InterviewSolution
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InterviewSolution
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For certain curves y= f(x) satisfying`[d^2y]/[dx^2]= 6x-4`, f(x) has local minimum value 5 whenx=1.9. Number of critical point for y=f(x) for x € [0,2](a) 0 (b)1. c).2 d) 310. Global minimum value y = f(x) for x € [0,2] is(a)5 (b)7 (c)8 d) 911 Global maximum value of y = f(x) for x € [0,2] is(a) 5 (b) 7 (c) 8 (d) 9 |
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Answer» Correct Answer - C 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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