MathMadeEasy

• $(9^{x-1})^{1/4} = 3$ (rewrite radical into fraction power format)
• Base 9 on the left, base 3 on the right (match the bases)
• $9 = 3^2$ (rewrite 9 with base 3)
• $(3^{2(x-1)})^{1/4} = 3^1$ (apply the power rule: multiply the exponents)
• $3^{\frac{2(x-1)}{4}} = 3^1$ (same base on both sides)
• $\dfrac{2(x-1)}{4} = 1$ (equality property: exponents must be equal)
• $x - 1 = 2$
• $x = 3$

Answer: C

• General percentile formula: $P_k = L + \left(\dfrac{\frac{kn}{100} - CF_b}{f}\right) \times w$ (L = lower boundary of the class, $CF_b$ = cumulative frequency before that class, f = frequency of that class, w = class width)
• Position $= \dfrac{kn}{100}$ (k = the percentile you want, n = total number of students)
• The position formula tells us which class the 70th percentile belongs to
• $n = 3+2+1+2+2+2 = 12$ (add up all frequencies to get total students)
• Cumulative frequencies: 12→3, 14→5, 16→6, 17→8, 21→10, 22→12 (running total as you move down the scores)
• $\dfrac{70 \times 12}{100} = 8.4$ (plug into the position formula: k = 70, n = 12)
• $8.4 > 8$, so the 70th percentile belongs to the 5th class (score 21)
• The 5th class has boundary 20.5–21.5, frequency 2, and cumulative frequency before it ($CF_b$) of 8
• $P_{70} = 20.5 + \left(\dfrac{8.4 - 8}{2}\right)(1) = 20.5+0.2 = 20.7$, closest to option A (20.6)

Answer: A

Watch this video for 50th percentile calculation (median)

General average rate of change formula for any function $f(x)$ in a closed interval $[a, b]$ is $\dfrac{f(b)-f(a)}{b-a}$.

Average rate of change $= \dfrac{f(1)-f(-2)}{1-(-2)}$.

$f(1) = 1+4-1 = 4$, and $f(-2) = 4-8-1 = -5$.

So the rate $= \dfrac{4-(-5)}{3} = \dfrac{9}{3} = 3$.

Answer: A

General volume formula for a frustum of any shape: $V = \dfrac{1}{3}h'\left(A_b + A'_b + \sqrt{A_b A'_b}\right)$, where $A_b$ is the lower base area, $A'_b$ is the upper base area, and $h'$ is the height. (Works for a frustum of a cone, a pyramid, or any other shape, just use that shape's own area formula for $A_b$ and $A'_b$.)

In our case it's circular, so we use $\pi r^2$ for the base areas.

Lower base area: $A_b = \pi R^2 = \pi(6)^2 = 36\pi$.

Upper base area: $A'_b = \pi r^2 = \pi(4)^2 = 16\pi$.

$\sqrt{A_b A'_b} = \sqrt{36\pi \times 16\pi} = \sqrt{576\pi^2} = 24\pi$.

$V = \dfrac{1}{3}(9)\left(36\pi+16\pi+24\pi\right) = 3(76\pi) = 228\pi \text{ cm}^3$.

Answer: B

Leading coefficient means the number in front of the highest-degree term, but we need to simplify first. The trap here: don't assume $x^5$ is the highest degree just because it appears in the unsimplified expression.

Expand: $p(x) = 2x^5-4x^3+2x^2-2x^5+x^4$.

Combine like terms: the $2x^5$ and $-2x^5$ cancel, leaving $p(x) = x^4-4x^3+2x^2$.

The highest-degree term is $x^4$, so the leading coefficient is $1$.

Answer: B

$420^\circ = 360^\circ + 60^\circ$.

$420^\circ$ is coterminal with $60^\circ$.

Coterminal angles have the same trigonometric values.

So $\cos 420^\circ = \cos 60^\circ = \dfrac12$.

Watch this to derive special angles yourself

Answer: D

Let $x$ = units of Product I, $y$ = units of Product II.

Tax collected: \$3 per unit of Product I, \$2 per unit of Product II.

Constraints given: 2 hours for Product I, 1 hour for Product II, max 40 hours/week $\rightarrow 2x + y \le 40$.

Product I cannot be manufactured more than 10 units $\rightarrow x \le 10$.

Since it's production, our units $x$ and $y$ are non-negative.

$x \ge 0$

$y \ge 0$

Our objective function is $Z = 3x + 2y$.

