(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)
Explanation
(a) The range of values of p must satisfy p(^2) – 4 x 4 x 1 (geq) 0 which will simplify to (p – 4)(p + 4) (geq) 0
Thus; p (geq) -4 or p (leq) 4
(b)(i) (1 + 3x)(^6) which is to be in ascending powers of x as ((^6_0))(3x)(^o) + ((^6_1))(3x)(^1) + ((^6_2))(3x)(^2) + ((^6_3))(3x)(^3) + ((^6_4))(3x)(^4) + ((^6_5))(3x)(^5) + ((^6_6))(3x)(^6)
= 1 + 18x + 135x(^2) + 540x(^3) + 1215x(^4) + 1458x(^5) + 729x(^6)
(b)(ii) Observe that 1 + 3x = 1.03
x = 0.01
Substituting for x in the expansion;
(1.03)(^6) = 1 + 18(0.01) + 135(0.01)(^2) + 540(0.01)(^3) + 1215(0.01)(^4) + 1458(0.01)(^5) + 729(0.01)(^6)
= 1 + 0.18 + 0.0135 + 0.00054 + 0.00001215 + …
= 1.19405215
= 1.194 correct to four significant figures.