OMPT-F mixed paper

Answer: $\frac{5 x + 27}{\left(x + 5\right) \left(x + 6\right)}$

The common denominator is $(x+6)(x+5)$. $\dfrac{3(x+5) + 2(x+6)}{(x+6)(x+5)} = \frac{5 x + 27}{\left(x + 5\right) \left(x + 6\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{5}{2 x + 11}$
• Used $(x+6+5)$ as the common denominator → $\frac{5}{x + 11}$

Answer: $9 x^{2} - 12 x + 4$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $9 x^{2} - 12 x + 4$.

Common mistakes, and the answer each one gives:

• Squared each term and left out the middle term → $9 x^{2} + 4$
• Gave the middle term the wrong sign → $9 x^{2} + 12 x + 4$
• Doubled the whole bracket instead of squaring it → $6 x - 4$

Answer: $\left(x + 2\right) \left(x + 8\right)$

Find two numbers whose PRODUCT is $16$ and whose SUM is $-10$: they are $-2$ and $-8$. So the factorisation is $\left(x + 2\right) \left(x + 8\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} - 10 x + 16$
• Matched the sum to the constant and the product to the middle term → $x^{2} + 9 x - 10$

Answer: $x^{11}$

Multiplying adds the exponents and dividing subtracts them: $9 + 7 - 5 = 11$, so the answer is $x^{11}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{58}$
• Divided the exponents instead of subtracting → $x^{\frac{16}{5}}$

Answer: $\frac{x - 4}{x + 4}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x - 4}{x - 2} \times \dfrac{x - 2}{x + 4}$. The $x - 2$ then cancels, leaving $\frac{x - 4}{x + 4}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} - 16}{x^{2} - 4 x + 4}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} - 4 x + 4}{x^{2} - 16}$

Answer: $\frac{5}{4}$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{5 - (0)}{4 - (0)} = \frac{5}{4}$.

Common mistakes, and the answer each one gives:

• Inverted the fraction (run over rise) → $\frac{4}{5}$
• Subtracted the coordinates in opposite orders → $- \frac{5}{4}$

Answer: $- \frac{7}{4}$

Expand the bracket: $2 x - 16 = 6 x - 9$. Collect the $x$ terms on one side and the numbers on the other, then divide: $x = - \frac{7}{4}$.

Common mistakes, and the answer each one gives:

• Multiplied only the $x$ by the number outside the bracket → $\frac{1}{4}$
• Moved a term across without changing its sign → $- \frac{17}{2}$

Answer: $43$

Cross-multiply to clear both denominators: $5\left(x - 7\right) = 4\left(x + 2\right)$. Expand, collect the $x$ terms and divide: $x = 43$.

Common mistakes, and the answer each one gives:

• Multiplied only the numerators by the other denominator → $37$
• Added the fractions instead of cross-multiplying → $3$

Answer: $-1$

$\left(x + 1\right) \left(x + 6\right) = 0$, so $x = -1$ or $x = -6$. The larger is $-1$.

Common mistakes, and the answer each one gives:

• Gave the smaller root → $-6$
• Read the roots off with the signs unchanged → $6$

Answer: $4 + \sqrt{17}$

The discriminant is $b^2 - 4ac = 68$, 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 $4 + \sqrt{17}$.

Common mistakes, and the answer each one gives:

• Used $+b$ in the numerator instead of $-b$ → $-4 + \sqrt{17}$
• Divided by $a$ instead of $2a$ → $8 + 2 \sqrt{17}$
• Gave the smaller root → $4 - \sqrt{17}$

Answer: $-3$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}$. Here $a = -2$ and $b = -12$, so $x = -\dfrac{-12}{2(-2)} = -3$. (Completing the square gives $-2\left(x + 3\right)^2 + 9$, which shows the same thing.)

Common mistakes, and the answer each one gives:

• Used $\dfrac{b}{2a}$ without the minus sign → $3$
• Gave the MAXIMUM VALUE instead of where it happens → $9$

Answer: $0$

Where the graphs meet, the two expressions are equal. Bring everything to one side: $x^{2} + 6 x = 0$, which factorises as $x \left(x + 6\right)$. The roots are $-6$ and $0$.

Common mistakes, and the answer each one gives:

• Gave the smaller root → $-6$
• Read the roots straight off the brackets with the signs unchanged → $6$

Answer: $-6$

The vertical asymptote sits where the denominator is zero: $x + 6 = 0$, so $x = -6$. (The horizontal one is $y = 2$, the ratio of the leading coefficients.)

Common mistakes, and the answer each one gives:

• Set the NUMERATOR to zero (that gives the x-intercept) → $- \frac{3}{2}$
• Gave the horizontal asymptote instead → $2$
• Forgot the sign when solving → $6$

Answer: $\frac{x + 8}{x - 7}$

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

Common mistakes, and the answer each one gives:

• Cancelled the $x^2$ terms, which are terms and not factors → $- \frac{15 x}{49} - \frac{8}{7}$
• Cancelled the wrong bracket → $\frac{x + 7}{x - 7}$

Answer: $12 x^{2} + 24 x + 16$

Square the bracket FIRST, then multiply by $3$, then add the constant — the order matters. $\left(2 x + 2\right)^2 = 4 x^{2} + 8 x + 4$, and the whole thing comes to $12 x^{2} + 24 x + 16$.

