A) \[f\left( x \right)>0\forall x\in R\]
B) \[f\left( x \right)<0\forall x\in R\]
C) \[f\left( x \right)=\sin \phi \forall x\in R\]
D) none of these
Correct Answer: A
Solution :
(a)| \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)+2xy-1\] |
| Put \[x=y=0\Rightarrow f\left( 0 \right)=2f\left( 0 \right)-1\Rightarrow f\left( 0 \right)=1\] |
| Also,\[f'\left( x \right)=\underset{h\,\to \,0}{\mathop{\operatorname{Lim}}}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}\]\[=\underset{h\,\to \,0}{\mathop{Lim}}\,\frac{f\left( x \right)+f\left( h \right)+2xh-1-f\left( x \right)}{h}\] |
| \[=2x+\underset{h\,\to \,0}{\mathop{Lim}}\,\frac{f\left( h \right)-1}{h}\]\[=2x+f'\left( 0 \right)=2x+\sin \phi \] |
| Integrating, we get \[f\left( x \right)={{x}^{2}}+x\sin \phi +C\]\[f\left( 0 \right)=1\Rightarrow 1=C\] |
| \[\therefore f\left( x \right)={{x}^{2}}+x\sin \phi +1>0\forall x\in R\] \[\left[ \because \Delta <0 \right]\]. |
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