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Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsFunctions3 Marks
Question
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$|x-4|+|x-2|=3$
✓

Answer

$|x-4|+|x-2|=3$
Case I: $x<2$
Equation (i) reduces to
$ 4-\mathrm{x}+2-\mathrm{x}=3 \ldots \ldots .[\mathrm{x}<2<4, \mathrm{x}-4<0, \mathrm{x}-2<0]$
$\therefore 6-3=2 \mathrm{x}$
$\therefore \mathrm{x}=\frac{3}{2} $
Case II: $2 \leq \mathrm{x}<4$
Equation (i) reduces to
$ 4-\mathrm{x}+\mathrm{x}-2=3$
$\therefore 2=3 \text { (absurd) } $
There is no solution in $[2,4)$
Case III: $x \geq 4$
Equation (i) reduces to
$ x-4+x-2=3$
$\therefore 2 x=6+3=9$
$\therefore x=\frac{9}{2}$
$\therefore x=\frac{3}{2}, \frac{9}{2} \text { are solutions. } $
The solution set $=\left\{\frac{3}{2}, \frac{9}{2}\right\}$

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