OMPT-B mixed paper

Answer: $\dfrac{4}{121}$

Multiply the numerators together and the denominators together, then cancel: $\dfrac{1}{11} \times \dfrac{4}{11} = \dfrac{4}{121}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Turned the second fraction over, as if dividing → $\frac{1}{4}$
• Multiplied the numerators but added the denominators → $\frac{2}{11}$

Answer: $\dfrac{256}{625}$

Raise the numerator and the denominator to the power $4\mathclose{:}$ $4^{4} = 256$ and $5^{4} = 625\mathclose{.}$ An even power of a negative number is positive, so the answer is $\dfrac{256}{625}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Kept the minus sign: an even power of a negative number is positive → $- \frac{256}{625}$
• Multiplied by the exponent instead of raising to it → $- \frac{16}{5}$

Answer: $x^{4}$

Multiplying adds the exponents and dividing subtracts them: $7 + 3 - 6 = 4\mathclose{,}$ so the answer is $x^{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{15}$
• Divided the exponents instead of subtracting → $x^{\frac{5}{3}}$

Answer: $\dfrac{x + 7}{x + 2}$

Factorise both: $\dfrac{\left(x + 4\right) \left(x + 7\right)}{\left(x + 2\right) \left(x + 4\right)}\mathclose{.}$ The factor $x + 4$ is common to both and cancels, leaving $\dfrac{x + 7}{x + 2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Cancelled the wrong bracket, keeping the common one → $\frac{x + 4}{x + 2}$
• Cancelled the $x^2$ terms, which are terms and not factors → $\frac{11 x + 28}{6 x + 8}$

Answer: $\dfrac{17 x + 69}{\left(x + 3\right)\left(x + 5\right)}$

The common denominator is $(x + 3)(x + 5)\mathclose{,}$ so multiply each numerator by the OTHER bracket: $\dfrac{9(x + 5) + 8(x + 3)}{\left(x + 3\right)\left(x + 5\right)} = \dfrac{17 x + 69}{\left(x + 3\right)\left(x + 5\right)}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the numerators and the denominators separately → $\dfrac{17}{2 x + 8}$
• Multiplied each numerator by its OWN denominator instead of the other one → $\dfrac{17 x + 67}{\left(x + 3\right)\left(x + 5\right)}$

Answer: $t = \dfrac{v - u}{a}$

Undo what is done to $t$ in reverse order, the same to both sides: take $u$ off both sides, $v - u = at\mathclose{,}$ then divide by $a\mathclose{.}$ So $t = \dfrac{v - u}{a}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added instead of subtracting $u$ → $t = \dfrac{v + u}{a}$
• Divided only $u$ by $a$ → $t = v - \dfrac{u}{a}$

Answer: $5$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{1 - 6}{-2 - (-1)} = 5\mathclose{.}$

Common mistakes, and the answer each one gives:

• Inverted the fraction (run over rise) → $\frac{1}{5}$
• Subtracted the coordinates in opposite orders → $-5$

Answer: $-4$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $x + \left(x + 6\right) = -2\mathclose{.}$ Expand: $2 x + 6 = -2\mathclose{,}$ so $2 x = -8$ and $x = -4\mathclose{.}$ (Then the first equation gives $y = 2\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $2$

Answer: $4$

No solution means parallel lines: the left sides must be multiples of each other. The $x$ coefficients are in the ratio $-2\mathclose{,}$ so $k = -2 \times (-2) = 4\mathclose{.}$ (The right sides are then $2$ against $4\mathclose{,}$ so the lines are parallel and distinct.)

Common mistakes, and the answer each one gives:

• Took the coefficient of $y$ from the first equation unchanged → $-2$
• Divided by the factor instead of multiplying → $1$

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

In that order: the stretch gives $y = 2x^{2}\mathclose{,}$ the shifts then give $y = 2\left(x - 1\right)^{2} + 2\mathclose{,}$ which is $y = 2 x^{2} - 4 x + 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 2 x^{2} - 4 x + 6$

Answer: $\dfrac{-4 + \sqrt{19}}{3}$

Here $a = 3\mathclose{,}$ $b = 8$ and $c = -1\mathclose{,}$ so the discriminant is $b^2 - 4ac = 64 - (-12) = 76$ — positive but not a perfect square, so the formula is needed: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-8 \pm \sqrt{76}}{6}\mathclose{.}$ Since $\sqrt{76} = 2 \sqrt{19}\mathclose{,}$ that is $\dfrac{-8 \pm 2 \sqrt{19}}{6} = \dfrac{-4 \pm \sqrt{19}}{3}\mathclose{.}$ The larger solution is $\dfrac{-4 + \sqrt{19}}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $+b$ in the numerator instead of $-b$ → $\dfrac{4 + \sqrt{19}}{3}$
• Divided by $a$ instead of $2a$ → $\dfrac{-8 + 2\sqrt{19}}{3}$
• Gave the smaller solution → $\dfrac{-4 - \sqrt{19}}{3}$

