Definite Integration Question 154
Question: 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
Options:
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}}} $
Show Answer
Answer:
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 $
BETA
