if we see the discriminant of the given exp. it is (a-2)²+8a=(a+2)² which is always +ve

so the its grph is definitely cutting the x axis...nw as we see f(0)= -2 i.e. the graph will hv to be an upward opening parabola to satisfy the given conditn as f(0) is -ve..

so..a>0 will be the ans..

(1+cosA)/sinA=11/2

2sin²(A/2)/2sin(A/2)cos(A/2)=cot(A/2)=11/2=>tanA/2=2/11

tanA=2tan(A/2)/1-tan²(A/2)

substituting the value of tanA/2

we get tanA=(44/105)

let the family of lines be 2x+6y=K

this can be written as x/(K/2) +y/(K/6) =1....(intercept form)

which gives the x & y intercept as K/2 & K/6 respectively....

as it is given that the intercept b/w the axes is 10 we can write

K²/4 + K²/36 = 100

it is a quadratic in K so we will get 2 lines.

If we integrate the given expression,i.e. 3ax²+2bx+c we get ax³+bx²+cx

applying limits frm 0 to 2 we find that the value of integeral is 2(4a+2b+c) which is 0

this means that the area under curve of the given expression is 0 b/w 0and2 so the curve of the given expression will have to intersect the x axis in this interval.....so there will be a root in the interval     (0,2)

ans is (c)