Test: Linear Inequalities- 1 - JEE MCQ

Given, 

6<3(x-5)-5(x-1)
          15
90<-2x-10
100<-2x
-50>x

check for interval (7/3, ∞ ) the whole would be +ve
check for interval (-∞,3/2 ) the whole would be +ve

(7x-5)/(8x+3) > 4
(7x-5)/(8x+3) - 4 >0
7x - 5 - 4 ( 8x + 3 ) / 8x + 3 > 0
- 25 x - 17 / 8x + 3 > 0
Now furthermore solving for general range :
x ∈ ( -17/ 25, - 3/8)

|4 − x| + 1 < 3
⇒ 4 − x + 1 < 3
Add −4 and −1 on both sides, we get
4 − x + 1 − 4 − 1 < 3 − 4 − 1
⇒ − x < −2
Multiply both sides by −1, we get
x > 2
Also,|4−x| + 1 < 3
⇒ −(4−x) + 1 < 3
⇒ − 4 + x + 1 < 3
Add 4 and −1 on both sides, we get
− 4 + x + 1 + 4 − 1 < 3 + 4 − 1
⇒ x < 6
Thus, x ∈ (2,6).

 |x-2|/(x-2) > 0
=> x - 2 > 0
x > 2
x denotes (2,∞)

x−13+4<x−55−2

Multiply by 15 both side we get

x−13×15+4×15<x−55×15−2×15

⇒5(x−1)+60<3(x−5)−30

⇒5x−5+60<3x−15−30

⇒5x+55<3x−45

Add −3x and −55 on both sides, we get

5x+55−3x−55<3x−45−3x−55

⇒5x−3x<−45−55

⇒2x<−100

Divided by 2 we get

x<−50

Then x is (−∞,−50)

 |(2x-1)/(x-1)| > 2
|x| > a
⇒ x > a
or x < -a
(2x-1)/(x-1) > 2 and (2x-1)/(x-1) < -2
(2x-2+1)/(x-1) > 2
⇒ (2(x-1) + 1)/(x-1) > 2
⇒ 2 + (1/(x-1)) > 2
1/(x-1) > 0
x-1 < 0
x < 1...........(1)
Now taking, (2x-1)/(x-1) < -2
2 + (1/(x-1) < -2
= 1/(x-1) < -4
x-1 > -1/4
x > -1/4 + 1
x > 3/4.......(2)
From (1) and (2)
x implies (3/4, 1)∪ (⁡1,∞)

Let marks in 5th paper be x
Average = (95 + 72 + 73 + 53 + x)/5
= (323 + x)/5
(75 ≤ 323 + x)/5 < 80
375 ≤ 323 + x < 400
52 ≤ x < 77
[52,77).

3 − 2x ≤ 9
-2x ≤ 6
-x ≤ 3
-3 ≤ x

To determine which statement matches the solutions x=4,5 and 6, let's analyze each option:

a) x > 4 and x < 7

• This means 4 < x < 7.
• The solutions x = 5,6 do not satisfy this condition because it does not contain 4.

b) x ≥ 4 and x ≤ 7

• This means 4 ≤ x ≤ 7.
• The solution x = 4,5,6,7  do not satisfy this condition as it contains 7 also .

c) x ≥ 4 and x < 7

• This means 4 ≤ x < 7.
• The solutions x=4,5,6 satisfy this condition because they are all greater than or equal to 4 and strictly less than 7.

d) x > 4 and x > 7

• This means x > 7.
• The solutions x = 4,5,6 do not satisfy this condition because none of them are greater than 7.

Therefore, after analyzing each option, we conclude that the solutions x=4,5,6 correspond to statement:

c) x ≥ 4 and x < 7