Linear Inequalities Test -2

Result

Linear Inequalities Test -2

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Solve inequality and show the graph of the solution, $$7x+3 < 5x+9$$

A

B

C

D

Solution

$$7x + 3 < 5x + 9$$
$$ \Rightarrow 7x - 5x < 9 - 3$$
$$ \Rightarrow 2x < 6$$
$$ \Rightarrow x < 3$$
Question 2
1 / -0

Area of the region {$$(x,y):{x}^{2}+{y}^{2}\le 1\le x+y$$} is

A
$$\dfrac{\pi}{4}+\dfrac{1}{2}$$

B
$$\dfrac{\pi}{4}-\dfrac{1}{2}$$

C
$$\dfrac{\pi}{4}+\dfrac{3}{4}$$

D
$$\pi+1$$

Solution

To find area $$x^2+y^2\leqslant |\leqslant x+y$$
1) Circle $$x^2+y^2\leqslant |$$
$$(x-0)^2+(y-0)^2\leqslant |$\Rightarrow $ centere $$(0,0)$$, Radius $$1$$
2) $$x+y\geqslant |$$
Required area is
$$A=\int_{1}^{0}\sqrt{1-x^2}-(1-x)$$
$$A\Rightarrow ^1_0\left[\dfrac{\sin ^{-1}(x)}{2}+\dfrac{x}{2}\sqrt{1-x^2}\right]^1_0\left[x-\dfrac{x^2}{2}\right]$$
$$\therefore A\Rightarrow \left[\dfrac{\pi}{4}+\dfrac{1}{2}\times 0-0-0\right]-\left[1-0-\dfrac{1}{2}+0\right]$$
$$\boxed{\therefore A\Rightarrow \dfrac{\pi}{4}-\dfrac{1}{2}}$$
Question 3
1 / -0

Number of integral solutions satisfy inequality $$\left| x-3 \right| -\left| 2x+5 \right| \ge \left| x+8 \right| $$ is

A
5

B
6

C
7

D
8

Solution

Case $$(1)\quad x\le -8\Rightarrow$$
$$-x+3+2x+5 \ge -x-8$$
$$2x \ge -8$$
$$x\ge -8\quad \therefore \boxed {x=-8}$$
Case $$(2)\quad x\in [-8, -5/2]$$
$$-x+3+2x+5\ \ge x+8$$
$$8\ge 8$$ true
$$\therefore \boxed {x\in [-8, -5/2]}$$
Case $$(3)\quad x\in [-5/2, 3]$$
$$-x+3-2x-5\ \ge x+8$$
$$-10\ge 4x$$
$$-5/2 \ge x$$
$$\therefore \boxed {x =-5/2}$$
Case $$(4)\quad x\ge 3$$
$$x-3-2x-5\ge x+8$$
$$-16\ge 2x$$
$$-8\ge x$$
$$\boxed {No\ solution}$$
$$\therefore $$
Finally $$x\in [-8, -5/2]$$
$$x\ \in \boxed {-8, -2.5}$$
$$\therefore $$
the integral solution of $$x$$ are
$$x=-8, -7, -6, -5, -4, -3$$
$$\therefore $$
$$B$$ Answer.
Question 4
1 / -0

The sum of four numbers in AP is $$20.$$ The numbers are such that the ratio of the product of first and fourth is to the product of second and third as $$2 : 3.$$ The greatest number is:-

A
$$8$$

B
$$7$$

C
$$14$$

D
$$4$$

Solution

Let the four terms be $$\;a-3d\;,a-d\;,a+d\;,and\;a+3d$$ with common difference $$2d$$.
$$sum=4a=20$$
$$a=5$$

$$\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}=\dfrac{2}{3}$$

$$\dfrac{25-9d^2}{25-d^2}=\dfrac{2}{3}$$

$$d^2=1$$$$\Rightarrow d=1$$

$$Largest \;term=a+3d$$$$=5+3(1)$$$$=8.$$
Question 5
1 / -0

The system of equation $$|x-1|+3y=4$$, $$x-|y-1|=2$$ has

A
No solution

B
A unique solution

C
Two solutions

D
More than two solutions

Solution

For, $$\mid x-1 \mid + 3y=4$$ ........(i)

When $$x \geq 1$$
$$\mid x-1 \mid + 3y=4$$
$$\implies (x-1) +3y = 4$$

When $$x<1$$
$$\mid x-1 \mid + 3y=4$$
$$\implies -(x-1) +3y=4$$

For, $$x - \mid y-1 \mid =2 $$ ........(ii)

When $$y \geq 1$$
$$x - \mid y-1 \mid =2 $$
$$\implies x-(y-1)=2$$

When $$y<1$$
$$x - \mid y-1 \mid =2 $$
$$\implies x-[-(y-1)]=2$$
$$\implies x+(y-1)=2$$

Now, we plot graphs of equations (i) and (ii)
From the graph, we can say that the given system of the equation has a unique solution.

Hence, option B is correct.
Question 6
1 / -0

If x $$\epsilon$$ I, the solution set of the inequation $$-2 \leq x < 3$$ is

A
$${-2, -1, 0, 1, 2}$$

B
$${-1, 0, 1, 2, 3}$$

C
$${0, 1, 2, 3}$$

D
$${0, 1, 2}$$

Solution

x $$\epsilon$$ I
Solution set $$-2 \leq x < 3= {-2, -1, 0, 1, 2}$$
Question 7
1 / -0

Which of the following is correct?

A
Every LPP has an optimal solution

B
A LPP has unique optimal solution

C
If LPP has two optimal solutions, then it has infinite number of optimal solutions

D
The set has two optimal solutions, then it has infinite number of optimal solutions
Question 8
1 / -0

The half plane represented by 3x + 2y < 8 contains the point

A
$$\left ( 1, \dfrac{5}{2} \right )$$

B
(2, 1)

C
(0, 0)

D
(5, 1)
Question 9
1 / -0

Region represented by the inequation system
$$x + y \leq 3$$
$$y \leq 6$$
and $$x \geq 0,y \geq 0$$
is :

A
Unbounded in the first quadrant

B
Unbounded in the first and second quadrant

C
Bounded in the first quadrant

D
None of the above

Solution

Converting the given inequations into equations
$$x + y = 3$$ …..(1)
$$y = 6$$ …..(2)
Region represented by $$x + y \leq 3$$ : The line $$x + y = 3$$ meets the coordinate axes are $$A(3,0)$$ and $$B(0,3)$$ respectively,
$$x + y = 3$$

Region represented by $$y \leq 6$$: The line $$y = 6$$ is parallel to the x-axis and its every point will satisfy the inquation in first quadrant, region containing the origin represents the solution set of this inequation.$$A(3, 0); B(0, 3)$$
Join points A and B to obtain the line. Clearly, $$(0, 0)$$ satisfies the inequation $$x + y \leq 3$$. So, the region containing the origin represent the solution set of the inequation.
Region represented by $$x \geq 0$$ and $$y \geq 0$$: Since every point in the first quadrant satisfy the inequations, so the first quadrant is the solution set of these inequations.

The shaded region is the common region of inequations. This is a feasible region of solution which is bounded and is in the first quadrant.
Hence the region is bounded and is in the first quadrant.
∴ Answer (c) is correct.
Question 10
1 / -0

One-half of a number is 17 more than one-third of that number. What is the number?

A
52

B
84

C
102

D
112

Solution