advanced inequations exercise - 1(b)

7/27/2019 Advanced Inequations Exercise - 1(B)

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INEQUATIONS & EQUATIONS

Exercise1 (B)

1. [C]

2

x 1 x 5x 7 x 1

2x 1 x 5x 7 x 1 0

2x 1 x 5x 7 1 0

2x 1 x 5x 5 0

(x1) (x3) (x2) < 0

by curve method we get x ,1 2,3

2. [D]

x 1 1 x , x R

Take square both side;

2 2

x 1 1 x

2 2

x 1 1 x 0

x 1 1 x x 1 1 x 0

x x 1 x x 0

L.H.S. of above inequality will become zero for x 0 , as | x |x = xx = 0 so x 0 is not the

feasible solution.

for x < 0

[x + x2] [xx] < 0

(2) (2x) < 0

4x < 0

1 2 3


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x < 0

so x < 0 is the solution of given inequality.

3. [A]

x 0.5 x4 7.2 4 ; x R

x 0.5

2 x2 7.2 4

2x 1 x2 7.2 4

2x x2.2 7.2 4 0

2x x x2.2 8.2 2 4 0

x x2.2 1 2 4 0

take x2 y

1y 4

2

x1 2 42

as x2 is always greater than1

2 .

So,x0 2 4

take log both sides

2x log 4

x 2

x 2

4. [A]

2

x 3x

y

2 x

1

4

x

1

2 4


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2

3 x 3x

2

3 xx

&2

x 3x

2

x 23 0

x

&

2x 2 3x0

x

2x 3x 2

0x

&

(x 2)(x 1)0

x

x 1 x 2

0x

& x , 0 1, 2 (2)

x 2, 1 0, .(1)

By taking intersection of (1) & (2)

x 2, 1 1, 2

5. [D]

xx2 2 2 2

case 1: x 0

x x2 2 2 2

x2 2

2x log 2

1

x2

case 2: x < 0

x x2 2 2 2 (as x2 > 0)

2 1 0

10 2


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2x x2 1 2 2 2

2x x2 2 2 2 1 0

take x2 y .(y > 0)

2y 2 2y 1 0

y 2 1 y 2 1 0

y 2 1 or y 2 1

x2 2 1 x2 2 1

2x log 2 1 2x log 2 1 0

As x < so but x < 0 for this solution is not true

2x , log 2 1 (2)

By taking union of (1) & (2)

2

1x , log 2 1 ,

2

6. [A]

213

log x x 1 1 0 here 2x x 1 0 x R

1 23log x x 1 1 0

231 log x x 1 1 k aa1

as log b log bk

23log x x 1 1

To hold true this inequality

2x x 1 3

3

1

1

y

x

2log 3


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2x x 2 0

x , 2 1,

7. [B]

2x x 2 x 0

case 1: x 2

2x x 2 x 0

2 2x 2 0 x 2

x , 2 2,

but x 2 so we get

x 2, 2 2, (1)

case 2: x


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9. [B]

22

3 3log x log x 10

2

1x

x

Take 3log x a ;ax 3

2

3 32 log x log x 10 2x 1

22a a 8x 1

22a a 8a

3 1

a a 4 a 23 1

So we get a (a + 4) (a2) = 0

a = 0,4, + 2 = 3log x

so, 4 2x 1, 3 , 3

1

x 1, ,981

10. [B]

a ax log bc x 1 1 log bc

a alog a log bc

ax 1 log abc

Similarly by log ca by 1 log abc

Similarly cz 1 log abc

So,a b c

1 1 1 1 1 1

x 1 y 1 z 1 log abc log abc log abc

abc abc abclog a log b log c


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abclog abc

= 1

11. [D]

x x3 1 3 9

take square both side

2 2

x x3 1 3 9 0

x x x x3 1 3 9 3 1 3 9 0

x2 3 10 8 0

x x3 5 0 3 5

3x log 5

12. [B]

x x25 5 4 16 (1)

Here x25 5 0 & x4 16 0

x5 25 & x 24 4

x 25 5 & x 2

x , 2 & x 2,

So only feasible region for given equation is x = 2

For x = 2, given equation is satisfied.

