Question

# If $$f(x)$$ be a continuously increasing function satisfying the condition that $$f(x)=\dfrac{1}{3}\left[f(x+6)+\dfrac{6}{f(x+7)}\right]$$ and $$f(x) \ge 0$$ for all $$x\ \epsilon \ \ R$$. If $$\displaystyle \lim _{ x\rightarrow \infty }{ f\left( x \right) =\sqrt { m } }$$ then value of $$m$$ is

A
3
B
4
C
6
D
5

Solution

## The correct option is A $$3$$We have, $$f\left( x \right)=\dfrac{1}{3}\left[ f\left( x+6 \right)+\dfrac{6}{f\left( x+7 \right)} \right]$$ and $$f\left( x \right)=\sqrt{m}$$ So, $$\sqrt{m}=\dfrac{1}{3}\left[ f\left( x+6 \right)+\dfrac{6}{f\left( x+7 \right)} \right]$$ Now, $$f\left( x+6 \right)=\sqrt{m+6}$$ $$f\left( x+7 \right)=\sqrt{m+7}$$ $$\sqrt{m}=\dfrac{1}{3}\left[ \sqrt{m+6}+\dfrac{6}{\sqrt{m+7}} \right]$$ $$\Rightarrow \sqrt{m}=\dfrac{1}{3}\left[ \dfrac{\sqrt{m+6}\sqrt{m+7}+6}{\sqrt{m+7}} \right]$$ $$\Rightarrow 3\sqrt{m}\sqrt{m+7}=\sqrt{m+6}\sqrt{m+7}+6$$ On squaring both side and we get, $$9m\left( m+7 \right)=\left( m+6 \right)\left( m+7 \right)+36+12\sqrt{\left( m+6 \right)\left( m+7 \right)}$$ $$\Rightarrow 9{{m}^{2}}+63m={{m}^{2}}+7m+6m+42+36+12\sqrt{\left( m+6 \right)\left( m+7 \right)}$$ $$\Rightarrow 8{{m}^{2}}+50m-78=12\sqrt{\left( m+6 \right)\left( m+7 \right)}$$ $$\Rightarrow 4{{m}^{2}}+25m-39=6\sqrt{\left( m+6 \right)\left( m+7 \right)}$$ On squaring both side and solve that, $$m=3$$ Hence, this is the answer.Mathematics

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