21if 2a+3b+6c=0, then atleast one root of the equation ax^2 +bx +c=0 lies in the interval:
1) 0,1
2)1,2
3)2,3
4)1,3
11BHAI 4 THE 2ND ONE
U HAVE TO INTEGRATE THIS FUNCTION TO FIND A NEW FUNCTION SO AS TO FIND THE INTERVAL
ON INTEGRATING U GET
ax3/3+bx2/2+cx=0
u get
2ax3+3bx2+6cx=0
now x=1 satisfies the given relation
and at x=0 c=0
soits 1 root must lie between (0,1)
1061=2/b+3/a ..
==> (2/b+3/a)/2 ≥ √6/ab (AM-GM)
==> 1/2 ≥ √6/ab
==> 1/4≥ 6/ab
==> ab ≥ 24
21and at x=0 c=0
soits 1 root must lie between (0,1)
manipal pls explain this sentence.[2]