Core Mathematics C12(GCE)Practice Answer 6

Standard 𝐀𝐀

𝑴𝑴𝑴𝑴.𝑺𝑺.𝑽𝑽.𝑺𝑺𝑺𝑺𝑺𝑺𝑴𝑴𝑺𝑺𝑺𝑺𝑴𝑴𝑺𝑺𝑺𝑺𝑺𝑺

𝑻𝑻𝑻𝑻𝑺𝑺𝑻𝑻 𝑳𝑳𝑻𝑻𝑺𝑺𝑳𝑳𝑻𝑻𝑴𝑴,𝑴𝑴𝑺𝑺𝑴𝑴𝑴𝑴𝑴𝑴𝑺𝑺𝑴𝑴 𝑬𝑬𝑬𝑬𝑺𝑺𝑻𝑻𝑴𝑴𝑺𝑺𝑻𝑻𝑴𝑴 & 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑴𝑴

𝑺𝑺𝑺𝑺𝑺𝑺. 𝒔𝒔𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝒔𝒔𝑨𝑨. 𝒄𝒄𝑨𝑨𝑻𝑻

Method 1

(a) 𝑥𝑥2 − 4𝑥𝑥 + 5 > 0

𝑓𝑓 𝑥𝑥 = 𝑥𝑥 − 2 2 + 1

For all 𝑥𝑥 values 𝑓𝑓(𝑥𝑥) is greater than zero

Therefore −∞ < 𝑥𝑥 < ∞

No part of this publication maybe reproduced in any form without the prior written permission fromMR.S.V.SWARNARAJA, (Team Leader, Marking Examiner & Author), email: swa@swanash.com

Method 2

(a) 𝑥𝑥2 − 4𝑥𝑥 + 5 > 0

discriminant

= 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎

= 16 − 4 × 1 × 5

= −4 < 0

No real solutions and

minimum curve 𝑎𝑎 < 0

Hence, for all 𝑥𝑥 values 𝑓𝑓(𝑥𝑥) is

greater than zero

Therefore, −∞ < 𝑥𝑥 < ∞

No part of this publication maybe reproduced in any form without the prior written permission fromMR.S.V.SWARNARAJA, (Team Leader, Marking Examiner & Author), email: swa@swanash.com

(b) 𝑥𝑥2 + 16 > 8𝑥𝑥

𝑥𝑥2 − 8𝑥𝑥 + 16 > 0

𝑓𝑓 𝑥𝑥 = 𝑥𝑥 − 4 2

Therefore 𝑥𝑥 < 4, 𝑥𝑥 > 4 ( 𝑥𝑥 ≠ 4)

No part of this publication maybe reproduced in any form without the prior written permission fromMR.S.V.SWARNARAJA, (Team Leader, Marking Examiner & Author), email: swa@swanash.com

• Linear Inequalities are solved in a similar manner to linearequations but If you are multiplying or dividing the inequalityby a negative number you must change the inequality signround.

−2𝑥𝑥 > 10

𝑥𝑥 <10−2

𝑥𝑥 < −5

−2𝑥𝑥 > 10

0 > 10 + 2𝑥𝑥

−10 > 2𝑥𝑥

−5 > 𝑥𝑥

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀: 1 (𝑤𝑤𝑤𝑤𝑀𝑀𝑀 𝑠𝑠𝑤𝑤𝑠𝑠𝑠𝑠 𝑎𝑎𝑀𝑎𝑎𝑠𝑠𝑠𝑠𝑀𝑀) 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀: 2 (𝑤𝑤𝑤𝑤𝑀𝑀𝑀𝑀𝑀𝑤𝑤𝑀𝑀 𝑠𝑠𝑤𝑤𝑠𝑠𝑠𝑠 𝑎𝑎𝑀𝑎𝑎𝑠𝑠𝑠𝑠𝑀𝑀)

𝑇𝑇𝑇𝑇𝑎𝑎𝑀𝑀𝑤𝑤𝑀𝑀𝑀𝑀𝑠𝑠𝑎𝑎𝑇𝑇 𝑀𝑀𝑇𝑇 𝑂𝑂𝑠𝑠𝑇𝑇𝑤𝑤𝑠𝑠𝑀𝑀 𝑎𝑎𝑇𝑇𝑎𝑎𝑠𝑠𝑠𝑠𝑀𝑀𝑠𝑠

𝑴𝑴𝑴𝑴.𝑺𝑺.𝑽𝑽.𝑺𝑺𝑺𝑺𝑺𝑺𝑴𝑴𝑺𝑺𝑺𝑺𝑴𝑴𝑺𝑺𝑺𝑺𝑺𝑺

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𝑤𝑤𝑤𝑤𝑤𝑤. 𝑠𝑠𝑤𝑤𝑎𝑎𝑠𝑠𝑎𝑎𝑠𝑠𝑀. 𝑎𝑎𝑀𝑀𝑇𝑇𝑀𝑀𝑇𝑇𝑎𝑎𝑤𝑤𝑇𝑇: 𝑠𝑠𝑤𝑤𝑎𝑎𝑠𝑠𝑠𝑤𝑤𝑎𝑎𝑠𝑠𝑎𝑎𝑠𝑠𝑀. 𝑎𝑎𝑀𝑀𝑇𝑇