to $\bar x^{2}-\dfrac{\sum x^{2}}{n}$, which gives $-9$. A negative variance is impossible and should stop you immediately.

Takeaway (📌):
Mean of the squares minus square of the mean — in that order. Variance is never negative, so the order is self-checking.

Correct Answer: E. $f(x) = x^{3}-1$

Fastest Approach (🚀):
Turning point or periodic $\Rightarrow$ many-to-one.
$x^{3}-1$ has neither.

Matches Option E.

Step-by-Step Breakdown:

1. The test

A mapping is one-to-one if distinct inputs always give distinct outputs — equivalently, if no horizontal line crosses the graph more than once.

2. Eliminate

$x^{2}$: $f(-2) = f(2) = 4$. Many-to-one.
$|x|$: $f(-3) = f(3) = 3$. Many-to-one.
$\sin x$: periodic, so $f(0) = f(2\pi) = 0$ — infinitely many inputs per output.
$x^{2}+2x$: a parabola with its vertex at $x = -1$, so $f(0) = f(-2) = 0$. Many-to-one.

3. Confirm the survivor

$f(x) = x^{3}-1$ has $f'(x) = 3x^{2}\ge 0$, and is zero only at the single point $x = 0$. The function is strictly increasing throughout, so it never revisits a value. One-to-one.

4. Why this matters

Only a one-to-one function has an inverse over its whole domain. $x^{2}$ acquires an inverse only once the domain is restricted to $x \ge 0$ — which is exactly why $\sqrt{\ }$ is defined as the positive root.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $f(x) = x^{2}$ — Many-to-One
  $f(-2) = f(2)$.
• B. $f(x) = |x|$ — Many-to-One
  $f(-3) = f(3)$.
• C. $f(x) = \sin x$ — Many-to-One
  Periodic, so every value recurs infinitely often.
• D. $f(x) = x^{2}+2x$ — Many-to-One
  A parabola: $f(0) = f(-2) = 0$.

Common Mistake (⚠️):
Testing only positive inputs. Every many-to-one option here is one-to-one on $x \ge 0$; the repetition appears only when negative values are included.

Takeaway (📌):
A turning point or a period makes a function many-to-one. One-to-one functions are exactly those with an inverse on their full domain.

Correct Answer: B. $x \lt -1$ or $x \gt 3$

Fastest Approach (🚀):
$f'(x) = 3x^{2}-6x-9 = 3(x-3)(x+1)$.
Positive quadratic $\Rightarrow$ positive outside its roots.

Matches Option B.

Step-by-Step Breakdown:

1. Translate the word into a condition

A function is strictly increasing where its gradient is positive, so solve $f'(x) \gt 0$.

2. Differentiate and factorise

$f'(x) = 3x^{2}-6x-9 = 3\left(x^{2}-2x-3\right) = 3(x-3)(x+1)$

3. Solve the quadratic inequality

The roots are $x = -1$ and $x = 3$. The coefficient of $x^{2}$ is positive, so the parabola opens upwards and is above the axis outside its roots:
$x \lt -1$ or $x \gt 3$

4. Read it back against the shape

$f$ is a positive cubic: it rises, turns at $x = -1$, falls through the middle, turns again at $x = 3$, and rises thereafter. The decreasing stretch is exactly $-1 \lt x \lt 3$ — which is Option A, and the answer to the opposite question.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $-1 \lt x \lt 3$ — Direction Reversed
  The interval where the function is decreasing — the inequality is the wrong way round.
• C. $x \gt 3$ only — Partial Region
  Only half the region; the cubic is also increasing to the left of the first turning point.
• D. $x \lt -1$ only — Partial Region
  The other half only.
• E. all real $x$ — Turning Points Ignored
  True only if $f'(x)$ had no real roots, which the discriminant rules out here.

Common Mistake (⚠️):
Giving the interval between the roots. A positive quadratic is negative between its roots and positive outside them; sketching the parabola takes two seconds and settles it.

Takeaway (📌):
Increasing means $f'(x) \gt 0$, decreasing means $f'(x) \lt 0$. Factorise the derivative and sketch it — do not solve $f'(x) = 0$ and stop.

Correct Answer: C. $2$

Fastest Approach (🚀):
$b\sin A = 10\sin 40^{\circ} \approx 6.43$, and $6.43 \lt 8 \lt 10$.
Side $a$ is longer than the height but shorter than $b$ — the two-triangle case.

Matches Option C.

Step-by-Step Breakdown:

1. Apply the sine rule

$\dfrac{\sin B}{b} = \dfrac{\sin A}{a} \implies \sin B = \dfrac{10\sin 40^{\circ}}{8} \approx 0.8035$

2. Take both solutions seriously

$\sin B = 0.8035$ has two solutions in $(0^{\circ},180^{\circ})$:
$B \approx 53.5^{\circ}$ and $B \approx 180^{\circ}-53.5^{\circ} = 126.5^{\circ}$

Your calculator returns only the acute one. The obtuse partner is a genuine solution because $\sin$ is positive in the second quadrant.

3. Test each against the angle sum

$40^{\circ}+53.5^{\circ} = 93.5^{\circ} \lt 180^{\circ}$ ✓
$40^{\circ}+126.5^{\circ} = 166.5^{\circ} \lt 180^{\circ}$ ✓

Both leave room for a positive third angle, so both give a valid triangle.

4. The test that skips the trigonometry

Drop a perpendicular from $C$ to $AB$; its length is $b\sin A \approx 6.43$. Then:
$a \lt b\sin A$: no triangle
$a = b\sin A$: exactly one, right-angled
$b\sin A \lt a \lt b$: two triangles
$a \ge b$: exactly one

Here $6.43 \lt 8 \lt 10$, so two.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $1$ — Obtuse Case Missed
  Taking only the acute solution the calculator returns.
• B. $0$ — Rearrangement Error
  Concluding $\sin B \gt 1$ from a mis-arranged sine rule.
• D. $3$ — Invalid Solution
  Counting the reflex solution, which cannot be an angle of a triangle.
• E. infinitely many — Under-determined
  Treating the data as insufficient — three pieces of information do constrain the triangle.

