OMPT-A mixed paper

Answer: $\frac{3}{11}$

Write both over the common denominator $11$, combine the numerators, then cancel: $\frac{3}{11}$.

Common mistakes, and the answer each one gives:

• Added (or subtracted) numerators and denominators separately → $3$

Answer: $\frac{8}{125}$

A negative exponent turns the fraction over, and the fraction $\frac{3}{2}$ means the square root raised to the power $3$: $\left(\dfrac{4}{25}\right)^{3/2} = \frac{8}{125}$.

Common mistakes, and the answer each one gives:

• Ignored the minus sign, so the fraction was never turned over → $\frac{125}{8}$
• Took the root but never applied the power → $\frac{2}{5}$
• Turned the fraction over and stopped there → $\frac{4}{25}$

Answer: $\frac{3}{4}$

The bars mean the inside is $5$ away from zero in EITHER direction, so there are two equations: $4 x + 2 = 5$ and $4 x + 2 = -5$. They give $\frac{3}{4}$ and $- \frac{7}{4}$; the larger is $\frac{3}{4}$.

Common mistakes, and the answer each one gives:

• Gave the smaller of the two solutions → $- \frac{7}{4}$
• Solved the positive branch, then negated the answer instead of solving the second branch → $- \frac{3}{4}$

Answer: $\frac{20}{11}$

Multiplication comes before addition, so work out $\dfrac{2}{11} \times \dfrac{3}{8} = \frac{3}{44}$ first, then add $\dfrac{7}{4}$: $\frac{20}{11}$.

Common mistakes, and the answer each one gives:

• Worked strictly left to right, adding before multiplying → $\frac{255}{352}$

Answer: $\frac{11 x + 54}{\left(x + 2\right) \left(x + 6\right)}$

The common denominator is $(x+2)(x+6)$. $\dfrac{8(x+6) + 3(x+2)}{(x+2)(x+6)} = \frac{11 x + 54}{\left(x + 2\right) \left(x + 6\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{11}{2 x + 8}$
• Used $(x+2+6)$ as the common denominator → $\frac{11}{x + 8}$

Answer: $25 x^{2} - 49$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $25 x^{2} - 49$.

Common mistakes, and the answer each one gives:

• Kept a middle term; in this product the two cancel → $25 x^{2} + 70 x - 49$
• Added the squares instead of subtracting → $25 x^{2} + 49$
• Squared only the first term → $25 x^{2} - 7$

Answer: $\left(x - 7\right) \left(x + 1\right)$

Find two numbers whose PRODUCT is $-7$ and whose SUM is $6$: they are $-1$ and $7$. So the factorisation is $\left(x - 7\right) \left(x + 1\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} + 6 x - 7$
• Matched the sum to the constant and the product to the middle term → $x^{2} - 7 x + 6$

Answer: $x^{10}$

Multiplying adds the exponents and dividing subtracts them: $7 + 5 - 2 = 10$, so the answer is $x^{10}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{33}$
• Divided the exponents instead of subtracting → $x^{6}$

Answer: $\frac{x + 5}{x - 5}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x + 5}{x + 2} \times \dfrac{x + 2}{x - 5}$. The $x + 2$ then cancels, leaving $\frac{x + 5}{x - 5}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} - 25}{x^{2} + 4 x + 4}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} + 4 x + 4}{x^{2} - 25}$

Answer: $-3$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{15 - (6)}{0 - (3)} = -3$.

Common mistakes, and the answer each one gives:

• Inverted the fraction (run over rise) → $- \frac{1}{3}$
• Subtracted the coordinates in opposite orders → $3$

Answer: $\frac{55}{3}$

Expand the bracket: $7 x - 63 = 4 x - 8$. Collect the $x$ terms on one side and the numbers on the other, then divide: $x = \frac{55}{3}$.

Common mistakes, and the answer each one gives:

• Multiplied only the $x$ by the number outside the bracket → $\frac{1}{3}$
• Moved a term across without changing its sign → $\frac{79}{3}$

Answer: $\frac{67}{3}$

Cross-multiply to clear both denominators: $7\left(x - 9\right) = 4\left(x + 1\right)$. Expand, collect the $x$ terms and divide: $x = \frac{67}{3}$.

