| Direction: The mean value of the continuous functions f(x) in the interval [a, b} is given by the formula |
| Mean value \[=\frac{\int\limits_{a}^{b}{f(x)\,dx}}{b-a}\] |
A) \[\frac{2}{\pi }\]
B) \[\frac{1}{\pi }\]
C) \[\frac{4}{\pi }\]
D) \[\frac{8}{\pi }\]
Correct Answer: C
Solution :
\[f(x)=(x+1)\,\,{{(\tan \,x)}^{x}}\cdot \,{{\sec }^{2}}x\] \[+{{\log }_{e}}\,(\tan \,x)\,{{(\tan \,x)}^{x+1}}\] ?(i) Let \[g(x)={{(\tan \,x)}^{x+1}}\] \[{{\log }_{e}}\,g(x)=(x+1)\,{{\log }_{e}}\,(\tan \,x)\] \[\frac{1}{g(x)}g'(x)=(x+1)\,\frac{1}{\tan \,x}{{\sec }^{2}}x\] \[+{{\log }_{e}}\,(\tan \,x).1\] \[\Rightarrow \] \[g'(x)={{(\tan \,x)}^{x+1}}\] \[\left\{ \frac{x+1}{\tan \,x}\cdot \,{{\sec }^{2}}x+{{\log }_{e}}\,(\tan \,x) \right\}\] \[\Rightarrow \] \[g'(x)=\,{{(\tan \,x)}^{x}}\,(x+1)\,{{\sec }^{2}}x\] \[+{{\log }_{e}}\,(\tan \,x)\,{{(\tan \,x)}^{x+1}}\] \[\Rightarrow \] \[g'(x)=f(x)\] [From Eq. (i)] \[\Rightarrow \] \[\int{f(x)dx=g(x)}\] So, Mean value \[=\,\frac{\int_{0}^{\pi /4}{f(x)dx}}{\frac{\pi }{4}-0}=\frac{[g(x)]_{0}^{\pi /4}}{\frac{\pi }{4}}\] \[=\frac{[{{(\tan \,x)}^{x+1}}]_{0}^{\pi /4}}{\frac{\pi }{4}}\] \[=\frac{{{(1)}^{\frac{\pi }{4}+1}}-{{(0)}^{0+1}}}{\frac{\pi }{4}}=\frac{4}{\pi }\]
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