If fx-x-1-x-1- then show thati f1-x-fxii f-1-x-1-fx

We have, f(x)=x 1/x+1 (i) f ( 1/x )=1/x 1/1/x+1=(1 x)/x/(1+x)/x=1 x/1+x = (x 1)/x+1= f(x) (ii) f ( 1/x )= 1/x 1/ 1/x+1 = 1 x/x/( 1+x)/x ⇒ f ( 1/x ) = (x+1) ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

If fx=x-1/x+1...

Question

If $f (x) = \frac{x - 1}{x + 1}$ , then show that
(i) $f (\frac{1}{x}) = - f (x)$
(ii) $f (- \frac{1}{x}) = \frac{- 1}{f (x)}$

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Solution

We have, $f (x) = \frac{x - 1}{x + 1}$

(i) $f (\frac{1}{x}) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1} = \frac{\frac{(1 - x)}{x}}{\frac{(1 + x)}{x}} = \frac{1 - x}{1 + x}$
$= \frac{- (x - 1)}{x + 1} = - f (x)$

(ii) $f (- \frac{1}{x}) = \frac{- \frac{1}{x} - 1}{- \frac{1}{x} + 1} = \frac{\frac{- 1 - x}{x}}{\frac{(- 1 + x)}{x}}$
$\Rightarrow f (\frac{- 1}{x}) = \frac{- (x + 1)}{x - 1}$
Now, $\frac{- 1}{f (x) = \frac{- 1}{\frac{x - 1}{x + 1}}} = \frac{- (x + 1)}{x - 1}$
$∴ f (- \frac{1}{x}) = - \frac{1}{f (x)}$

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Similar questions

Q. If

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

, then show that

(i)

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

(ii)

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

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

, then

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

f (x)

II.

f (- x)

III.

- f (x)

IV.

\frac{1}{f (x)}

Q. If

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

the show that

f (x) + f (\frac{1}{x}) = 0

If $f (x) = \frac{x + 1}{x - 1},$ show that $f [f (x)] = x .$

Q. Assertion :If

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

then

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

Reason: If

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

then

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

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If f(x)=x−1x+1, then show that (i) f(1x)=−f(x) (ii) f(−1x)=−1f(x)

We have, f(x)=x−1x+1 (i) f(1x)=1x−11x+1=(1−x)x(1+x)x=1−x1+x =−(x−1)x+1=−f(x) (ii) f(−1x)=−1x−1−1x+1=−1−xx(−1+x)x ⇒f(−1x)=−(x+1)x−1 Now, −1f(x)=−1x−1x+1=−(x+1)x−1 ∴f(−1x)=−1f(x)

If $f (x) = \frac{x - 1}{x + 1}$ , then show that
(i) $f (\frac{1}{x}) = - f (x)$
(ii) $f (- \frac{1}{x}) = \frac{- 1}{f (x)}$