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CBSE Questions for Class 11 Engineering Maths Linear Inequalities Quiz 2 - MCQExams.com
CBSE
Class 11 Engineering Maths
Linear Inequalities
Quiz 2
Which of the following inequations represents the shaded region?
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$$\displaystyle 2x+y\leq 4$$
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$$\displaystyle 2x+y\geq 4$$
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$$\displaystyle x+2y\leq 4$$
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$$\displaystyle x+2y\geq 4$$
Explanation
The line passes through $$(0,4)$$ and $$(2,0)$$
Hence equation of the line
$$\dfrac{y}{4}+\dfrac{x}{2}=1$$
$$2x+y=4$$
Now in the above inequality graph the shading is done towards the origin.
Hence by substituting
$$x=0,y=0$$ we get an inequality which is true.
Hence the possible inequality is $$2x+y<4$$
Now the straight line is dark and not dotted. This indicates that the points on the line are a part of the inequality.
Hence the inequality can be written as
$$2x+y=4$$ or $$2x+y<4$$
$$2x+y\leq 4$$.
The locus represented by equation |z i| + |z + i| $$=$$ 1 on Argand plane
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ellipse
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hyperbola
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straight line
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no locus
The shaded region is represented by the inequation:
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$$\displaystyle y\geq x$$
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$$\displaystyle y\geq- x$$
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$$\displaystyle y\geq \left | x \right |$$
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$$\displaystyle y\leq \left | x \right |$$
Explanation
The equations of both the lines in the above graph are
$$y=-x$$ and $$y=x$$
Hence if we put them together we get
$$y=|x|$$
Now
Let us take a point inside the shaded region.
Let it be $$(0,2)$$
Now
$$2>0$$
$$y>|x|$$
Hence the required inequality is
$$y\geq |x|$$.
The table shown here contains pricing information for the chopper (a kitchen tool) that is advertised on television The price depends on how soon a customer calls after the end of the advertisement Approximately, how much more would a customer who orders 3 choppers, 23 minutes after the end of the television advertisement pay than a customer who orders 2 chopper, 17 minutes after the advertisement ?
Minutes (m) after television
advertisement ends
Price per chopper
(In Rs.)
$$\displaystyle 0\leq m\leq10$$
34.95
$$\displaystyle 10< m\leq20$$
39.95
$$\displaystyle 20< m\leq30$$
44.95
$$\displaystyle 30< m\leq40$$
49.95
$$\displaystyle m> 40$$
59.95
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Rs 15
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Rs 35
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Rs 40
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Rs 55
The inequalities in which terms compared are never equal to each other is classified as
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strict equality
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strict inequality
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strict quadres
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differential quadrates
Explanation
The inequalities in which the terms are never equal to each other are termed as $$strict$$ $$inequality$$
The inequality symbol in the strict inequality is either $$\gt$$ or $$\lt$$ i.e., it doesn't have any equality conditions.
The point which does not belong to the feasible region of the LPP:
Minimize: $$Z=60x+10y$$
subject to $$3x+y \ge 18$$
$$2x+2y \ge 12$$
$$x+2y\ge 10$$
$$x,y \ge 0$$ is
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$$(0,8)$$
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$$(4,2)$$
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$$(6,2)$$
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$$(10,0)$$
Explanation
We test whether the inequalitiies are satisfied or not
$$(0,8)$$, $$3(0)+8 \ge 8$$ $$8\ge 8$$ is true.
$$2(0)+2(8)=16\ge 12$$ is true.
$$0+2(8)=16 \ge 10$$ is true.
$$\therefore$$ $$(0,8)$$ is in the feasible region.
$$(4,2)$$, $$3(4)+2=14 \ge 8$$
$$2(4)+2(2)=16\ge 12$$
$$4+2(2)=8\ge 10$$ is not true
$$\therefore$$ $$(4,2)$$ is not a point in the feasible region
$$\therefore$$ (2) is correct
Which of the following number line represents the solution of the inequality
$$-6x + 12 > -7x + 17$$ ?
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Explanation
Given, $$-6x+12>-7x+17$$
$$\Rightarrow -6x+7x>17-12$$
$$\Rightarrow x > 5$$
Hence, option A is correct.
