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maths p32 answers

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yeah and there were 3 stationary points! XD pi/2 was one of them!!!

Rightly said.

y = 3 sin x + 4 cos^3 x
dy/dx = 3 cos x - 12 cos^2 x sin x
0 = 3 cos x ( 1 - 4 cos x sin x)

From here, we get the first stationary point.

3 cos x = 0
cos x = 0
x = pi/2

0 = 1 - 4 cos x sin x
0 = 1 - 2 (2 cos x sin x)
0 = 1 - 2 (sin 2x)
2 sin 2x = 1
sin 2x = 1/2

The range '0<x<pi' needs to be modified over here as we are dealing with '2x'.

0<x<pi
0<2x<2pi

sin 2x = 1/2
2x = pi/6 and 2x = 5pi/6
x = pi/12 and x = 5pi/12

Therefore, there were 3 stationary points; pi/2, pi/12 and 5pi/12.

As far as finding the nature of the smallest 'x' value was concerned, we had to double differentiate 'dy/dx' and then replace 'x' with 'pi/12'.

dy/dx = 3 cos x - 12 cos^2 x sin x
d^2 y/dx^2 = -3 sin x - 12 cos^3 x + 24 cos x sin^2 x

Put 'x=pi/12'.

d^2 y/dx^2 = -10.04

Therefore, the stationary point is a maximum.
 
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Rightly said.

y = 3 sin x + 4 cos^2 x
dy/dx = 3 cos x - 12 cos^2 x sin x <-----
0 = 3 cos x ( 1 - 4 cos x sin x)

From here, we get the first stationary point.

3 cos x = 0
cos x = 0
x = pi/2

0 = 1 - 4 cos x sin x
0 = 1 - 2 (2 cos x sin x)
0 = 1 - 2 (sin 2x)
2 sin 2x = 1
sin 2x = 1/2

The range '0<x<pi' needs to be modified over here as we are dealing with '2x'.

0<x<pi
0<2x<2pi

sin 2x = 1/2
2x = pi/6 and 2x = 5pi/6
x = pi/12 and x = 5pi/12

Therefore, there were 3 stationary points; pi/2, pi/12 and 5pi/12.

As far as finding the nature of the smallest 'x' value was concerned, we had to double differentiate 'dy/dx' and then replace 'x' with 'pi/12'.

dy/dx = 3 cos x - 12 cos^2 x sin x
d^2 y/dx^2 = -3 sin x - 12 cos^3 x + 24 cos x sin^2 x

Put 'x=pi/12'.

d^2 y/dx^2 = -10.04

Therefore, the stationary point is a maximum.
See red coloured mark...
Isnt the differentiation of cos^2 x like this:
mzda.png
 
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I'm losing a mark in the last vector question. Another mark in a trigo question. And maybe 1 or 2 more here and there. Hoping for 70+
 
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@Prisonbreak - Why don't you just make a new topic and upload the paper out there? Better than sending it to soo many people right?
 
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aditiya. i think the answer to your lasr part is wrong. can you pls disscus it with me.
 
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