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For a certain curve `y=f(x)` satisfying `(d^(2)y)/(dx^(2))=6x-4,f(x)` has a local minimum value `5` when `x=1`. The global maximum value of `f(x)` if `0lexle2`, is |
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Answer» Correct Answer - 7 Integrating both sides w.r.t `x` `(dy)/(dx)=3x^(2)-4x+c` at `x=1 (dy)/(dx)=0impliesc=1` `implies(dy)/(dx)=3x^(2)-4x+1`…………(1) Integeratig both sides w.r.t `x` `y=x^(3)-2x^(2)+x+c_(1)` at `x=1, y=5impliesc_(1)=5` `impliesy=x^(2)-2x^(2)+x+5` from equation (1) we get the critical points `x=1/3,1` `f(1)=5` `f(0)=5,f(2)=7 1(1/3)=112/27` Hence the global maximum value `=7` |
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