Which region is described by the shade in the graph given?
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$$2x+3y=3$$
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$$2x+3y>3$$
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$$2x+3y<3$$
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None of these
Explanation
The graph of the function $$2x+3y=3$$ is shown in the figure.
It is clear from the graph, $$(0,0)$$ does not lie on the line.
Now, put $$(0,0)$$ in the equation of line.
$$2(0)+3(0)=3$$
$$0\neq 3$$ hence, $$0<3$$ is true
Hence, region which
lies
below the graph is denoted by $$2x+3y<3$$
It is clear from the graph, $$(0,2)$$ does not lie on the line.
N
ow, put $$(0,2)$$ in the equation of line,
$$2(0)+3(2)=3$$
$$0\neq 6$$ hence, $$6>3$$ is true.
Hence, the region which lies above the graph is denoted by $$2x+3y>3$$.
Sketch the solution to system of ineqalities
$$x\le -3$$
$$y < \frac{5}{3}x+2$$
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Explanation
Given, $$x\leq -3$$ and $$y<\dfrac{5}{3}x+2$$
first, draw the graph for equations
$$x= -3$$ and $$y=\dfrac{5}{3}x+2$$
$$x=-3$$ is the line which is parallel to the y-axis and passes through (-3,0).
Hence, $$x\leq -3$$ includes the left side region of the line.
similarly, for
$$y=\dfrac{5}{3}x+2$$
substitute y=0 we get, $$\dfrac{5}{3}x=-2 \implies x=-\dfrac{6}{5}=-1.2$$
substitute x=0 we get, $$y=2$$
therefore,
$$y=\dfrac{5}{3}x+2$$
line passes through $$(-1.2,0)$$ and $$(0,2)$$.
Hence,
$$y<\dfrac{5}{3}x+2$$
includes the region below the line.
Option D satisfies the above conditions.
A graph and the system of inequalities are shown above. Which region of the graph could represent the solution for the system of in equations?
$$y > x$$
$$3y≤-4x+6$$
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$$A$$
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$$B$$
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$$C$$
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$$D$$
Explanation
given $$y>x$$
and $$3y\leq -4x+6$$
first, draw the graph for equations
$$y =x$$ and
$$3y= -4x+6$$
$$y=x$$ is the line which passes through the origin as shown in the above fig
Hence, $$y>x$$ includes the above region of the line.
similarly, for
$$3y= -4x+6$$
substitute y=0 we get, $$-4x+6=0 \implies x=1.5$$
substitute x=0 we get, $$3y=6 \implies y=2$$
therefore,
$$3y= -4x+6$$ line passes through (1.5,0) and (0,2) as shown in fig.
Hence,
$$3y\leq -4x+6$$ includes the region below the line.
the intersection region is the D region as shown in above figure.
If the solution set for the system $$\begin{cases} y>x \\ y\le -\dfrac { 3 }{ 7 } x+5 \end{cases}$$ is given by the above figure, then which of the following is NOT a solution to the system?
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$$(0,3)$$
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$$(1,2)$$
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$$(2,4)$$
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$$(3,3)$$
Explanation
As per the given figure, $$(0,3)$$ and $$(1,2)$$ lie in the desired region, hence these are solutions of the system.
Whereas $$(2,4)$$ and $$(3,3)$$ lie on the lines but $$(3,3)$$ does not satisfy the equation $$y>x$$.
Hence, it is not a solution to the system.
State true or false:
The statement $$0>1$$ $$\rightarrow$$ $$\sin x = 2$$.
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True
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False
Explanation
$$\mathrm P$$
$$\mathrm Q$$
$$\mathrm P \rightarrow Q$$
True
True
True
True
False
False
False
True
True
False
False
True
$$0>1 $$ is False
$$\sin x=2$$ is False
(since, $$\sin x$$ range is $$[-1,1]$$)
From the above truth table, $$\mathrm {False} \rightarrow \mathrm {False}$$ is
$$\mathrm {True}$$
Which equation has the solution shown on the number line?