Common mistakes, and the answer each one gives:

• Multiplied only the numerators by the other denominator → $\frac{64}{3}$
• Added the fractions instead of cross-multiplying → $\frac{59}{11}$

Answer: $-6$

Eliminate $y$ by scaling and adding, or substitute. The solution is $x = -6$, $y = 4$.

Common mistakes, and the answer each one gives:

• Solved for y instead of x → $4$

Answer: $4$

The first equation already gives $y$, so put it straight into the second: $5x + 5\left(21 - 4 x\right) = 45$. That leaves one unknown, and $x = 4$.

Common mistakes, and the answer each one gives:

• Gave the value of $y$ rather than of $x$ → $5$
• Substituted into the equation it came from, which says nothing → $9$

Answer: $6$

$x \left(x - 6\right) = 0$, so $x = 0$ or $x = 6$. The larger is $6$.

Common mistakes, and the answer each one gives:

• Gave the smaller root → $0$
• Read the roots off with the signs unchanged → $0$

Answer: $\frac{3}{2} + \frac{3 \sqrt{3}}{2}$

The discriminant is $b^2 - 4ac = 108$, which is positive but not a perfect square, so the roots are irrational and the formula is needed: $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$. The larger root is $\frac{3}{2} + \frac{3 \sqrt{3}}{2}$.

Common mistakes, and the answer each one gives:

• Used $+b$ in the numerator instead of $-b$ → $- \frac{3}{2} + \frac{3 \sqrt{3}}{2}$
• Divided by $a$ instead of $2a$ → $3 + 3 \sqrt{3}$
• Gave the smaller root → $\frac{3}{2} - \frac{3 \sqrt{3}}{2}$

Answer: $6$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}$. Here $a = -2$ and $b = 24$, so $x = -\dfrac{24}{2(-2)} = 6$. (Completing the square gives $-2\left(x - 6\right)^2 - 9$, which shows the same thing.)

Common mistakes, and the answer each one gives:

• Used $\dfrac{b}{2a}$ without the minus sign → $-6$
• Gave the MAXIMUM VALUE instead of where it happens → $-9$

Answer: $-3$

Where the graphs meet, the two expressions are equal. Bring everything to one side: $x^{2} + 10 x + 21 = 0$, which factorises as $\left(x + 3\right) \left(x + 7\right)$. The roots are $-7$ and $-3$.

Common mistakes, and the answer each one gives:

• Gave the smaller root → $-7$
• Read the roots straight off the brackets with the signs unchanged → $7$

Answer: $-4$

The vertical asymptote sits where the denominator is zero: $2 x + 8 = 0$, so $x = -4$. (The horizontal one is $y = -1$, the ratio of the leading coefficients.)

Common mistakes, and the answer each one gives:

• Set the NUMERATOR to zero (that gives the x-intercept) → $4$
• Gave the horizontal asymptote instead → $-1$
• Forgot the sign when solving → $4$

Answer: $\frac{x - 6}{x + 5}$

Factorise both: $\dfrac{\left(x - 6\right) \left(x + 1\right)}{\left(x + 1\right) \left(x + 5\right)}$. The factor $x + 1$ is common to both and cancels, leaving $\frac{x - 6}{x + 5}$.

Common mistakes, and the answer each one gives:

• Cancelled the $x^2$ terms, which are terms and not factors → $\frac{- 5 x - 6}{6 x + 5}$
• Cancelled the wrong bracket → $\frac{x + 1}{x + 5}$

Answer: $100 x^{2} - 200 x + 109$

Square the bracket FIRST, then multiply by $4$, then add the constant — the order matters. $\left(5 x - 5\right)^2 = 25 x^{2} - 50 x + 25$, and the whole thing comes to $100 x^{2} - 200 x + 109$.

