Questions (1 Viewer)

[taggs-sasuke]

Can I please have help with this question? (Thank you!)

1. 2004 3(b)

Question:

P(x) = (x+1)(x-3)Q(x) + 3(x+1) - 11

What is the remainder when P(x) is divided by (x+1)(x-3)?

Answer:

The remainder is 3(x+1) - 11 - but why?

 

[Trebla]

The Remainder Theorem

It should be obvious that for some polynomial P(x) divided by A(x), you get a quotient Q(x) and a remainder R(x), so P(x) = A(x)Q(x) + R(x).

 

[Timothy.Siu]

i dont really get the rest of the question...

but thats the answer because

P(x)=Q(x)+Remainder.

 

[taggs-sasuke]

Oh right.

Thank you.

 

[taggs-sasuke]

2.

Question:

Graph y = xlnx - 1 , x>0

Answer:

• dy / dx = inx + 1

• Minimum turning point ( 1/e , -1/e-1)

• For x>0, the second derivative > 0

My question is how do we know there is a discontinuity at (0,-1)?

 

[taggs-sasuke]

3.

(I've done part i)

i. Show that the point (6p,3p^2) lies on the parabola x^2=12y

ii. The chord of the parabola x^2=12y passes through the point c(8,0). If A = (6p,3p^2) and B = (6q,3q^2) show that 4(p+q) = 3pq.

Can I please have help with part ii? Thank you.

 

[jet]

2.

Question:

Graph y = xlnx - 1 , x>0

Answer:

• dy / dx = inx + 1

• Minimum turning point ( 1/e , -1/e-1)

• For x>0, http://latex.codecogs.com/gif.latex?\frac{d^2y}{dx^2} > 0

My question is how do we know there is a discontinuity at (0,-1)?

Because, since ln (0) is undefined, the derivative is undefined at that point.

EDIT: And, because it says x > 0

 

[untouchablecuz]

2.

Question:

Graph y = xlnx - 1 , x>0

Answer:

• dy / dx = inx + 1

• Minimum turning point ( 1/e , -1/e-1)

• For x>0, http://latex.codecogs.com/gif.latex?\frac{d^2y}{dx^2} > 0

My question is how do we know there is a discontinuity at (0,-1)?

f(0) = 0Ln(0) - 1, which is undefined

however, to find the y coordinate we take limits

as x --> 0^+ (i.e as x approaches 0 from the positive side), y ---> -1, since xLnx --> 0 (the graph of y = Ln x has an asymptote at x = 0)

 

[jet]

 

[taggs-sasuke]

Thanks!

 

[taggs-sasuke]

Can I please have help again?

4.

RS^2 = 74 + 70cos(theta - 36degrees52minutes)

Find the maximum length of RS and the value of theta for which this occurs.

Thanks!

 

[Dumbledore]

Can I please have help again?

4.

RS^2 = 74 + 70cos(theta - 36degrees52minutes)

Find the maximum length of RS and the value of theta for which this occurs.

Thanks!

the maximum value of cos@ is 1 so RS^2=74+70(1); RS = 12

the max value of @ is when @-36'52'=0, because cos(0) = 1
so @ = 36'52'

 

[taggs-sasuke]

Thank you.