A Level Maths : Common Mistakes/misconceptions Watch

Sir Cumference
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I thought it would be useful for A Level maths students if we compile a list of common mistakes and misconceptions that occur in pure, stats and mechanics. Please post anything that you can think of.

Here are two to start things off and I'll add more at some point:

- The standard trig derivatives that you've learnt e.g. \frac{d}{dx}\left(\sin x\right) = \cos x are only true if the angle (x in this case) is measured in radians. So as a general rule it's best to always work in radians (especially if the question involves calculus) unless a question specifically mentions degrees. Of course if you haven't learnt radians yet then everything will be in degrees!

- For trig equations you should only divide the equation by a trig function if you have considered whether the function could be equal to 0. For example, if you have \sin x \cos x + \cos x = 0 and you divide by \cos x you will get \sin x + 1 = 0 but you will lose the solutions for when \cos x =0. Normally it's best to factorise instead of dividing, so here you would get \cos x\left(\sin x + 1\right) = 0. This doesn't just apply to trig equations e.g. x^2 + x = 0 \Rightarrow x + 1 = 0 but it's common for the mistake to occur with trig equations.
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Laboromniavincit
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(Original post by Sir Cumference)
I thought it would be useful for A Level maths students if we compile a list of common mistakes and misconceptions that occur in pure, stats and mechanics. Please post anything that you can think of.

Here are two to start things off and I'll add more at some point:

- The standard trig derivatives that you've learnt e.g. \frac{d}{dx}\left(\sin x\right) = \cos x are only true if the angle (x in this case) is measured in radians. So as a general rule it's best to always work in radians (especially if the question involves calculus) unless a question specifically mentions degrees. Of course if you haven't learnt radians yet then everything will be in degrees!

- For trig equations you should only divide the equation by a trig function if you have considered whether the function could be equal to 0. For example, if you have \sin x \cos x + \cos x = 0 and you divide by \cos x you will get \sin x + 1 = 0 but you will lose the solutions for when \cos x =0. Normally it's best to factorise instead of dividing, so here you would get \cos x\left(\sin x + 1\right) = 0. This doesn't just apply to trig equations e.g. x^2 + x = 0 \Rightarrow x + 1 = 0 but it's common for the mistake to occur with trig equations.
More please please please
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Sir Cumference
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- The force that an object exerts on the ground is not the same as the weight of the object - they are two different forces. They may have the same magnitude e.g. if the object is stationary on the ground.

- Speed = distance / time can only be used if there is no acceleration. SUVAT equations can only be used if there is constant acceleration. For non-constant acceleration you must use calculus.
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ghostwalker
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The solution to x^2-3x+2=0 is; x= 2 OR x=1; NOT x=2 AND x=1 (x can't be both at the same time).
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Deggs_14
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It’s bloody hard 😂
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RDKGames
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\sqrt{a^2 + b^2} \neq a+b

:facepalm:
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Laboromniavincit
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(Original post by ghostwalker)
The solution to x^2-3x+2=0 is; x= 2 OR x=1; NOT x=2 AND x=1 (x can't be both at the same time).
I am taking down the points
Thank you so much
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ghostwalker
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(Original post by RDKGames)
\sqrt{a^2 + b^2} \neq a+b

:facepalm:
And its other form, which I've seen on here a few times.

(a+b)^2\not=a^2+b^2
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4D Chess
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\ln (x^2) \neq (\ln x)^2
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Sir Cumference
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If you are asked to calculate something like

\displaystyle \sum_{k=1}^{15} 3 \times 4^k

Then this is clearly a geometric series but a common mistake is to assume that a=3 and apply the geometric series formula using that. This is incorrect and you can see that if you write out the series:

(12) + (12 \times 4) + (12 \times 4^2), ...

So a=12, r=4 and n=15 and you can plug these into the formula. If in doubt, always write out the series so you can see it more clearly.
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_gcx
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(Original post by ghostwalker)
The solution to x^2-3x+2=0 is; x= 2 OR x=1; NOT x=2 AND x=1 (x can't be both at the same time).
This is more of a linguistic thing. I think "x = 2 and x = 1 (are solutions)" is an acceptable enough answer to "what are the solutions of x^2 - 3x + 2 = 0". Obviously if the question is asking "if x^2 - 3x + 2 = 0 what's x" (or even "solve" that) then giving the answer x=2 and x=1 is problematic.
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_gcx
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There is a lot of confusion with square roots.

\sqrt{x^2} = x

only holds for x \ge 0. The identity \sqrt{x^2} = |x| holds for all real x.

