CSCA sample maths questions with answers

Answer: B, $\varnothing$

1. The left side is a perfect square: $x^2-6x+9=(x-3{)}^2$.
2. A square is $\ge 0$ for every $x$, and equals $0$ only at $x=3$.
3. A square is never negative, so it is never below $0$: no solution.
4. So the solution set is $\varnothing$.

Why the other options are wrong

• A That set solves $(x-3{)}^2>0$, not $(x-3{)}^2<0$.
• C That set solves $(x-3{)}^2\le 0$, not $(x-3{)}^2<0$.
• D That set solves $(x-3{)}^2\ge 0$, not $(x-3{)}^2<0$.

Answer: C, $(-\infty ,-5)\cup (-5,4]$

1. $(x+5{)}^0$ needs a non-zero base: $x\ne -5$ ($0^0$ is undefined).
2. $\sqrt{4-x}$ needs $4-x\ge 0$: $x\le 4$.
3. So the domain is $(-\infty ,-5)\cup (-5,4]$.

Why the other options are wrong

• A $0^0$ is undefined, so $x=-5$ (where the base is $0$) is excluded.
• B $\sqrt{0}$ is defined, so $x=4$ is in the domain.
• D $4-x\ge 0$ means $x\le 4$.

Answer: A, $-\frac{120}{169}$

1. $\cos \alpha =-\frac{12}{13}$ (from ${\sin }^2\alpha +{\cos }^2\alpha =1$, sign from the second quadrant).
2. $\sin 2\alpha =2\sin \alpha \cos \alpha =2\cdot \frac{5}{13}\cdot (-\frac{12}{13})$.
3. So $\sin 2\alpha =-\frac{120}{169}$.

Why the other options are wrong

• B $\sin 2\alpha$ is not $2\sin \alpha$.
• C $\sin 2\alpha =2\sin \alpha \cos \alpha$: don't forget the $2$.
• D The cosine is negative in the second quadrant; its sign carries into the answer.

Answer: D, $\frac{35}{13}$

1. $d=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$.
2. $|5\cdot (-1)-12\cdot (-4)-8|=35$ and $\sqrt{25+144}=13$.
3. So $d=\frac{35}{13}$.

Why the other options are wrong

• A Divide by $\sqrt{A^2+B^2}$, not by $A^2+B^2$.
• B Keep the constant term's sign: $|A{x}_0+B{y}_0+C|$.
• C Divide by $\sqrt{A^2+B^2}$.

Answer: D, $(-5,5)$ and $4$

1. Complete the squares in $x$ and in $y$.
2. $(x+5{)}^2+(y-5{)}^2=16$.
3. The centre is $(-5,5)$ and the radius is $4$.

Why the other options are wrong

• A $r^2=16$: take the square root.
• B The centre is $\left(-\frac{D}{2},-\frac{E}{2}\right)$: mind the signs.
• C Halve the coefficients: the centre is $\left(-\frac{D}{2},-\frac{E}{2}\right)$.

Answer: A, $2$

1. Change base: ${\log }_{a}b=\frac{\lg b}{\lg a}$.
2. ${\log }_{3}11\cdot {\log }_{11}5\cdot {\log }_{5}9=\frac{\lg 11}{\lg 3}\cdot \frac{\lg 5}{\lg 11}\cdot \frac{\lg 9}{\lg 5}=\frac{\lg 9}{\lg 3}$.
3. $\frac{\lg 9}{\lg 3}=\frac{2\lg 3}{\lg 3}=2$.

Why the other options are wrong

• B Change everything to base $10$: the middle numbers cancel and $\frac{\lg 9}{\lg 3}$ is left.
• C The chain leaves $\frac{\lg 9}{\lg 3}={\log }_{3}9$, not ${\log }_{9}3$.
• D The logs don't cancel to $1$: the first base and the last number remain.