If the solution set of -2 x -1-3 x -4-5 x -5- is x - p - -4-3- - then the value of p isA- 1B- 0 C- 1-2 D- 2

The correct option is C 12|2x 1|+|3x 4|=|5x 5| Here 5x 5=(2x 1)+(3x 4) So, (2x 1)(3x 4)≥0 (∵ |x+y|=|x|+|y|, then xy≥0) ⇒(x 12 )(x 4 3) ≥ 0 ∴ x∈( ∞,12]∪[43 ...

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Standard XII

Mathematics

Absolute Value Function

If the soluti...

Question

If the solution set of $| 2 x - 1 | + | 3 x - 4 | = | 5 x - 5 |$ is $x \in (- \infty, p] \cup [\frac{4}{3}, \infty)$ , then the value of $p$ is

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\frac{1}{2}

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Solution

The correct option is B $\frac{1}{2}$
$| 2 x - 1 | + | 3 x - 4 | = | 5 x - 5 |$
Here $5 x - 5 = (2 x - 1) + (3 x - 4)$
So,
$(2 x - 1) (3 x - 4) \geq 0$
$(∵ | x + y | = | x | + | y |, then x y \geq 0)$
$\Rightarrow (x - \frac{1}{2}) (x - \frac{4}{3}) \geq 0 ∴ x \in (- \infty, \frac{1}{2}] \cup [\frac{4}{3}, \infty)$
Hence the value of $p$ is $\frac{1}{2}$

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If the solution set of |2x−1|+|3x−4|=|5x−5| is x∈(−∞,p ]∪[43,∞), then the value of p is

The correct option is B 12|2x−1|+|3x−4|=|5x−5| Here 5x−5=(2x−1)+(3x−4) So, (2x−1)(3x−4)≥0 (∵|x+y|=|x|+|y|, then xy≥0) ⇒(x−12)(x−43)≥0∴x∈(−∞,12]∪[43,∞) Hence the value of p is 12

If the solution set of $| 2 x - 1 | + | 3 x - 4 | = | 5 x - 5 |$ is $x \in (- \infty, p] \cup [\frac{4}{3}, \infty)$ , then the value of $p$ is