yr 11 revision - questions (1 Viewer)

[-pari-]

hey, i'm revising yr 11 stuff, and came up with a few problems....would be great if someone could take a look n help out

1)
|x + 1|
x^2 - 1

can someoe just check if i've got the right jist of things:

x cannot= 1 or -1 [from the denominator, x^2 - 1 =/= 0]

when |x + 1| >0
x>-1

.'. for x > -1

x + 1
x^2 - 1

= x+1
(x+1)(x-1)

= 1
(x-1)

when |x +1| <0
-x-1 < 0
-x <1
x > -1 .........wait...but thats the same thing as before? :S

how do i get

1
1 - x

when x < - 1 ??

2) -4 (smaller.than.or.equal.to) 3
(x+3)

i never get this..
book says you multiply both sides by (x+3)^2...

...why? why can't you just multiply it by (x+3) like you normally would?


P.S: Happy new year all

 

[Riviet]

2) We don't know whether x+3 is positive or negative and this means that if we multiply boths sides by x+3, we don't know whether to keep the inequality sign the same or to switch it to the other way. However (x+3)² is always positive (or zero) so we can safely multiply by that on both sides of the inequality.

 

[-pari-]

ahh...right! cheers

 

[jyu]

hey, i'm revising yr 11 stuff, and came up with a few problems....would be great if someone could take a look n help out

1)
|x + 1|
x^2 - 1

can someoe just check if i've got the right jist of things:

x cannot= 1 or -1 [from the denominator, x^2 - 1 =/= 0]

when |x + 1| >0
x>-1

.'. for x > -1

x + 1
x^2 - 1

= x+1
(x+1)(x-1)

= 1
(x-1)

when |x +1| <0
-x-1 < 0
-x <1
x > -1 .........wait...but thats the same thing as before? :S

how do i get

1
1 - x

when x < - 1 ??

2) -4 (smaller.than.or.equal.to) 3
(x+3)

i never get this..
book says you multiply both sides by (x+3)^2...

...why? why can't you just multiply it by (x+3) like you normally would?


P.S: Happy new year all

when |x +1| <0 .......... This is wrong.
| | is always greater than or equal to 0, and since x =/= -1,
so |x + 1|>0.

For x < -1, |x + 1|/[(x - 1)(x + 1)] = -(x +1)/[(x - 1)(x + 1)]

= -1/(x - 1)

= 1/(1 - x)

Happy 2007

:wave:

 

[bos1234]

how do u do it this way??...

denominator cannot = 1 or -1

so on the number line... A-1-------------0B-----------1C +++(-2,-1,0,1,2,3)

so for A when x is < -1... solve etc etc and i got as above

and for C when x > 1 solve 1/x-1 --- is this bit right?

how about the middle bit.. if i put x=0 i get x+1/x^2+1

 

[jyu]

how do u do it this way??...

denominator cannot = 1 or -1

so on the number line... A-1-------------0B-----------1C +++(-2,-1,0,1,2,3)

so for A when x is < -1... solve etc etc and i got as above

and for C when x > 1 solve 1/x-1 --- is this bit right?

how about the middle bit.. if i put x=0 i get x+1/x^2+1

|x+1|/(x^2 - 1) = |x+1|/[(x-1)(x+1)] is defined for all x values except -1 and 1.

To simplify the expression, consider the two cases: x < -1 and x > -1.

For x < -1, |x+1|/(x^2 - 1) = 1/(1-x)

For x > -1, |x+1|/(x^2 - 1) = 1/(x-1).
This second case covers x =0, |x+1|/(x^2 - 1) = 1/(0-1)=-1

:wave:

 

[bos1234]

ok thanks!

 

[-pari-]

doing revision of calculus:

find the acute angle between the curve y= x^2 and the line y= 2x +3 at each point of intersection

say what??

dont understand the question let alone what i'm meant to do.....

 

[jyu]

doing revision of calculus:

find the acute angle between the curve y= x^2 and the line y= 2x +3 at each point of intersection

say what??

dont understand the question let alone what i'm meant to do.....

Meaning the angle between the line y =2x + 3 and the tangent to the curve y = x^2 at an intersecting point.

Step 1: Find the x-coordinates of the intersecting points.

Step 2: Use calculus to find the gradient of the tangent at each intersecting point.

Step 3: At each intersecting point use the gradients of the line and the tangent to find the angle that each makes with the x-axis (positive direction).

angle with x-axis = tan^(-1) gradient

Step 4: Required angle = Larger angle - smaller angle


:wave:

 

[Riviet]

Basically you're finding the angle made by the tangents drawn to the point of intersection for each curve. So find the point of intersection, differentiate and find the gradient of the tangent for each curve, which allows you to use the formula tan@=|(m1-m2)/(1+m1m2)| to find @ or you can find and subtract the angles as jyu has explained.

 

[-pari-]

hmm....yeah i initially thought it was asking for the angle between the two tangents

but then i figured that wasn't right...

thnks 4 the help

 

[bos1234]

how do i do this qn without multiplying by the square...

(x^2 - 9) / x > 0

x^2 - 9 divide by x > 0

denom. =/= 0??

 

[jyu]

how do i do this qn without multiplying by the square...

(x^2 - 9) / x > 0

x^2 - 9 divide by x > 0

denom. =/= 0??

For (x^2 - 9) / x > 0, either both x^2 - 9 and x are negative, or both are positive, i.e.

either x^2 - 9 < 0 and x < 0

or x^2 - 9 > 0 and x > 0.

First case yields -3 < x < 0, second case x > 3.

Solution set is {x: -3 < x < 0} U {x: x > 3}.


:wave:

 

[Slidey]

ALWAYS DRAW A SMALL SKETCH FOR INEQUALITIES!!!!!!!!!!!!111

Please.

(In general, not to anybody specific in this thread)

 

[~shinigami~]

Whoa! Slidey, what happened to your orange colour which signifies your awesome as a mod?

 

[-pari-]

questions

1)

y^2 - 6 (smaller.than.or.equal.to) 1
y

i get...

y^3 - 6y - y^2 (smallerthanorequalto) 0

are we meant to use polynomials for this?

same with ...

2) Factorise & solve x^5 - 9x^3 - 8x^2 + 72 = 0

3) solve simultaneous equations y= x^3 = x^2 & y = x + 1

 

[Riviet]

y³-y²-6y=y(y²-y-6)
=y(y-3)(y+2)

Now sketch this and find the points that are on the x-axis OR below the x-axis. These are your solutions.

2) Substitute numbers that are factors of 72, use the factor theorem and long divide if necessary after you've found a few solutions.

3) y=x³+x²=x+1
x³+x²-x-1=0
Substitute x=-1 and then use long division to find the quadratic and use the quad formula if necessary to find the other two roots.

 

[-pari-]

1) would that be a parabola? coz there is an extra y at the front... y(y-3)(y+2)? wudn't that change the shape of the curve into something else?

2) & 3) are those polynomial functions? or are you solving them using methods you'd use to solve polynomial functions...?


[coz i'm pretty sure thats 3u, so just so i knw whether i need to worry about it or not ]

 

[Riviet]

Ah sorry, I didn't realise you were only doing 2u. =\

1) It's a cubic (expand it and see for yourself).

2) & 3) Yes they are polynomials, the one in 2 is a polynomial equation and the ones in 3 are polynomial functions.

This is the 2u method that I'd use which happens to be easier/quicker than 3u methods:

x³+x²-x-1=0
x³-x+x²-1=0
x(x²-1) + 1(x²-1) = 0
(x²-1)(x+1)=0
x=-1,1

 

[-pari-]

hmm...smart


thanks