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Answer 1 · 2017-06-03T05:41:34

We have $a = 1$ $a=1$ and $b = 2$ $b=2$ . For details please see below.

Explanation:

We have $3 x^{2} + 6 x + 5$ $3x^2+6x+5$

= $3 (x^{2} + 2 x) + 5$ $3(x^2+2x)+5$

= $3 (x^{2} + 2 \times x \times 1 + 1^{2}) - 3 \times 1^{2} + 5$ $3(x^2+2xx x xx1+1^2)-3xx1^2+5$

= $3 {(x + 1)}^{2} - 3 + 5$ $3(x+1)^2-3+5$

= $3 {(x + 1)}^{2} + 2$ $3(x+1)^2+2$

Hence $a = 1$ $a=1$ and $b = 2$ $b=2$

As ${(x + 1)}^{2}$ $(x+1)^2$ is a perfect square, and minimum value of $3 x^{2} + 6 x + 5$ $3x^2+6x+5$ is $2$ $2$ and hence it is always non-zero and positive and hence $\frac{2}{3 x^{2} + 6 x + 5} \geq 0$ $2/(3x^2+6x+5) >=0$

$3 {(x + 1)}^{2} \geq 0$ $3(x+1)^2>=0$ or $3 {(x + 1)}^{2} + 2 \geq 2$ $3(x+1)^2+2>=2$

i.e. $3 x^{2} + 6 x + 5 \geq 2$ $3x^2+6x+5 >=2$ and dividing by $3 x^{2} + 6 x + 5$ $3x^2+6x+5$ , we get

$0 \leq \frac{2}{3 x^{2} + 6 x + 5} \leq 1$ $0 <=2/(3x^2+6x+5) <= 1$
graph{2/(3x^2+6x+5) [-10, 10, -5, 5]}

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