From where did that a a come from?

Algebra Level 4

3 x 2 + 2 x + a 2 x 2 + 3 x + 2 ≤ 3 \frac {3x^{2}+2x+a}{2x^{2}+3x+2}\leq3 for all real x x

Then which one of the following values a a cannot take?

1.082 0 -1 2

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Sabhrant Sachan
May 29, 2016

We have an expression : 3 x 2 + 2 x + a 2 x 2 + 3 x + 2 ≤ 3 The Quadratic equation in the denominator is always greater than 0 as, D = 9 − 4 ⋅ 2 ⋅ 2 ⟹ − 7 < 0 and the coefficient of x 2 ⟹ 2 > 0 , Now 3 x 2 + 2 x + a − 6 x 2 − 9 x − 6 2 x 2 + 3 x + 2 ≤ 0 3 x 2 + 2 x + a − 6 x 2 − 9 x − 6 ≤ 0 3 x 2 + 7 x + ( 6 − a ) ≥ 0 The Discriminant of the equation should be less than or equal to 0 D ≤ 0 49 − 4 ⋅ 3 ⋅ ( 6 − a ) ≤ 0 a ≤ 23 12 < 2 Our Answer is 2 \text{We have an expression : }\dfrac{3x^2+2x+a}{2x^2+3x+2}\le3 \\ \text{The Quadratic equation in the denominator is always greater than 0} \\ \text{ as, D }= 9-4\cdot2\cdot2 \implies -7<0 \text{ and the coefficient of } x^2 \implies 2>0 \text{, Now } \\ \dfrac{3x^2+2x+a-6x^2-9x-6}{2x^2+3x+2}\le0 \\ 3x^2+2x+a-6x^2-9x-6\le0 \\ 3x^2+7x+(6-a)\ge0 \\ \text{The Discriminant of the equation should be less than or equal to 0} \\ D \le 0 \\ 49-4\cdot3\cdot(6-a)\le0 \\ a\le \dfrac{23}{12}<2 \\ \text{Our Answer is }\color{#3D99F6}{\boxed{2}}

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