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$$-3 > x > -1$$
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$$-5 < x < -2$$
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$$-2 > x > 0$$
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$$-1 > x > -6$$
Explanation
Since shaded region on the graph is from $$-5$$ to $$-2$$
So solution is $$-5<x<-2$$
Hence, option B is correct.
Which of the following is the correct graph for $$x\ge 3$$ or $$x\le -2$$?
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Line A
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Line B
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Line C
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Line D
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Line E
Explanation
It is stated $$x \ge 3$$ that means including $$3$$ and till infinity (positive) $$= \{3, 4, 5, 6, ....\}$$
And, it also stated $$x \le -2$$ that means including $$-2$$ and till infinity (negative) $$= \{3, 4, 5, 6, ... \}$$
So, the line D perfectly defines the two stated i.e. $$x \ge 3$$ and $$x \le -2$$
$$\therefore $$ Option D is correct.
The point $$(2,1)$$ lies in the _____ region of the solution of linear inequations $$x-2y \leq 0, $$ and $$x-y>0$$
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feasible
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non-feasible
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unbounded
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bounded
Explanation
Given, $$x-2y\leq 0$$ and
$$x-y>0\implies x>y \implies y<x$$
First draw the graph for the equations,
$$x-2y= 0$$ and
$$y=x$$
$$x-2y= 0$$ and
$$y=x$$ are
the lines which pass through the origin as shown in the above fig.
Hence,
$$y<x$$ includes the below region of the line and
$$x-2y\leq 0$$ also includes the below
region of the line.
Therefore, the blue shaded region is the feasible region which is unbounded.
Given point (2,1). Substituting in the given inequations,
$$x-2y\leq 0\implies 2-2*1\leq 0 \implies 0\leq 0$$ True
$$x-y>0 \implies 2-1>0 \implies 1>0$$ True
Therefore, (2,1) belongs to the feasible region which is unbounded.
Which quadrant does the solution lie in?
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$$I$$
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$$II$$
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$$III$$
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$$IV$$
Explanation
Given, $$2x+3y>2$$
$$x-y<0 \implies y>x$$
$$y\geq 0$$
$$x\leq 3$$
first, draw the graph for equations
$$2x+3y=2$$
$$x-y=0 \implies x=y$$
$$y= 0 \implies x-axis$$
$$x= 3$$
$$x=y$$ is the line which passes through the origin as shown in the above fig
Hence, $$y>x$$ includes the above region of the line.
$$y= 0 \implies x-axis$$. Hence,
$$x\leq 3$$
includes the left region of the line.
$$x= 3 \implies $$line parallel to x-axis. Hence,
$$x\leq 3$$
includes the left region of the line.
for
$$2x+3y=2$$
substituting y=0, we get $$2x=2\implies x=1$$
substituting x=0, we get $$3y=2\implies y=\dfrac{2}{3}$$
Therefore, line
$$2x+3y=2$$ passes through (1,0) and (0,2/3) as shown in the above figure
Hence,
$$2x+3y>2$$
includes the above region of the line.
Therefore, the blue shaded region is the feasible region which is present in I quadrant as shown in the figure.
How many solutions does this set include?
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$$5$$
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$$10$$
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$$20$$
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Infinite
Explanation
Given, $$2x+3y>2$$
$$x-y<0 \implies y>x$$
$$y\geq 0$$
$$x\leq 3$$
first, draw the graph for equations
$$2x+3y=2$$
$$x-y=0 \implies x=y$$
$$y= 0 \implies x-axis$$
$$x= 3$$
$$x=y$$ is the line which passes through the origin as shown in the above fig
Hence, $$y>x$$ includes the above region of the line.
$$y= 0 \implies x-axis$$. Hence, $$y \geq 0$$
includes the region above the x axis.
$$x= 3 \implies $$line parallel to x-axis. Hence,
$$x\leq 3$$
includes the left region of the line.
for
$$2x+3y=2$$
substituting y=0, we get $$2x=2\implies x=1$$
substituting x=0, we get $$3y=2\implies y=\dfrac{2}{3}$$
Therefore, line
$$2x+3y=2$$ passes through (1,0) and (0,2/3) as shown in the above figure
Hence,
$$2x+3y>2$$
includes the above region of the line.
