OMPT-D mixed paper

Answer: $\dfrac{49}{132}$

Write both fractions over the lowest common denominator, then combine the numerators: $\dfrac{5}{11} - \dfrac{1}{12} = \dfrac{60}{132} - \dfrac{11}{132} = \dfrac{49}{132}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added (or subtracted) the numerators and the denominators separately → $-4$
• Wrote both over $132$ but did not change the numerators → $\frac{1}{33}$

Answer: $-\dfrac{27}{64}$

Raise the numerator and the denominator to the power $3\mathclose{:}$ $3^{3} = 27$ and $4^{3} = 64\mathclose{.}$ An odd power of a negative number is negative, so the answer is $-\dfrac{27}{64}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Dropped the minus sign: an odd power of a negative number is negative → $\frac{27}{64}$
• Multiplied by the exponent instead of raising to it → $- \frac{9}{4}$

Answer: $\dfrac{103}{15}$

Multiplication comes before addition, so work out $\dfrac{6}{5} \times \dfrac{7}{2} = \dfrac{21}{5}$ first, then add $\dfrac{8}{3}\mathclose{:}$ $\dfrac{103}{15}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Worked strictly left to right, adding before multiplying → $\frac{203}{15}$

Answer: $\left(x - 4\right)\left(x + 4\right)$

$x^{2} - 16 = x^{2} - 4^{2}$ is a difference of two squares, $p^2 - q^2 = (p - q)(p + q)\mathclose{,}$ so the answer is $\left(x - 4\right)\left(x + 4\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Wrote it as a perfect square → $\left(x - 4\right)^{2}$

Answer: $\dfrac{10}{3}$

Expand the bracket: $4 x + 12 = 7 x + 2\mathclose{.}$ Put the $x$ terms on the left and the numbers on the right, changing the sign of each term that crosses over: $- 3 x = -10\mathclose{,}$ so $x = \dfrac{10}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied only the $x$ by the number outside the bracket → $\frac{1}{3}$
• Moved the number across without changing its sign → $- \frac{14}{3}$

Answer: $-26$

Multiply both sides by $20$ to clear the denominators: $4\left(x - 9\right) = 5\left(x - 2\right)\mathclose{.}$ Expand: $4 x - 36 = 5 x - 10\mathclose{.}$ Collect: $- x = 26\mathclose{,}$ so $x = -26\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied out $5\left(x - 2\right)$ as $5 x - 2\mathclose{,}$ multiplying only the $x$ → $-34$
• Cross-multiplied the wrong way, writing $5\left(x - 9\right) = 4\left(x - 2\right)$ → $37$

Answer: $2$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $4 x + 4\left(4 x - 10\right) = 0\mathclose{.}$ Expand: $20 x - 40 = 0\mathclose{,}$ so $20 x = 40$ and $x = 2\mathclose{.}$ (Then the first equation gives $y = -2\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $-2$
• Multiplied only the first term of the bracket by $4$ → $\frac{1}{2}$

Answer: $\text{infinitely many solutions}$

The second equation's left side is $2$ times the first's. So is its right side $(2 \times (-8) = -16)\mathclose{:}$ it is the SAME line, and every point on it is a solution.

Common mistakes, and the answer each one gives:

• Saw two different equations and concluded the lines never meet → $\text{no solution}$
• Assumed two lines always cross exactly once → $\text{exactly one solution}$

Answer: $4$

Half the coefficient of $x$ is $-4\mathclose{,}$ so $x^{2} - 8 x + 20 = \left(x - 4\right)^{2} - (-4)^{2} + 20 = \left(x - 4\right)^{2} + 4\mathclose{.}$ So $q = 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the square of half the coefficient instead of subtracting it → $36$
• Subtracted the square of the WHOLE coefficient → $-44$

Answer: $9$

Exactly one solution means $D = b^2 - 4ac = 0\mathclose{:}$ $(-6)^{2} - 4c = 0\mathclose{,}$ so $c = \dfrac{36}{4} = 9\mathclose{.}$ (Then the left side is $\left(x - 3\right)^{2}\mathclose{.})$

Common mistakes, and the answer each one gives:

• Left the 4 out: set $b^2 - c = 0$ → $36$
• Halved $b$ but did not square it → $3$

Answer: $y = x - 1$

Divide: $x^{2} - 5 x - 4 = \left(x - 4\right)\left(x - 1\right) - 8\mathclose{,}$ so $f(x) = x - 1 - \dfrac{8}{x - 4}\mathclose{.}$ The fraction tends to $0$ as $x$ grows, so the asymptote is $y = x - 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the vertical asymptote instead → $x = 4$
• Divided only the leading terms, $x^2 \div x$ → $y = x$
• Divided by $x$ instead of by $x - 4$ → $y = x - 5$

Answer: $\dfrac{5}{6}$

Divide every term by $x^2\mathclose{,}$ the highest power: the fraction is $\dfrac{5 + \frac{6}{x} + \frac{6}{x^2}}{6 - \frac{5}{x} + \frac{1}{x^2}}\mathclose{,}$ and every term with $x$ underneath tends to $0\mathclose{.}$ The limit is $\dfrac{5}{6}\mathclose{,}$ the graph's horizontal asymptote.