Common Mistake (⚠️):
Taking only the calculator's acute answer and reporting one triangle. That is the single most common error in sine-rule work, and the specification names the ambiguous case explicitly.

Takeaway (📌):
With two sides and a non-included angle, compare $a$ with $b\sin A$ and with $b$. Strictly between them means two triangles.

Correct Answer: C. $2$

Fastest Approach (🚀):
$\displaystyle\sum_{r\ge 1} r t^{r} = \frac{t}{(1-t)^{2}}$ with $t = \tfrac12$.
$\dfrac{1/2}{(1/2)^{2}} = \dfrac{1/2}{1/4} = 2$.

Matches Option C.

Step-by-Step Breakdown:

1. Check the distribution is valid

$\sum_{r\ge1}\left(\tfrac12\right)^{r} = \dfrac{1/2}{1-1/2} = 1$ — a geometric series summing to $1$, so this is a genuine probability distribution.

2. Write the expectation as a series

$E(X) = \sum_{r=1}^{\infty} r\left(\dfrac12\right)^{r}$

3. Sum it

Differentiating the geometric series $\sum_{r\ge0}t^{r} = \dfrac{1}{1-t}$ and multiplying by $t$ gives
$\sum_{r\ge1}rt^{r} = \dfrac{t}{(1-t)^{2}}$

At $t = \tfrac12$:
$\dfrac{\tfrac12}{\left(\tfrac12\right)^{2}} = \dfrac{\tfrac12}{\tfrac14} = 2$

4. Sanity check by partial sums

$1(0.5)+2(0.25)+3(0.125)+4(0.0625) = 0.5+0.5+0.375+0.25 = 1.625$, and the remaining terms carry it towards $2$. Convergent, and to a finite value.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $1$ — Mode vs Mean
  Giving the most likely single value rather than the expectation.
• B. $\dfrac{3}{2}$ — Truncation Error
  Summing only the first two or three terms.
• D. $4$ — Formula Misuse
  Using $\dfrac{1}{(1-t)^{2}}$ without the factor $t$.
• E. The expectation is infinite — Convergence Error
  Assuming an infinite range implies an infinite expectation.

Common Mistake (⚠️):
Assuming that infinitely many outcomes force an infinite expectation. The probabilities decay geometrically, which is fast enough for the series to converge.

Takeaway (📌):
An expectation over a countable range is a series. If it is geometric in form, the sum-to-infinity toolkit applies unchanged.

Correct Answer: A. $x+3y+1 = 0$

Fastest Approach (🚀):
$m = -\tfrac13$; $y+1 = -\tfrac13(x-2)$.
$3y+3 = -x+2 \implies x+3y+1 = 0$.

Matches Option A.

Step-by-Step Breakdown:

1. Find the gradient

The given line has gradient $3$. Perpendicular gradients multiply to $-1$:
$m\times 3 = -1 \implies m = -\dfrac13$

2. Use the point

$y-y_1 = m(x-x_1)$ with $(2,-1)$:
$y-(-1) = -\dfrac13(x-2)$
$y+1 = -\dfrac13(x-2)$

3. Rearrange

Multiply through by $3$:
$3y+3 = -(x-2) = -x+2$
$x+3y+1 = 0$

4. Verify by substituting the point

$2+3(-1)+1 = 2-3+1 = 0$ ✓

That check is decisive and takes two seconds. Option E differs only in the constant, and fails it: $2-3-1 = -2 \neq 0$.

5. The two conditions

Parallel: same gradient, $m_1 = m_2$.
Perpendicular: $m_1m_2 = -1$, so each is the negative reciprocal of the other.

Option D keeps the original gradient, so it is parallel rather than perpendicular — the error of doing the point step and forgetting the gradient step.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $3x+y-5 = 0$ — Reciprocal Omitted
  Using $m = -3$, the negative but not the reciprocal.
• C. $x-3y-5 = 0$ — Sign Omitted
  Using $m = +\tfrac13$, the reciprocal without the sign.
• D. $y = 3x-7$ — Parallel Not Perpendicular
  Keeps the original gradient, so this line is parallel.
• E. $x+3y-1 = 0$ — Constant Error
  Right gradient, wrong constant — the point does not satisfy it.

Common Mistake (⚠️):
Using the negative of the gradient, $-3$, rather than the negative reciprocal, $-\tfrac13$. Both the sign and the inversion are required.

Takeaway (📌):
Perpendicular gradients multiply to $-1$. Find the gradient, use the point, then substitute the point back to check.

Correct Answer: C. It crosses the $x$-axis at $x = 0$.

Fastest Approach (🚀):
$2^{x} \gt 0$ always $\Rightarrow$ no $x$-intercept.
At $x = 0$, $y = 1$, not $0$.

Matches Option C.

Step-by-Step Breakdown:

1. Check the y-intercept

$2^{0} = 1$, so the curve passes through $(0,1)$. Every graph $y = a^{x}$ does, whatever the value of $a$ — a useful anchor point.

2. Check the behaviour at each end

As $x\to-\infty$, $2^{x}\to 0$ but never reaches it, so $y = 0$ is a horizontal asymptote.
As $x\to+\infty$, $2^{x}$ grows without limit.

3. Check monotonicity

Since $2 \gt 1$, each unit step to the right doubles $y$, so the function is increasing everywhere. Had the base been between $0$ and $1$ it would be decreasing everywhere instead.

4. Identify the false statement

A positive number raised to any real power is positive, so $2^{x} \gt 0$ for all $x$: the curve lies entirely above the $x$-axis and never crosses it. Option C is false — and it is contradicted by Option A, since the curve cannot be at both $(0,0)$ and $(0,1)$.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. It passes through $(0,1)$. — True Statement
  True: $2^{0} = 1$.
• B. It has the line $y = 0$ as an asymptote. — True Statement
  True: $2^{x}\to 0$ as $x\to-\infty$.
• D. It is increasing for every value of $x$. — True Statement
  True: the base exceeds $1$.
• E. It lies entirely above the $x$-axis. — True Statement
  True, and the reason Option C is false.

Common Mistake (⚠️):
Confusing the $y$-intercept with an $x$-intercept. The curve meets the $y$-axis at $1$ and meets the $x$-axis nowhere.

