OMPT-F mixed paper

Answer: $2 x^{2}$

Only LIKE terms combine — the same letters to the same powers: the $x^2$ terms give $2 x^{2}\mathclose{,}$ the $x$ terms $0$ and the $y$ terms $0\mathclose{,}$ so the answer is $2 x^{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Combined the $x^2$ terms with the $x$ terms as if they were alike → $2 x$

Answer: $\dfrac{169}{112}$

Multiplication comes before addition, so work out $\dfrac{9}{8} \times \dfrac{5}{6} = \dfrac{15}{16}$ first, then add $\dfrac{4}{7}\mathclose{:}$ $\dfrac{169}{112}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Worked strictly left to right, adding before multiplying → $\frac{475}{336}$

Answer: $5\left(x - 9\right)\left(x - 7\right)$

Take out the common factor $5$ first: $5 x^{2} - 80 x + 315 = 5\left(x^{2} - 16 x + 63\right)\mathclose{.}$ Then look for two numbers whose PRODUCT is $63$ and whose SUM is $-16\mathclose{:}$ they are $-9$ and $-7\mathclose{.}$ So $x^{2} - 16 x + 63 = \left(x - 9\right)\left(x - 7\right)\mathclose{,}$ and the answer is $5\left(x - 9\right)\left(x - 7\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Found the right two numbers but put the opposite signs in the brackets → $5\left(x + 7\right)\left(x + 9\right)$
• Used two numbers with the right product but the wrong sum → $5\left(x - 21\right)\left(x - 3\right)$
• Left out the common factor $5$ → $\left(x - 9\right)\left(x - 7\right)$

Answer: $\dfrac{x + 7}{x + 8}$

Factorise both: $\dfrac{\left(x - 2\right) \left(x + 7\right)}{\left(x - 2\right) \left(x + 8\right)}\mathclose{.}$ The factor $x - 2$ is common to both and cancels, leaving $\dfrac{x + 7}{x + 8}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Cancelled the wrong bracket, keeping the common one → $\frac{x - 2}{x + 8}$
• Cancelled the $x^2$ terms, which are terms and not factors → $\frac{5 x - 14}{6 x - 16}$

Answer: $3$

The point is on the graph, so its coordinates fit the formula: $21 = 2p^2 + 3\mathclose{,}$ so $p^2 = 9$ and, as $p > 0\mathclose{,}$ $p = 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to take the square root → $9$
• Gave the negative root → $-3$

Answer: $-\dfrac{3}{2}$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-1 - 2}{1 - (-1)} = -\dfrac{3}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Inverted the fraction (run over rise) → $- \frac{2}{3}$
• Subtracted the coordinates in opposite orders → $\frac{3}{2}$

Answer: $38$

Multiply both sides by $35$ to clear the denominators: $5\left(x + 4\right) = 7\left(x - 8\right)\mathclose{.}$ Expand: $5 x + 20 = 7 x - 56\mathclose{.}$ Collect: $- 2 x = -76\mathclose{,}$ so $x = 38\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied out $7\left(x - 8\right)$ as $7 x - 8\mathclose{,}$ multiplying only the $x$ → $14$
• Cross-multiplied the wrong way, writing $7\left(x + 4\right) = 5\left(x - 8\right)$ → $-34$

Answer: $x > -4$

Collect the $x$ terms on the left and the numbers on the right: $- 3 x < 12\mathclose{.}$ Divide by $-3$ — a NEGATIVE number, so the inequality sign turns round: $x > -4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not reverse the inequality when dividing by a negative number → $x < -4$
• Moved the number across without changing its sign → $x > -2$

Answer: $2$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}\mathclose{.}$ Here $a = -2$ and $b = -16\mathclose{,}$ so $x = -\dfrac{-16}{2 \cdot (-2)} = -4\mathclose{.}$ The maximum value is $f(-4) = 2\mathclose{.}$ (Completing the square gives $f(x) = -2\left(x + 4\right)^{2} + 2\mathclose{,}$ which shows the same thing.)

Common mistakes, and the answer each one gives:

• Gave the $x$-coordinate of the turning point instead of the value → $-4$
• Substituted $x = \dfrac{b}{2a}\mathclose{,}$ without the minus sign → $-126$

Answer: $y = 2 x^{2} - 8 x + 4$

In that order: the stretch gives $y = 2x^{2}\mathclose{,}$ the shifts then give $y = 2\left(x - 2\right)^{2} - 4\mathclose{,}$ which is $y = 2 x^{2} - 8 x + 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 2 x^{2} - 8 x$

Answer: $2$

Where the graphs meet, the two expressions are equal. Bring everything to one side: $x^{2} + 5 x - 14 = 0\mathclose{,}$ which factorises as $\left(x + 7\right)\left(x - 2\right) = 0\mathclose{.}$ The solutions are $-7$ and $2\mathclose{;}$ the larger is $2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $-7$
• Read the solutions off the brackets with the signs unchanged → $7$