$\begin{cases} 2x+y\le 40 \\ x\le 10 \\ x\ge 0 \\ y\ge 0 \end{cases}$

In linear programming (LP), the maximum (or minimum) occurs at a corner point of the feasible region.

Corner points: $(0,0)$, $(10,0)$, $(10,20)$, $(0,40)$.

$Z(0,0) = 0$

$Z(10,0) = 30$

$Z(10,20) = 30+40 = 70$

$Z(0,40) = 0+80 = 80$

The largest value is at $(0,40)$: $Z=80$.

Answer: B

A universal set is defined relative to a particular discussion: it is the set containing all elements relevant to that discussion, not necessarily infinite or restricted to real numbers.

Answer: C

Total ratio parts $= 5+4+3+2 = 14$. One part $= \dfrac{56}{14} = 4$ kg.

Wheat $= 5(4)=20$ kg ✓, Barley $=4(4)=16$ kg ✓, Oat $=3(4)=12$ kg ✓, Maize $=2(4)=8$ kg.

Option D claims maize is 4 kg, but it is actually 8 kg, so D is the false statement.

Answer: D

• Class interval: the class itself, e.g. $10-19$
• Class limit: the two boundary values of that interval, if the class is $10-19$, then $10$ is the lower limit and $19$ is the upper limit
• Class width: upper limit minus lower limit, $19-10=9$
• Class mark: average of the lower and upper limits, $\dfrac{10+19}{2}=14.5$
• A is false: that's actually describing class mark, not class width
• B is false: that's also describing class mark, not class limit
• C is true: this matches the class mark definition exactly
• D is false: class interval is the grouping itself (like $10-19$), not a frequency count

Answer: C

What we'll use:

• A tangent line is always perpendicular to the radius drawn to the point of tangency
• The product of the slope of two perpendicular lines: $m_1 \times m_2 = -1$
• Completing the square method: rewrite the circle's equation in standard form $(x-h)^2+(y-k)^2=r^2$ to find center $(h, k)$
• Point-slope form: once you have a point and a slope, you can write the equation of ANY line (Lines masterclass: watch this for more equation-of-a-line past exam questions)

Now the computation:

• $x^2+y^2-4x-6y+3=0 \Rightarrow (x-2)^2+(y-3)^2=10$
• Center $(2,3)$, $r^2=10$
• Calculate the slope using center $(2,3)$ and point of tangency $(-1,2)$
• $m_1 = \dfrac{2-3}{-1-2} = \dfrac13$
• $\dfrac13 \times m_2 = -1 \Rightarrow m_2=-3$ (slope of the tangent line)
• $\dfrac{y-2}{x-(-1)} = -3$
• Cross multiply: $y-2=-3(x+1)$
• $y-2=-3x-3$
• $3x+y=-1$

Answer: C

• A: $A\cap B=B\cap A$ true, this is the commutative property
• B: $(A\cap B)\cap C=A\cap(B\cap C)$ true, this is the associative property
• C: $A\cap D=A$ true, $A\subseteq D$ means every element of $A$ is already in $D$
• D: $A\cap B=A$ not true, this would require $A\subseteq B$, which is not given anywhere in the problem

Answer: D

• 13,900 is in income bracket number 7
• The rate for that bracket is 35%, but there is a deduction of 1,500
• Tax $= 13{,}900 \times 0.35 - 1{,}500 = 4{,}865 - 1{,}500 = 3{,}365$ Birr

Answer: D

Shortcut method (fast, exam-friendly):

• $A \times A^{-1}$ must give the identity matrix $I = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$
• Pick row 2 of $A$, $[3, 0, 0]$, since it makes the multiplication easy (mostly zeros)
• Multiply row 2 of $A$ with column 2 of each choice, we need to get $1$ (matching the identity's row 2, column 2 entry)
• A: $[3,0,0]\cdot\begin{pmatrix}1\\-1\\1\end{pmatrix}$
  $= 3(1)+0(-1)+0(1)=3$ ✗
• B: $[3,0,0]\cdot\begin{pmatrix}1\\0\\0\end{pmatrix}$
  $= 3(1)+0(0)+0(0)=3$ ✗
• C: $[3,0,0]\cdot\begin{pmatrix}\frac13\\-1\\\frac23\end{pmatrix}$
  $= 3\left(\frac13\right)+0(-1)+0\left(\frac23\right)=1$ ✓
• D: $[3,0,0]\cdot\begin{pmatrix}-1\\0\\0\end{pmatrix}$
  $= 3(-1)+0(0)+0(0)=-3$ ✗

Answer: C

Formal method:

