InterviewSolution
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InterviewSolution
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A differentiable function `f(x)`has a relative minimum at `x=0.`Then the function `f=f(x)+a x+b`has a relative minimum at `x=0`forall `a`and all`b`(b) all `b`if `a=0`all `b >0`(d) all `a >0`A. all a and all bB. all b if a =0C. all b `gt` 0D. all a `gt` 0 |
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Answer» Correct Answer - 2 Since f(x) has a relative minimum at x =0 f(X)=0 and `f(0)gt0` (assuming `f(X)ne 0)` If the funciton y =f(X) ax+b has a relative minimum at x=0 then `(dy)/(dx)=0 at x =0 or f(X) + a =0 for x =0` F(0)+a=0 or a =0 or a=0 Now `(d^(2)y)/(dx^(2))=f'(x)or (d^(2)y)/(dx^(2))_(x=0)=f'(0)gt0 [therefore f'(0)gt0]` Hence y has a relative minimum at x= 0 if a =0 and b can attain anyreal value. |
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