Takeaway (📌):
$y = a^{x}$ for $a \gt 1$: through $(0,1)$, asymptote $y = 0$, always positive, always increasing. For $0 \lt a \lt 1$ only the last changes.

Correct Answer: D. $\dfrac{3}{5}$

Fastest Approach (🚀):
$x_2 = \dfrac{1}{2}$, $x_3 = \dfrac{1}{1+\tfrac12} = \dfrac{2}{3}$, $x_4 = \dfrac{1}{1+\tfrac23} = \dfrac{3}{5}$.

Matches Option D.

Step-by-Step Breakdown:

1. Apply the rule term by term

$x_{2} = \dfrac{1}{1+1} = \dfrac{1}{2}$
$x_{3} = \dfrac{1}{1+\tfrac12} = \dfrac{1}{\tfrac32} = \dfrac{2}{3}$
$x_{4} = \dfrac{1}{1+\tfrac23} = \dfrac{1}{\tfrac53} = \dfrac{3}{5}$

2. Invert, do not divide

$\dfrac{1}{\tfrac53} = \dfrac{3}{5}$. Each step is a reciprocal of $1+x_n$, so writing the denominator as a single fraction first keeps the arithmetic to one line.

3. The pattern hiding underneath

$1,\ \tfrac12,\ \tfrac23,\ \tfrac35,\ \tfrac58,\ \ldots$ — the numerators and denominators are consecutive Fibonacci numbers. The sequence converges to $\dfrac{\sqrt5-1}{2}\approx 0.618$, the reciprocal of the golden ratio, which is the positive solution of $x = \dfrac{1}{1+x}$.

4. The index check

Three applications take $x_1$ to $x_4$. Counting four would give $x_5 = \tfrac58$ — Option E, and the reason that distractor is there.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{2}{3}$ — Off by One
  Stopping at $x_3$.
• B. $\dfrac{1}{2}$ — Off by One
  Stopping at $x_2$.
• C. $\dfrac{5}{3}$ — Reciprocal Missed
  Giving $1+x_3$ rather than its reciprocal.
• E. $\dfrac{8}{5}$ — Off by One
  Going one step too far, to $x_5$.

Common Mistake (⚠️):
Applying the rule one time too many or too few, or converting to decimals and rounding $\tfrac23$ to $0.67$ before the next step.

Takeaway (📌):
For $x_{n+1} = f(x_n)$, generate the terms in exact form and label the index at every step. Any limit satisfies $L = f(L)$.

Correct Answer: D. $12$

Fastest Approach (🚀):
$\sigma_y = |3|\times 4 = 12$.
(The mean would be $3(20)-5 = 55$, but that is not what was asked.)

Matches Option D.

Step-by-Step Breakdown:

1. Separate location from spread

For $y = ax+b$:
$\bar y = a\bar x+b$ but $\sigma_y = |a|\,\sigma_x$

2. Why the constant disappears from the spread

Standard deviation is built from deviations $x-\bar x$. Under the transformation:
$y-\bar y = (ax+b)-(a\bar x+b) = a(x-\bar x)$

The $b$ cancels in the subtraction, so only the multiplier survives.

3. Substitute

$\sigma_y = 3\times 4 = 12$

4. Note the modulus

If $a$ were negative the spread would still be positive — $y = -3x-5$ would also give $\sigma_y = 12$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $7$ — Spread Shift Error
  Applying the full transformation $3(4)-5$ to the standard deviation.
• B. $-5$ — Misread Question
  Giving the additive constant.
• C. $55$ — Misread Question
  Giving the new mean rather than the new standard deviation.
• E. $4$ — Scaling Ignored
  Leaving the spread unchanged, ignoring the multiplier as well.

Common Mistake (⚠️):
Applying the whole transformation to the standard deviation, giving $3(4)-5 = 7$. Shifting a data set moves it without stretching it.

Takeaway (📌):
$\bar y = a\bar x+b$, $\sigma_y = |a|\sigma_x$. Location feels both operations; spread feels only the multiplication.

Correct Answer: D. the third

Fastest Approach (🚀):
Take $y = -x+2$: passes $(1,1)$, $(-1,3)$, $(3,-1)$.
For $x \lt 0$, $y = -x+2 \gt 2 \gt 0$ — never in the third.

Matches Option D.

Step-by-Step Breakdown:

1. Take a concrete example

Any line with $m \lt 0$ and $c \gt 0$ behaves the same way, so use $y = -x+2$.

2. Test each quadrant

first ($x \gt 0$, $y \gt 0$): $(1,1)$ ✓
second ($x \lt 0$, $y \gt 0$): $(-1,3)$ ✓
fourth ($x \gt 0$, $y \lt 0$): $(3,-1)$ ✓
third ($x \lt 0$, $y \lt 0$): impossible

3. Prove the third case in general

For $x \lt 0$ and $m \lt 0$, the product $mx$ is positive. Adding a positive $c$ keeps $y$ positive:
$y = mx+c \gt 0$ whenever $x \lt 0$

So no point with $x \lt 0$ has $y \lt 0$, and the third quadrant is unreachable.

4. The general rule

A line misses a quadrant only when it is not horizontal or vertical, and the missed quadrant is determined by the two signs. With $m \lt 0$ and $c \lt 0$ it would miss the first instead; with $m \gt 0$ the line always meets three quadrants and misses either the second or the fourth.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. the first — Quadrant Misread
  The line does reach the first quadrant, for small positive $x$.
• B. the second — Quadrant Misread
  The line rises to the left, so it reaches the second quadrant.
• C. the fourth — Quadrant Misread
  For large enough $x$ the line drops below the axis, reaching the fourth.
• E. it passes through all four — Over-general
  A non-horizontal line always misses one quadrant.

Common Mistake (⚠️):
Reasoning only from the intercept. A positive $y$-intercept places the line in the upper half, but which quadrants it reaches depends on the gradient as well.

Takeaway (📌):
Sketch one representative line rather than arguing abstractly. Two sign conditions on $m$ and $c$ determine the quadrants exactly.

Correct Answer: C. $28$

Fastest Approach (🚀):
$F'(x) = x^{3}+1$.
$F'(3) = 27+1 = 28$.

Matches Option C.