• $A^{-1} = \dfrac{\text{Adj}(A)}{|A|}$

Watch this for the formal matrix inverse method

• Group the logs to one side and the number to the other: $\log_2(x+3)-\log_2 x = 1$
• Quotient rule: $\log\left(\dfrac{a}{b}\right) = \log a - \log b$
• We're going the other way here, combining two subtracted logs back into one: $\log_2(x+3)-\log_2 x = \log_2\left(\dfrac{x+3}{x}\right)$
• So $\log_2\left(\dfrac{x+3}{x}\right)=1$
• Convert to exponential form to solve: $2^1 = \dfrac{x+3}{x}$
• $2x = x+3$
• $x=3$
• Before picking your answer, always check the result against the domain
• For a log to be defined, what's inside it must be greater than 0, it can't even be zero
• $x+3>0$ and $x>0$
• To satisfy both, the domain is $\{x: x>0\}$

Answer: D

• A: $(a^r)^s=a^{rs}$ true (power rule)
• B: $(a+b)^r=a^r+b^r$ not true (exponents don't distribute over addition)
• C: $\left(\dfrac{a}{b}\right)^r=\dfrac{a^r}{b^r}$ true (quotient rule)
• D: $(ab)^r=a^rb^r$ true (product rule)

Answer: B

Arrange the data in order (increasing or decreasing): 1, 2, 3, 4, 5, 6, 8, 8, 9, 9, 10, 10 (12 values).

With an even count, the median is the average of the 6th and 7th values: $\dfrac{6+8}{2}=7$.

Answer: C

• During the exam, you don't need to solve or derive the inequalities from scratch
• Just pick a convenient point from inside the shaded region, like $(-3, 0)$
• Plug that point into each choice and see which one it satisfies
• A: $0\le 2(-3)+4=-2$? False, eliminate A
• B: $0\ge -2(-3)+4=10$? False, eliminate B
• C: $0\ge 2(-3)+4=-2$? True.
  $0\le -(-3)+2=5$? True. Both hold
• D: $0\le -2(-3)+4=10$? True.
  $0\ge -(-3)+2=5$? False, eliminate D

Answer: C

• Total surface area = Lateral surface area + area of base
• Lateral surface area $=\dfrac12 \times (\text{perimeter of base}) \times \text{slant height} = \dfrac12 Pl$
• Area of base = area of a circle $=\pi r^2$
• To find lateral surface area we need the slant height $l$, use the Pythagorean theorem to find it
• $l=\sqrt{r^2+h^2}=\sqrt{25+144}=\sqrt{169}=13$
• Perimeter of base (a circle) $=2\pi r = 2\pi(5)=10\pi$
• Lateral surface area $=\dfrac12(10\pi)(13)=65\pi$
• Area of base $=\pi(5)^2=25\pi$
• Total surface area $=65\pi+25\pi=90\pi\text{ cm}^2$

Answer: A

• The rationalizing factor is the denominator with the opposite sign (we call it the conjugate)
• For $\sqrt3-\sqrt2$, the conjugate is $\sqrt3+\sqrt2$
• We multiply both the top and bottom by the conjugate: $\dfrac{5}{\sqrt3-\sqrt2}\times\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}$

Answer: D

By the binomial theorem, the coefficients for power 5 come from row 5 of Pascal's triangle: $1,5,10,10,5,1$.

So $(x+y)^5 = x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$.

Answer: D

$\det = \left(\dfrac35\right)\left(-\dfrac35\right) - \left(\dfrac45\right)\left(\dfrac45\right)$

$= -\dfrac{9}{25}-\dfrac{16}{25} = -\dfrac{25}{25}=-1$.

Answer: B

• General formula: Area $=|\vec u \times \vec v|$ (the area of a parallelogram spanned by two vectors is the magnitude of their cross product)
• The cross product is computed the same way as a $3\times3$ determinant, with $\hat\imath, \hat\jmath, \hat k$ as the top row: $\vec u\times\vec v = \begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 1 & 2 & 3\\ 2 & -1 & 1\end{vmatrix}$
• For the $\hat\imath$ component, cross out its row and column, leaving the 2×2 piece behind:

• $\hat\imath$ component: $(2)(1)-(3)(-1)=2+3=5$
• For the $\hat\jmath$ component, cross out its row and column:

• $\hat\jmath$ component: $-\left[(1)(1)-(3)(2)\right]=-(1-6)=5$
• For the $\hat k$ component, cross out its row and column:

