Test Index

Linear Inequalities

Question : 10 of 54

Marks: +1, -0

$L_{1}: x_{1} \geq 0$
$L_{2}: x_{2} \geq 0$
$L_{3}: x_{1}+x_{2} \leq 1$
$L_{4}: x_{1}-x_{2} \geq 1$
$L_{5}:-x_{1}+2 x_{2} \leq 0$
$L_{1}: x_{1} \leq 0$
$L_{2}: x_{2} \geq 0$
$L_{3}: x_{1}+x_{2} \geq 1$
$L_{4}: x_{1}-x_{2} \geq 1$
$L_{5}:-x_{1}+2 x_{2} \leq 0$
$L_{1}: x_{1} \geq 0$
$L_{2}: x_{2} \geq 0$
$L_{3}: x_{1}+x_{2} \geq 1$
$L_{4}: x_{1}-x_{2} \leq 1$
$L_{s}:-x_{1}+2 x_{2} \leq 0$
None of these

Solution:

Concept:

Identify the set of inequalities that bound the shaded region in the graph.

Explanation:

The common region lies in the first quadrant, so

x_{1} \geq 0

and

x_{2} \geq 0

The region is above the line

x_{1}+x_{2}=1

, so

x_{1}+x_{2} \geq 1

The region is below the line

x_{1}-x_{2}=1

, so

x_{1}-x_{2} \leq 1

The region is below the line

-x_{1}+2x_{2}=0

(i.e.,

x_{2} \leq \frac{1}{2}x_{1}

), so

-x_{1}+2x_{2} \leq 0

These match option C.

Answer:

Option C is correct.

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