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Find the range of value of p for which 4x(^2) – px + 1…

(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.