• $\hat k$ component: $(1)(-1)-(2)(2)=-1-4=-5$
• $\vec u\times\vec v = 5\hat\imath+5\hat\jmath-5\hat k$
• Since we are asked for area, we need the magnitude of the cross product
• $|\vec u\times\vec v| = \sqrt{5^2+5^2+(-5)^2}=\sqrt{75}=5\sqrt3$

Answer: C

• Since $4.4$ is rounded to 1 decimal place, we add and subtract half of that place value ($0.1 \div 2 = 0.05$) to find its bounds
• $4.4-0.05=4.35$, this is our lower bound. $4.4+0.05=4.45$ is the upper bound, so $[4.35,\,4.45)$
• Same logic for $2.6$: bounds are $[2.55,\,2.65)$
• Lower bound of product $= 4.35\times2.55 = 11.0925$
• Upper bound of product $= 4.45\times2.65 = 11.7925$

Answer: C

• $x=2.137777\ldots$ (let $x$ equal the repeating decimal)
• $100x=213.7777\ldots$ (multiply by 100 to move past the two non-repeating digits)
• $1000x=2137.7777\ldots$ (multiply by 1000 to move one full repeat past the decimal)
• $1000x-100x=2137.7777\ldots-213.7777\ldots$ (subtract to cancel the repeating part)
• $900x=1924$
• $x=\dfrac{1924}{900}$
• $\dfrac{1924}{900} = \dfrac{4\times481}{4\times225}$
• $x=\dfrac{481}{225}$

Answer: D

• This is indirectly asking for the derivative of $x^2$. Using $g$ instead of the more familiar $\dfrac{f(x+h)-f(x)}{h}$ is meant to test whether you understand the concept, not just recognize a memorized formula
• The derivative of $x^2$ is $2x$

Answer: B

• A geometric series $\sum ar^n$ diverges if $|r|>1$
• A: $4^n\left(-\frac13\right)^n = \left(-\frac43\right)^n$, so $r=-\frac43$, and $|r|=\frac43>1$, so it diverges
• B: $r=-\frac23$, $|r|<1$, converges
• C: $r=0.\overline{27}\approx0.27$, $|r|<1$, converges
• D: $r=\frac{\sqrt3}{2}\approx0.87$, $|r|<1$, converges

Answer: A

Watch more challenging past exam questions

The expression is undefined where the denominator is zero: $x^2+3x+2=0 \Rightarrow (x+1)(x+2)=0$.

$\Rightarrow x=-1,-2$.

So the domain is $\{x\in\mathbb R: x\ne -1,-2\}$.

Answer: C

The tangent function repeats every $\pi$ radians. Sine and cosine repeat every $2\pi$.

Answer: D

See sine and cosine past entrance exam questions

• Case 1: $\dfrac{d_1}{d_2}=\dfrac{e_1}{e_2}=\dfrac{f_1}{f_2}$ → infinite
• Case 2: $\dfrac{d_1}{d_2}=\dfrac{e_1}{e_2}\ne\dfrac{f_1}{f_2}$ → no solution
• Case 3: $\dfrac{d_1}{d_2}\ne\dfrac{e_1}{e_2}$ → unique (one solution)

Answer: B

• Write each equation in slope form ($y=mx+b$) and compare slopes
• A: $3x-6y+1=0 \Rightarrow y=\frac12 x+\frac16$; $-x+2y-1=0 \Rightarrow y=\frac12 x+\frac12$. Same slope ($\frac12$), different intercepts, so these lines are parallel
• B: $x-y+1=0 \Rightarrow y=x+1$ (slope $1$); $x+y+1=0 \Rightarrow y=-x-1$ (slope $-1$). Different slopes, not parallel
• C: $x-3y+2=0 \Rightarrow y=\frac13x+\frac23$ (slope $\frac13$); $-3x+y+4=0 \Rightarrow y=3x-4$ (slope $3$). Different slopes, not parallel
• D: $2x-4y+2=0 \Rightarrow y=\frac12x+\frac12$ (slope $\frac12$); $-2x+y+5=0 \Rightarrow y=2x-5$ (slope $2$). Different slopes, not parallel

Answer: A

Watch this for more challenging past exam questions

• The pattern: $270, 180, 120, 80, \ldots$
• $\dfrac{180}{270}=\dfrac{120}{180}=\dfrac{80}{120}=\dfrac23$. All consecutive term ratios are equal, so this is a geometric sequence, and our common ratio is $\dfrac23$
• First term $a=270$, common ratio $r=\dfrac23$
• Since $|r|<1$, the sum to infinity is $S_\infty = \dfrac{a}{1-r}$
• $S_\infty = \dfrac{270}{1-\frac23} = \dfrac{270}{\frac13} = 810$ kg

