A) all a and All B
B) all b, if \[a=0\]
C) all \[b>0\]
D) all \[a>0\]
Correct Answer: B
Solution :
Since, f(x) has a relative minimum at x = 0. Therefore, \[f'(0)=0\] and\[f''(0)>0\]. If the function \[y=f(x)+ax+b\]has a relative minimum at \[x\text{ }=\text{ }0\], then \[\frac{dy}{dx}=0\text{ }at\text{ }x\text{ }=\text{ }0\] \[\Rightarrow \] \[f'(x)+a=0\] for \[x=0\] \[\Rightarrow \] \[f'(0)+a=0\] \[\Rightarrow \] \[0+a=0\] \[\Rightarrow \] \[a=0\] \[[\because f'(0)=0]\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=f''(x)\] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=0}}=f''(0)>0\] \[[\because f''(0)>0]\] Hence, y has relative minimum at x = 0, if a = 0 and b can attain any real value.
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