Question

Show 2ae^1/2a - 4e^1/2a = 2 as x=2+e^-1/2x

Answer 1

Could you use brackets to clarify the expression?

Answer 2

\[2ae ^{a/2} - 4e^{a/2} =2\] as \[x = 2+ e^{-x/2}\]

Answer 3

can you post a copy of the original question?

Answer 4

okayy.
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w08_qp_3.pdf Question 9i

Answer 5

I was doing it halfway, then i couldnt show that x = ______ equation.

Answer 6

ok.
\[ 2a e^\frac{a}{2}- 4e^\frac{a}{2}=2 \\
2e^\frac{a}{2}(a-2) = 2 \\
e^\frac{a}{2}(a-2) =1 \\
a-2 = e^{-\frac{a}{2}}
\]

Answer 7

and finally
a= 2 + exp(-a/2)

Answer 8

thankss. could you help me with ques 5ii?

Answer 9

did you find "a" ?

Answer 10

oh yes a =3.

Answer 11

so (2x+1) ( 2x^2-3x+3) <0

Answer 12

I assume you mean
\[ (2x+1) ( 2x^2-3x+3) <0 \]
for that expression to be negative we need either
1. 2x+1 < 0 and (2x^2-3x+3) is positive
or
2. vice versa

Answer 13

Okay.

Answer 14

< 0 means expression has to be negative, is it?

Answer 15

yes, stuff < 0 means stuff is less than 0 (i.e. negative)
the roots of (2x^2-3x+3) are complex
that means this parabola does not cross the x-axis
the 2 in front of x^2 makes it concave up (smile shape)
and as a check, the vertex is at -(-3)/(2*2) = 3/4. that means it's min y value is at x=3/4
and it is 2*9/16 - 9/4 + 3 = 15/8
so the parabola is always positive

that means we need the other factor to always be negative
because we need minus time plus to get a minus

Answer 16

I seee. so we take 2x+1 <0? is it?

Answer 17

yes

Answer 18

nicee thanks!!another ques. no 4!!!

Answer 19

just want you to check my answer.

Answer 20

can you see?

Answer 21

.

Answer 22

as a check to Q5, here is a plot, showing the curve goes negative for x< -0.5

Answer 23

oh cool! okayy!! thanks!

Answer 24

there was once you showed me a similar working to that ques no. 4. I was quite confused about this. how is sin2x/ (cos2x +1) = sinx /cosx ? D:

Answer 25

yes, it looks ok until you "jump" to the answer
use:
sin(2x) = 2 sinx cosx
cos(2x) = cos^2 x - sin^2(x)
which can be rewritten using cos^2 + sin^2 = 1
i.e. cos^2 x = 1 - sin^2 x, so
cos(2x) = (1-sin^2 x) - sin^2 x = 1 - 2 sin^2x

Answer 26

now you can work on
\[ \frac{\sin 2x}{1 - \cos 2x } \]

Answer 27

you should get cos x / sin x = cot x

Answer 28

I'm lost! :(
So,
dy/dx = sin 2x / (1-cos2x)
dy/dx = 2 sinxcosx / (1 - sin^2 x) is it?

Answer 29

cos2x = 1 - 2 sin^2x

so you second line should read
dy/dx = 2 sinxcosx / (1 -( 1 - 2 sin^2 x))

Answer 30

\[ \frac{dy}{dx} = \frac{2 \sin x \cos x}{1 - 1 + 2 \sin^2 x} \]

Answer 31

what should i do next? i kept going back to the previous pattern.

Answer 32

1-1 = 0
\[ \frac{dy}{dx} = \frac{2 \sin x \cos x}{1 - 1 + 2 \sin^2 x} \\
= \frac{2 \sin x \cos x}{2 \sin^2 x}
\]

Answer 33

the denominator starts as
\[ 1 - \cos 2x \]
we have the identity
\[ \cos 2x = 1 - 2\sin^2 x \]
so the bottom becomes
\[ 1 - ( 1 - 2\sin^2 x) \\
= 1 - 1 + 2\sin^2 x
\\ = 2 \sin^2 x
\]

Answer 34

okay what's next?

Answer 35

simplify
\[ \frac{dy}{dx}= \frac{2 \sin x \cos x}{2 \sin^2 x} \]

Answer 36

cannot do. D: I kept getting back sin2x/ 1- cox2x

Answer 37

if the problem was
\[ \frac{dy}{dy}= \frac{2 w \ z}{2 w^2} \]
can you simplify that ?

Answer 38

cos x / sinx?

Answer 39

yes.
\[ \frac{2 \sin x \cos x}{2 \sin x \sin x} = \frac{\cos x }{\sin x } \]

Answer 40

OHH. I see it now! thankss!

Answer 41

Could Pls check my solution for me. The question is (1/ 1-x) + (2 / 1+2x) - (4 / 2+x)