help with questions from practise paper (1 Viewer)

jjuunnee

Active Member
Joined
Mar 16, 2016
Messages
169
Location
Sydney
Gender
Female
HSC
2018
Solve 2sin2x = sinx for 0 degrees <= x <= 360 degrees

Given equation x2 - (k + 2)x + (2k + 4) = 0, determine the values of k such that one root is twice the other

Is the curve y = x2 - 4 / x3 + 8 differentiable for all x? Give a reason for your answer.



Sorry if this is hard to read
 

jjuunnee

Active Member
Joined
Mar 16, 2016
Messages
169
Location
Sydney
Gender
Female
HSC
2018
sorry there's more

The line l is tangent to the curve y = 3x2 at the point (3, 27). What is the gradient of line l?

Simplify |x + 2| / x2 - 4 for x does not equal to + or - 2
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Solve 2sin2x = sinx for 0 degrees <= x <= 360 degrees

Given equation x2 - (k + 2)x + (2k + 4) = 0, determine the values of k such that one root is twice the other

Is the curve y = x2 - 4 / x3 + 8 differentiable for all x? Give a reason for your answer.



Sorry if this is hard to read




 

kashkow

Active Member
Joined
Oct 29, 2015
Messages
177
Location
Right here.
Gender
Male
HSC
2016
For the first question:

> Simplify down to: sinx(2sinx-1) = 0

> then solve: sinx = 0 and 2sinx-1 = 0 for x

For the second question:

> let (alpha) be one root and (beta) be another root such that (beta) = 2(alpha)

> Thus roots are (alpha) and 2(alpha)

> now (2k+4) = product of roots = (alpha)*2(alpha)= 2(alpha)^2

> Also (k+2) = sum of roots = (alpha) + 2(alpha) = 3(alpha)

> Solve simultaneous equations to solve for k (substitute alpha).

For the third question:

> Recall that differentiation is finding the tangent to the curve

> As f'(x) = 2x +12/x^4 is not continuous you cannot find tangent to the curve for ALL x (ie. x=0).

> :. answer is No; reason because the original function is a non-continuous function

For the fourth question:

> Recall that finding gradient of the tangent at a point is the same as finding the gradient of the curve of which the tangent belongs at that point, (3, 27)

> thus simply differentiate the curve y=3x^2

> Then use the gradient function and substitute for f'(3)

For your fifth question:

|x + 2| can be + or -

ie x+2 > 0
or x+2 < 0

Thus x>-2 or x<-2

When x>-2
x+2 / (x+2)(x-2)

1/(x-2)

When x<-2
(-x-2)/(x+2)(x-2)

-(x+2)/(x+2)(x-2)
-1/(x-2)

:. expression becomes +- 1/(x-2)


I think the fifth one is right but I'm not too sure; I'm not the best with absolute values... sorry :S Also not 100% on the third question... I'm not too sure what "differentiable for all x" means exactly. You can differentiate it and find the gradient function; however you cannot use all x within the gradient function to find the tangent. Thus that is why I think it's not "differentiable" for all x.

Rats; i was too slow :p
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top