Common mistakes, and the answer each one gives:

• Multiplied the bracket by the outside number before squaring → $36 x^{2} + 72 x + 40$
• Squared each term inside the bracket separately → $12 x^{2} + 16$

Answer: $\frac{29}{5}$

Multiply both sides by $x - 3$: $2 x + 8 = 7\left(x - 3\right)$. Expand, collect and divide: $x = \frac{29}{5}$. (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 → $- \frac{1}{2}$
• Set the denominator to zero, which is the EXCLUDED value → $3$

Answer: $0$

A cube is undone by a cube root, and unlike a square root it keeps the sign: $\sqrt[3]{-64} = -4$. So $x - 4 = -4$ and $x = 0$.

Common mistakes, and the answer each one gives:

• Divided by 3 instead of taking the cube root → $- \frac{52}{3}$
• Took the cube root but lost the sign → $8$

Answer: $3 x^{4}$

The power rule turns $4\log_{10}(x)$ into $\log_{10}(x^{4})$, and the product rule combines the two into $\log_{10}(3 x^{4})$.

Common mistakes, and the answer each one gives:

• Added the arguments instead of multiplying them → $x^{4} + 3$
• Multiplied by the power instead of raising to it → $12 x$

Answer: $4$

Equal logarithms with the same base have equal arguments, so $x + 8 = 12$ and $x = 4$. Check the domain: $x + 8 > 0$ holds.

Common mistakes, and the answer each one gives:

• Subtracted the shift from the wrong side → $20$
• Treated the log as a multiplier and divided → $\frac{3}{2}$

Answer: $3$

Write the right-hand side as a power of $5$: $3125 = 5^{5}$. With the same base on both sides the exponents must be equal, so $2 x - 1 = 5$ and $x = 3$.

Common mistakes, and the answer each one gives:

• Set the exponent equal to 3125 instead of to 5 → $1563$
• Forgot to move the constant in the exponent across → $\frac{5}{2}$

Answer: $2$

Write both numbers as powers of $5$: $25 = 5^{2}$ and $625 = 5^{4}$. Then $\log_{25}625 = \dfrac{4}{2} = 2$, 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 → $25$
• Turned the fraction of exponents upside down → $\frac{1}{2}$

Answer: $72 x \left(9 x^{2} + 1\right)^{3}$

Outer power 4, inner $9x^2 + 1$, so $f' = 4(9x^2+1)^{3}\cdot 18x$. This gives $f'(x) = 72 x \left(9 x^{2} + 1\right)^{3}$.

Common mistakes, and the answer each one gives:

• Differentiated each part separately and combined them → $4 \left(9 x^{2} + 1\right)^{3}$

Answer: $-6$

$f'(x) = 12 x^{2} + 12 x - 360 = 0$ at $x = -6$ and $x = 5$. $f''(x) = 24 x + 12$, and $f''(-6) = -132 < 0$, so $x = -6$ 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 → $1512$
• Solved f(x) = 0 instead of f'(x) = 0 → $0$

Answer: $\frac{- 6 x^{2} - 6 x + 42}{x^{4} + 14 x^{2} + 49}$

The quotient rule is $\dfrac{u'v - uv'}{v^2}$ — and the ORDER of the two products matters, unlike the product rule. With $u = 6 x + 3$ and $v = x^{2} + 7$: $u' = 6$ and $v' = 2x$, giving $\frac{- 6 x^{2} - 6 x + 42}{x^{4} + 14 x^{2} + 49}$.

Common mistakes, and the answer each one gives:

• Wrote $uv' - u'v$ in the numerator, the two products swapped → $\frac{6 x^{2} + 6 x - 42}{x^{4} + 14 x^{2} + 49}$
• Differentiated top and bottom separately → $\frac{3}{x}$

Answer: $4 x^{3} - 32 x$

Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside: $2\left(x^{2} - 8\right)^{1} \times 2 x$, which expands to $4 x^{3} - 32 x$.

Common mistakes, and the answer each one gives:

• Forgot to multiply by the derivative of the inside → $2 x^{2} - 16$
• Differentiated the inside and raised THAT to the power → $4 x^{2}$

Answer: $\frac{10 x + 29}{\left(x + 2\right) \left(x + 5\right)}$

The common denominator is $(x+5)(x+2)$. $\dfrac{7(x+2) + 3(x+5)}{(x+5)(x+2)} = \frac{10 x + 29}{\left(x + 2\right) \left(x + 5\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{10}{2 x + 7}$
• Used $(x+5+2)$ as the common denominator → $\frac{10}{x + 7}$

Answer: $25 x^{2} - 4$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $25 x^{2} - 4$.

Common mistakes, and the answer each one gives:

• Kept a middle term; in this product the two cancel → $25 x^{2} + 20 x - 4$
• Added the squares instead of subtracting → $25 x^{2} + 4$
• Squared only the first term → $25 x^{2} - 2$

Answer: $\left(x - 1\right) \left(x + 3\right)$

Find two numbers whose PRODUCT is $-3$ and whose SUM is $-2$: they are $-3$ and $1$. So the factorisation is $\left(x - 1\right) \left(x + 3\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} - 2 x - 3$
• Matched the sum to the constant and the product to the middle term → $x^{2} + x - 2$

Answer: $x^{7}$

Multiplying adds the exponents and dividing subtracts them: $7 + 4 - 4 = 7$, so the answer is $x^{7}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{24}$
• Divided the exponents instead of subtracting → $x^{\frac{11}{4}}$

Answer: $\frac{x + 1}{x + 7}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x + 1}{x - 1} \times \dfrac{x - 1}{x + 7}$. The $x - 1$ then cancels, leaving $\frac{x + 1}{x + 7}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} + 8 x + 7}{x^{2} - 2 x + 1}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} - 2 x + 1}{x^{2} + 8 x + 7}$