Question

If $f\left(x\right)$ be a continuously increasing function satisfying the condition that $f\left(x\right)=\frac{1}{3}\left[f\right(x+6)+\frac{6}{f(x+7)}]$ and $f\left(x\right)\ge 0$ for all $xϵR$. If $\underset{x\to \infty }{lim}f\left(x\right)=\sqrt{m}$ then value of $m$ is

• A. $3$
• B. $4$
• C. $6$
• D. $5$

Answer 1

The correct option is A $3$

We have,

$f\left(x\right)=\frac{1}{3}[f(x+6)+\frac{6}{f(x+7)}]$ and $f\left(x\right)=\sqrt{m}$

So,

$\sqrt{m}=\frac{1}{3}[f(x+6)+\frac{6}{f(x+7)}]$

Now,

$f(x+6)=\sqrt{m+6}$

$f(x+7)=\sqrt{m+7}$

$\sqrt{m}=\frac{1}{3}[\sqrt{m+6}+\frac{6}{\sqrt{m+7}}]$

$\Rightarrow \sqrt{m}=\frac{1}{3}\left[\frac{\sqrt{m+6}\sqrt{m+7}+6}{\sqrt{m+7}}\right]$

$\Rightarrow 3\sqrt{m}\sqrt{m+7}=\sqrt{m+6}\sqrt{m+7}+6$

On squaring both side and we get,

$9m(m+7)=(m+6)(m+7)+36+12\sqrt{(m+6)(m+7)}$

$\Rightarrow 9m^2+63m=m^2+7m+6m+42+36+12\sqrt{(m+6)(m+7)}$

$\Rightarrow 8m^2+50m-78=12\sqrt{(m+6)(m+7)}$

$\Rightarrow 4m^2+25m-39=6\sqrt{(m+6)(m+7)}$

On squaring both side and solve that,

$m=3$

Hence, this is the answer.