If for all x different from both 1 and 0 we have f1x-x-x-1- f2x-1-1-x- and for all integers n- 1- we have f n-2 x- f n-1 f 1 x if n is odd f n-1 f 2 x if n is even - then f4 equals

The correct option is C f_1(x) f_3 =f_2(f_1(x)) =11 x/x 1 =x 1x 1 x =x 1 1 =1 x .Hence f_4 =f_3(f_2(x))=1 11 x=1 x 1 ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

If for all ...

Question

If for all $x$ different from both $1$ and $0$ we have $f_{1} (x) = \frac{x}{x - 1}, f_{2} (x) = \frac{1}{1 - x}$ , and for all integers $n \geq 1$ , we have $f_{n + 2} (x) = {\begin{matrix} f_{n + 1} (f_{1} (x)) i f n i s o d d f_{n + 1} (f_{2} (x)) i f n i s e v e n \end{matrix}$ then $f_{4}$ equals

x

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x - 1

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f_{1} (x)

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f_{2} (x)

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Solution

The correct option is C $f_{1} (x)$
$f_{3}$
$= f_{2} (f_{1} (x))$
$= \frac{1}{1 - \frac{x}{x - 1}}$
$= \frac{x - 1}{x - 1 - x}$
$= \frac{x - 1}{- 1}$
$= 1 - x$ .
Hence
$f_{4}$
$= f_{3} (f_{2} (x)) = 1 - \frac{1}{1 - x} = \frac{1 - x - 1}{1 - x}$
$= \frac{- x}{1 - x}$
$= \frac{x}{x - 1}$
$= f_{1} (x)$

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If for all x different from both 1 and 0 we have f1(x)=xx−1,f2(x)=11−x, and for all integers n≥1, we have fn+2(x)={fn+1(f1(x))ifnisoddfn+1(f2(x))ifniseven then f4 equals

The correct option is C f1(x)f3=f2(f1(x))=11−xx−1=x−1x−1−x=x−1−1=1−x.Hencef4=f3(f2(x))=1−11−x=1−x−11−x=−x1−x=xx−1=f1(x)