Let fx-Kxx-1x- -1 then the value of K for which fofx-x is

The correct option is C 1 f ( f ( x ) )=f ( kxx+1 ) =k ( kxx+1 )kxx+1=1=k^2x ( k+1 )x+1 ∴ for f ( f ( x ) )=x ⇒ k+1=0 ...

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Multiplication with Vectors

Let fx=Kxx+...

Question

Let $f (x) = \frac{K x}{x + 1} (x \neq - 1)$ then the value of $K$ for which $(f o f) (x) = x$ is

1

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- 1

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2

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\sqrt{2}

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f (x) = {\begin{matrix} x & - 1 \leq x \leq 1 x^{2} & 1 < x \leq 2 \end{matrix}

h^{- 1} (x)

h (x) = f o f (x)

k

f (x) = k x + 5, w h e n x \leq 2

f (x) = x - 1, w h e n x > 2

x = 2

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

(f o f) (x)

k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} & i f - 1 \leq x < 0 \frac{2 x + 1}{x - 1} & i f 0 \leq x < 1 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

x = 0

f (x) = k x + 5, w h e n x \leq 2

x - 1, w h e n x > 2

x = 2

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