Q78 / 200 · Manipal (MET) 2013 · Mathematics

Figure

  • A
    $0$
  • Correct answer
    B
    $1$
  • C
    $2$
  • D
    $1 / 2$

Solution

Let $y=\frac{x}{x+\frac{\sqrt[3]{x}}{x+\sqrt[3]{x} \ldots \text { infinity }}}$ $=\frac{x}{x+\frac{1}{x^{2 / 3}} \cdot \frac{x}{x+\sqrt[3]{x} \ldots \infty}}=\frac{x}{x+\frac{y}{x^{2 / 3}}}$ $\begin{aligned} & \Rightarrow y=\frac{x^{5 / 3}}{x^{5 / 3}+y} \\ & \Rightarrow y^2+\left(x^{5 / 3}\right) \cdot y-x^{5 / 3}=0 \\ & \therefore \quad y=\frac{-x^{5 / 3} \pm \sqrt{x^{10 / 3}+4 x^{5 / 3}}}{2}\end{aligned}$ $=\frac{-x^{5 / 3}+\sqrt{x^{10 / 3}+4 x^{5 / 3}}}{2}$ $(\because y\gt0)$ $\begin{aligned} & =\frac{4 x^{5 / 3}}{2\left(\sqrt{x^{10 / 3}+4 x^{5 / 3}}+x^{5 / 3}\right)}=\frac{2}{\sqrt{\left(1+\frac{4}{x^{5 / 3}}\right)+1}} \\ & \begin{aligned} & \therefore \quad \lim _{x \rightarrow \infty} y=\frac{2}{\sqrt{1+0}+1} \\ & \quad=\frac{2}{2}=1 .\end{aligned}\end{aligned}$

Hence, the correct answer is option: 'B'.

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