Q42 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$f$ is real valued function satisfying the relation $f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}$. If $f\left(x+\frac{1}{x}\right)=1$ then $x=$

  • A
    $\pm 2$
  • Correct answer
    B
    $\pm 1$
  • C
    $\pm 3$
  • D
    $\pm 6$

Solution

Given that $f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}$ $f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}+2 \cdot 3 x \cdot \frac{1}{2 x}-3$ $\Rightarrow f\left(3 x+\frac{1}{2 x}\right)=\left(3 x+\frac{1}{2 x}\right)^2-3$ $\therefore f(x)=x^2-3$ $\text {Also } f\left(x+\frac{1}{x}\right)=1 \Rightarrow\left(x+\frac{1}{x}\right)^2-3=1 \Rightarrow\left(x+\frac{1}{x}\right)^2=4$ $\Rightarrow x+\frac{1}{x}= \pm 2 \Rightarrow x^2-2 x+1=0 \text { or } x^2+2 x+1=0$ $\Rightarrow(x-1)^2=0 \text { or }(x+1)^2=0 \Rightarrow x=1 \text { or } x=-1$

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

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