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This is one of our many quizzes on Graphing compound inequalities. Click on this to access more such competitive quizzes on Linear Inequalities in One Variable.

Graphing compound inequalities Quiz 2

mathematics

 Which of the following inequality belongs to the given graph? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/81f8b4c6-2cdc-4ef6-9a4f-c63491c7df7f8792197264840513212.png"/>

 Write the equation of the given graph in slope intercept form

From the given graph find its slope intercept form, since two of its intercepts are given, $ \\ \left( {0,\;3} \right)\;{\text{and}}\;\left( {6,\;0} \right) \\ $ Therefore its slope intercept form is $ \\ y = mx + c \\ m = \dfrac{{0 - 3}}{{6 - 0}} =  - \dfrac{1}{2}\;{\text{and}}\;c = 3 \\ $ That is $ \\   y =  - \dfrac{1}{2}x + 3 \\   \Rightarrow x + 2y = 6 \\  $  According to the graph, inequality is $x + 2y < 6$ 

 Evaluate the compound inequality and give its solution in the interval form $ \\ x - 2 >  - 1\;{\text{and}}\;x + 3 < 5$ 

 Find solution for each individually and then see what operation is there in between them and do accordingly

$ \Rightarrow x - 2 >  - 1\;{\text{and}}\;x + 3 < 5 \\   \Rightarrow x >  - 1 + 2\;{\text{and}}\;x < 5 - 3 \\   \Rightarrow x > 1\;{\text{and}}\;x < 2 \\  $ Since there is “and” between the two inequalities, so there intersection will be taken as the solution, That is  $  x \in \left( {1,\;\infty } \right) \cap \left( { - \infty ,\;2} \right) \\  x \in \left( {1,\;2} \right) \\  $ 

 The given graph represents which of the following inequality? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/dc408c13-ebab-4231-a04a-2f4569593f473075020409026715184.png"/>

 Observe the dotted line in the graph and check for greater than or less than sign by putting a point in it.

Since there are four options of same equality that is $y = 2x$ $ \\ $ We can see there is a dotted line, which tells that either $y \leqslant 2x\;{\text{or}}\;y \geqslant 2x$ are not the correct options, $ \\ $ Now, testing the point $\left( {0,\; - 2} \right)$ which gives $ \\ $ $0 <  - 2$ and holds true only if we take the first inequality $y < 2x$ $ \\ $ Therefore $y < 2x$ that is option (a) is the correct option

 Double inequality gives two set of solutions, and the final solution can be found by intersection of both the solutions

 $ - 4 < 3x - 7 < 8 \\ $ Can be written as $ \\  - 4 < 3x - 7\;{\text{and}}\;3x - 7 < 8 \\   \Rightarrow  - 4 + 7 < 3x\;{\text{and}}\;3x < 8 + 7 \\   \Rightarrow 3 < 3x\;{\text{and}}\;3x < 15 \\   \Rightarrow 1 < x\;{\text{and}}\;x < 5 \\  $ Its interval notation is $x \in \left( {1,\;5} \right)$ 

 Which of the following inequality$ / $ies will have a finite solution for the value of $x$?

 Check for each and every option, for the quadratic inequality try to factorize it to solve and for the modulus inequality split it into two parts positive and negative $\left| {x + a} \right| > b \Rightarrow x + a > b\;{\text{or}}\;x + a <  - b$

Check for each Options, $ \\ $ Option (b) $ \\  \Rightarrow {x^2} - 6x + 8 < 0 \\   \Rightarrow {x^2} - 2x - 4x + 8 < 0 \\   \Rightarrow x\left( {x - 2} \right) - 4\left( {x - 2} \right) < 0 \\   \Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) < 0 \\  $ Here there are two cases for this, $ \\ $ Case I:  $ \\ \Rightarrow \left( {x - 2} \right) < 0\;{\text{and}}\;\left( {x - 4} \right) > 0 \\   \Rightarrow x < 2\;{\text{and}}\;x > 4 \\  $  Such number does not exist $ \\ $ Case II: $ \\   \Rightarrow \left( {x - 2} \right) > 0\;{\text{and}}\;\left( {x - 4} \right) < 0 \\   \Rightarrow x > 2\;{\text{and}}\;x < 4 \\  $ That is $x \in \left( {2,\;4} \right)$ which is finite