Q2: Solutions (1 Viewer)

[MarsBarz]

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.
T_k+1 = ⁿC_k.a^n-k.b^k
T_k+1 = ¹²C_k.(2x)^12-k.(-1/x²)^k
T_k+1 = ¹²C_k.(2)^12-k.x^12-k.(-1)^k.x^-2k
T_k+1 = ¹²C_k.(2)^12-k.(-1)^k.x^12-3k
For the independent term, x^12-3k = 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 e^3x(cosx-3sinx)
d/dx[e^3x(cosx-3sinx)] = 3e^3x(cosx-3sinx)+e^3x(-sinx-3cosx)
= -10e^3xsinx



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...

 

[Slidey]

Part a needs a root sign.

 

[Kutay]

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.
T_k+1 = ⁿC_k.a^n-k.b^k
T_k+1 = ¹²C_k.(2x)^12-k.(-1/x²)^k
T_k+1 = ¹²C_k.(2)^12-k.x^12-k.(-1)^k.x^-2k
T_k+1 = ¹²C_k.(2)^12-k.(-1)^k.x^12-3k
For the independent term, x^12-3k = 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 e^3x(cosx-3sinx)
d/dx[e^3x(cosx-3sinx)] = 3e^3x(cosx-3sinx)+e^3x(-sinx-3cosx)
= -10e^3xsinx



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 ?