WJEC C3/C4 8th/17th June 2016

There was an integration-only section. The one with a = 2.3 was the definite integration, but there were three indefinite integration questions before that.

There was an integration-only section. The one with a = 2.3 was the definite integration, but there were three indefinite integration questions before that.

Hey man, do you remember the 3 indefinite integral questions? or their answers? thanks

Hey man, do you remember the 3 indefinite integral questions? or their answers? thanks Previous years there were only two indefinites, are you sure about 3?

Some answers:

1. 0.74379 to 5dp
Multiply this by e: 2.02183 to 5dp.

2. 30, 150, 196.6, 343.4 degrees

146.3,326.3 degrees

3. $\dfrac{2}{3}$

4. $-\frac{1}{2}\tan 3t$
$-\frac{1}{8}\sec^3 3t$
$-\dfrac{1}{y^3}$

5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle $\dfrac{\theta}{2}$.
Usual stuff with iterations etc; for function use $f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5$

6. An easy counterexample is $a=b=1, c=2, d=3$

7. $x=-\dfrac{1}{3}$ ($x=2$ doesn't work)

8. $a=-3,5$
$b=-\frac{2}{3}$

9. $f^{-1}(x)=12-3\ln(x-8)$
Domain is $[9,+\infty)$

10. To show $hh(x)=x$ simply expand out the left.
So it is clear that h is its own inverse and so $h^{-1}(-1)=\frac{1}{9}$

That's all I can remember, please correct anything that's wrong .

EDIT: Just realised I left out diff/int questions.

Differentiation: $\ln(\cos x)$ gives $-\tan x$,

$\tan^{-1}(\frac{x}{3})$ gives $\dfrac{3}{9+x^2}$

$e^{6x}(3x-2)^4$ gives $18xe^{6x}(3x-2)^3$

Integration: $a=2.3$

I don't remember exactly, sorry. I vaguely remember some of the types, though: one had an 8 in the numerator (which you had to take out as a constant) with a polynomial in the denominator raised to the power of 4, I think, which had to be flipped (i.e. brought into the numerator with a minus sign on the power). The chain rule one was ln(cosx), I think, and that came out to -sinx/cosx, simplified to -tanx. The third one was 1/polynominal not raised to a power (or raised to the power of 1), so it simply integrated to 1/differential of polynominal x ln|polynomial| + c.

Hope that helps.


Fairly sure, yeah.

Could you explain Q2? Also could there be other answers for 4bii? Could you tell me what y is

i got that!!

The chain rule one is a differentiation one lol, the other two i remember as well. :/

how is it over 8??

Some answers:

1. 0.74379 to 5dp
Multiply this by e: 2.02183 to 5dp.

2. 30, 150, 196.6, 343.4 degrees

146.3,326.3 degrees

3. $\dfrac{2}{3}$

4. $-\frac{1}{2}\tan 3t$
$-\frac{1}{8}\sec^3 3t$
$-\dfrac{1}{y^3}$

5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle $\dfrac{\theta}{2}$.
Usual stuff with iterations etc; for function use $f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5$

6. An easy counterexample is $a=b=1, c=2, d=3$

7. $x=-\dfrac{1}{3}$ ($x=2$ doesn't work)

8. $a=-3,5$
$b=-\frac{2}{3}$

9. $f^{-1}(x)=12-3\ln(x-8)$
Domain is $[9,+\infty)$

10. To show $hh(x)=x$ simply expand out the left.
So it is clear that h is its own inverse and so $h^{-1}(-1)=\frac{1}{9}$

That's all I can remember, please correct anything that's wrong .

EDIT: Just realised I left out diff/int questions.

Differentiation: $\ln(\cos x)$ gives $-\tan x$,

$\tan^{-1}(\frac{x}{3})$ gives $\dfrac{3}{9+x^2}$

$e^{6x}(3x-2)^4$ gives $18xe^{6x}(3x-2)^3$

Integration: $a=2.3$

I don't remember exactly, sorry. I vaguely remember some of the types, though: one had an 8 in the numerator (which you had to take out as a constant) with a polynomial in the denominator raised to the power of 4, I think, which had to be flipped (i.e. brought into the numerator with a minus sign on the power). The chain rule one was ln(cosx), I think, and that came out to -sinx/cosx, simplified to -tanx. The third one was 1/polynominal not raised to a power (or raised to the power of 1), so it simply integrated to 1/differential of polynominal x ln|polynomial| + c.

Hope that helps.


Fairly sure, yeah.

I swear the grade boundaries are going to be lower... How the hell should we know that x=2 shouldnt work?!

@IrrationalRoot, some questions about these:

gives

gives


For the first one, would I still get the marks if I only partially simplified it? I initially got a 1/3 in the numerator and 1+(1/9)x² in the denominator and I multiplied top and bottom by 3 instead of 9. Its the equivalent, but this would mean I left a 1/3 in the denominator...
The second one - was this actually on the paper? I've no recollection of it. The product rule question was a different one, I feel...

Some answers:

1. 0.74379 to 5dp
Multiply this by e: 2.02183 to 5dp.

2. 30, 150, 196.6, 343.4 degrees

146.3,326.3 degrees

3. $\dfrac{2}{3}$

4. $-\frac{1}{2}\tan 3t$
$-\frac{1}{8}\sec^3 3t$
$-\dfrac{1}{y^3}$

5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle $\dfrac{\theta}{2}$.
Usual stuff with iterations etc; for function use $f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5$

6. An easy counterexample is $a=b=1, c=2, d=3$

7. $x=-\dfrac{1}{3}$ ($x=2$ doesn't work)

8. $a=-3,5$
$b=-\frac{2}{3}$

9. $f^{-1}(x)=12-3\ln(x-8)$
Domain is $[9,+\infty)$

10. To show $hh(x)=x$ simply expand out the left.
So it is clear that h is its own inverse and so $h^{-1}(-1)=\frac{1}{9}$

That's all I can remember, please correct anything that's wrong .

EDIT: Just realised I left out diff/int questions.

Differentiation: $\ln(\cos x)$ gives $-\tan x$,

$\tan^{-1}(\frac{x}{3})$ gives $\dfrac{3}{9+x^2}$

$e^{6x}(3x-2)^4$ gives $18xe^{6x}(3x-2)^3$

Integration: $a=2.3$

Are you sure about your answer for inverse tan of x/3. Could there be a different definition for 'simplest form'?

Hmm, not sure exactly how strict the marking is. You'd definitely only lose one mark at most, but I don't know whether they'd take a mark off. I think I'd lean towards yes since I recall there being a fair few marks for these, but I can't remember what the exact marks where.

With the second one I'm pretty sure this was on the paper? I certainly did that derivative anyway, unless I misread the question or something .
What do you think the product rule Q was?

I hope the boundaries are lower than last year.



I remember a product rule question having e^6x as one of the functions, but I think the polynomial next to it is from an integration question where it was in the denominator and and you had to flip it (raise it to the negative power), and then divide by the product of the differential of and the new power, which was -3. I distinctly remember that.