Therefore, the blue shaded region is the feasible region which is not a closed region or it is not bounded. Hence, there are infinite number of solutions.
Which of the following points lie in the solution set?
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$$(1,1)$$
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$$(1,2)$$
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$$(2,1)$$
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$$(3,2)$$
Explanation
By option verification,
i.e., substituting the options in given linear inequations and verifying
Given, $$2x+3y>2$$
$$x-y<0$$
$$y\geq 0$$
$$x\leq 3$$
substituting option A (i.e.,) $$(x,y)=(1,1)$$
$$2x+3y>2\implies 2\times 1+3\times 1>2 \implies 5>2$$ True
$$x-y<0 \implies 1-1<0 \implies 0<0$$ False
$$y\geq 0 \implies 1\geq 0$$ True
$$x\leq 3 \implies 1\leq 3$$ True
substituting option B (i.e.,) $$(x,y)=(1,2)$$
$$2x+3y>2\implies 2\times 1+3\times 2>2 \implies 8>2$$ True
$$x-y<0 \implies 1-2<0 \implies -1<0$$ True
$$y\geq 0 \implies 2\geq 0$$ True
$$x\leq 3 \implies 1\leq 3$$ True
substituting option C (i.e.,) $$(x,y)=(2,1)$$
$$2x+3y>2\implies 2\times 2+3\times 1>2 \implies 7>2$$ True
$$x-y<0 \implies 2-1<0 \implies 1<0$$ False
$$y\geq 0 \implies 1\geq 0$$ True
$$x\leq 3 \implies 2\leq 3$$ True
substituting option D (i.e.,) $$(x,y)=(3,2)$$
$$2x+3y>2\implies 2\times 3+3\times 2>2 \implies 12>2$$ True
$$x-y<0 \implies 3-2<0 \implies 1<0$$ False
$$y\geq 0 \implies 2\geq 0$$ True
$$x\leq 3 \implies 3\leq 3$$ True
Therefore option B satisfies the above linear inequalities.
Consider the linear inequations and solve them graphically:
$$3x-y-2 > 0;\,\,\, x+y \leq 4;\,\,\, x >0;\,\,\, y \geq 0$$.
The solution region of these inequations is a convex polygon with _____ sides.
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$$3$$
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$$4$$
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$$5$$
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$$7$$
Explanation
Given, $$3x-y-2>0\implies y<3x-2$$
$$x+y\leq 4$$
$$x>0$$ and $$y\geq 0$$
First, draw the graph for equations
$$3x-y-2=0$$
$$x+y= 4$$
$$x=0$$ and $$y= 0$$
$$x=0$$ is the Y-axis.
Hence, $$x>0$$ includes the right side region of the line.
$$y=0$$ is the X-axis.
Hence, $$y\geq 0$$ includes the upper side region of the line.
Similarly, for
$$y=3x-2$$
Substitute y=0 we get, $$3x=2 \implies x=\dfrac{2}{3}$$
Substitute x=0 we get, $$y=-2$$
Therefore,
$$y=3x-2$$
line passes through (2/3,0) and (0,-2).
Hence,
$$y<3x-2$$
includes the region below the line.
Similarly, for
$$x+y= 4$$
Substitute y=0 we get, $$x=4$$
Substitute x=0 we get, $$y=4$$
Therefore,
$$x+y= 4$$
line passes through (2/3,0) and (0,-2).
Hence,
$$x+y\leq 4$$
includes the region below the line.
As shown in the above figure, the blue shaded region is the feasible region which is the intersection of all 4 solution sets. Feasible region is the polygon with $$3 \space sides$$ as shown in the figure.
Find solution of following inequality also show it graphically.
$$x+3<5,x\in R$$
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Explanation
Given, $$x+3<5\implies x<2$$ and $$x\in R$$
The real numbers includes all the integers(e.g., -4), fractions
(e.g., 3/4)
and irrational numbers(e.g., $$\sqrt{2}$$).
Option C, represents only integers.
Option B, represents all the numbers $$<2$$
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