(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) ...
Question 1 Report
(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) find, correct to four significant figures, the value of (1.03)\(^6\)
Answer Details
(a) For \(4x^2 - px + 1 = 0\) to have real roots, the discriminant must be non-negative:
\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-p)^2 - 4(4)(1) \ge 0 \;\Rightarrow\; p^2 - 16 \ge 0\] \[(p-4)(p+4) \ge 0 \;\Rightarrow\; p \le -4 \ \text{ or } \ p \ge 4\](b)(i) Using \((1+3x)^6 = \displaystyle\sum_{r=0}^{6}\binom{6}{r}(3x)^r\):
\[(1+3x)^6 = 1 + 18x + 135x^2 + 540x^3 + 1215x^4 + 1458x^5 + 729x^6\](ii) Put \(3x = 0.03\), i.e. \(x = 0.01\), so \((1+3x)^6 = (1.03)^6\):
\[(1.03)^6 \approx 1 + 18(0.01) + 135(0.01)^2 + 540(0.01)^3 + \dots\] \[= 1 + 0.18 + 0.0135 + 0.00054 + 0.0000122 + \dots = 1.19405\]To four significant figures, \((1.03)^6 \approx \mathbf{1.194}\).
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