Step-by-Step Breakdown:

1. State the theorem

If $F(x) = \displaystyle\int_{a}^{x} f(t)\,\mathrm{d}t$ then $F'(x) = f(x)$, for any constant lower limit $a$.

2. Apply it

Here $f(t) = t^{3}+1$, so $F'(x) = x^{3}+1$ and
$F'(3) = 3^{3}+1 = 28$

3. Confirm the long way

$F(x) = \left[\tfrac14 t^{4}+t\right]_{2}^{x} = \tfrac14 x^{4}+x-8$
$F'(x) = x^{3}+1$ ✓

The constant $-8$ carries the lower limit, and it differentiates away — which is exactly why the value of $a$ never affects $F'$.

4. What the theorem is saying

Integrating builds up area as $x$ moves right; the rate at which that area grows is the height of the curve at $x$. That is the whole content of the Fundamental Theorem, and the reason integration can be done by reversing differentiation at all.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $27$ — Term Omitted
  Dropping the $+1$ from the integrand.
• B. $10$ — Wrong Operation
  Substituting into $\tfrac14 x^{4}+x$ or integrating first.
• D. $9$ — Extra Differentiation
  Differentiating the integrand as well, giving $3t^{2}$ at $t\ldots$
• E. $30$ — Limit Misused
  Including the lower limit as a contribution.

Common Mistake (⚠️):
Integrating first and then evaluating $F(3)$, which answers a different question — or substituting into $t^{3}$ and forgetting the $+1$.

Takeaway (📌):
$\dfrac{\mathrm{d}}{\mathrm{d}x}\displaystyle\int_{a}^{x} f(t)\,\mathrm{d}t = f(x)$. The lower limit is irrelevant to the derivative.

Correct Answer: A. $3$

Fastest Approach (🚀):
$\dfrac{\sqrt3}{1/\sqrt3} = \sqrt3\times\sqrt3 = 3$.

Matches Option A.

Step-by-Step Breakdown:

1. Recall the two values

From the $30$–$60$–$90$ triangle with sides $1$, $\sqrt3$, $2$:
$\tan 60^{\circ} = \dfrac{\sqrt3}{1} = \sqrt3$
$\tan 30^{\circ} = \dfrac{1}{\sqrt3}$

2. Divide

$\dfrac{\sqrt3}{\frac{1}{\sqrt3}} = \sqrt3\times\dfrac{\sqrt3}{1} = 3$

Dividing by a fraction is multiplying by its reciprocal, which turns this into a product rather than a nested fraction.

3. The two triangles worth drawing

$30$–$60$–$90$: an equilateral triangle of side $2$ cut in half, giving sides $1$, $\sqrt3$, $2$.
$45$–$45$–$90$: a unit square cut along its diagonal, giving sides $1$, $1$, $\sqrt2$.

Every exact value on the specification reads off one of these two, so nothing needs memorising as a list.

4. The complete table

$\sin$: $0$, $\tfrac12$, $\tfrac{1}{\sqrt2}$, $\tfrac{\sqrt3}{2}$, $1$
$\cos$: $1$, $\tfrac{\sqrt3}{2}$, $\tfrac{1}{\sqrt2}$, $\tfrac12$, $0$
$\tan$: $0$, $\tfrac{1}{\sqrt3}$, $1$, $\sqrt3$, undefined

for $0^{\circ}$, $30^{\circ}$, $45^{\circ}$, $60^{\circ}$, $90^{\circ}$.

Note that $\tan 90^{\circ}$ is undefined, since $\cos 90^{\circ} = 0$ and $\tan = \dfrac{\sin}{\cos}$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\sqrt3$ — Division Error
  Dividing $\sqrt3$ by $1$ rather than by $\tfrac{1}{\sqrt3}$.
• C. $\tfrac13$ — Inverted
  The quotient inverted.
• D. $2$ — Wrong Ratio
  Using $\tan 60^{\circ} = 2$ from the hypotenuse.
• E. $\tfrac{1}{\sqrt3}$ — Wrong Quantity
  Quoting $\tan 30^{\circ}$ itself.

Common Mistake (⚠️):
Inverting one of the two values, giving $\tfrac13$. The tangent increases with the angle in this range, so $\tan 60^{\circ}$ must exceed $\tan 30^{\circ}$ and the quotient must exceed $1$.

Takeaway (📌):
Draw the two standard triangles rather than memorising a table. Dividing by a fraction is multiplying by its reciprocal.

Correct Answer: A. $50$

Fastest Approach (🚀):
$\dfrac{n+1}{2} = 25.5 \implies n+1 = 51 \implies n = 50$.

Matches Option A.

Step-by-Step Breakdown:

1. Write the mean in terms of n

$\sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}$, so
$\bar x = \dfrac{n(n+1)/2}{n} = \dfrac{n+1}{2}$

The $n$ cancels — the mean of the first $n$ integers is just the midpoint of $1$ and $n$.

2. Solve

$\dfrac{n+1}{2} = 25.5 \implies n+1 = 51 \implies n = 50$

3. Check

The integers $1$ to $50$ are symmetric about $25.5$, and their sum is $\dfrac{50\times 51}{2} = 1275$, giving $\dfrac{1275}{50} = 25.5$. Correct.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $51$ — Incomplete Answer
  Stopping at $n+1 = 51$.
• C. $25$ — Operation Error
  Halving the mean.
• D. $26$ — Rounding Error
  Rounding the mean up and halving.
• E. $100$ — Formula Misuse
  Doubling $50$, or using $\bar x = \tfrac{n}{4}$.

Common Mistake (⚠️):
Answering $51$ by stopping at $n+1$, or assuming the mean must be a member of the set — for even $n$ it never is.

Takeaway (📌):
For any evenly spaced set the mean equals the median equals the midpoint of the extremes. The sum formula is only needed to confirm it.

Correct Answer: C. $9$

Fastest Approach (🚀):
$\sum x = 24+66 = 90$ over $10$ values.
$\bar x = 9$.

Matches Option C.

Step-by-Step Breakdown:

1. Convert each mean to a total

$\sum x_1 = 4\times 6 = 24$
$\sum x_2 = 6\times 11 = 66$

2. Combine

$\sum x = 24+66 = 90$ over $n = 4+6 = 10$ values.

