13x2+2ax+b=0
By Sridhacharya method to find its roots √4a2-3b this has to be positve
6x+2a≥0 → x=-a/3→ -3x=a therefore a is also positive (c)
341Fact: It attains an extremum
Implication: f'(x) = 3x2+2ax+b = 0 has real roots
And so?: This means discriminant 4(a2-3b)>0. So the 1st quantity is positive (funnily, in all your choices, its +, so this was not needed. still...)
Fact the extremum is a minima. Also the point of minimum is negative.
Implication f"(x0)>0 with x0<0 where x0 is the minimum point
And so? f"(x0) >0 means 6x0+2a>0. or x_0> -\frac{a}{3} . You can see that if a<0, we are in serious problems as that would mean x0>0. So a>0
Fact 3x2+2ax+b = 0 has a negative root. I hope that means only one negative root
Implication Not much really to add to previous sentence.
And so? so the other roots is +ve and hence product of roots is -ve. so that b<0.
otherwise both options c and d are correct
1f(x)=x3 + ax2 + bx +c ,
we have f'(x)=3x2+2ax+b
now we see that this is an upward concave curve,
if at all this eqn has roots, then the first one wud be a pt of maxima of
f(x) and the second root wud be the minima of f(x),
so we realise that, if the 'minima x' > 'maxima x'
or if the minima x is <0, even the maxima x<0
so, we have sum of the roots of the eqn f'(x)=0
= -2a/3 <0....1)
we hve product of roots= b/3>0.....2)
we assume the eqn to actually have maxima and minima points,
so D>0, or a2-3b>0....3)
so D)
cheers!!
