Functions 2 Question 4
5. For $x \in R-{0,1}$, let $f _1(x)=\frac{1}{x}, f _2(x)=1-x \quad$ and $f _3(x)=\frac{1}{1-x}$ be three given functions. If a function, $J(x)$ satisfies $\left(f _2 \circ J \circ f _1\right)(x)=f _3(x)$, then $J(x)$ is equal to
(2019 Main, 9 Jan I)
(a) $f _2(x)$
(b) $f _3(x)$
(c) $f _1(x)$
(d) $\frac{1}{x} f _3(x)$
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Answer:
Correct Answer: 5. (b)
Solution:
- We have,
$$ f _1(x)=\frac{1}{x}, f _2(x)=1-x \text { and } f _3(x)=\frac{1}{1-x} $$
Also, we have $\left(f _2 \circ J\right.$ o $\left.f _1\right)(x)=f _3(x)$
$$ \begin{array}{cc} \Rightarrow & f _2\left(\left(J o f _1\right)(x)\right)=f _3(x) \\ \Rightarrow & f _2\left(J\left(f _1(x)\right)=f _3(x)\right. \\ \Rightarrow & 1-J\left(f _1(x)\right)=\frac{1}{1-x} \\ {\left[\because f _2(x)=1-x \text { and } f _3(x)=\frac{1}{1-x}\right]} \\ \Rightarrow & 1-J \frac{1}{x}=\frac{1}{1-x} \quad\left[\because f _1(x)=\frac{1}{x}\right] \\ \Rightarrow & J \frac{1}{x}=1-\frac{1}{1-x} \\ =\frac{1-x-1}{1-x}=\frac{-x}{1-x} \end{array} $$
Now, put $\frac{1}{x}=X$, then
$$ \begin{array}{rlr} J(X) & =\frac{\frac{-1}{X}}{1-\frac{1}{X}} \\ & =\frac{-1}{X-1}=\frac{1}{1-X} \end{array} $$
$$ \Rightarrow \quad J(X)=f _3(X) \text { or } J(x)=f _3(x) $$
