Maths AS Level exam question (Binomial Expansion)?

a) (1 + px)^9 ≈ 1^9 + 9C1(px) + 9C2(px)²

=> 1 + 9px + 36p²x²

b) 36x = 9px => p = 4

qx² = 36p²x² = 576x² => q = 576

:)>

The first three terms of a binomial expansion (q+p)^n are

nC0*q^(n-0)*p^0 , nC1*q^(n-1)*p^1 and nC2*q^(n-2)*p^2

In this question q = 1, p = px and n = 9

Second term is 9C1*1^8*(px)^1 = 36x

9*1*px = 36x

9px = 36x

p = 36x/9x

p = 4

Third term is 9C2*1^7*(px)^2 = qx^2

36*1*p^2x^2 = qx^2

36p^2x^2 = qx^2

36p^2 = q

q = 36*4^2 = 36*16 = 576

a million. (x^2 + 4x + 9) might properly be written (x + 2)^2 + 5. [multiply out and spot] You halve the coefficient of x first then upload it to x and sq. the effect. So a million / (x^2 + 4x + 9) = a million / [(x+2)^2 + 5] optimal fee of the fraction will correspond to the minimum fee of (x+2)^2 + 5. positioned y = (x+2)^2 + 5 dy/dx = 2(x+2). while dy/dx = 0, x = -2 positioned x = -2 interior the fraction and get a million / 5, this is the optimal fee of the fraction. {the different fee of x will supply a larger denominator and consequently a smaller fraction} 2. For the quadratic equation to haven't any real roots, you have sixteen < 4(ok+2)(ok+5), which boils right down to ok^2 + 7k + 6 > 0 as you're saying. (ok+6)(ok+a million) > 0 via factorising it. it is effective if the two the two factors are unfavourable or the two are effective. ok > -a million ensures the two are effective [examine and spot] ok < -6 ensures the two are unfavourable [lower back examine] that's while ok lies between -6 and -a million (inclusive of equalling -6 and -a million) which you get a unfavourable or 0 result. consequently the values of ok are the two decrease than -6 or extra suitable than -a million. 3. For the binomial enlargement you get (a million + n.(2x) + nC2. (2x)^2 + nC3.(2x)^3 + etc ) So coefficient of x^2 is 4.nC2 coefficient of x is 2n Equation is 4nC2 = 4n 4n(n-a million) / 2 = 4n n(n-a million) = 2 n = 2. the different(s) I circulate away to you.

First 3 terms are:

(9 choose 0)p^0x^0 + (9 choose 1)p^1x^1 + (9 choose 2)p^2x^2

(9 choose 0) = 9!/(9!0!) = 1

(9 choose 1) = 9!/(8!1!) = 9

(9 choose 2) = 9!/(7!2!) = 9*8/2 = 36

So the first 3 terms are:

1, 9px, 36p^2x^2

9px = 36x

9p = 36

p = 4

36p^2x^2 = qx^2

36p^2 = q

Since we know that p = 4:

36*16 = q

q = 576

2a)

1+9(1)(px)+36(1)(px)^2

b)

36=9p

p=4

q=36p^2

q=576