The minimum value of `y=5x^(2)-2x+1` is

Options

A. `1/5`
B. `2/5`
C. `4/5`
D. `3/5`

Answer

C

Explanation

For maximum / minimum value `(dy)/(dx)=0 implies 5(2x)-2(1)+0=0impliesx=(1)/5`
Now at ` x=(1)/5,(d^2y)/(dx^2)=10` which is positive so y has minimum value at `x=(1)/5` .
Therefore `y_("min")=5((1)/(5))^(2)-2((1)/5)+1=(4)/5`