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

The minimum/minimum value `(dy)/(dx)=0rArr5(2x)-2(1)+0=0=0rArrx=(1)/(5)`
Now at `x=(1)/(5)m(d_(y^2))/(dx^(2))=10 x=(1)/(5)m(d_(y^2))/(dx^(2))=10` which is positive so minima at `x=(1)/(5)`.
Therefore `y_("min")=5((1)/(5))^(2)-2((1)/(5))+1=(4)/(5)`