If `ax^(2)+bx+c=0` have two distinct roots lying int eh interval `(0,1),a,b,c ` ` epsilon N` The least value of `b` is

Options

A. 5
B. 6
C. 7
D. 8

Answer

A

Explanation

Let `f(x)=ax^(2)-bx+c` has two distinct roots `alpha` and `beta`. Then `f(x)=a(x-alpha)(x-beta)`. Since `f(0)` and `f(1)` are of same sign.
Therefore `c(a-b+c)gt0`
`impliesc(a-b+c)ge1`
`:.a^(2) alpha beta(1-apha)(1-beta)ge1`
But `alpha(1-alpha)=1/4=(1/2-alpha)^(2)le1/4`
`:.a^(2) alpha beta(1-alpha)(1-beta)lt(a^(2))/16`
`implies(a^(2))/16gt1impliesagt4 [ :' alphq!=beta]`
`impliesage5` as `a epsilonI`
Also `b^(2)-4acge0`
`impliesb^(2)ge4acge20`
`impliesbge5`
Next `age5, bge5` we get `cge1` ltbr `:.abcge25`
`log_(5)abcgelog_(5)25=2`
Least value of `b` is 5