SATHEE: Functions 3 Question 4

Functions 3 Question 4

4. If $f : [1, \infty) \to [2, \infty)$ 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) $\frac{x}{1 + x^{2}}$

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

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Answer:

Correct Answer: 4. (a)

Solution:

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 $\geq G M$ , we get $x + \frac{1}{x} \geq 2$

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

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

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

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

$∵ x \in R$ , so $D \geq 0$

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

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

Functions 3 Question 4

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

Answer:

IIIrd Method

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Functions 3 Question 4

4. If f:[1,∞)→[2,∞) is given by f(x)=x+1x, then f−1(x) equals

Answer:

IIIrd Method

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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