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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
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B
4
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C
6
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D
5
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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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