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.

Start

Congratulations - you have completed *Core 4 basic facts quiz*.

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1 |

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

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

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

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

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

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

Question 1 Explanation:

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

Question 2 |

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

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

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

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

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

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

Question 3 |

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

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

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

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

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

Question 4 |

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

A | 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 $t$. |

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

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

D | 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 $t$. |

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

F | 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. |

Question 5 |

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

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

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

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

Question 6 |

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

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

B | $4xy\frac{dy}{dx}$ 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 | $2x^2 y \frac{dy}{dx}$ Hint: Good effort - but you need to use product rule. |

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

Question 7 |

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

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

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

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

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

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

Question 8 |

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

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

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

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

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

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

Question 9 |

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

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

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

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

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

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

Question 10 |

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

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

B | $\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. |

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

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

Question 10 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.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
Get Results

There are 10 questions to complete.

← |
List |
→ |

Return

Shaded items are complete.

1 | 2 | 3 | 4 | 5 |

6 | 7 | 8 | 9 | 10 |

End |

Return

You have completed

questions

question

Your score is

Correct

Wrong

Partial-Credit

You have not finished your quiz. If you leave this page, your progress will be lost.

Correct Answer

You Selected

Not Attempted

Final Score on Quiz

Attempted Questions Correct

Attempted Questions Wrong

Questions Not Attempted

Total Questions on Quiz

Question Details

Results

Date

Score

Hint

Time allowed

minutes

seconds

Time used

Answer Choice(s) Selected

Question Text

All done

Need more practice!

Keep trying!

Not bad!

Good work!

Perfect!

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

TIL Pierre Bézier (who patented, but apparently didn't discover, the curves), lived much more recently than I imagined (1910-1999).

Ask Uncle Colin: integrating a trigonometric product www.flyingcoloursmaths.co.uk/a…

A Puzzle Full of Nines www.flyingcoloursmaths.co.uk/a…

He headed home after an abrupt 180. twitter.com/IMcMillan/status/1…

Oh dear, Asda. cc @standupmaths pic.twitter.com/voi2yPcxdc