Let $f (x) = | x - 1 | + | x - 2 | + | x - 3 | + | x - 4 |$ then

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The least value of $f (x)$ is 4

The least value is not attained at a unique point

The number of integral solution of f(x) = 4 is 2

The value of $\frac{f (π - 1) + f (e)}{2 f (12 / 5)}$ is 1

answer is A, B, C, D.

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Detailed Solution

$f (x) = {\begin{matrix} 10 - 2 x & i f x < 1 \\ 8 - 2 x & i f 1 \leq x \leq 2 \\ 4 & i f 2 < x \leq 3 \\ 2 x - 2 & i f 3 < x \leq 4 \\ 4 x - 10 & i f x > 4 \end{matrix}$ Question Image

Clearly the least value of f(x) is 4
The no. of integral solutions of f(x) = 4 are two {2, 3}
Since $π - 1, e, \frac{12}{5} \in [2, 3]$ $∴ f (π - 1) = f (e) = f (\frac{12}{5}) = 4$ $∴ \frac{f (π - 1) + f (e)}{2 f (12 / 5)} = 1$