How to solve this ???

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Answer 1 · 2017-06-03T05:41:34

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

We have #3x^2+6x+5#

= #3(x^2+2x)+5#

= #3(x^2+2xx x xx1+1^2)-3xx1^2+5#

= #3(x+1)^2-3+5#

= #3(x+1)^2+2#

Hence #a=1# and #b=2#

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

#3(x+1)^2>=0# or #3(x+1)^2+2>=2#

i.e. #3x^2+6x+5 >=2# and dividing by #3x^2+6x+5#, we get

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