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Question 74 of 104

Q.If f:R→Rf: R \to R be defined by f(x)={x,x<1x2,x≥1f(x) = \begin{cases} x, & x < 1 \\ x^2, & x \ge 1 \end{cases} then f−1(x)f^{-1}(x) is:

(a) {x,x<1x,x≥1\begin{cases} x, & x<1 \\ \sqrt{x}, & x \ge 1 \end{cases}
(b) {x,x≤12x,x>1\begin{cases} x, & x\le 1 \\ 2x, & x > 1 \end{cases}
(c) {x,x<1x,x≥1\begin{cases} \sqrt{x}, & x<1 \\ x, & x \ge 1 \end{cases}
(d) {1,x<1x,x≥1\begin{cases} 1, & x<1 \\ x, & x \ge 1 \end{cases}
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2018MCQ· 1mImportance★★★★★
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On x<1x<1, f(x)=xf(x)=x is its own inverse; on x≥1x\ge1, f(x)=x2f(x)=x^2 inverts to x\sqrt{x}. Piecing these together gives option (a).

Given f(x)={x,x<1x2,x≥1f(x) = \begin{cases} x, & x<1 \\ x^2, & x\ge 1\end{cases}.

For the branch x<1x<1: as xx ranges over (−∞,1)(-\infty,1), f(x)=xf(x)=x ranges over (−∞,1)(-\infty,1) too (identity map). So on this range of outputs y<1y<1, the inverse is simply x=yx=y, i.e. f−1(y)=yf^{-1}(y)=y.

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