If -4 x -5-6 x -12-2 x -7- then x belongs toA- -5-4- 2-B- 2- -C- -1-3-D- - 2-

The correct option is A [54,2]|4x 5|+|6x 12|=|2x 7| We know that |a b| = |a| +|b|, then ab ≤ 0 Here, 2x 7=(6x 12) (4x 5) So, (6x 12)(4x 5)≤0 ⇒ (x 2)(x 5 4)≤ ...

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Byju's Answer

Standard XII

Mathematics

Inequalities Involving Modulus Function

If |4 x -5|+|...

Question

If $| 4 x - 5 | + | 6 x - 12 | = | 2 x - 7 |$ , then $x$ belongs to

[\frac{5}{4}, 2]

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(- \infty, 2]

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(2, \infty)

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(- 1, 3]

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Solution

The correct option is A $[\frac{5}{4}, 2]$
$| 4 x - 5 | + | 6 x - 12 | = | 2 x - 7 |$
We know that $| a - b | = | a | + | b |$ , then $a b \leq 0$
Here, $2 x - 7 = (6 x - 12) - (4 x - 5)$
So,
$(6 x - 12) (4 x - 5) \leq 0 \Rightarrow (x - 2) (x - \frac{5}{4}) \leq 0$
$∴ x \in [\frac{5}{4}, 2]$

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Q. If

| 4 x - 5 | + | 6 x - 12 | = | 2 x - 7 |

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- 4 \leq | x | \leq 2

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If $| x - 5 | + | x - 10 | ⩽ 20$ , then x belongs to

Question 3
Which of the following equations has 2 as a root?
(a) $x^{2} - 4 x + 5 = 0$
(b) $x^{2} + 3 x - 12 = 0$
(c) $2 x^{2} - 7 x + 6 = 0$
(d) $3 x^{2} - 6 x - 2 = 0$

Q. Simplify:

{(3 x^{2} - 6 x + 5) + (2 x - 2 x^{2} + 5)} - (x^{2} - 4 x + 10)

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If |4x−5|+|6x−12|=|2x−7|, then x belongs to

The correct option is A [54,2]|4x−5|+|6x−12|=|2x−7| We know that |a−b|=|a|+|b|, then ab≤0 Here, 2x−7=(6x−12)−(4x−5) So, (6x−12)(4x−5)≤0⇒(x−2)(x−54)≤0 ∴x∈[54,2]

If $| 4 x - 5 | + | 6 x - 12 | = | 2 x - 7 |$ , then $x$ belongs to

The correct option is A $[\frac{5}{4}, 2]$
$| 4 x - 5 | + | 6 x - 12 | = | 2 x - 7 |$
We know that $| a - b | = | a | + | b |$ , then $a b \leq 0$
Here, $2 x - 7 = (6 x - 12) - (4 x - 5)$
So,
$(6 x - 12) (4 x - 5) \leq 0 \Rightarrow (x - 2) (x - \frac{5}{4}) \leq 0$
$∴ x \in [\frac{5}{4}, 2]$