Practice Trial HSC Questions HELP (1 Viewer)

yanujw

Well-Known Member
Joined
May 23, 2020
Messages
343
Location
57 Mount Pleasant St.
Gender
Male
HSC
2022
c) The way I did it was using the general term as follows;
1620898337862.png
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
 
  • Like
Reactions: csi

Qeru

Well-Known Member
Joined
Dec 30, 2020
Messages
368
Gender
Male
HSC
2021
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

Member
Joined
Nov 10, 2019
Messages
94
Gender
Undisclosed
HSC
2021
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
 

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
583
Gender
Male
HSC
2017
di)

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

The graph will look like this

1620918468580.png

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

Well-Known Member
Joined
Dec 30, 2020
Messages
368
Gender
Male
HSC
2021
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:
1621003825761.png
from i we found that the poi is (-2,2) hence the region where the blue graph is above the red graph is:
 

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
583
Gender
Male
HSC
2017
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

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
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:





 

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

Top