Answer: D

• A is true, the two graphs are mirror images along the $x$-axis
• B is false, the graphs intersect at $(1,0)$ not $(0,1)$
• C is false: log functions have a vertical asymptote at $x=0$ (the $y$-axis), not a horizontal one
• D is false: the graph of $f$ is increasing and $g$ is decreasing

Answer: A

Watch this for past exam questions on exponential graphs

• General formula: mean deviation $=\dfrac{\sum|x_i-\bar x|}{n}$
• Calculate the mean: $\bar x=\dfrac{40+60+60+90+100}{5}=\dfrac{350}{5}=70$
• Find the deviations: $|40-70|=30,\ |60-70|=10,\ |60-70|=10,\ |90-70|=20,\ |100-70|=30$
• $\dfrac{\sum|x_i-\bar x|}{n} = \dfrac{30+10+10+20+30}{5}=\dfrac{100}{5}=20$

Answer: A

Watch this for deviation from mode past exam

• In cluster sampling, each cluster (subgroup) is meant to be a small-scale representation of the whole population, meaning internally heterogeneous but similar to the overall group, not "different" from it
• Option A describes the opposite of what cluster sampling actually is, so it is incorrect
• B, C, and D are all standard, correct definitions

Answer: A

Watch this for sampling technique past exam questions to understand the different sampling techniques. This question appears in a different format every year.

• A $180^\circ$ rotation about the origin maps $(x,y)\to(-x,-y)$
• $A=(2,2)\to(-2,-2)$
• $B=(2,6)\to(-2,-6)$
• $C=(5,6)\to(-5,-6)$

Answer: A

• Only row 3 changes: new $R_3 = R_3 - 2R_1 = (-1,4,2) - 2(1,2,-1)$
• $= (-1-2,\ 4-4,\ 2+2) = (-3,0,4)$
• Rows 1 and 2 stay the same, giving $\begin{pmatrix}1 & 2 & -1\\ 3 & 2 & 4\\ -3 & 0 & 4\end{pmatrix}$

Answer: D

• By the Location Theorem, check the sign of $f$ at integer points
• A: $f(0)=-3$ (negative), $f(1)=1-2+3+3-3=2$ (positive), sign change, so there exists a number between $0$ and $1$ that makes $f(x)$ zero
• B: $f(-2)=-32-32+12-6-3=-61$ (negative), $f(-1)=-1-2+3-3-3=-6$ (negative), no sign change
• C: $f(1)=2$ (positive), $f(2)=32-32+12+6-3=15$ (positive), no sign change
• D: $f(-1)=-6$ (negative), $f(0)=-3$ (negative), no sign change

Answer: A

Watch this video for the Location Theorem

• Students remember the derivative of $\sqrt{x}$ as $\dfrac{1}{2\sqrt x}$
• General chain rule for a square root: if $f(x)=\sqrt{u}$ where $u$ is a function of $x$, then $f'(x)=\dfrac{1}{2\sqrt u}\cdot u'$
• Let $u=2x^2+2x$
• $u'=4x+2$
• $f'(x) = \dfrac{1}{2\sqrt u}\cdot u' = \dfrac{4x+2}{2\sqrt{2x^2+2x}}$
• Simplify: $\dfrac{4x+2}{2\sqrt{2x^2+2x}} = \dfrac{2x+1}{\sqrt{2x^2+2x}}$

Answer: C

Watch this for more challenging chain rule past exam questions

Order of arrangement doesn't matter when selecting two cars, Car A, B is the same as Car B, A. So we use combination: $C(n,r)=\dfrac{n!}{r!(n-r)!}$.

$C(7,2) = \dfrac{7!}{2!\,5!} = \dfrac{7\times6}{2} = 21$.

Answer: B

• General formula: variance $\sigma^2=\dfrac{\sum f_i(x_i-\bar x)^2}{\sum f_i}$
• Total $n=\sum f_i=11+8+5+1=25$
• Find the mean $\bar x$
• $\bar x = \dfrac{4(11)+7(8)+8(5)+10(1)}{25} = \dfrac{44+56+40+10}{25} = \dfrac{150}{25}=6$
• $(x_i-\bar x)^2 \times f_i$
• $(4-6)^2(11)=44$
• $(7-6)^2(8)=8$
• $(8-6)^2(5)=20$
• $(10-6)^2(1)=16$
• $\sum(x_i-\bar x)^2 f_i=44+8+20+16=88$
• $\sigma^2 = \dfrac{88}{25} = 3.52$

Answer: C

• A power function is defined as $f(x)=ax^r$, $a\ne0$, $r\in\mathbb R$
• A is an exponential function
• B is a power function (correctly defined)
• C is a power function, but it wrongly allows $a$ to be zero
• D is a constant function, $f(x)=1$ for any $x$

Answer: B

Shortcut:

• The graph crosses the $y$-axis at $(0,-2)$. Substitute $x=0$ into each of the four choices and see which one gives $f(0)=-2$, this eliminates most of the choices right away
• You can also use the $x$-intercept $(-2,0)$ to double check: substitute $x=-2$ and see which choice gives $f(-2)=0$
• Or use the vertical dotted line ($x=\frac12$): substituting $\frac12$ into $x$ should make the denominator zero for the correct choice

Answer: D

Formal method: the graph shows a vertical asymptote at $x=\frac12$ and a horizontal asymptote at $y=\frac12$.

• A: $f(x)=\dfrac{x-2}{x+\frac12}$, vertical asymptote at $x=-\frac12$, horizontal asymptote $y=1$, matches neither.
• B: $f(x)=\dfrac{2x-3}{x-\frac12}$, vertical asymptote at $x=\frac12$ (matches), horizontal asymptote $y=2$ (does not match).
• C: $f(x)=\dfrac{x+2}{2x}$, vertical asymptote at $x=0$ (does not match), horizontal asymptote $y=\frac12$ (matches).
• D: $f(x)=\dfrac{x+2}{2x-1}$, vertical asymptote at $x=\frac12$ (matches), horizontal asymptote $y=\frac12$ (matches).

• Notice the multiple choice angles and the given table: $145+35=180$, $162+18=180$, $136+44=180$, such two angles are called reference angles
• Reference angles have the same trig value but may differ in sign, follow ASTC (All Students Take Calculus) to help you remember which trig value is positive or negative depending on which quadrant the original angle belongs to

• Use reference angles
• $145^\circ=180^\circ-35^\circ$
• $162^\circ=180^\circ-18^\circ$
• $136^\circ=180^\circ-44^\circ$
• A: $\tan 145^\circ=-\tan 35^\circ=-0.7002$, not $+0.7002$, so this is false
• B: $\cos 145^\circ=-\cos 35^\circ=-0.8192$, not $-0.5736$, so this is false
• C: $\cos 162^\circ=-\cos 18^\circ=-0.9511$, not $+0.9511$, so this is false
• D: $\sin 136^\circ=\sin 44^\circ=0.6947$, which is true

Answer: D

• Split the fraction: $\dfrac{x^2+1}{x^{3/4}} = \dfrac{x^2}{x^{3/4}} + \dfrac{1}{x^{3/4}}$
• Put them in separate integral signs (integration sum property): $\displaystyle\int \dfrac{x^2+1}{x^{3/4}}\,dx = \int \dfrac{x^2}{x^{3/4}}\,dx + \int \dfrac{1}{x^{3/4}}\,dx$
• Simplify each term: $\dfrac{x^2}{x^{3/4}}=x^{2-3/4}=x^{5/4}$, and $\dfrac{1}{x^{3/4}}=x^{-3/4}$
  $= \displaystyle\int x^{5/4}\,dx + \int x^{-3/4}\,dx$
• $\dfrac{4}{9}x^{9/4} + 4x^{1/4}$
• Sum: $\dfrac49 x^{9/4} + 4x^{1/4} + c = 4x^{1/4}\left(\dfrac19 x^2+1\right)+c$

Answer: B

• Drop: $20$ m (hits ground 1st time, total $=20$)
• Bounce 1: up $10$, down $10$ (hits 2nd time, total $=20+20=40$)
• Bounce 2: up $5$, down $5$ (hits 3rd time, total $=40+10=50$)
• Bounce 3: up $2.5$, down $2.5$ (hits 4th time, total $=50+5=55$)
• Bounce 4: up $1.25$, down $1.25$ (hits 5th time, total $=55+2.5=57.5$ m)

Answer: C

• A is true: a central angle ($\theta$) is formed by two radii, vertex at the center
• B is true: an inscribed angle ($\alpha$) is formed by two chords, vertex on the circle
• C is true: inscribed angle $=\dfrac12$ arc length. Semicircle arc $=180^\circ$, so $\dfrac12(180^\circ)=90^\circ$
• D is false: $\theta = $ arc length, $\alpha = \dfrac12$ arc length, the two are not the same

