Questions from everywhere (1 Viewer)

[steve001]

Hey i just got a couple of questions,

1. Of the three roots of the cubic equation x^3 - 15x + 4 = 0, two are reciprocals.
(i) Find the other root. (i got -4)
(ii) Find all the roots and verify that two of them are reciprocals.

2. A solid cylinder, radius r and height h, is to be constructed under the condition that the sum of its height and circumference is S, where S is constant.
Prove that
(i) if it is given its maximum possible volume, h = S/3
(ii) if it is given its maximum possible total surface area, r+ h = S/2

3. (i) Express sinx - cosx in the form Asin(x-y), with A > 0 and 0 < y < pi/2. (i got (rt2)sin(x-pi/4).)
(ii) Determine lim x->pi/4 (sinx-cosx)/(x-pi/4)

4. AB is a chord of a circle of radius 1 metre that subtends an angle x at the centre, where 0<x<pi. the perimeter of the minor segment cut off by AB is equal to the diameter of the circle.
(i) Show that x +2sin(0.5x) - 2 = 0
(ii) Show that the value of x is such that 1<x<2.

5. (i) Write down the binomial expression of (1+x)^n in ascending powers of x.
(ii) Show that n(sigma)r+1 nCr = 2^n -1 (i thought that you will use induction but its only 1 mark).
(iii) Use integration and the answer to part (i) to show that
(1/n+1)(n+1 sigma r=1) n+1Cr = n sigma r=0 (nCr/r+1)

Sorry for the setting out especially for the last question but any help will be greatly appreciated.

Thanks in advanced.

 

[scardizzle]

1. ii) i just used long division you get the polynomial x^2 -4x + 1
using the GQF i get x = 2+/- 3^1/2 and (2+root3) x (2-root3) = 1

 

[scardizzle]

3. (sinx - cos)/(x-pi/4) = rt2sin (x - pi/4)/(x-pi/4)

using the small angle theory

sin (x-pi/4)/ x-pi/4 -> 1 as x -> pi/4

therefore its equal to rt2

 

[steve001]

Thanks for the help
Your answers are right coz i put them in the equation and they both equalled zero.

Now just the other 4 to go.

 

[Revacious]

the last one, you just sub in x=1

(1+x)^n = sum (r=0~n) ncr x^r

put x=1

2^n = sum (r=0~n) ncr


then integrate both sides to get the last aprt, its not induction.