Functions 3 Question 4

4. If $f:[1, \infty) \rightarrow[2, \infty)$ is given by $f(x)=x+\frac{1}{x}$, then $f^{-1}(x)$ equals

(2001, 1M)

(a) $\frac{x+\sqrt{x^{2}-4}}{2}$

(b) $\frac{x}{1+x^{2}}$

(c) $\frac{x-\sqrt{x^{2}-4}}{2}$

(d) $1+\sqrt{x^{2}-4}$

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Answer:

Correct Answer: 4. (a)

Solution:

  1. We have, $f(x)=\frac{x}{1+x^{2}}, x \in R$

Ist Method $f(x)$ is an odd function and maximum occur at $x=1$

From the graph it is clear that range of $f(x)$ is

$$ -\frac{1}{2}, \frac{1}{2} $$

IInd Method $f(x)=\frac{1}{x+\frac{1}{x}}$

If $x>0$, then by AM $\geq GM$, we get $x+\frac{1}{x} \geq 2$

$$ \Rightarrow \quad \frac{1}{x+\frac{1}{x}} \leq \frac{1}{2} \quad \Rightarrow \quad 0<f(x) \leq \frac{1}{2} $$

If $x<0$, then by AM $\geq$ GM, we get $x+\frac{1}{x} \leq-2$

$$ \Rightarrow \quad \frac{1}{x+\frac{1}{x}} \geq-\frac{1}{2} \quad \Rightarrow-\frac{1}{2} \leq f(x)<0 $$

If $x=0$, then $f(x)=\frac{0}{1+0}=0$

Thus,

$$ -\frac{1}{2} \leq f(x) \leq \frac{1}{2} $$

Hence, $f(x) \in-\frac{1}{2}, \frac{1}{2}$

IIIrd Method

Let $y=\frac{x}{1+x^{2}} \Rightarrow y x^{2}-x+y=0$

$\because x \in R$, so $D \geq 0$

$\Rightarrow \quad 1-4 y^{2} \geq 0$ $\Rightarrow(1-2 y)(1+2 y) \geq 0 \Rightarrow y \in-\frac{1}{2}, \frac{1}{2}$

So, range is $-\frac{1}{2}, \frac{1}{2}$.



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