Answer: $1$

Where the graphs meet, the two expressions are equal. Bring everything to one side: $x^{2} + 2 x - 3 = 0\mathclose{,}$ which factorises as $\left(x + 3\right)\left(x - 1\right) = 0\mathclose{.}$ The solutions are $-3$ and $1\mathclose{;}$ the larger is $1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $-3$
• Read the solutions off the brackets with the signs unchanged → $3$

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

Multiply both sides by $x + 4\mathclose{:}$ $6 x + 2 = 3\left(x + 4\right)\mathclose{,}$ which is $6 x + 2 = 3 x + 12\mathclose{.}$ Collect: $3 x = 10\mathclose{,}$ so $x = \dfrac{10}{3}\mathclose{.}$ That is not $x = -4\mathclose{,}$ the value the equation forbids, so it stands.

Common mistakes, and the answer each one gives:

• Ignored the denominator and solved $6 x + 2 = 3$ → $\frac{1}{6}$
• Multiplied out $3\left(x + 4\right)$ as $3 x + 4\mathclose{,}$ multiplying only the $x$ → $\frac{2}{3}$

Answer: $(-4, 3)$

Factorise the numerator: $x^{2} + 11 x + 28 = \left(x + 4\right)\left(x + 7\right)\mathclose{.}$ The factor $\left(x + 4\right)$ cancels, so $f(x) = x + 7$ everywhere EXCEPT at $x = -4\mathclose{,}$ where the original is $\frac{0}{0}$ and undefined. The hole is at $x = -4\mathclose{,}$ at the height the simplified formula gives there: $y = -4 + 7 = 3\mathclose{.}$ So it is $(-4, 3)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Read the hole off the factor $\left(x + 4\right)$ with the wrong sign → $(4, 11)$
• Gave the point where the graph meets the $x$-axis → $(-7, 0)$

Answer: $\dfrac{- 6 x - 4}{2 x + 4}$

Write $y = \dfrac{- 4 x - 4}{2 x + 6}$ and solve for $x\mathclose{:}$ $y\left(2 x + 6\right) = - 4 x - 4\mathclose{,}$ so $2 x y + 6 y = - 4 x - 4\mathclose{.}$ Gather the $x$ terms: $x\left(2 y + 4\right) = - 6 y - 4\mathclose{,}$ so $x = \dfrac{- 6 y - 4}{2 y + 4}\mathclose{.}$ Swap the letters: $f^{-1}(x) = \dfrac{- 6 x - 4}{2 x + 4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the reciprocal $\frac{1}{f(x)}$ instead of the inverse → $\frac{2 x + 6}{- 4 x - 4}$
• Swapped $a$ and $d$ but did not change the signs of $b$ and $c$ → $\frac{6 x - 4}{2 x - 4}$

Answer: $-54$

The leading term of a product is the product of the leading terms: $\left(3 x\right)^{3} \cdot (-2x) = 27x^{3} \cdot (-2x) = -54x^{4}\mathclose{.}$ So the leading coefficient is $-54\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not raise the coefficient to its power → $-6$
• Took the leading term of $\left(2 - 2x\right)$ as $+2x$ → $54$

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

Write both numbers as powers of $5\mathclose{:}$ $25 = 5^{2}$ and $125 = 5^{3}\mathclose{.}$ Then $\log_{25}(125) = \dfrac{3}{2}\mathclose{,}$ because $25$ raised to that power is $5^{3} = 125\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided the two numbers → $5$
• Turned the fraction of exponents upside down → $\frac{2}{3}$

Answer: $x = 3 \text{ or } x = 9$

Put $u = \log_{3}(x)\mathclose{:}$ $u^{2} - 3 u + 2 = (u - 1)(u - 2) = 0\mathclose{,}$ so $u = 1$ or $u = 2\mathclose{.}$ Then $x = 3^{u}\mathclose{:}$ $x = 3^{1} = 3$ or $x = 3^{2} = 9\mathclose{.}$ Both are positive, so both logarithms exist.