Common mistakes, and the answer each one gives:

• Divided the constant terms → $6$
• Divided the coefficients of $x$ → $-\dfrac{6}{5}$

Answer: $y = 3\sqrt{x - 4} + 2$

In that order: the stretch gives $y = 3\sqrt{x}\mathclose{,}$ and the shifts then give $y = 3\sqrt{x - 4} + 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 3\left(\sqrt{x - 4} + 2\right)$
• Shifted the wrong way horizontally → $y = 3\sqrt{x + 4} + 2$

Answer: $\left(x - 6\right)^{2} + 5 \text{ for } x \ge 6$

Write $y = \sqrt{x - 5} + 6$ and undo the steps in reverse order: take $6$ off, $y - 6 = \sqrt{x - 5}\mathclose{;}$ square, $\left(y - 6\right)^{2} = x - 5\mathclose{;}$ then $x = \left(y - 6\right)^{2} + 5\mathclose{.}$ Swap the letters: $f^{-1}(x) = \left(x - 6\right)^{2} + 5\mathclose{.}$ A square root is never negative, so $f$ only takes values $y \ge 6$ — and those are the $x$ the inverse is defined for: $x \ge 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Squared both sides before taking the $6$ off → $x^{2} - 1$
• Undid the steps in the original order instead of the reverse order → $\left(x + 5\right)^{2} - 6$

Answer: $2 x^{3}$

The power rule turns $3\log_{3}(x)$ into $\log_{3}(x^{3})\mathclose{,}$ and the product rule combines the two into $\log_{3}\left(2 x^{3}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the arguments instead of multiplying them → $x^{3} + 2$
• Multiplied by the power instead of raising to it → $6 x$

Answer: $20$

$\log_{5}(\text{something}) = 2$ means $\text{something} = 5^{2} = 25\mathclose{.}$ So $x + 5 = 25$ and $x = 20\mathclose{.}$ (Check: $x + 5$ is then positive, as a logarithm needs.)

Common mistakes, and the answer each one gives:

• Multiplied the base by $2$ instead of raising it to the power $2$ → $5$
• Forgot to subtract $5$ at the end → $25$
• Raised the exponent to the power of the base, $2^{5}$ → $27$

Answer: $16 x^{3} + 6 x + 5$

Term by term (the sum rule): $x^n$ becomes $nx^{n-1}\mathclose{,}$ a number in front stays in front, and the constant $6$ becomes $0\mathclose{.}$ So $f'(x) = 16 x^{3} + 6 x + 5\mathclose{.}$

Common mistakes, and the answer each one gives:

• Kept the constant $6$ instead of making it $0$ → $16 x^{3} + 6 x + 11$
• Multiplied by each power but did not lower it → $16 x^{4} + 6 x^{2} + 5 x$

Answer: $3x^{2}\left(\sin(3x) + x\cos(3x)\right)$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{3}$ and $v = \sin(3x)\mathclose{:}$ $u' = 3x^{2}$ and $v' = 3\cos(3x)\mathclose{.}$ So $f'(x) = 3x^{2}\sin(3x) + x^{3} \cdot 3\cos(3x)\mathclose{,}$ which factorises as $3x^{2}\left(\sin(3x) + x\cos(3x)\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $9x^{2}\cos(3x)$
• Forgot the factor $3$ that the chain rule gives inside $\sin(3x)$ → $x^{2}\left(3\sin(3x) + x\cos(3x)\right)$

Answer: $-5$

$f'(x) = 6 x^{2} - 12 x - 210 = 6\left(x + 5\right)\left(x - 7\right)\mathclose{,}$ which is zero at $x = -5$ and $x = 7\mathclose{.}$ $f''(x) = 12 x - 12\mathclose{,}$ and $f''(-5) = -72 < 0\mathclose{,}$ so $x = -5$ is the local maximum.

