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2019 WAEC Further Mathematics Theory (a) Find the range of value of p for which 4x\(^2\) - px + 1...

Further Mathematics
WAEC 2019

(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)\(^6\)

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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. 

 


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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995