Let `f(x) =4x^2-4ax+a^2-2a+2` be a quadratic polynomial in x,a be any real number. If x-coordinate ofd vertex of parabola y =f(x) is less thna 0 and f(x) has minimum value 3 for `x in [0,2]` then value of a is

Options

A. 1
B. 2
C. 3
D. 0

Answer

2

Explanation

`f(x) = 4x^(2) - 4ax + a^(2)-2a + 2` represents a parabola whose vertex is at `((a)/(2), 2 - 2a)`
Case-I : `0 lt(a)/(2) lt 2`
In this case , `f(x)` attains the global minimum value at `x = (a)/(2)` .
Thus , `f((a)/(2)) = 3`
`rArr 3 = -2a + 2`
`rArr a = - (1)/(2)` (Rejected)
Case II : `(a)/(2)le2`
In this cases, f(x) attains the global minimun value at x = 0 .
Thus, `f(0) = 3`
`rArr 3 = a^(2) - 2a + 2`
`rArr a = 5 pm sqrt(10)`
So, `a = 5 + sqrt(10)`
Case III: `(a)/(2) le 0`
In case,f(x) attains the global minimum value at x = 0.
`rArr 3=a^(2) - 2a + 2 `
`rArr a = 1 p, sqrt(2)`.
So, `a= 1 - sqrt(2)`.
Hence, premissible values of a are `1-sqrt(2)` and `5+ sqrt(10)`.
`f(x) = 4x^(2)-4ax + a^(2) -2a+ 2` is monotonic in `[0,2]`
Hence, point of minima of funciton should not lie in `[0,2]`.
Now, `f(x) = 0`
`rArr 8x - 4a =0`
If `(a)/(2)in [0,2]`, then `a in [0,4]`.
For f(x) to be montomic in `[0,2], a notin [0,4]`. So , `a le 0 or a ge 4`.