3. Divide

$\bar x = \dfrac{90}{10} = 9$

4. Check the position

The unweighted midpoint of $6$ and $11$ is $8.5$, but the larger sample sits at $11$, so the combined mean must be pulled above $8.5$. It is.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $8.5$ — Weighting Error
  Taking the unweighted average of the two means.
• B. $8$ — Weighting Error
  Weighting towards the smaller sample.
• D. $17$ — Additive Error
  Adding the two means.
• E. $9.5$ — Weighting Error
  Over-weighting the second sample.

Common Mistake (⚠️):
Averaging the two means to $8.5$, which would only be right if the samples were the same size.

Takeaway (📌):
Combined mean $= \dfrac{\sum \text{all values}}{\text{total count}}$, and it always leans towards the larger group.

Correct Answer: E. $\tfrac18$

Fastest Approach (🚀):
$\sqrt[4]{16} = 2$; $2^{3} = 8$; reciprocal $= \tfrac18$.

Matches Option E.

Step-by-Step Breakdown:

1. Read the index in three parts

$16^{-3/4}$:
the denominator $4$ means a fourth root
the numerator $3$ means cube it
the minus means take the reciprocal

2. Do the root first

$\sqrt[4]{16} = 2$, because $2^{4} = 16$.

Rooting first keeps the numbers small. Powering first would need $16^{3} = 4096$ and then its fourth root — the same answer, far more work.

3. Then the power, then the reciprocal

$2^{3} = 8$
$8^{-1} = \dfrac18$

4. A negative index never makes the answer negative

It signals a reciprocal, not a sign change. $16^{-3/4}$ is positive, and any answer below zero has confused the two — which is exactly what Options B and D offer.

5. The general form

$a^{-m/n} = \dfrac{1}{\left(\sqrt[n]{a}\right)^{m}}$

Root, power, reciprocal. The order is a choice made for arithmetic convenience, not a rule — but it is always the easier choice.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\tfrac{1}{12}$ — Index Multiplied
  Computing $\tfrac34\times 16 = 12$ and reciprocating.
• B. $-8$ — Sign Misread
  Reading the minus sign as a sign change rather than a reciprocal.
• C. $8$ — Reciprocal Omitted
  Ignoring the negative index entirely.
• D. $-12$ — Index Multiplied
  Computing $-\tfrac34\times 16$.

Common Mistake (⚠️):
Treating the negative index as a negative answer. A negative index means one over the quantity, and the result here is positive.

Takeaway (📌):
Denominator roots, numerator powers, minus sign reciprocates. Take the root first to keep the numbers manageable.

Correct Answer: C. $44$

Fastest Approach (🚀):
$3\times 12 = 36$ and $-2\times(-4) = +8$.
$36+8 = 44$.

Matches Option C.

Step-by-Step Breakdown:

1. Use linearity

Over the same range,
$\displaystyle\int_{a}^{b}\left[pf(x)+qg(x)\right]\mathrm{d}x = p\int_{a}^{b} f(x)\,\mathrm{d}x+q\int_{a}^{b} g(x)\,\mathrm{d}x$

2. Substitute the given values

$3\displaystyle\int_{2}^{5} f-2\int_{2}^{5} g = 3(12)-2(-4)$

3. Watch the double negative

$-2\times(-4) = +8$, so the total is $36+8 = 44$.

That sign is where the question is lost: subtracting a negative integral increases the answer.

4. The limits matter

This works only because both integrals run over exactly the same range. Over different ranges the values cannot be combined this way — for contiguous ranges you would instead use $\displaystyle\int_{a}^{b}+\int_{b}^{c} = \int_{a}^{c}$.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $28$ — Sign Error
  Treating $-2\times(-4)$ as $-8$.
• B. $36$ — Term Omitted
  Dropping the $g$ term entirely.
• D. $40$ — Coefficient Dropped
  Using $3(12)+(-4)$ without the factor of $2$.
• E. $32$ — Coefficient Error
  Using $2(12)+2(4)$ or another mis-collected combination.

Common Mistake (⚠️):
Computing $36-8 = 28$ by treating $-2\times(-4)$ as negative. Both minus signs are real: one from the expression, one from the value of the integral.

Takeaway (📌):
Integrals are linear: constants factor out and sums split, provided the range is shared. Contiguous ranges join end to end instead.

Correct Answer: A. $x = 120^{\circ}$ and $x = 240^{\circ}$

Fastest Approach (🚀):
Acute angle: $\cos^{-1}\tfrac12 = 60^{\circ}$.
Cosine negative in quadrants 2 and 3: $120^{\circ}$, $240^{\circ}$.

Matches Option A.

Step-by-Step Breakdown:

1. Take the acute angle from the magnitude

Ignore the sign for a moment: $\cos 60^{\circ} = \tfrac12$, so the reference angle is $60^{\circ}$.

2. Decide which quadrants

Cosine is positive in the first and fourth quadrants and negative in the second and third. The equation asks for a negative cosine, so both solutions lie there.

3. Place the reference angle in each

Second quadrant: $180^{\circ}-60^{\circ} = 120^{\circ}$
Third quadrant: $180^{\circ}+60^{\circ} = 240^{\circ}$

4. Check both

$\cos 120^{\circ} = -\tfrac12$ ✓
$\cos 240^{\circ} = -\tfrac12$ ✓

Both lie within $0^{\circ}$ to $360^{\circ}$, so both are kept.

5. Why a calculator alone is not enough

$\cos^{-1}\left(-\tfrac12\right)$ returns $120^{\circ}$ and stops. The second solution has to be constructed from the symmetry of the cosine curve, which is why Option C is there for anyone who takes the single value and moves on.

6. The quadrant rule

All positive in the first, Sine only in the second, Tangent only in the third, Cosine only in the fourth. Reading it backwards tells you where each function is negative, which is what a negative right-hand side needs.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $x = 60^{\circ}$ and $x = 300^{\circ}$ — Sign Ignored
  The quadrants where cosine is **positive**.
• C. $x = 120^{\circ}$ only — Solution Missed
  Only the value the calculator returns; the second solution is missing.
• D. $x = 210^{\circ}$ and $x = 330^{\circ}$ — Wrong Function
  Solves $\sin x = -\tfrac12$ instead.
• E. $x = 60^{\circ}$ and $x = 120^{\circ}$ — Reference Angle Kept
  Mixes the acute angle with one genuine solution.

