If -3 x -4-2 x -3- x -2-1-2- then the number of solutions is

We know that |a|+|b|+|c|≥ |a+b+c| ∀ a,b,c ∈ℝ Let a=3x+4, b= (2x+3) and c= (x+2) Then |3x+4|+|2x+3|+|x+2|≥ |3x+4 2x 3 x 2| nbsp; ⇒ |3x+4|+|2x+3|+|x+2|≥1 Hence, ...

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Standard VI

Mathematics

Equal Sets

If |3 x +4|+|...

Question

If $| 3 x + 4 | + | 2 x + 3 | + | x + 2 | = \frac{1}{2}$ , then the number of solutions is

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Solution

We know that $| a | + | b | + | c | \geq | a + b + c | \forall a, b, c \in R$
$Let a = 3 x + 4, b = - (2 x + 3) and c = - (x + 2)$
$Then | 3 x + 4 | + | 2 x + 3 | + | x + 2 | \geq | 3 x + 4 - 2 x - 3 - x - 2 |$
$\Rightarrow | 3 x + 4 | + | 2 x + 3 | + | x + 2 | \geq 1$
Hence, $| 3 x + 4 | + | 2 x + 3 | + | x + 2 | = \frac{1}{2}$ has no solution.

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Q. If

| 3 x + 4 | + | 2 x + 3 | + | x + 2 | = \frac{1}{2}

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∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 {2 x}^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12,

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∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

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∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = A x - 12

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If |3x+4|+|2x+3|+|x+2|=12, then the number of solutions is

We know that |a|+|b|+|c|≥|a+b+c| ∀ a,b,c∈R Let a=3x+4, b=−(2x+3) and c=−(x+2) Then |3x+4|+|2x+3|+|x+2|≥|3x+4−2x−3−x−2| ⇒|3x+4|+|2x+3|+|x+2|≥1 Hence, |3x+4|+|2x+3|+|x+2|=12 has no solution.

If $| 3 x + 4 | + | 2 x + 3 | + | x + 2 | = \frac{1}{2}$ , then the number of solutions is