Common mistakes, and the answer each one gives:

• Gave the local MINIMUM instead → $7$
• Gave the $y$-value rather than the $x$-coordinate → $650$

Answer: $0$

$\frac{3\pi}{2}$ is a special angle, so the value is exact. Its reference value is $0\mathclose{,}$ and the quadrant fixes the sign: $\cos\left(\frac{3\pi}{2}\right) = 0\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the sine instead → $-1$

Answer: $\frac{2\pi}{3}$

Adding or taking off whole turns of $2\pi$ does not move a point on the unit circle. $\frac{20\pi}{3} - 6\pi = \frac{2\pi}{3}\mathclose{,}$ which lies in $[0, 2\pi)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took off one turn too few → $\frac{8\pi}{3}$
• Took off a multiple of $\pi$ that is not a whole number of turns → $-\frac{\pi}{3}$

Answer: $\frac{\pi}{6}$

$\arcsin$ gives the ONE angle $y$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ whose sine is $\frac{1}{2}\mathclose{:}$ that is $\frac{\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave a solution of $\sin\,y = \frac{1}{2}$ that lies outside the range of $\arcsin$ → $\frac{5\pi}{6}$
• Gave the angle in degrees → $30$

Answer: $\frac{44}{3}$

Integrate term by term: $\int \left(5 x^{2} + 3\right) dx = \frac{5 x^{3}}{3} + 3 x\mathclose{.}$ Then substitute the upper limit and subtract the lower: $\left[\frac{5 x^{3}}{3} + 3 x\right]_{-2}^{-1} = - \frac{14}{3} - \left(- \frac{58}{3}\right) = \frac{44}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round → $- \frac{44}{3}$
• Substituted the limits into the ORIGINAL function, not its integral → $-15$

Answer: $\dfrac{256}{3}$

They meet where $x^{2} = 2 x + 15\mathclose{,}$ that is $x^{2} - 2 x - 15 = \left(x + 3\right)\left(x - 5\right) = 0\mathclose{:}$ at $x = -3$ and $x = 5\mathclose{.}$ Between them the line is on top, so the area is $\displaystyle\int_{-3}^{5} \left(- x^{2} + 2 x + 15\right) dx = \dfrac{256}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round, parabola minus line → $- \frac{256}{3}$
• Integrated only the parabola between the meeting points → $\frac{152}{3}$

Answer: $\frac{\pi}{2} - 1$

By parts with $u = x$ and $dv = \cos(x)\,dx\mathclose{,}$ so $v = \sin(x)\mathclose{:}$ $\int x\cos(x)\,dx = x \cdot \sin(x) - \int \sin(x)\,dx = x\sin(x) + \cos(x)\mathclose{.}$ From $0$ to $\frac{\pi}{2}$ that is $\frac{\pi}{2} - 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $\frac{\pi}{2}$
• Got the sign of the second term wrong → $\frac{\pi}{2} + 1$

Answer: $13$

Pythagoras: the square on the hypotenuse equals the sum of the squares on the legs. $5^2 + 12^2 = 169\mathclose{,}$ and $\sqrt{169} = 13\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the legs instead of their squares → $17$
• Subtracted the squares, which finds a LEG, not the hypotenuse → $\sqrt{119}$

Answer: $\frac{25 \pi}{6}$

The arc is $\dfrac{150}{360} = \dfrac{5}{12}$ of the circumference $2\pi r\mathclose{:}$ $\dfrac{5}{12} \times 2\pi \times 5 = \frac{25 \pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the area of the sector instead → $\frac{125 \pi}{12}$
• Used the diameter in place of the radius → $\frac{25 \pi}{3}$

Answer: $y = - \frac{x}{3} - \frac{16}{3}$

The given line has slope $3\mathclose{,}$ so a perpendicular one has slope $-\dfrac{1}{3}\mathclose{.}$ Through $(2, -6)\mathclose{:}$ $y = -\dfrac{1}{3}\left(x - 2\right) - 6\mathclose{,}$ which is $y = - \frac{x}{3} - \frac{16}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the same slope, which gives a PARALLEL line → $y = 3 x - 12$
• Used $-3$ as the slope instead of $-\frac{1}{3}$ → $y = - 3 x$

Answer: $\dfrac{35}{13}$

For $ax + by + c = 0$ the distance is $\dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\mathclose{.}$ Here the top is $|12 \cdot (-4) + 5 \cdot 3 - 2| = 35$ and the bottom $\sqrt{169} = 13\mathclose{,}$ so the distance is $\dfrac{35}{13}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by $\sqrt{169}$ → $35$
• Divided by $169$ instead of by its square root → $\frac{35}{169}$
• Left out the absolute value → $- \frac{35}{13}$

Answer: $\dfrac{29}{4}$

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3 t^{2} + 2}{2 t - 2}\mathclose{.}$ At $t = 3$ that is $\dfrac{29}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided $dx/dt$ by $dy/dt\mathclose{,}$ upside down → $\frac{4}{29}$
• Gave $dy/dt$ alone → $29$