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If x satisfies |3x – 2| + |3x – 4| + |3x – 6| >= 12 to what interval does x belong? |...| represents modulus function. [BITSAT 2013]

If x satisfies |3x – 2| + |3x – 4| + |3x – 6| >= 12 to what interval does x belong? |...| represents modulus function. [BITSAT 2013]

Grade:12

1 Answers

Samyak Jain
333 Points
4 years ago
|3x – 2| + |3x – 4| + |3x – 6| \geq 12.
First we find critical values of x by equating each modulus expression to zero.
Thus critical values of x are 2/3 , 4/3 and 6/3 = 2.
Case 1.  x \geq 2        ...(1)
3x – 2 + 3x – 4 + 3x – 6  \geq 12   \Rightarrow   9x – 12  \geq  12
x \geq 24 / 9      ...(2)
Taking intersection of (1) and (2), we get x \geq 24 / 9          ...(i)
 
Case 2.  4/3  \leq  x 
3x – 2 + 3x – 4 – (3x – 6) \geq 12   \Rightarrow   3x  \geq 12
\geq 4        ...(4)
Intersection of (3) and (4) gives x  \epsilon  \phi          ...(ii)
 
Case 3. 2/3  \leq  x 
3x – 2 – (3x – 4) – (3x – 6)  \geq  12   \Rightarrow   – 3x + 8  \geq  12   \Rightarrow   3x \leq – 4
\leq – 4/3    ...(6)
Taking intersection of (5) and (6), we get x  \epsilon  \phi           ...(iii)
 
Case 4.  x
– (3x – 2) – (3x – 4) – (3x – 6)  \geq  12   \Rightarrow   – 9x + 12  \geq  12
x  \leq  0            ...(8)
Intersection of (7) and (8) gives x  \leq  0            ...(iv)
Union of (i), (ii), (iii) & (iv) gives \geq 24/9  \cup  x \leq 0   i.e.   x  \epsilon  (– \infty , 0]  \cup  [24 / 9 , \infty).

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