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ate maximum or minimum value of the function ` y= 25x ^(2) +5- 10x `A. `y _(min)=4`B. ` y _(max) =8`C. `y _(min) =8`D. `y _(max) =4` |
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Answer» Correct Answer - A `(dy)/(dx) =50x-10 =0` `x-1/5implies (d^(2)y)/( dx ^(2)) =50gt 0` So y is minimum at `x=1//5` `y_(""(min))=25xx(1)/(25)+ 5-10 xx1/5 =4` |
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