This short quiz is to make sure you know the basic stuff you need for Core 4 maths. It's not going to guarantee you a good grade, but it'll give you a good base to work from.

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Question 1 |

What does $\frac{3x + 2}{x(x^2 + 1)}$ look like in partial fractions?

A | $\frac{A}{x} + \frac{Bx + C}{x^2+1}$ Hint: Yes - a quadratic factor needs a linear thing on top. |

B | $\frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}$ Hint: Good effort - bad factorising. |

C | $\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$ Hint: No. That would be for $(x+1)^2$ on the bottom. |

D | $\frac{Ax + B}{x} + \frac{Cx + D}{x^2+1}$ Hint: Overkill. |

E | $\frac{A}{x} + \frac{C}{x^2+1}$ Hint: *QI klaxen* Something missing. |

Question 2 |

How would you work out $(2 + x)^{-2}$?

A | Use the $(1+x)^n$ formula with $n=-2$, $x = {2x}$ and divide by 2 at the end. |

B | Use the $(1+x)^n$ formula with $n=-2$ and double it at the end. Hint: No. |

C | Use the $(1+x)^n$ formula with $n=-2$, $x = \frac{x}{2}$ and multiply by 4 at the end. Hint: No. |

D | Use the $(1+x)^n$ formula with $n=-2$, $x = \frac{x}{2}$ and divide by 2 at the end. |

E | Use the $(1+x)^n$ formula with $n=-2$, $x = \frac{x}{2}$ and divide by 4 at the end. Hint: Yep |

Question 2 Explanation:

$(2+x)^{-2}$ = \left[2 (1 + \frac{x}{2})\right]^{-2} = \frac{1}{4}\left(1+\frac{x}{2}\right)^{-2}.

Question 3 |

What's the formula for a volume of revolution around the $x$-axis?

A | $V = \int x^2 ~dy$ Hint: No - that would be the $y$-axis. And there's something missing |

B | $V = \int y^2 ~dx$ Hint: Nope - something missing! |

C | $V = \pi \int y^2 ~dx$ Hint: Correct! |

D | $V = \int y ~dx$ Hint: No - that gives the area |

E | $V = \pi \int x^2 ~dy$ Hint: No, that would be the $y$-axis. |

Question 4 |

How would you integrate $\int x \ln(x)~dx$?

A | By substitution, with $u = x$. Hint: That really doesn't help. |

B | By parts, with $u = x$ and $v' = \ln(x)$. Hint: No - it's not easy to integrate $\ln(x)$. |

C | By inspection, it's $\frac{1}{2}x^2 + c$. Hint: It isn't, you know. |

D | By substitution, with $u = \ln(x)$. Hint: You can actually do it that way - although you'll need to use parts later. |

E | By parts, with $u = \ln(x)$ and $v' = x$. Hint: The classical solution. |

Question 5 |

What do you get if you differentiate $x^2y^2$ with respect to $x$?

A | $4xy\frac{dy}{dx}$ Hint: Good effort - but you need to use product rule. |

B | $2x^2 y \frac{dy}{dx}$ Hint: Good effort - but you need to use product rule. |

C | $2xy^2 + 2x^2y \frac{dy}{dx}$ Hint: Correct! Not tidy, but correct! |

D | $2xy\left(y + x \frac{dy}{dx}\right)$ Hint: Good factorising! |

E | $4xy$ Hint: No, you might want to look up implicit differentiation. |

Question 6 |

Which of these is the way to get the gradient of a parametric curve?

A | Work out $\frac{dy}{dt} \times \frac{dx}{dt}$ and substitute the correct value of $t$. |

B | Work out $\frac{dy}{dt} \div \frac{dx}{dt}$ and substitute the correct value of $x$. Hint: You shouldn't have any $x$s knocking around when everything is in terms of $t$. |

C | Work out $\frac{dy}{dt} \div \frac{dx}{dt}$ and substitute the correct value of $t$. Hint: Correct! |

D | Work out $\frac{dx}{dt} \div \frac{dy}{dt}$ and substitute the correct value of $t$. Hint: That would give you $\frac{dx}{dy}$! |

E | Work out the Cartesian form and differentiate that. Hint: Good luck with that. It's possible to make that work, sometimes, but it's not the way they're looking for. |

F | Work out $\frac{dy}{dt} \div \frac{dx}{dt}$ and substitute the correct value of $y$. Hint: You shouldn't have any $y$s knocking around when everything is in terms of $t$. |

Question 7 |

If you're given $x \frac{dy}{dx} = y^2$, what do you end up integrating?

A | $\int x~dx = \int y^2~dy$ Hint: Careful with your algebra! |

B | $\int x~dy = \int y^2~dx$ Hint: Nope - you can't mix your $x$s and $y$s up in an integral. |

C | $\int \frac{dy}{dx} = \int \frac{y^2}{x} $ Hint: Careful! You need to integrate with respect to something (there needs to be a $dx$ or $dy$ or similar at the end of an integral. |

D | $\int \frac{dy}{dx} = \int \frac{y^2}{x} $ Hint: Careful! You need to integrate with respect to something (there needs to be a $dx$ or $dy$ or similar at the end of an integral. |

E | $\int y^{-2}~dy = \int x^{-1} dx$ Hint: There you go. |

Question 7 Explanation:

To separate a differential equation, you need to get all of the $x$s (for instance) on one side of the equation, and all of the $y$s on the other.

Question 8 |

How would you work out $\int \frac{x}{\sqrt{x^2 + 1}} dx$?

A | By substitution, with $u = x^2 + 1$ Hint: That'll work just fine - although it's not the easiest way. |

B | Using the quotient rule. Hint: There's no such thing for integration! |

C | By parts, with $u = x$ and $v' = (x^2 + 1)^{\frac{1}{2}}$. Hint: No. |

D | By parts, with $u = x$ and $v' = (x^2 + 1)^{-\frac{1}{2}}$. Hint: Yikes, you're braver than me. |

E | By substitution, with $u^2 = x^2 + 1$ Hint: Yep - this drops out really easily. |

Question 9 |

How do you work out $\int \sin^2(x)~dx$?

A | By parts, with $u = \sin(x)$ and $v' = \sin(x)$. Hint: You end up going in circles unless you're really careful! |

B | Use the identity $\sin^2(x) = 1 - \cos^2(x)$, rearrange and integrate. Hint: This goes around in circles. |

C | Use the identity $\cos(2x) = 1 - 2 \sin^2(x)$, rearrange and integrate. Hint: Yep - $\cos(2x)$ is relatively easy to integrate. |

D | Use the substitution $u = \sin(x)$. Hint: This doesn't come out as nicely as you'd hope. |

E | Use the identity $\cos(2x) = 1 - \sin^2(x)$, rearrange and integrate. Hint: That's not an identity. |

Question 10 |

How do you find the angle between a line $\mathbf{a} + \lambda \mathbf{b}$ and a vector $\mathbf{c}$?

A | Use a protractor. Hint: Nice try. |

B | Work out $\cos^{-1}(\frac{\mathbf{a \cdot c}}{|\mathbf{a}||\mathbf{c}|)}$. Hint: No - the point on the line (a) is irrelevant. |

C | Work out $\sin^{-1}(\mathbf{a \cdot c})$. Hint: No - that's about as wrong as you can get. |

D | Work out $\cos^{-1}(\frac{\mathbf{b \cdot c}}{|\mathbf{b}||\mathbf{c}|)}$. Hint: Yep! |

E | Work out $\cos^{-1}(\mathbf{b \cdot c})$. Hint: No - that would only work with unit vectors. |

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