Answer: D

Watch this for past exams and more on this topic

• A is not true (this misstates associative property below)
• B is true (identity property)
• C is true (commutative property)
• D is true (associative property, correctly stated)

Answer: A

• Position vector formula: $\vec u = $ terminal $-$ initial
• $\vec u = (3-2,\ 5-(-4)) = (1,9)$
• $\vec u = \hat\imath+9\hat\jmath$

Answer: D

Watch this for more entrance questions on vectors

• A is true (by definition, the mode is the value that occurs most often)
• B is false (mode applies to both quantitative and qualitative data, not just quantitative)
• C is false (unlike the mean, mode is not affected by extreme values)
• D is false (a data set can have more than one mode, or none, so it's not always unique)

Answer: A

Watch this for more exam questions on mode

• General distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
• $d = \sqrt{(4-1)^2+(-2-2)^2} = \sqrt{9+16} = \sqrt{25} = 5$

Answer: D

• Splitting the pentagon into 5 triangles from the center, notice a right triangle formed by the apothem (dotted line)
• In that right triangle: $\sin 36^\circ = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{s/2}{r}$
• Solve for $s$: $s = 2r\sin 36^\circ$
• $s = 2(3)(0.5878) = 3.5268$ cm
• Perimeter $= 5s = 5(3.5268) \approx 17.63$ cm

Answer: D

Alternative method: the general formula for any regular $n$-gon with circumradius $r$ is $s=2r\sin\left(\dfrac{180^\circ}{n}\right)$.

• $s = 2(3)\sin(36^\circ) = 2(3)(0.5878) = 3.5268$ cm
• Perimeter $= 5(3.5268) \approx 17.63$ cm, the same result.

Watch this for hexagon and other related past exams

• The graph shows $(-5,0)$ and $(1,0)$ as roots (x-intercepts)
• Substitute the simplest one, $x=1$, and check if you get zero
• A: $-1+4+5=8$, not zero
• B: $-1-4+5=0$ ✓
• C: $-1+2+5=6$, not zero
• D: $-1-2+5=2$, not zero

Answer: B

Alternative method: for any quadratic $ax^2+bx+c$, the vertex's $x$-coordinate is $-\dfrac{b}{2a}$. The graph's bending point (vertex) is $(-2,9)$. Check against the multiple choices:

• A: $a=-1,b=4$, vertex $x=-\dfrac{4}{-2}=2$, not $-2$
• B: $a=-1,b=-4$, vertex $x=-\dfrac{-4}{-2}=-2$ ✓
• C: $a=-1,b=2$, vertex $x=-\dfrac{2}{-2}=1$, not $-2$
• D: $a=-1,b=-2$, vertex $x=-\dfrac{-2}{-2}=-1$, not $-2$

• A is false, only a perpendicular segment bisects a chord, not any segment

• B is false, only equal-length chords subtend equal central angles

• C is true (standard circle theorem): two equal chords are always the same perpendicular distance from the center

• D is false, same issue as A, passing through center doesn't guarantee perpendicularity

Answer: C

• For any quadratic $ax^2+bx+c=0$ with roots $r_1$ and $r_2$, the sum of the roots is $r_1+r_2=-\dfrac{b}{a}$ and the product of the roots is $r_1r_2=\dfrac{c}{a}$, this works without ever needing to solve for $r_1$ and $r_2$ individually
• Here $a=2$, $b=3$, $c=-4$, so $r_1+r_2=-\dfrac{3}{2}$ and $r_1r_2=\dfrac{-4}{2}=-2$
• $\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{r_1^2+r_2^2}{(r_1r_2)^2}=\dfrac{(r_1+r_2)^2-2r_1r_2}{(r_1r_2)^2}$
• $(r_1+r_2)^2=\left(-\dfrac32\right)^2=\dfrac94$, and $2r_1r_2=2(-2)=-4$, so $r_1^2+r_2^2=\dfrac94-(-4)=\dfrac{25}{4}$
• $(r_1r_2)^2=(-2)^2=4$
• Putting it together: $\dfrac{25/4}{4}=\dfrac{25}{16}$

Answer: B


Watch this for quadratic-related past exam questions

• $m(\angle A)=\dfrac12\big(m(\text{arc }BC)-m(\text{arc }CD)\big)$
• $15=\dfrac12\big(70-m(\text{arc }CD)\big)$
• $30=70-m(\text{arc }CD)$
• $m(\text{arc }CD)=40^\circ$

