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Standard XII

Mathematics

Inequality

Let x>0 and...

Question

Let $x > 0$ and $| x - 1 | + | x - 2 | = | 2 x - 3 |$ , then the least integral value of $^{'} x^{'}$ is :

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Solution

The correct option is B $1$
$\begin{matrix} | x - 1 | + | x - 2 | = | 2 x - 3 | f o r 1 \leq x < \frac{3}{2} x - 1 + 2 - x = 3 - 2 x 1 = 3 - 2 x 2 x = 2 \Rightarrow x = 1 f o r x < 1 1 - x + 2 - x = 3 - 2 x 3 - 2 x = 3 - 2 x B u t f r o m g i v e n o p t i o n B i s c o r r e c t . \end{matrix}$

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Let x>0 and |x−1|+|x−2|=|2x−3| , then the least integral value of ′x′ is :

The correct option is B 1|x−1|+|x−2|=|2x−3|for1≤x<32x−1+2−x=3−2x1=3−2x2x=2⇒x=1forx<11−x+2−x=3−2x3−2x=3−2xButfromgivenoptionBiscorrect.

Let $x > 0$ and $| x - 1 | + | x - 2 | = | 2 x - 3 |$ , then the least integral value of $^{'} x^{'}$ is :

The correct option is B $1$
$\begin{matrix} | x - 1 | + | x - 2 | = | 2 x - 3 | f o r 1 \leq x < \frac{3}{2} x - 1 + 2 - x = 3 - 2 x 1 = 3 - 2 x 2 x = 2 \Rightarrow x = 1 f o r x < 1 1 - x + 2 - x = 3 - 2 x 3 - 2 x = 3 - 2 x B u t f r o m g i v e n o p t i o n B i s c o r r e c t . \end{matrix}$