Solution of Past Paper of CAIE Math 9709/12 Feb/Mar 2020 Question # 9
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CAIE AS & A Level Mathematics P1 (9709/1)
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CAIE AS & A Level Mathematics P1 (9709/1)
This document presents most carefully worked out and thoroughly explained complete Solution of Past Paper of Cambridge AS/A Level Mathematics (9709), Pure Mathematics 1, Feb/March 2020
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CIE/GCE/AS/Math/P1/20/Mar/12/Q#9
Question
a) Express 2𝑥2 + 12𝑥 + 11 in the form 2(𝑥 + 𝑎)2 + 𝑏, where 𝑎 and 𝑏 are constants.
The function f is defined by 𝑓 (𝑥 ) = 2𝑥 2 + 12𝑥 + 11 for 𝑥 ≤ −4.
b) Find an expression for 𝑓 −1 (𝑥 ) and state the domain of 𝑓 −1 (𝑥 ).
The function 𝑔 is defined by 𝑔(𝑥 ) = 2𝑥 − 3 for 𝑥 ≤ −4.
c) For the case where k = −1, solve the equation 𝑓𝑔(𝑥 ) = 193.
d) State the largest value of 𝑘 possible for the composition 𝑓𝑔 to be defined.
Solution
a)
We are given that;
2𝑥 2 + 12𝑥 + 11
We use method of “completing square” to obtain the desired form. We complete the
square for the terms which involve 𝑥.
2(𝑥 2 + 6𝑥 ) + 11
We have the algebraic formula;
(𝑎 − 𝑏)2 = (𝑎)2 − 2(𝑎)(𝑏) + (𝑏)2
(𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏2
For the given case we can compare the given terms with the formula as below;
𝑎2 = 𝑥 2
2𝑎𝑏 = 6𝑥 = 2(𝑥 )(3)
Therefore, we can deduce that;
, CIE/GCE/AS/Math/P1/20/Mar/12/Q#9
𝑏2 = (3)2
Hence, we can write;
2[(𝑥 )2 + 2(𝑥 )(3)] + 11
To complete the square, we can add and subtract the deduced value of 𝑏2 ;
2[(𝑥 )2 + 2(𝑥 )(3) + (3)2 − (3)2 ] + 11
2[{(𝑥 )2 + 2(𝑥 )(3) + (3)2} − (3)2 ] + 11
2[{(𝑥 )2 + 2(𝑥 )(3) + (3)2} − 9] + 11
2[{(𝑥 + 3)2 } − 9] + 11
2{(𝑥 + 3)2 } − 18 + 11
2 (𝑥 + 3 )2 − 7
b)
We are given that;
𝑓(𝑥 ) = 2𝑥 2 + 12𝑥 + 11 for 𝑥 ≤ −4
To find the inverse of a given function 𝑦 we need to write it in terms of 𝑦 rather than in
terms of 𝑥.
We have found in (a) that the given function can be written as;
𝑓 (𝑥 ) = 2 ( 𝑥 + 3 )2 − 7
𝑦 = 2 (𝑥 + 3 )2 − 7
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