Answer: D


Watch this for related past exam questions, this has more challenging questions

• A tangent line needs two things: a point it passes through, and a slope. First find the point by plugging $x=2$ into $f$ itself: $f(2)=\dfrac{2(2)+1}{2-1}=\dfrac{5}{1}=5$, so the line passes through $(2,5)$
• The slope at any point on the curve is given by $f'(x)$, so differentiate using the quotient rule: $f'(x)=\dfrac{(x-1)(2)-(2x+1)(1)}{(x-1)^2}=\dfrac{2x-2-2x-1}{(x-1)^2}=\dfrac{-3}{(x-1)^2}$
• Evaluate the slope at $x=2$: $f'(2)=\dfrac{-3}{(2-1)^2}=\dfrac{-3}{1}=-3$
• Point-slope form: $\dfrac{y-y_1}{x-x_1}=m$. With $(x_1,y_1)=(2,5)$ and $m=-3$: $\dfrac{y-5}{x-2}=-3$
• Cross-multiply: $y-5=-3(x-2)$
• $y=-3x+6+5=-3x+11$
• Rearranged into standard form: $3x+y=11$

Answer: A

Watch this video for more on equation of line for absolute beginners

Shortcut method (fast, exam-friendly):

• Domain check first: $x^2+3x\ge0 \Rightarrow x(x+3)\ge0 \Rightarrow x\le-3$ or $x\ge0$
• The domain is $\{x: x\le-3 \text{ or } x\ge0\}$
• $-1$ fails this, so eliminate any option containing $-1$: C and D are out
• Between A $\{1,-4\}$ and B $\{1,4\}$, test $x=-4$: $\sqrt{16-12}=\sqrt4=2$ ✓; test $x=4$: $\sqrt{16+12}=\sqrt{28}\ne2$ ✗

Answer: A

Formal method:

• Square both sides: $x^2+3x=4$
• $x^2+3x-4=0 \Rightarrow (x+4)(x-1)=0 \Rightarrow x=-4 \text{ or } x=1$
• Both satisfy the domain found above, so both belong in the solution set: $\{1,-4\}$

• $S_n=\dfrac{n}{2}(a_1+a_n)$, sum of $n$ terms of an arithmetic sequence
• $a_n=a_1+(n-1)d$, $n$th term of an arithmetic sequence
• Here $a_1=11$, $d=-13$, and $n=7$
• $a_7=11+(7-1)(-13)=11+6(-13)=11-78=-67$
• $S_7=\dfrac{7}{2}\big(a_1+a_7\big)=\dfrac{7}{2}\big(11+(-67)\big)$
• $=\dfrac{7}{2}(-56)$
• $=7\times(-28)=-196$

Answer: A

• The denominator $(x+2)(x+1)^2$ has a repeated linear factor, so the decomposition needs one term for the simple factor and two terms for the repeated one: $\dfrac{3x^2+5x-1}{(x+2)(x+1)^2}=\dfrac{A}{x+2}+\dfrac{B}{x+1}+\dfrac{C}{(x+1)^2}$
• Multiply both sides by $(x+2)(x+1)^2$ to clear all the denominators: $3x^2+5x-1=A(x+1)^2+B(x+2)(x+1)+C(x+2)$
• Choosing convenient values of $x$ makes terms vanish and solves for the constants directly, no system of equations needed. Let $x=-2$ (this zeroes out the $B$ and $C$ terms): $3(4)+5(-2)-1=1=A(1)^2 \Rightarrow A=1$
• Let $x=-1$ (this zeroes out the $A$ and $B$ terms): $3(1)+5(-1)-1=-3=C(1) \Rightarrow C=-3$

Exam shortcut:

• Substitute $A=1$ and $C=-3$ into the decomposition from step 1: $\dfrac{3x^2+5x-1}{(x+2)(x+1)^2}=\dfrac{1}{x+2}+\dfrac{B}{x+1}-\dfrac{3}{(x+1)^2}$
• Only option C matches this, no need to solve for $B$ at all

Answer: C

Formal method (solve for $B$ too):

• Substituting $A=1$ and $C=-3$ back into the cleared equation above: $3x^2+5x-1=(x+1)^2+B(x+2)(x+1)-3(x+2)$, now $B$ is the only unknown left
• Match the coefficient of $x^2$ on both sides: on the right side it's $1+B$, so $3=1+B \Rightarrow B=2$
• Substituting the constants back in: $\dfrac{3x^2+5x-1}{(x+2)(x+1)^2}=\dfrac{1}{x+2}+\dfrac{2}{x+1}-\dfrac{3}{(x+1)^2}$