Common Mistake (⚠️):
Giving the acute angle $60^{\circ}$ itself, or stopping at the single value the calculator returns. A negative cosine has no solution in the first quadrant.

Takeaway (📌):
Reference angle from the magnitude, quadrants from the sign, then place the angle in each and keep those inside the interval.

Correct Answer: E. $2.1$

Fastest Approach (🚀):
$0.2+1.0+0.9 = 2.1$.

Matches Option E.

Step-by-Step Breakdown:

1. Check the distribution is valid

$0.2+0.5+0.3 = 1$. Good — the probabilities are exhaustive.

2. Multiply each value by its probability

$1\times 0.2 = 0.2$
$2\times 0.5 = 1.0$
$3\times 0.3 = 0.9$

3. Add

$E(X) = 0.2+1.0+0.9 = 2.1$

4. Why there is no divisor

In a frequency mean you divide by $\sum f$. Here the weights are probabilities that already total $1$, so the division is by $1$ and disappears.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $2$ — Weighting Error
  Taking the unweighted mean of $1, 2, 3$.
• B. $0.5$ — Misread Question
  Giving the largest probability.
• C. $1.9$ — Pairing Error
  Pairing the probabilities with the values in the wrong order.
• D. $6$ — Weighting Error
  Summing the values without weighting.

Common Mistake (⚠️):
Averaging the three values to get $2$, ignoring that $X = 2$ is two and a half times as likely as $X = 1$.

Takeaway (📌):
$E(X)$ is the mean you would approach over many repetitions. It need not be a value $X$ can actually take.

Correct Answer: B. $2\sqrt3$

Fastest Approach (🚀):
$\dfrac{6\sqrt3}{3} = 2\sqrt3$.

Matches Option B.

Step-by-Step Breakdown:

1. Multiply top and bottom by the surd

$\dfrac{6}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3} = \dfrac{6\sqrt3}{\left(\sqrt3\right)^{2}} = \dfrac{6\sqrt3}{3}$

Multiplying by $\dfrac{\sqrt3}{\sqrt3}$ is multiplying by $1$, so the value is unchanged — only its form.

2. Simplify

$\dfrac{6}{3} = 2$, so the result is $2\sqrt3$.

3. Check numerically

$\sqrt3 \approx 1.732$, so $\dfrac{6}{1.732}\approx 3.46$, and $2\times 1.732 = 3.46$ ✓

4. Why rationalise at all

A surd in the denominator is awkward to evaluate by hand and awkward to combine with other fractions. Clearing it puts every expression in a comparable form, which is why answers are conventionally given that way.

5. When the denominator is a sum

For $\dfrac{1}{2+\sqrt3}$, multiply by the conjugate $\dfrac{2-\sqrt3}{2-\sqrt3}$. The difference of two squares then clears the surd: $(2+\sqrt3)(2-\sqrt3) = 4-3 = 1$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $6\sqrt3$ — Value Changed
  Only the numerator multiplied, so the value has tripled.
• C. $3\sqrt3$ — Arithmetic Error
  Dividing $6$ by $2$ rather than by $3$.
• D. $\dfrac{\sqrt3}{2}$ — Inverted
  The fraction inverted.
• E. $\sqrt{2}$ — Surd Error
  Cancelling inside the root incorrectly.

Common Mistake (⚠️):
Multiplying only the numerator by $\sqrt3$, giving $6\sqrt3$. Both parts of the fraction must be multiplied, or the value changes.

Takeaway (📌):
Multiply top and bottom by the denominator's surd; use the conjugate when the denominator is a sum or difference.

Correct Answer: B. $-6$

Fastest Approach (🚀):
$\displaystyle\int_{0}^{5} f = 8-2 = 6$.
$\displaystyle\int_{5}^{0} f = -6$.

Matches Option B.

Step-by-Step Breakdown:

1. Join the contiguous ranges

The ranges $[0,3]$ and $[3,5]$ meet at $x = 3$, so
$\displaystyle\int_{0}^{5} f(x)\,\mathrm{d}x = \int_{0}^{3} f+\int_{3}^{5} f = 8+(-2) = 6$

2. Reverse the limits

$\displaystyle\int_{b}^{a} f(x)\,\mathrm{d}x = -\int_{a}^{b} f(x)\,\mathrm{d}x$
so $\displaystyle\int_{5}^{0} f(x)\,\mathrm{d}x = -6$.

3. Why the sign flips

From $F(b)-F(a)$, swapping $a$ and $b$ gives $F(a)-F(b)$ — the same magnitude with the opposite sign. Integrating right to left counts the same area backwards.

4. What the negative value on $[3,5]$ means

$\displaystyle\int_{3}^{5} f = -2$ says the curve is below the axis over most of that stretch. The integral is $6$; the total area enclosed would be $8+2 = 10$, since area ignores sign. Option C is that value, and answers a different question.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $6$ — Reversal Missed
  Joining the ranges but not reversing the limits.
• C. $10$ — Sign Ignored
  Adding the magnitudes, which gives the total area rather than the integral.
• D. $-10$ — Sign Ignored
  Both the area error and the reversal.
• E. $2$ — Combination Error
  Subtracting $8-2$ then $-2-8$, or another mis-combination.

Common Mistake (⚠️):
Stopping at $6$ after joining the ranges, or subtracting the second integral instead of adding it because its value is negative.

Takeaway (📌):
$\displaystyle\int_{a}^{b}+\int_{b}^{c} = \int_{a}^{c}$, and $\displaystyle\int_{b}^{a} = -\int_{a}^{b}$. Signed integrals add; unsigned areas do not.

Correct Answer: D. $0.7$

Fastest Approach (🚀):
$E(X^{2}) = 1(0.2)+4(0.5)+9(0.3) = 0.2+2.0+2.7 = 4.9$.
$\text{Var} = 4.9-4.41 = 0.49 \implies \sigma = 0.7$.

