AP EAMCET · Maths · Functions
If \(f(x)=|x-1|+|x-2|+|x-3|, 2 < x < 3\), then \(f\) is
- A an onto function but not one-one
- B one-one function but not onto
- C a bijection
- D neither one-one nor onto
Answer & Solution
Correct Answer
(C) a bijection
Step-by-step Solution
Detailed explanation
We have, \(\begin{aligned} & f(x)=|x-1|+|x-2|+|x-3| \\ & \Rightarrow f(x)=\left\{\begin{array}{cc} 6-3 x, & x 3 \end{array}\right. \end{aligned}\) For \(2 < x < 3\), then \(f(x)=x\) So, \(f(x)\) is one-one and onto function. It implies that \(f(x)\) is a bijection.
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