Answer: $x < 4 \text{ or } x > 6$

Factorise: $\left(x - 4\right)\left(x - 6\right) > 0\mathclose{.}$ The parabola opens upwards and cuts the axis at $4$ and $6\mathclose{,}$ so it is above the axis outside them: $x < 4 \text{ or } x > 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the region on the wrong side of the roots → $4 < x < 6$
• Used the roots with their signs swapped → $x < -6 \text{ or } x > -4$

Answer: $(-\infty, 1]$

Complete the square: $f(x) = -\left(x + 1\right)^{2} + 1\mathclose{.}$ The vertex is $(-1, 1)$ and the parabola opens downwards, so its highest value is $1\mathclose{,}$ and every value below it is taken: the range is $(-\infty, 1]\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the $x$-coordinate of the vertex as the bound → $(-\infty, -1]$
• Took the parabola to open the other way → $[1, \infty)$

Answer: $[2, 5) \cup (5, \infty)$

The root needs $x - 2 \ge 0\mathclose{,}$ so $x \ge 2\mathclose{,}$ and the denominator is zero at $x = 5\mathclose{,}$ which is left out: the domain is $[2, 5) \cup (5, \infty)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot that $x = 5$ makes the denominator zero → $[2, \infty)$
• Required the inside of the root to be strictly positive → $(2, 5) \cup (5, \infty)$

Answer: $y = \dfrac{3}{2}$

For large $x$ only the leading terms matter: $f(x) \approx \dfrac{3 x}{2 x} = \dfrac{3}{2}\mathclose{.}$ So the horizontal asymptote is $y = \dfrac{3}{2}\mathclose{.}$ (The vertical one is $x = -\dfrac{5}{2}\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the vertical asymptote instead → $x = -\dfrac{5}{2}$
• Divided the constant terms instead of the leading coefficients → $y = \dfrac{4}{5}$

Answer: $\dfrac{25}{2}$

Multiply both sides by $x - 3\mathclose{:}$ $4 x + 7 = 6\left(x - 3\right)\mathclose{,}$ which is $4 x + 7 = 6 x - 18\mathclose{.}$ Collect: $- 2 x = -25\mathclose{,}$ so $x = \dfrac{25}{2}\mathclose{.}$ That is not $x = 3\mathclose{,}$ the value the equation forbids, so it stands.

Common mistakes, and the answer each one gives:

• Ignored the denominator and solved $4 x + 7 = 6$ → $- \frac{1}{4}$
• Multiplied out $6\left(x - 3\right)$ as $6 x - 3\mathclose{,}$ multiplying only the $x$ → $5$

Answer: $y = 3\sqrt{x + 1} + 3$

In that order: the stretch gives $y = 3\sqrt{x}\mathclose{,}$ and the shifts then give $y = 3\sqrt{x + 1} + 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 3\left(\sqrt{x + 1} + 3\right)$
• Shifted the wrong way horizontally → $y = 3\sqrt{x - 1} + 3$

Answer: $x = 9$

Put $u = \sqrt{x}\mathclose{,}$ so that $x = u^2\mathclose{:}$ $u^2 - u - 6 = (u + 2)(u - 3) = 0\mathclose{,}$ so $u = -2$ or $u = 3\mathclose{.}$ A square root is never negative, so $\sqrt{x} = -2$ is impossible; $\sqrt{x} = 3$ gives $x = 9\mathclose{.}$

Common mistakes, and the answer each one gives:

• Kept the negative value of $\sqrt{x}$ and squared it → $x = 4 \text{ or } x = 9$
• Gave the values of $u = \sqrt{x}$ and stopped → $x = -2 \text{ or } x = 3$

Answer: $\dfrac{3 x + 9}{4}$

Write $y = \dfrac{4 x - 9}{3}$ and solve for $x\mathclose{:}$ $3y = 4 x - 9\mathclose{,}$ so $x = \dfrac{3 y + 9}{4}\mathclose{.}$ Swap the letters: $f^{-1}(x) = \dfrac{3 x + 9}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the reciprocal $\frac{1}{f(x)}$ instead of the inverse → $\frac{3}{4 x - 9}$
• Undid each step, but in the original order instead of the reverse order → $\frac{3 x}{4} + 27$

Answer: $x = -2,\ x = -1,\ x = 1 \text{ or } x = 2$

Put $u = x^2\mathclose{:}$ $u^2 - 5u + 4 = (u - 1)(u - 4) = 0\mathclose{,}$ so $x^2 = 1$ or $x^2 = 4\mathclose{.}$ Each gives TWO values of $x\mathclose{:}$ $x = -2,\ x = -1,\ x = 1 \text{ or } x = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $u = x^2$ and stopped → $x = 1 \text{ or } x = 4$
• Kept only the positive square roots → $x = 1 \text{ or } x = 2$

