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Polynomial Functions

If fx=x-1x+...

Question

If $f (x) = \frac{x - 1}{x + 1}$ , which of the following statements is /are correct?

A

f (\frac{1}{x}) = - f (x)

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B

f (\frac{1}{x}) = f (x)

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C

f (\frac{1}{x}) = \frac{1}{f (x)}

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D

f (\frac{1}{x}) = - \frac{1}{f (x)}

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Solution

The correct option is A $f (\frac{1}{x}) = - f (x)$
Here we have,
$f (x) = \frac{x - 1}{x + 1}$
Then, $f (\frac{1}{x}) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1}$
$\Rightarrow f (\frac{1}{x}) = \frac{\frac{1 - x}{x}}{\frac{1 + x}{x}} = \frac{1 - x}{1 + x} = - f (x)$

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Similar questions

f (x) = \frac{4^{x}}{4^{x} + 2} for all x \in R .

Which of the following is/are true?

If $f : R \to R$ be given by $f (x) = \frac{4^{x}}{4^{x} + 2}$ for all $x ϵ R$ . Then,

Q. If f : R → R be given by for all

f (x) = \frac{4^{x}}{4^{x} + 2}

x ∈ R, then

(a) f(x) = f(1 − x)
(b) f(x) + f(1 − x) = 0
(c) f(x) + f(1 − x) = 1
(d) f(x) + f(x − 1) = 1

Q. Assertion :Statement 1:

F (x) = \int_{1}^{x} \frac{log t}{1 + t + t^{2}} d t

F (x) = - F (1 / x)

Reason: Statement 2: If

F (x) = \int_{1}^{x} \frac{log t}{t + 1} d t

F (x) + F (1 / x) = (1 / 2) (log x)^{2}

Q. Assertion :If

f (x) = \int_{1}^{x} \frac{l n t d t}{1 + t + t^{2}} (x > 0)

f (x) = - f (\frac{1}{x})

f (x) = \int_{1}^{x} \frac{l n t d t}{t + 1},

f (x) + f (\frac{1}{x}) = \frac{1}{2} {(l n x)}^{2}

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