If x satisfies x-1-x-2-x-3-6- then the values of x lie in the range

Let nbsp;f(x)=x 1+x 2+x 3→If nbsp;x≤1f(x)= (x 1) (x 2) (x 3)⇒f(x)= x+1 x+2 x+3⇒f(x)= 3x+6Now, nbsp;f(x)≥6⇒ 3x+6≥6⇒x≤0∴x≤0→If nbsp;1≤x≤2f(x)=(x 1) (x 2) ...

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Byju's Answer

Standard XIII

Mathematics

Modulus Function

If x satisfie...

Question

If x satisfies $(x - 1 | + (x - 2 | + (x - 3 | \geq 6$ , then the values of x lie in the range

0 \leq x \leq 4

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x \leq - 2 o r x \geq 4

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x \leq 0 o r x \geq 4

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x \leq 1 o r x \geq 4

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Solution

The correct option is C $x \leq 0 o r x \geq 4$
$Let f (x) = (x - 1 | + (x - 2 | + (x - 3 | \to If x \leq 1 f (x) = - (x - 1) - (x - 2) - (x - 3) \Rightarrow f (x) = - x + 1 - x + 2 - x + 3 \Rightarrow f (x) = - 3 x + 6 Now, f (x) \geq 6 \Rightarrow - 3 x + 6 \geq 6 \Rightarrow x \leq 0 ∴ x \leq 0 \to If 1 \leq x \leq 2 f (x) = (x - 1) - (x - 2) - (x - 3) \Rightarrow f (x) = x - 1 - x + 2 - x + 3 \Rightarrow f (x) = - x + 4 Now, f (x) \geq 6 \Rightarrow - x + 4 \geq 6 \Rightarrow - x \geq 2 which is not possible as 1 \leq x \leq 2 \to If 2 < x \leq 3 f (x) = (x - 1) + (x - 2) - (x - 3) \Rightarrow f (x) = x - 1 + x - 2 - x + 3 \Rightarrow f (x) = x Now, f (x) \geq 6 \Rightarrow x \geq 6 which is not possible as x \leq 3 \to If x > 3 f (x) = (x - 1) + (x - 2) + (x - 3) \Rightarrow f (x) = x - 1 + x - 2 + x - 3 \Rightarrow f (x) = 3 x - 6 Now, f (x) \geq 6 \Rightarrow 3 x - 6 \geq 6 \Rightarrow x \geq 4 ∴ x \geq 4 Range of x is x \leq 0 or x \geq 4$

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If x satisfies (x−1|+(x−2|+(x−3|≥6, then the values of x lie in the range

If x satisfies $(x - 1 | + (x - 2 | + (x - 3 | \geq 6$ , then the values of x lie in the range