InterviewSolution
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InterviewSolution
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For certain curve `y=f(x)` satisfying `(d^(2)y)/(dx^(2))=6x-4, f(x)` has local minimum value 5 when `x=1` Global maximum value of `y=f(x)` for `x in [0,2]` isA. 5B. 7C. 8D. 9 |
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Answer» Correct Answer - B Integrating `(d^(2)y)/(dx^(2))=6x-4`, we get `(dy)/(dx) = 3x^(2)-4x+A` When `x=1, (dy)/(dx)=0`, so that `A=1`,Hence, `(dy)/(dx) = 3x^(2)-4x+1` Integrating, we get `y=x^(3)-2x^(2)+x+5`. From equation (1), we get the cricitical points `x=1//3, x=1`. At the critical point `x=1/3, (d^(2)y)/(dx^(2))` is negative Therefore, at `x=1//3, y` has a local maximum. At `x=1, (d^(2)y)/(dx^(2))` is positive. Therefore, at `x=1, y` has a local minimum. Also, `f(1) =5, f(1/3)=139/7, f(0) =5, f(2)=7` Hence, the global maximum value =7 and the global minimum value =5 |
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