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If $m n=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$, then the value of $0.1+0.1^{\frac{1}{m}}+0.1^{\frac{1}{n}}$ is
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The correct answer is:
$0.2+0.1^{\frac{1}{3}}$
We have, $m n=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$
$\begin{aligned} & \Rightarrow \quad m n=3 \text { and } \frac{m+n}{m n}=\frac{4}{3} \\ & \Rightarrow \quad m n=3 \text { and } m+n=4\end{aligned}$
Solving, we get
$m=1, n=3$ or $m=3, n=1$
Now, $0 \cdot 1+(0 \cdot 1)^{\frac{1}{m}}+(0 \cdot 1)^{\frac{1}{n}}$
Put $m=1, n=3$, we get
$0 \cdot 1+(0 \cdot 1)+(0 \cdot 1)^{1 / 3}=0 \cdot 2+(0 \cdot 1)^{1 / 3}$
$\begin{aligned} & \Rightarrow \quad m n=3 \text { and } \frac{m+n}{m n}=\frac{4}{3} \\ & \Rightarrow \quad m n=3 \text { and } m+n=4\end{aligned}$
Solving, we get
$m=1, n=3$ or $m=3, n=1$
Now, $0 \cdot 1+(0 \cdot 1)^{\frac{1}{m}}+(0 \cdot 1)^{\frac{1}{n}}$
Put $m=1, n=3$, we get
$0 \cdot 1+(0 \cdot 1)+(0 \cdot 1)^{1 / 3}=0 \cdot 2+(0 \cdot 1)^{1 / 3}$
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