A) \[f(x)=\int_{1}^{x}{\frac{\log \,t}{1+t}}dt\]
B) \[\frac{1}{2}\]
C) \[f:R\to R\]
D) \[f(x)=\]
Correct Answer: C
Solution :
Let temperature at the interface be T. For part AB, Rate of heat transmission, \[\pm 1\] For part BC, Rate of heat transmission, \[F(x)=f(x)+f\left( \frac{1}{x} \right),\] Here, A is area of cross-sections. At equilibrium, \[f(x)=\int_{1}^{x}{\frac{\log \,t}{1+t}}dt\] \[\frac{1}{2}\] \[f:R\to R\] \[f(x)=\] \[min\{x+1,|~x|+1\}\]You need to login to perform this action.
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