# Q2: Solutions (1 Viewer)

#### MarsBarz

##### Member
Question 2:

a. Find d/dx(2sin-15x).
d/dx(2sin-15x = [2/(1-25x2)1/2] . 5
= 10/(1-25x2)1/2

b. Use the binomial theorem to find the term independent of x in the expansion of [2x-(1/x2)]12.
Tk+1 = nCk.an-k.bk
Tk+1 = 12Ck.(2x)12-k.(-1/x2)k
Tk+1 = 12Ck.(2)12-k.x12-k.(-1)k.x-2k
Tk+1 = 12Ck.(2)12-k.(-1)k.x12-3k
For the independent term, x12-3k = 1
12-3k = 0
12 = 3k
k = 4
Hence the independent term is, T5 = 12C4.(2)12-4.(-1)4
.:. T5 = 126720

c.
(i) Differentiate e3x(cosx-3sinx)
d/dx[e3x(cosx-3sinx)] = 3e3x(cosx-3sinx)+e3x(-sinx-3cosx)
= -10e3xsinx

I'll edit this post and do the rest later...only got 4 hours of sleep last night and I'm starting to feel it. Someone else has made full solutions anyway...

Last edited:

#### Slidey

##### But pieces of what?
Part a needs a root sign.

#### Kutay

##### University
MarsBarz said:
Question 2:

a. Find d/dx(2sin-15x).
d/dx(2sin-15x = [2/(1-25x2)] . 5
= 10/(1-25x2)

b. Use the binomial theorem to find the term independent of x in the expansion of [2x-(1/x2)]12.
Tk+1 = nCk.an-k.bk
Tk+1 = 12Ck.(2x)12-k.(-1/x2)k
Tk+1 = 12Ck.(2)12-k.x12-k.(-1)k.x-2k
Tk+1 = 12Ck.(2)12-k.(-1)k.x12-3k
For the independent term, x12-3k = 1
12-3k = 0
12 = 3k
k = 4
Hence the independent term is, T5 = 12C4.(2)12-4.(-1)4
.:. T5 = 126720

c.
(i) Differentiate e3x(cosx-3sinx)
d/dx[e3x(cosx-3sinx)] = 3e3x(cosx-3sinx)+e3x(-sinx-3cosx)
= -10e3xsinx

I'll edit this post and do the rest later...only got 4 hours of sleep last night and I'm starting to feel it. Someone else has made full solutions anyway...
For the binomial one i put -answer stupid mistake how much would i loose ?