Matches Option D.

Step-by-Step Breakdown:

1. Compute E(X²)

Square the values, keep the same probabilities:
$1^{2}(0.2)+2^{2}(0.5)+3^{2}(0.3) = 0.2+2.0+2.7 = 4.9$

Note this is not $\left[E(X)\right]^{2} = 4.41$ — the two differ by exactly the variance.

2. Apply the formula

$\text{Var}(X) = E(X^{2})-\left[E(X)\right]^{2} = 4.9-4.41 = 0.49$

3. Take the square root

$\sigma = \sqrt{0.49} = 0.7$

4. Sanity check

$X$ only ranges over $1$ to $3$ and is concentrated near $2$, so a spread of $0.7$ is plausible. A standard deviation larger than the range would be impossible.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $0.49$ — Incomplete Answer
  Giving the variance rather than the standard deviation.
• B. $2.1$ — Misread Question
  Giving $E(X)$.
• C. $4.9$ — Incomplete Answer
  Giving $E(X^{2})$ without subtracting.
• E. $0.3$ — Misread Question
  Giving a probability from the table.

Common Mistake (⚠️):
Squaring $E(X)$ and calling it $E(X^{2})$. Those two are equal only when the variable is constant — their difference is the variance.

Takeaway (📌):
$E(X^{2}) \ne \left[E(X)\right]^{2}$. Square the values first, then weight; never weight first and square after.

Correct Answer: A. $x = 3$ and $x = -2$

Fastest Approach (🚀):
$x^{2}-5 = x+1 \implies x^{2}-x-6 = 0$.
$(x-3)(x+2) = 0 \implies x = 3, -2$.

Matches Option A.

Step-by-Step Breakdown:

1. Eliminate y

Both equations are already solved for $y$, so set the right-hand sides equal:
$x^{2}-5 = x+1$

2. Rearrange to zero

$x^{2}-x-6 = 0$

Collect everything on the side that keeps the $x^{2}$ term positive — it makes the factorisation easier to spot.

3. Factorise

Two numbers multiplying to $-6$ and adding to $-1$: those are $-3$ and $+2$.
$(x-3)(x+2) = 0$
$x = 3$ or $x = -2$

4. Complete the solution

The question asks only for the $x$-coordinates, but a full answer pairs them:
$x = 3 \implies y = 4$
$x = -2 \implies y = -1$

The solutions are $(3,4)$ and $(-2,-1)$ — the two points where the line crosses the parabola.

5. What the number of solutions means

Two real solutions means the line cuts the curve twice. One repeated root would mean it is a tangent, and no real roots would mean it misses entirely — which is the discriminant question in another guise.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $x = -3$ and $x = 2$ — Sign Reversed
  Signs taken directly from the brackets without reversing.
• C. $x = 6$ and $x = -1$ — Rearrangement Error
  Factorising $x^{2}-5x-6$ instead.
• D. $x = 3$ only — Incomplete
  Only one root given; a quadratic here has two.
• E. no real solutions — Contradicts Discriminant
  The discriminant is $1+24 = 25 > 0$, so two real roots exist.

Common Mistake (⚠️):
Reading the factors' signs straight off the brackets. $(x-3)(x+2) = 0$ gives $x = 3$ and $x = -2$, with the signs reversed from those inside.

Takeaway (📌):
Substitute the linear into the quadratic, rearrange to zero, factorise. Pair each $x$ with its $y$ unless only one is asked for.

Correct Answer: A. $\tfrac{25}{2}$

Fastest Approach (🚀):
Antiderivative $= 2x^{3/2}+\dfrac{2}{x}$.
At $4$: $16+\tfrac12$. At $1$: $2+2 = 4$.
$16.5-4 = 12.5$.

Matches Option A.

Step-by-Step Breakdown:

1. Rewrite in index form

$3\sqrt{x} = 3x^{1/2}$ and $\dfrac{2}{x^{2}} = 2x^{-2}$, so the integrand is
$3x^{1/2}-2x^{-2}$

2. Integrate term by term

$\displaystyle\int 3x^{1/2}\,\mathrm{d}x = 3\cdot\dfrac{x^{3/2}}{3/2} = 2x^{3/2}$

$\displaystyle\int -2x^{-2}\,\mathrm{d}x = -2\cdot\dfrac{x^{-1}}{-1} = 2x^{-1} = \dfrac{2}{x}$

So the antiderivative is $2x^{3/2}+\dfrac{2}{x}$.

3. Evaluate between the limits

At $x = 4$: $2(4)^{3/2}+\dfrac{2}{4} = 2(8)+\dfrac12 = 16\tfrac12$
At $x = 1$: $2(1)+2 = 4$

$16\tfrac12-4 = \tfrac{25}{2}$

4. The two signs to watch

Integrating $x^{-2}$ raises the index to $-1$ and divides by $-1$, so the minus in front of the $2$ becomes a plus. Two negatives, and losing either gives $\tfrac{33}{2}$ or $\tfrac{15}{2}$ — both offered.

5. Why $4^{3/2} = 8$

Root first, then power: $\sqrt{4} = 2$, then $2^{3} = 8$. Taking the power first would need $\sqrt{64}$, which is the same answer by a slower route.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\tfrac{33}{2}$ — Sign Error
  Sign error on the second term, adding rather than subtracting at the limits.
• C. $16$ — Term Omitted
  Omitting the second term entirely.
• D. $\tfrac{15}{2}$ — Sign Error
  Sign error in the antiderivative of $x^{-2}$.
• E. $\tfrac{41}{2}$ — Limits Reversed
  Evaluating the limits the wrong way round on one term.

Common Mistake (⚠️):
Integrating $\dfrac{2}{x^{2}}$ as $2\ln x$. That rule belongs to $\dfrac{1}{x}$ alone; $x^{-2}$ obeys the ordinary power rule.

Takeaway (📌):
Rewrite roots and reciprocals as powers of $x$ before integrating, then apply the power rule term by term and watch the double negative.

Correct Answer: B. $8.3$

Fastest Approach (🚀):
$3(2.1)+2 = 6.3+2 = 8.3$.

Matches Option B.

