Maths Q's (1 Viewer)

[sle3pe3bumz]

A few questions from my maths test which im curious about.



1.Let c(x)= (3^x + 3^-x)/2

Show that (c(x))^2 = 1/2(c(2x) + 1 )

do we just do all the working out, cos when i done it got muddled up so i gave up. ><"




2. y=-1/4*b^2*x + b has intercepts at A & B.

i) Find coordinates of P, the midpoint of AB.

ii) Show that P lies on the hyperbola y=1/x

iii) Show that the area of triangle OAB, where O is the origin, is independent of the value of b.

i could do i) but i couldnt do the others. ><" gosh i do feel stupid thoughh .. lol ..

 

[watatank]

1.Let c(x)= 0.5(3^x + 3^-x]

Show that 0.5[ (3^x + 3^-x]² = 0.5(c(2x) + 1 )

LHS = 0.5[ (3^x + 3^-x]²
= 0.5^2 * [ 3^x + 3^-x)²
=0.25 * [ 3^2x + 2 + 3^(-2x) ]


RHS = 1/2(c(2x) + 1 )
= 0.5* ( 0.5 *[ (3^2x + 3^-2x ])
= 0.5*0.5 ( [ (3^2x + 3^-2x ])
= 0.25 ( [ (3^2x + 3^-2x )
=LHS

For 2. ii) If it is going to satisfy y = 1/x, then the coordinates of P must be reciprocals. If they are it works. If not check your working for the first question.

Hope that helps! :wave:

 

[z600]

For the first one

c(2x)=(9^x+9^-x)/2

So (c(x))^2=(9^x+9^-x)/4+1/2

Square c(x) and get

(9^x+9^-x+2)/4

=(9^x+9^-x)/4+1/2

 

[sle3pe3bumz]

1.Let c(x)= 0.5(3^x + 3^-x]

Show that 0.5[ (3^x + 3^-x]² = 0.5(c(2x) + 1 )

LHS = 0.5[ (3^x + 3^-x]²
= 0.5^2 * [ 3^x + 3^-x)²
=0.25 * [ 3^2x + 2 + 3^(-2x) ]


RHS = 1/2(c(2x) + 1 )
= 0.5* ( 0.5 *[ (3^2x + 3^-2x ])
= 0.5*0.5 ( [ (3^2x + 3^-2x ])
= 0.25 ( [ (3^2x + 3^-2x )
=LHS

For 2. ii) If it is going to satisfy y = 1/x, then the coordinates of P must be reciprocals. If they are it works. If not check your working for the first question.

Hope that helps! :wave:

um they're not the same ? is there spose to be no 2 ? but besides that i get what your doing .. ^^"

& could somebody show me all the working out for the 2nd Q ? thanks =] much appreciated !

 

[sle3pe3bumz]

For the first one

c(2x)=(9^x+9^-x)/2

So (c(x))^2=(9^x+9^-x)/4+1/2

Square c(x) and get

(9^x+9^-x+2)/4

=(9^x+9^-x)/4+1/2

& i dont really get what your trying to say .. ><"

 

[watatank]

um they're not the same ? is there spose to be no 2 ? but besides that i get what your doing .. ^^"

& could somebody show me all the working out for the 2nd Q ? thanks =] much appreciated !

Opps! Sorry, forgot to do 1/2(c(2x) + 1 )

LHS = 0.5[ (3^x + 3^-x]²
= 0.5^2 * [ 3^x + 3^-x)²
=0.25 * [ 3^2x + 2 + 3^(-2x) ]


RHS = 1/2(c(2x) + 1 )
= 0.5* ( 0.5 *[ (3^2x + 3^-2x] +1 )
= 0.5 * ( [ (3^2x + 3^-2x ]) + 0.5 * 1
= 0.25 ( [ (3^2x + 3^-2x ) + 0.5
= 0.25 ( [ (3^2x + 3^-2x + 2 )
=LHS

NB: in the very last step, I factorised...

Fixed now, all good?

Oh yeah regarding the other questions, if you can do the first part of the second question you just have to show that that the coordinates of point P are reciprocals for the second part...if you could do the first question it should be right, if not youve got it wrong. I fair failed to do that part. No idea how to do the third part. I failed. lol.

 

[sle3pe3bumz]

Opps! Sorry, forgot to do 1/2(c(2x) + 1 )

LHS = 0.5[ (3^x + 3^-x]²
= 0.5^2 * [ 3^x + 3^-x)²
=0.25 * [ 3^2x + 2 + 3^(-2x) ]


RHS = 1/2(c(2x) + 1 )
= 0.5* ( 0.5 *[ (3^2x + 3^-2x] +1 )
= 0.5 * ( [ (3^2x + 3^-2x ]) + 0.5 * 1
= 0.25 ( [ (3^2x + 3^-2x ) + 0.5
= 0.25 ( [ (3^2x + 3^-2x + 2 )
=LHS

NB: in the very last step, I factorised...

Fixed now, all good?

Oh yeah regarding the other questions, if you can do the first part of the second question you just have to show that that the coordinates of point P are reciprocals for the second part...if you could do the first question it should be right, if not youve got it wrong. I fair failed to do that part. No idea how to do the third part. I failed. lol.

hm, call me stupid but can when you have

= 0.5* ( 0.5 *[ (32x + 3-2x] +1 )

can you bring the .5 and times it by 1 to get

= 0.5 * ( [ (32x + 3-2x ]) + 0.5 * 1 ? how ? ><" lol ..

& also how did the 0.5 become a 0.25 ?

= 0.5 * ( [ (32x + 3-2x ]) + 0.5 * 1
= 0.25 ( [ (32x + 3-2x ) + 0.5

lol, i feel like an idiot .. ^^" & also for the 2nd Q, I'm pretty sure i got that wrongg .. ><"

 

[watatank]

You simply can't. Sorry, I made a slight mistake. Again.

Damn my incompetence!

LHS = ( 0.5*(3^x + 3^-x )²
= 0.5^2 * [ 3^x + 3^-x)²
=0.25 * [ 3^2x + 2 + 3^(-2x) ]


RHS = 1/2(c(2x) + 1 )
= 0.5* ( 0.5 *[ (3^2x + 3^-2x] +1 )
= 0.5 * (0.5 * [ (3^2x + 3^-2x ]) + 0.5 * 1
= 0.25 ( [ (3^2x + 3^-2x ) + 0.5
= 0.25 ( [ (3^2x + 3^-2x + 2 )
=LHS

 

[nelses]

For the second question:

i. when x = 0, y = b
when y = 0, x = 4/b

Therefore letting A be the y intercept is (0,b) and B being the x intercept (4/b,0)

The x coordinate of P = (0+4/b) /2 = 2/b
The y coordinate of P = (0+b)/2 = b/2

Therefore P(2/b,b/2)

ii. y=1/x => subbing P in
L.H.S. = b/2
R.H.S = 1/(2/b)
= b/2
= L.H.S.

iii. Area of OAB = 1/2 *OA *OB (area of this right angled triangle)
= 0.5*2/b *b/2
= 0.5 u^2

Therefore the area of OAB is a constant and independent of the value of b

 

[sle3pe3bumz]

oh okayy .. i see where everything goes now .. ahah .. thanks for that you guys ^^"