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The minimum value of `y=5x^(2)-2x+1` isA. `1/5`B. `2/5`C. `4/5`D. `3/5`

Answer» Correct Answer - C
For maximum/ minimum value `(dy)/(dx)=0rArr 5(2x)-2(1)+0=0rArr x=1/5`
Now at `x=1/5, (d^(2)y)/(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`


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