Common mistakes, and the answer each one gives:

• Gave the values of $u$ and stopped → $x = 1 \text{ or } x = 2$
• Took $x = 3u$ instead of $x = 3^u$ → $x = 3 \text{ or } x = 6$

Answer: $x^{2}\left(3\sin(5x) + 5x\cos(5x)\right)$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{3}$ and $v = \sin(5x)\mathclose{:}$ $u' = 3x^{2}$ and $v' = 5\cos(5x)\mathclose{.}$ So $f'(x) = 3x^{2}\sin(5x) + x^{3} \cdot 5\cos(5x)\mathclose{,}$ which factorises as $x^{2}\left(3\sin(5x) + 5x\cos(5x)\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $15x^{2}\cos(5x)$
• Forgot the factor $5$ that the chain rule gives inside $\sin(5x)$ → $x^{2}\left(3\sin(5x) + x\cos(5x)\right)$

Answer: $20 x^{2} - 26$

Put $f(x)$ in place of every $x$ in the outer function: $g(f(x)) = 4\left(5 x^{2} - 7\right) + 2\mathclose{.}$ Altogether $g(f(x)) = 20 x^{2} - 26\mathclose{.}$

Common mistakes, and the answer each one gives:

• Composed the wrong way round, working out $f(g(x))$ → $80 x^{2} + 80 x + 13$
• Multiplied $f(x)$ by $g(x)$ instead of composing them → $20 x^{3} + 10 x^{2} - 28 x - 14$

Answer: $-5$

$f'(x) = 6 x^{2} + 24 x - 30 = 6\left(x + 5\right)\left(x - 1\right)\mathclose{,}$ which is zero at $x = -5$ and $x = 1\mathclose{.}$ $f''(x) = 12 x + 24\mathclose{,}$ and $f''(-5) = -36 < 0\mathclose{,}$ so $x = -5$ is the local maximum.

Common mistakes, and the answer each one gives:

• Gave the local MINIMUM instead → $1$
• Gave the $y$-value rather than the $x$-coordinate → $200$

Answer: $6$

At an inflection point $f''$ changes sign. $f''(x) = 6 x - 36\mathclose{,}$ which is zero at $x = 6$ and changes sign there, from negative to positive.

Common mistakes, and the answer each one gives:

• Solved $f'(x) = 0$ and gave the smaller root → $3$
• Solved $f''(x) = 0$ with the sign turned round → $-6$

Answer: $104$

The three angles of any triangle add to $180^\circ\mathclose{,}$ so the third is $180 - 24 - 52 = 104\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $360^\circ\mathclose{,}$ the angles round a POINT → $284$
• Subtracted only one of the two given angles → $156$

Answer: $-1$

$-\frac{\pi}{4}$ is a special angle, so the value is exact. Its reference value is $1\mathclose{,}$ and the quadrant fixes the sign: $\tan\left(-\frac{\pi}{4}\right) = -1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the right size but the wrong sign for that quadrant → $1$

Answer: $5$

The amplitude is the size of the number in front, $|5| = 5\mathclose{:}$ the graph swings $5$ above and below its middle line.

Common mistakes, and the answer each one gives:

• Gave the vertical shift instead → $-1$
• Gave the coefficient of $x$ instead → $\frac{1}{3}$

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: $\dfrac{23}{3}$

The graph crosses the axis at $x = 2\mathclose{:}$ below it on $[0, 2]$ and above it on $[2, 3]\mathclose{.}$ An area is never negative, so integrate the two parts separately and add their sizes: $\displaystyle\int_{0}^{2} \left(4 - x^{2}\right) dx + \int_{2}^{3} \left(x^{2} - 4\right) dx = \dfrac{16}{3} + \dfrac{7}{3} = \dfrac{23}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the plain integral from $0$ to the end, so the part below the axis counted negative → $-3$
• Gave only the part below the axis → $\dfrac{16}{3}$

Answer: $8 \pi$

About the $y$-axis the slices are discs of radius $x\mathclose{,}$ with $x^{2} = y\mathclose{:}$ $V = \pi\displaystyle\int_{0}^{4} x^{2} \, dy = \pi\int_{0}^{4} y \, dy = 8 \pi\mathclose{.}$

Common mistakes, and the answer each one gives:

• Rotated about the $x$-axis instead → $\frac{32 \pi}{5}$
• Left out the $\pi$ → $8$

Answer: $1$

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

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $0$
• Got the sign of the second term wrong → $-1$

Answer: $\dfrac{1}{4}\ln\left(\frac{5}{3}\right)$

Split into partial fractions: $\dfrac{1}{\left(x - 1\right)\left(x + 3\right)} = \dfrac{1}{4}\left(\dfrac{1}{x - 1} - \dfrac{1}{x + 3}\right)\mathclose{.}$ The antiderivative is $\dfrac{1}{4}\ln\left|\dfrac{x - 1}{x + 3}\right|\mathclose{;}$ from $2$ to $3$ that is $\dfrac{1}{4}\ln\left(\frac{5}{3}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot the factor $\dfrac{1}{4}$ from the partial fractions → $\ln\left(\frac{5}{3}\right)$
• Added the two logarithms instead of subtracting them → $\dfrac{1}{4}\ln\left(\frac{12}{5}\right)$