The area bounded by the curve \[y=f(x),y=x\]and the lines \[x=1,x=t\] is \[(t+\sqrt{1+{{t}^{2}}})-\sqrt{2}-1\]sq. unit, for all t > 1. If f(x) satisfying f(x)>x for all x>1, then f(x) is equal to

\[x+\frac{x}{\sqrt{1+{{x}^{2}}}}\]

\[x+1+\frac{x}{\sqrt{1+{{x}^{2}}}}\]

\[1+\frac{x}{\sqrt{1+{{x}^{2}}}}\]

\[\frac{x}{\sqrt{1+{{x}^{2}}}}\]

The area bounded by the curve \[y=f(x),y=x\]and the lines \[x=1,x=t\]

1.	The area bounded by the curve \[y=f(x),y=x\]and the lines \[x=1,x=t\] is \[(t+\sqrt{1+{{t}^{2}}})-\sqrt{2}-1\]sq. unit, for all t > 1. If f(x) satisfying f(x)>x for all x>1, then f(x) is equal to
A.	\[x+1+\frac{x}{\sqrt{1+{{x}^{2}}}}\]
B.	\[x+\frac{x}{\sqrt{1+{{x}^{2}}}}\]
C.	\[1+\frac{x}{\sqrt{1+{{x}^{2}}}}\]
D.	\[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
Answer» B. \[x+\frac{x}{\sqrt{1+{{x}^{2}}}}\]

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The area bounded by the curve \[y=f(x),y=x\]and the lines \[x=1,x=t\] is \[(t+\sqrt{1+{{t}^{2}}})-\sqrt{2}-1\]sq. unit, for all t > 1. If f(x) satisfying f(x)>x for all x>1, then f(x) is equal to

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