Step-by-Step Breakdown:

1. Use linearity of expectation

$E(aX+b) = aE(X)+b$

The distribution itself is not needed — only $E(X)$.

2. Substitute

$E(3X+2) = 3(2.1)+2 = 6.3+2 = 8.3$

3. Contrast with the variance

Variance is not linear: $\text{Var}(aX+b) = a^{2}\text{Var}(X)$, with the $b$ dropping out entirely. Expectation feels both constants; variance feels only the multiplier, squared.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $6.3$ — Omitted Term
  Dropping the $+2$.
• C. $4.1$ — Omitted Factor
  Adding $2$ without multiplying by $3$.
• D. $2.1$ — Transformation Ignored
  Giving $E(X)$ unchanged.
• E. $15.3$ — Formula Confusion
  Squaring the multiplier as though this were a variance.

Common Mistake (⚠️):
Forgetting the $+2$, or applying the transformation as though it were a variance and squaring the $3$.

Takeaway (📌):
$E(aX+b) = aE(X)+b$ but $\text{Var}(aX+b) = a^{2}\text{Var}(X)$. Location takes both constants; spread takes the multiplier squared.

Correct Answer: C. $y = x^{3}-3x$

Fastest Approach (🚀):
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}-3 = 0 \implies x = \pm 1$: two turning points.
$x = 0 \implies y = 0$. ✓

Matches Option C.

Step-by-Step Breakdown:

1. Test the turning points

$y = x^{2}$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x = 0$ at $x = 0$ — one turning point.

$y = x^{3}$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} = 0$ at $x = 0$, but the second derivative is also zero there and the gradient does not change sign. It is a point of inflection, not a turning point — so this curve has none.

$y = x^{3}-3x$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}-3 = 0$ at $x = \pm 1$ — two turning points, a maximum at $x = -1$ and a minimum at $x = 1$.

$y = \dfrac1x$: the gradient $-\dfrac{1}{x^{2}}$ is never zero — no turning points, and it is undefined at $x = 0$ so it cannot pass through the origin either.

$y = 2^{x}$: always increasing, no turning points, and it passes through $(0,1)$ rather than the origin.

2. Test the origin

$x^{3}-3x$ at $x = 0$ gives $0$. ✓

3. Why $x^{3}$ is the sharp distractor

A cubic can have two turning points but need not. $x^{3}$ is the case where the two coincide into an inflection. Only when the derivative — a quadratic — has two distinct real roots does the cubic actually turn twice.

4. The general rule

A polynomial of degree $n$ has at most $n-1$ turning points. A quadratic has exactly one; a cubic has two or none.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $y = x^{2}$ — Wrong Count
  One turning point only.
• B. $y = x^{3}$ — Inflection Not Turning
  A point of inflection at the origin, not a turning point.
• D. $y = \dfrac{1}{x}$ — Wrong Family
  No turning points, and undefined at $x = 0$.
• E. $y = 2^{x}$ — Wrong Family
  No turning points, and passes through $(0,1)$.

Common Mistake (⚠️):
Assuming every cubic has two turning points. It has two only when its derivative has two distinct real roots, which $y = x^{3}$ does not.

Takeaway (📌):
Turning points come from $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ with a sign change. A degree-$n$ polynomial has at most $n-1$ of them.

Correct Answer: E. $0$

Fastest Approach (🚀):
$\sum x_i = n\bar x$ by definition.
$\sum(x_i-\bar x) = n\bar x-n\bar x = 0$.

Matches Option E.

Step-by-Step Breakdown:

1. Split the summation

$\sum_{i=1}^{n}\left(x_i-\bar x\right) = \sum_{i=1}^{n}x_i - \sum_{i=1}^{n}\bar x$

2. Evaluate each part

$\sum x_i = n\bar x$ — that is the definition of the mean, rearranged.
$\sum \bar x = n\bar x$ — the constant $\bar x$ added $n$ times.

3. Subtract

$n\bar x-n\bar x = 0$

4. Why this matters

The deviations always cancel, so their plain sum measures nothing. That is precisely why variance squares them first — squaring removes the cancellation and leaves a genuine measure of spread.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $n\bar x$ — Incomplete Expansion
  Giving $\sum x_i$ without subtracting the mean terms.
• B. It depends on the data — Identity Missed
  Missing that the cancellation is exact for every data set.
• C. $n\sigma^{2}$ — Measure Confusion
  Confusing the sum of deviations with the sum of *squared* deviations.
• D. $\sum x_i^{2}$ — Measure Confusion
  Giving the sum of squares.

Common Mistake (⚠️):
Answering 'it depends on the data'. The result is an identity: it holds for every data set, which is exactly what makes it useful.

Takeaway (📌):
$\sum(x-\bar x) = 0$ always. It is the reason variance squares the deviations rather than simply adding them.

Correct Answer: B. mean $53$, median $48$

Fastest Approach (🚀):
Mean: $+\dfrac{60}{20} = +3 \implies 53$.
Median: the ordering is unchanged, so it stays at $48$.

Matches Option B.

Step-by-Step Breakdown:

1. Effect on the mean

The total increases by $60$, and the mean is the total over $n$:
$\Delta\bar x = \dfrac{60}{20} = 3 \implies \bar x = 53$

2. Effect on the median

The median of $20$ values is the average of the 10th and 11th in order. Only the largest value changed, and it remained the largest, so the values in positions 1 to 19 are undisturbed.
median $= 48$, unchanged.

3. What this demonstrates

The mean is sensitive to every value; the median is sensitive only to the middle. This is the robustness that makes the median the right average for skewed data.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. mean $53$, median $51$ — Robustness Ignored
  Assuming the median shifts with the mean.
• C. mean $110$, median $48$ — Divisor Ignored
  Adding the whole $60$ to the mean.
• D. mean $50$, median $48$ — Conceptual Error
  Assuming a single value cannot move the mean.
• E. mean $52$, median $48$ — Arithmetic Error
  Dividing $60$ by a wrong count.

Common Mistake (⚠️):
Moving the median along with the mean, or adding the whole $60$ to the mean rather than sharing it across all $20$ values.

Takeaway (📌):
Changing one extreme value shifts the mean by $\dfrac{\text{change}}{n}$ and leaves the median alone. That difference is robustness.