Common mistakes, and the answer each one gives:

• Multiplied the bracket by the outside number before squaring → $400 x^{2} - 800 x + 409$
• Squared each term inside the bracket separately → $100 x^{2} + 109$

Answer: $-10$

Multiply both sides by $x - 3$: $3 x + 4 = 2\left(x - 3\right)$. Expand, collect and divide: $x = -10$. (Check it is not $x = 3$, which the original forbids.)

Common mistakes, and the answer each one gives:

• Multiplied only the numerator by the right-hand side → $- \frac{2}{3}$
• Set the denominator to zero, which is the EXCLUDED value → $3$

Answer: $-3$

A cube is undone by a cube root, and unlike a square root it keeps the sign: $\sqrt[3]{-1} = -1$. So $x + 2 = -1$ and $x = -3$.

Common mistakes, and the answer each one gives:

• Divided by 3 instead of taking the cube root → $- \frac{7}{3}$
• Took the cube root but lost the sign → $-1$

Answer: $8 x^{2}$

The power rule turns $2\log_{3}(x)$ into $\log_{3}(x^{2})$, and the product rule combines the two into $\log_{3}(8 x^{2})$.

Common mistakes, and the answer each one gives:

• Added the arguments instead of multiplying them → $x^{2} + 8$
• Multiplied by the power instead of raising to it → $16 x$

Answer: $5$

Equal logarithms with the same base have equal arguments, so $x + 8 = 13$ and $x = 5$. Check the domain: $x + 8 > 0$ holds.

Common mistakes, and the answer each one gives:

• Subtracted the shift from the wrong side → $21$
• Treated the log as a multiplier and divided → $\frac{13}{8}$

Answer: $0$

Write the right-hand side as a power of $3$: $27 = 3^{3}$. With the same base on both sides the exponents must be equal, so $3 x + 3 = 3$ and $x = 0$.

Common mistakes, and the answer each one gives:

• Set the exponent equal to 27 instead of to 3 → $8$
• Forgot to move the constant in the exponent across → $1$

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

Write both numbers as powers of $2$: $4 = 2^{2}$ and $8 = 2^{3}$. Then $\log_{4}8 = \dfrac{3}{2} = \frac{3}{2}$, because the base has to be raised to that power to reach the argument.

Common mistakes, and the answer each one gives:

• Divided the two numbers → $2$
• Turned the fraction of exponents upside down → $\frac{2}{3}$

Answer: $64 x \left(8 x^{2} + 1\right)^{3}$

Outer power 4, inner $8x^2 + 1$, so $f' = 4(8x^2+1)^{3}\cdot 16x$. This gives $f'(x) = 64 x \left(8 x^{2} + 1\right)^{3}$.

Common mistakes, and the answer each one gives:

• Differentiated each part separately and combined them → $4 \left(8 x^{2} + 1\right)^{3}$

Answer: $-4$

$f'(x) = 9 x^{2} - 9 x - 180 = 0$ at $x = -4$ and $x = 5$. $f''(x) = 18 x - 9$, and $f''(-4) = -81 < 0$, so $x = -4$ is the maximum.

Common mistakes, and the answer each one gives:

• Gave the local MINIMUM instead → $5$
• Gave the y-value rather than the x-coordinate → $456$
• Solved f(x) = 0 instead of f'(x) = 0 → $0$

Answer: $\frac{- 4 x^{2} + 16 x + 32}{x^{4} + 16 x^{2} + 64}$

The quotient rule is $\dfrac{u'v - uv'}{v^2}$ — and the ORDER of the two products matters, unlike the product rule. With $u = 4 x - 8$ and $v = x^{2} + 8$: $u' = 4$ and $v' = 2x$, giving $\frac{- 4 x^{2} + 16 x + 32}{x^{4} + 16 x^{2} + 64}$.

Common mistakes, and the answer each one gives:

• Wrote $uv' - u'v$ in the numerator, the two products swapped → $\frac{4 x^{2} - 16 x - 32}{x^{4} + 16 x^{2} + 64}$
• Differentiated top and bottom separately → $\frac{2}{x}$