Find all real values of m for which the inequalty mx2- 4x + 3m + 1 >0 is satisfied for all positive x.

Hint : THE GIVEN EXPRESSION IS +VE ONLY AND ONLY FOR POSTIVE x.

Hello Student, Please find the solution

The value of this quadratic is always positive (>0)
So graph will always be above x-axis
therefore the condition will be :-
D>0 and a>0 (parabola upwards and not touching the x-axis)

Solve both the conditions nw
1) D>0
16 – 4m(3m+1) >0
16 – 12m2 – 4m >0
multiplying by -1 both side
12m2+ 4m -16<0
Divide by 4
3m2+m – 4 <0
3m2+4m – 3m – 4 <0
m(3m+4) – 1 (3m + 4) <0
(3m+4)(m – 1) <0
[m \epsilon [-4/3,1]]

2) a>0
=>m>0

Now you can intersection of both the cases
[m \epsilon (0,1]]