Question

If f:{x | x≥1,x ∈ R} —> {x | x≥2,x ∈ R},f(x)=x+1/x, f⁻¹(x)=.......,Select Proper option from the given options.
(a) x+√x²-4/2
(b) x-√x²-4/2
(c) x²+1/x
(d) √x²-4

Answer 1

Answer:

The proper answer is (a) $f^{-1}(x) = \frac{x+\sqrt{x^2-4}}{2}$

Step-by-step explanation:

Let $f(x) = y$. Then

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

We can find x in function of y by quadratic formula, that is,

$x = \frac{y \pm \sqrt{y^2-4\cdot 1 \cdot 1}}{2} = \frac{y \pm \sqrt{y^2-4}}{2}$

Once we have $x\geq 1$, we have to choose the signal to be $+$, otherwise x would be very small if y is great (for example, if y = 100, x = 0.01)

Then the correct answer is (a).

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{A \: function \: f : \{x : x \geqslant 1 \: ,x \in \mathbb{R} \} \to \: \{x : x \geqslant 2\: ,x \in \mathbb{R} \}}$

defined as

$\displaystyle \sf{ f(x) = x + \frac{1}{x} \: }$

TO DETERMINE

$\sf{ {f}^{ - 1}(x) \: \: }$

EVALUATION

Here

$\sf{{f}^{ - 1} : \{x : x \geqslant 2 \: ,x \in \mathbb{R} \} \to \: \{x : x \geqslant 1\: ,x \in \mathbb{R} \}}$

$\sf{Let \: \: y = {f}^{ - 1} (x)}$

$\implies \sf{f(y) = x \: \: }$

$\implies \: \displaystyle \sf{ x = y + \frac{1}{y} \: }$

$\implies \: \displaystyle \sf{ {y}^{2} - xy + 1 = 0\: }$

$\implies \: \displaystyle \sf{ y = \frac{x \: \pm \: \sqrt{ {x}^{2} - 4 } }{x} \: }$

$\: \displaystyle \sf{Now \: if \: \: \: {f}^{ - 1} (x) = \frac{x \: - \: \sqrt{ {x}^{2} - 4 } }{x} \: }$

$\: \displaystyle \sf{Then \: \: \: {f}^{ - 1} (3) = \frac{3 - \sqrt{5} }{2} \notin \: \: \{x : x \geqslant 1\: ,x \in \mathbb{R}\} }$

$\therefore \: \: \displaystyle \sf{ {f}^{ - 1} (x) = \frac{x \: + \: \sqrt{ {x}^{2} - 4 } }{x} \: }$

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LEARN MORE FROM BRAINLY

f: R—>R,f(x)=x²+2x+3

Select Proper option from the given options.

(a) a bijection

(b) one-one but not onto

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