Practice Trial HSC Questions HELP (1 Viewer)

[csi]

Hey guys,

The attached are a few questions I need help with. Any help is appreciated. Thanks!

 

[yanujw]

c) The way I did it was using the general term as follows;

1) Find the term that expresses each term by indexing a variable k.
2) Find the value of x that represents x^2 (in this case the term indexed by k=4)
3) Sub back in to the expression

Let me know if you don't fully understand how to get the first line

 

[Qeru]

Hey guys,

The attached are a few questions I need help with. Any help is appreciated. Thanks!

general term is:
We want: so coefficient is

 

[Qeru]

c) The way I did it was using the general term as follows;
View attachment 30637
1) Find the term that expresses each term by indexing a variable k.
2) Find the value of x that represents x^2 (in this case the term indexed by k=4)
3) Sub back in to the expression

Let me know if you don't fully understand how to get the first line

super nitpicky but its 'coefficient' not 'term'

 

[csi]

c) The way I did it was using the general term as follows;
View attachment 30637
1) Find the term that expresses each term by indexing a variable k.
2) Find the value of x that represents x^2 (in this case the term indexed by k=4)
3) Sub back in to the expression

Let me know if you don't fully understand how to get the first line

thanks

 

[csi]

general term is:
We want: so coefficient is

thanks

 

[cossine]

di)

(x+4) > 2 / (x+3)

The graph will look like this



So if you can find the intersection points you can determine the values that satisfy the inequality

(x+4) = 2 / (x+3) (Note x cannot equal -3)

=> (x+3)(x+4) = 2

=> x^2 +7x +12 - 2 =0

=> x^2 +7x +10 =0

=> (x+5)(x+2) = 0

=> x= -5, -2

From this we can determine y = -1, 2 respectively.

The inequality is satisfied when " -5 < x < -3 or x>-2"

ii)
Based on the graph we can determine the inequality is satisfied when x<-5 or (x>-2 and x !=3


Alternative for d i)

(x+4)(x+3)^2 > 2(x+3) (We need to multiply by (x+3)^2 as x+3 could be negative and result in change of the sign of the inequality)

=> (x+4)(x+3)^2 -2(x+3) > 0

=> (x+3) * [ (x+4)(x+3) -2 ] >0

=> (x+3) * [x^2 +7x + 10] > 0

=> (x+3) * (x+5) * (x+2) > 0

Sketching this we can get the answer.

Alt for dii)

(x+4)|x+3| -2 >0

Then consider the case when x < -3

=> (x+4)(-x-3) -2 >0

or the case when x>= -3

(x+4)(x+3) -2 >0

 

[Qeru]

di)

(x+4) > 2 / (x+3)

The graph will look like this

View attachment 30646

So if you can find the intersection points you can determine the values that satisfy the inequality

(x+4) = 2 / (x+3) (Note x cannot equal -3)

=> (x+3)(x+4) = 2

=> x^2 +7x +12 - 2 =0

=> x^2 +7x +10 =0

=> (x+5)(x+2) = 0

=> x= -5, -2

From this we can determine y = -1, 2 respectively.

The inequality is satisfied when " -5 < x < -3 or x>-2"

ii)
Based on the graph we can determine the inequality is satisfied when x<-5 or (x>-2 and x !=3


Alternative for d i)

(x+4)(x+3)^2 > 2(x+3) (We need to multiply by (x+3)^2 as x+3 could be negative and result in change of the sign of the inequality)

=> (x+4)(x+3)^2 -2(x+3) > 0

=> (x+3) * [ (x+4)(x+3) -2 ] >0

=> (x+3) * [x^2 +7x + 10] > 0

=> (x+3) * (x+5) * (x+2) > 0

Sketching this we can get the answer.

Alt for dii)

(x+4)|x+3| -2 >0

Then consider the case when x < -3

=> (x+4)(-x-3) -2 >0

or the case when x>= -3

(x+4)(x+3) -2 >0

I don't think your right for ii. If we have a graph of and want to sketch all parts of the graph below the y axis is flipped along the x axis. Hence the left branch of the hyperbola now becomes completely positive as shown:

from i we found that the poi is (-2,2) hence the region where the blue graph is above the red graph is:

 

[cossine]

I don't think your right for ii. If we have a graph of and want to sketch all parts of the graph below the y axis is flipped along the x axis. Hence the left branch of the hyperbola now becomes completely positive as shown:
View attachment 30656
from i we found that the poi is (-2,2) hence the region where the blue graph is above the red graph is:

I was thinking of |x+4| < 2/ (x+3) for some reason. Thank you

 

[CM_Tutor]

I don't think your right for ii. If we have a graph of and want to sketch all parts of the graph below the y axis is flipped along the x axis. Hence the left branch of the hyperbola now becomes completely positive as shown:
View attachment 30656
from i we found that the poi is (-2,2) hence the region where the blue graph is above the red graph is:

Qeru is correct on the solution. One little piece of advice that I would add, though...

If drawing this sketch of

​

in an exam, I would be careful to include the vertical asymptote at . Otherwise, had the question been

​

and the asymptote is missing, it would be easy to make a mistake and think that the solution was and miss that the domain of is the same as the domain of - namely, - and thus that the solution must exclude the value where the RHS is undefined, thus being or . Similarly, the solution of

​

is

​

or, in interval notation, .

Also, remember with solving inequalities questions that the form in which the question is written matters. Consider the questions







They look the same, but the solutions are: