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) \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
Correct Answer: A
Solution :
[a] It is given that, \[f(x)>x\], for all \[x>1\]. So, area bounded by \[y=f(x),y=x\] and the lines \[x=1,x=t\] is given by \[\int_{1}^{t}{\{f(x)-x\}dx}\] But this area is given equal to \[(t+\sqrt{1+{{t}^{2}}}-\sqrt{2}-1)\] sq. unit. Therefore, \[\int_{1}^{t}{\{f(x)-x\}dx=t+\sqrt{1+{{t}^{2}}}-\sqrt{2}-1,}\] for all t>1 On differentiating both sides w.r.t. t, we get \[f(t)-t=1+\frac{t}{\sqrt{1+{{t}^{2}}}}\] for all \[t>1\] \[\Rightarrow f(t)=t+1+\frac{t}{\sqrt{1+{{t}^{2}}}}\] for all \[t>1\] Hence, \[f(x)=x+1+\frac{x}{\sqrt{1+{{x}^{2}}}}\] for all \[x>1\]
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