Some students write \sqrt {x^2} = \pm x too! We usually define \sqrt x to have a single value, for calculus and stuff to make sense, but you often see students say something like \sqrt 4 = \pm 2. My teacher in secondary school (who is an examiner lol, iirc the author of an A-level textbook also did this) was also convinced that this was the case. We usually pick the positive square root of x but you can pick the negative one if you want to. (you can adapt calculus stuff relatively painlessly but idk why you'd want to do this) It's just convention. The vast majority of the time we'll pick the positive square root, making \sqrt 4 = 2.

If you define \sqrt x as a set, ie. \sqrt x = \{y \in \mathbb R \mid y^2 = x\} you do have \sqrt{x^2} = \{x, -x\} (which you might informally write as \pm x) for all real x, and this is called a multifunction but these are only really studied properly deep into complex analysis, and this is not what is normally meant by the square root.
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_gcx
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Indefinite integrals can be confusing particularly the constant of integration.

For example, look at the integral:

\displaystyle \int \frac 3 {3 x} \mathrm dx

You can cancel the 3 and find that this is "equal to" \ln|x| + C. (let's write the constant C_1 instead, for a reason that'll be clear in a second) But if you decide, instead to do a substitution u = 3 x you might be confused that you instead get \ln|3x| + C_2. It might appear that you've got two different answers from working out the integral two different ways. This actually isn't what's happened.

It helps to understand an indefinite integral giving the set of functions that differentiate to give the function you were integrating. (the integrand) You can notice that \ln|3x| + C_2 = \ln|x| + (C_2 + \ln 3). This actually describes the same set of functions as \ln|x| + C_1, and the sets are related by C_2 + \ln 3 = C_1, so there is no problem here. I've seen a lot of students on here get confused about this, hope this is clear.

It gets even worse if you for some reason wanted to work out:

\displaystyle \int \frac 1 x \mathrm dx

by integration by parts. Writing:

\displaystyle \frac 1 x = \left(\frac 1 x\right) \times \frac {\mathrm d} {\mathrm dx}(x)

Then using IBP you'll get:

\displaystyle \int \frac 1 x \mathrm dx = \frac x x - \int \left(-\frac x {x^2}\right) \mathrm dx

giving:

\displaystyle \int \frac 1 x \mathrm dx = 1 + \int \frac 1 x \mathrm dx

At which point you might conclude that:

0 = 1

This basically shows that indefinite integrals, how they're usually handled, make no sense. (if you try this with bounds, you'll encounter no such problem) But you can kind of rationalise this for A-level's sake. We're basically saying, like last time, that f + C and f + 1 + C describe the same set of functions, so there is no problem here. (and IBP has done precisely nothing to help us with our problem )

Again, I've seen students do stuff like this on Reddit (though with more complicated/less obvious examples) and confuse themselves. (similarly, you can get 0 = 0 with IBP which is similarly frustrating)
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Sir Cumference
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It's common for students to mix up the log laws and say things like log(A + B) = log(A) x log(B) or similar. Also I see equations like this being solved

e^{2x} + e^x = 3

by "logging every term" to give 2x + x = \ln 3. This confusion probably arises from students getting used to multiplication and assuming that other functions share the same properties (associativity in this case).
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Laboromniavincit
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Unrelated post
Desperate to find solution for following exercise
No idea which book
Could you help please?
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the bear
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f -1(x) is not the reciprocal of f(x)

it is the INVERSE of f(x)
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the bear
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(Original post by Laboromniavincit)
Unrelated post
Desperate to find solution for following exercise
No idea which book
Could you help please?
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if you post your answers we can check them.
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deathbySTEP
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The fact that the 2nd derivative is 0 does not automatically imply a point of inflexion (e.g. y=x^4)
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deathbySTEP
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(Original post by Laboromniavincit)
Unrelated post
Desperate to find solution for following exercise
No idea which book
Could you help please?
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Looks like the CGP A-Level one to me
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Sir Cumference
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The dreaded "reverse chain rule": some students take it literally and say things like this

\displaystyle \int (x^2+3)^3 \ dx = \frac{1}{8x}(x^2+3)^4 + c

\displaystyle \int e^{x^2} \ dx = \frac{1}{2x}e^{x^2}+ c

These are both wrong. The only time you can do this kind of thing is if the inner function is linear e.g.

\displaystyle \int (2x+3)^3 \ dx = \frac{1}{8}(2x+3)^4 + c

If in doubt, use a substitution.
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