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Rational Function

Let f :-∞,+1]...

Question

Let $f : (- \infty, + 1] \to R, g : [- 1, \infty) \to R$ be such that $f (x) = \sqrt{1 - x}$ and $g (x) = \sqrt{1 + x}$ , then $f (x) + \frac{1}{g (x)}$ exist if $x \in$

A

[- 1, 1]

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B

(- \infty, - 1)

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C

(1, \infty)

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D

(- 1, 1]

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Solution

The correct option is D $(- 1, 1]$
$1 - x \geq 0, 1 + x > 0 1 \geq x \Rightarrow x \leq 1 x > - 1 ∴ x \in (- 1, 1]$

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Similar questions

Q. The distinct linear functions that map [−1, 1] onto [0, 2] are
(a)

f (x) = x + 1, g (x) = - x + 1

f (x) = x - 1, g (x) = x + 1

f (x) = - x - 1, g (x) = x - 1

(d) None of these

f (x) = m a x {1 + sin x, 1, 1 - cos x}, x \in [0, 2 π]

g (x) = m a x {1, | x - 1 |}, x \in R

f (x) = max {1 + sin x, 1, 1 - cos x}, \forall x \in [0, 2 π]

g (x) = max {1, x - 1} \forall x \in R

f (x) = sin (\frac{π}{6} sin (\frac{π}{2} sin x))

x \in R

g (x) = \frac{π}{2} sin x

x \in R .

(f \circ g) (x)

f (g (x))

(g \circ f) (x)

g (f (x)) .

Then which of the following is (are) true?

f : R - {1} \to R

f (x) = x + 1

g (x) : R - {1} \to

g (x) = \frac{x^{2} - 1}{x - 1}

f = g

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