Let f(x) = `4x^(2)-4ax+a^(2)-2a+2` and the golbal minimum value of f(x) for x in [0,2] is equal to 3
The number of values of a for which the global minimum value equal to 3 for x in [0,2] occurs at the endpoint of interval[0,2] is

Options

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

Answer

4

Explanation

`f(x)=4x^(2)-4ax+a^(2)-2a+2`
vertex of this parabola is `((a)/(2),2-2a)`
caseI:`0lta//2lt2`
In this case f(x) will attain the minium value at `x=a/2` .Thus `f(a/2)^(3)`

or `3=-2a+2 or a =-1/2 (rejected)`
CaseII: `(a)/(2)ge2`
In this f(x) attains global minimum value at x =2 thus f(2)=3
`3=16-8a+a^(2)-2a+2 or a =5pm sqrt(10)`
thus `a=5+sqrt(10)`
CaseIII:` (a)/(2)ge0`
In this case f(x) attains the global minimum value at x =0 thus f(0)=3
`therefore 3=a^(2)-2a+2 or a =1 pm`
Thus a =1-`sqrt(2)`
Hecne the permissibel values of a are `1-sqrt(2) and 2+sqrt(10)`
f(x) =`4x^(2)-49x+a^(2)-2a+2` is monotonic in [0,2]
Hence the point of minima of funciton should not lie in [0,2]
Now f(x) =0 or 8x-4a=0 or x `=a//2`
`(a)/(2)in [0,2] or a in [0,4]`
For f(x) to be monotomic [0,2], and [0,4] i/e `ale0` or age4