Answer: $y = -\log_{2}\left(x - 3\right) - 2$

Reflecting in the $x$-axis changes the sign of $y\mathclose{:}$ $y = -\log_{2}(x)\mathclose{.}$ Shifting 3 units to the right replaces $x$ by $x - 3\mathclose{,}$ and 2 units down adds $-2\mathclose{:}$ $y = -\log_{2}\left(x - 3\right) - 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Reflected in the $y$-axis instead → $y = \log_{2}\left(3 - x\right) - 2$
• Shifted the wrong way horizontally → $y = -\log_{2}\left(x + 3\right) - 2$

Answer: $4 x^{6}$

The power rule turns $6\log_{6}(x)$ into $\log_{6}(x^{6})\mathclose{,}$ and the product rule combines the two into $\log_{6}\left(4 x^{6}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the arguments instead of multiplying them → $x^{6} + 4$
• Multiplied by the power instead of raising to it → $24 x$

Answer: $3$

Write both numbers as powers of $2\mathclose{:}$ $4 = 2^{2}$ and $64 = 2^{6}\mathclose{.}$ Then $\log_{4}(64) = \dfrac{6}{2} = 3\mathclose{,}$ because $4$ raised to that power is $2^{6} = 64\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided the two numbers → $16$
• Turned the fraction of exponents upside down → $\frac{1}{3}$

Answer: $11$

The difference quotient is the change in $f$ divided by the change in $x\mathclose{:}$ $\dfrac{f(4) - f(1)}{4 - 1} = \dfrac{37 - 4}{3} = 11\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not divide by the length of the interval → $33$
• Subtracted the $x$-values the other way round → $-11$

Answer: $\dfrac{5}{\cos^2(x)}$

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}\mathclose{,}$ and the quotient rule gives $\dfrac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \dfrac{1}{\cos^2(x)}\mathclose{.}$ So $f'(x) = \dfrac{5}{\cos^2(x)}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided the derivatives of the sine and the cosine → $-\dfrac{5\cos(x)}{\sin(x)}$
• Gave $\dfrac{1}{\sin^2(x)}$ in place of $\dfrac{1}{\cos^2(x)}$ → $\dfrac{5}{\sin^2(x)}$

Answer: $4x^{3}\left(\cos(4x) - x\sin(4x)\right)$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{4}$ and $v = \cos(4x)\mathclose{:}$ $u' = 4x^{3}$ and $v' = -4\sin(4x)\mathclose{.}$ So $f'(x) = 4x^{3}\cos(4x) + x^{4} \cdot \left(-4\sin(4x)\right)\mathclose{,}$ which factorises as $4x^{3}\left(\cos(4x) - x\sin(4x)\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $-16x^{3}\sin(4x)$
• Forgot the factor $4$ that the chain rule gives inside $\cos(4x)$ → $x^{3}\left(4\cos(4x) - x\sin(4x)\right)$
• Took the derivative of $\cos$ to be $+\sin$ → $x^{3}\left(4\cos(4x) + 4x\sin(4x)\right)$

Answer: $20x\left(2 x^{2} - 2\right)^{4}$

Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside: $5\left(2 x^{2} - 2\right)^{4} \cdot 4x = 20x\left(2 x^{2} - 2\right)^{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to multiply by the derivative of the inside → $5\left(2 x^{2} - 2\right)^{4}$
• Differentiated the inside and raised THAT to the power → $1024 x^{5}$

Answer: $y = 10 x - 3$

The slope is the derivative there: $f'(x) = 2 x + 4\mathclose{,}$ so $m = f'(3) = 10\mathclose{.}$ The point is $(3, f(3)) = (3, 27)\mathclose{.}$ So $y = 10\left(x - 3\right) + 27\mathclose{,}$ which is $y = 10 x - 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the $y$-value $f(3)$ as the slope → $y = 27 x - 54$
• Wrote $y = f'(a)\,x + f(a)\mathclose{,}$ leaving out the shift to $x = a$ → $y = 10 x + 27$

Answer: $4$

$f'(x) = 12 x^{2} + 12 x - 240 = 12\left(x + 5\right)\left(x - 4\right)\mathclose{,}$ which is zero at $x = -5$ and $x = 4\mathclose{.}$ $f''(x) = 24 x + 12\mathclose{,}$ and $f''(4) = 108 > 0\mathclose{,}$ so $x = 4$ is the local minimum.

Common mistakes, and the answer each one gives:

• Gave the local MAXIMUM instead → $-5$
• Gave the $y$-value rather than the $x$-coordinate → $-608$

Answer: $53$

On a closed interval the extremes are at a stationary point or at an end. $f'(x) = 3\left(x - 1\right)\left(x - 5\right)$ is zero at $x = 1$ and $x = 5\mathclose{;}$ compare $f(-1) = -28\mathclose{,}$ $f(1) = 4\mathclose{,}$ $f(5) = -28$ and $f(8) = 53\mathclose{.}$ The maximum is $53\mathclose{,}$ at the end $x = 8\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the local maximum, without checking the ends → $4$
• Gave the $x$-coordinate $x = 8$ instead of the value → $8$