(a) Find the range of value of p for which 4x\(^2\) - px + 1 = 0 (b)(i) Expand (1 + 3x)\(^6\) in ascending powers of x (ii) Using the expression in 10 (ii) ...

Answer Details

a) To find the range of values of p for which 4x\(^2\) - px + 1 = 0, we need to use the discriminant of the quadratic equation.

The discriminant is given by the expression b\(^2\) - 4ac, where a = 4, b = -p and c = 1.

So, the discriminant is:

b\(^2\) - 4ac = (-p)\(^2\) - 4(4)(1) = p\(^2\) - 16

For the quadratic equation to have real roots, the discriminant must be greater than or equal to zero.

So, we have:

p\(^2\) - 16 \(\geq\) 0

Solving for p, we get:

p \(\leq\) -4 or p \(\geq\) 4

Therefore, the range of values of p for which 4x\(^2\) - px + 1 = 0 has real roots is p \(\leq\) -4 or p \(\geq\) 4.

b)

i) To expand (1 + 3x)\(^6\) in ascending powers of x, we can use the binomial theorem. The general term in the expansion is given by:

C(n, r) a\(^r\) b\(^{n-r}\)

where C(n, r) is the binomial coefficient, n is the power of the binomial, a is the first term, and b is the second term.

In this case, we have:

n = 6, a = 1, b = 3x

So, the expansion is:

(1 + 3x)\(^6\) = C(6,0) 1\(^6\) (3x)\(^0\) + C(6,1) 1\(^5\) (3x)\(^1\) + C(6,2) 1\(^4\) (3x)\(^2\) + C(6,3) 1\(^3\) (3x)\(^3\) + C(6,4) 1\(^2\) (3x)\(^4\) + C(6,5) 1\(^1\) (3x)\(^5\) + C(6,6) 1\(^0\) (3x)\(^6\)

Simplifying and collecting like terms, we get:

(1 + 3x)\(^6\) = 1 + 18x + 135x\(^2\) + 540x\(^3\) + 1215x\(^4\) + 1458x\(^5\) + 729x\(^6\)

ii) To find the value of (1.03)\(^6\) correct to four significant figures, we can substitute x = 0.03 into the expression we obtained in part (i):

(1 +