Consider the quadratic equation ` ax^(2) - bx + c =0,a,b,c in N` which has two distinct real roots belonging to the interval (1,2).
The least value of a is Option 1: 4 Option 2: 6 Option 3: 7 Option 4: 5

Options

A. 4
B. 6
C. 7
D. 5

Answer

4

Explanation

`az^(2) - bx + c =0`
Let `f(x) = ax^(2) - bx + c `be
the correspoding quadratic expression and `alpha,beta` be the roots of `f(x) = 0`.

Then, `f(x) = a(x-alpha) (x -beta)`
Now, `af(1) gt 0, af(2) gt 0,1lt(b)/(2a) lt2,b^(2) - 4ac gt 0`
`rArr a(1-alpha) (1-beta) gt 0, a(2-alpha)(2-beta) gt 0, 2a lt b lt 4a, b^(2) - 4ax gt 0`
`rArr a^(2) (1-alpha)(1-beta)(2-alpha) (2-beta) gt 0`
`rArr a^(2) (1-alpha)(1-beta)(2-alpha)(2-beta) gt 0`
As `f(x)` and `f(2)` both are integers and `f(1) gt 0`, and `f(2) gt 0`, so
`f(1)f(2) gt 0`
`rArr f(1) f(2) gt 1`
`rArr 1ge a^(2)(alpha -1)(2-alpha)(beta-1)(2-beta)" "(1)`
Now, `((alpha-1)+(2-alpha))/(2)gt ((alpha -1)(2-alpha))^(1//2)`
`rArr (alpha -1)(2-alpha) le (1)/(4)`
Similarly,
`(beta -1)(2-beta) le (1)/(4)`
`rArr (alpha -1)(2-alpha) (beta- 1)(2-beta)lt (1)/(16)`
As ` alpha ne beta`, so `alpha^(2) gt 1" "("Using (1)")`
`rArr a gt 5`
`rArr b^(2) gt 20c and b gt 10 rArr b ge 11`
Also, `b^(2) gt 100 rArr c gt 5 rArr c ge 6`