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If ax^(2)-bx+c=0 have two distinct roots...

If `ax^(2)-bx+c=0` have two distinct roots lying int eh interval `(0,1),a,b,c epislonN` The least value of `log_(5)abc` is

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
B
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 `log_(b)abc` is 2.
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