The solution set of |x-1|+|x-2|<3 is - Vidyadip

MCQ

The solution set of $|x-1|+|x-2|<3$ is

✓
$(0,3)$
B
$(1,2)$
C
$(0,2)$
D
$(2,3)$

✓

Answer

Correct option: A.

$(0,3)$

a

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 11 - 5. linear inequalities→STD 11 - 5. linear inequalities — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

$\lim \tan\text{x} = \text{x}\rightarrow \frac{\pi }{2}$

If $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{\text{x}}=1,$ then:

If $\omega $ is a complex cube root of unity, then $225 + $${(3\omega + 8{\omega ^2})^2}$$ + {(3{\omega ^2} + 8\omega )^2} = $

What is the slope of the tangent line drawn to the hyperbola $xy = a\,(a \ne 0)$ at the point $(a, 1)$

$2 \sin A \sin B$ is equal to :

If $10^\text{n} + 3 \times 4^{\text{n}+2}+\lambda$ is divisible by $9$ for all $\text{n}\in\text{N},$ then the least positive integer value of $\lambda$ is

The set of values of $x$ for which the expression $\frac{{\tan 3x - \tan 2x}}{{1 + \tan 3x\tan 2x}} = 1$, is

Two sets $A$ and $B$ are as under:

$A = \{ \left( {a,b} \right) \in R \times R:\left| {a - 5} \right| < 1 \,\,and\,\,\left| {b - 5} \right| < 1\} $; $B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a - 6} \right)}^2} + 9{{\left( {b - 5} \right)}^2} \le 36} \right\}$ then : . . . . .

If $1 + \cot \theta = {\rm{cosec}}\theta $, then the general value of $\theta $ is

For any odd integer $n \ge 1$, ${n^3} - {(n - 1)^3} + ........... + {( - 1)^{n - 1}}{1^3} = $