The number of positive integral values of $k$ for which the equation $k = | x + | 2 x - 1 | | - | x - | 2 x - 1 | |$ has exactly three real solution is

0

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2

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3

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5

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Solution

The correct option is A $0$

For $x < | 2 x - 1 |$ we can see that even if $x$ becomes negative $| 2 x - 1 |$ will be a bigger positive as it is increasing with double rate.
So, $k = x + | 2 x - 1 | + x - | 2 x - 1 | = 2 x$
When $x > | 2 x - 1 |$ then
$k = x + | 2 x - 1 | - x + | 2 x - 1 | = 2 | 2 x - 1 |$
When $x = | 2 x - 1 |$
$\Rightarrow x = 2 x - 1 \Rightarrow x = 1$ or $x = - 2 x + 1 \Rightarrow x = \frac{1}{3}$
$∴ | x + | 2 x - 1 | | - | x - | 2 x - 1 | | = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} 2 x; & x < \frac{1}{3} 2 | 2 x - 1 |; & \frac{1}{3} \leq x \leq 1 2 x; & x > 1 \end{matrix} = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 2 x; & x \leq \frac{1}{3} - 4 x + 2; & \frac{1}{3} < x < \frac{1}{2} 4 x - 2; & \frac{1}{2} \leq x \leq 1 2 x; & x > 1 \end{matrix}$

Hence no integral value of $k$ exists for which the equation has three real solutions.

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