So no of solutions = 1

13. [D]

4x 1 7 x 0

As square root is always positive so given equation is feasible only if,


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4x + 1 = 0 & 7x = 0

1

x4

& x = 7

So no common solution.

14. [C]

2 2

x 2 x 24 9 2 8 0

2 22x 4 2 x 82 9 2 2 0

2 22x x16 2 36 2 8 0

2 2

2x x4 2 9 2 2 0

put2

x2 a

so,24a 9a 2 0

(4a1) (a2) = 0

1

a , 24

2

x 12 , 24

2x 2,1

2x 2 is not possible ; 2x 1

x 1

15. [A]

3

2 44 3 1x 1 2x 1 x

2

; x 0,1

1

x 1 2x 1 x2

(as 1 + 2x > 0 for x 0,1 )


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1

x 1 2x 1 x2

5

2

16. [A]

x 1

x 1

2 12

2 1

We can cross multiply x 12 1 as x 12 1 0 for x R

x 1 x 12 1 2 2 1

x

x21 4 2 4

2

x7

2 52

x 10

27

This is true for x R

17. [C]

2x x2 7 2 1 ..(1)

Here2x2 7 0

2x2 7

22x log 7

21x log 72

(2) (feasible region)

form feasible region it is clear thatx2 1 0

so by taking square of (1)


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2

2x x2 7 2 1

2x 2x x2 7 2 1 2 2

x

2 2 8

x2 4

x < 2 ..(2)

by taking intersection of (1) & (2)

we get 21

x log 7, 22

4x log 7, 2

18. [A]

3x 1 1 x

case 1: 3x 1 0

x 1

so

3x 1 1 x

3x x 2 0

2x 1 x x 2 0

x 1, .(1)

case 2:3x 1 0

3

x 1 x 1

So, 3x 1 1 x

3x x 0

3x x 0

101


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x x 1 x 1 0

x , 1 0, 1 ..(2)

So take union of (1) & (2)

x , 1 0,

19. [B]

x 2 x

2x

case 1 : x 2 0

x 2 ..(i)

x 2 x

2x

2

2 0x

1 x

0x

x 1 0x

x , 0 1,

Taking intersection of (i) & (ii)

x 2, 0 1, ..(iii)

case 2:

x + 2 < 0 x , 2 (A)

x 2 x

2x

0 1


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2x 2 2x

0x

2x 1

0x

1

x , 0,2

..(B)

Taking intersection of A & B

x , 2 ...(C)

So by taking union of (iii) & (c)

x , 0 1,

20. [A]

2

2

x 5x 41

x 4

2

2

x 5x 41 1

x 4

case 1:

2

2

x 5x 4

1x 4

2 2

2

x 5x 4 x 40

x 4

x 2x 50

x 2 x 2

5

x , 2 0, 2 ,2

(1)

case 2:2

2

x 5x 41

x 4

2 2 4

2

x 5x 4 x0

x 4

01

2

5

2

02 2


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8 5x0

x 2 x 2

5x 80

x 2 x 2

8

x 2, 2,5

.(2)

By taking intersection of (1) & (2)

8 5

x 0, ,5 2

21. [C]

2x x 12

2xx 3

case 1 : x 0

2x x 12

2xx 3

2x x 12

2x 0

x 3

2 2x x 12 2x 6x

0x 3

2x 5x 12

0x 3

2

x 5x 120

x 3

as2x 5x 12 is always greater than zero for x R

so,1

0x 3

x ,3

85

2 2


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case 2 : x < 0

2x x 12

2x 0x 3

2 2

x x 12 2x 6x 0x 3

2x 7x 12

0x 3

2

x 7x 120

x 3

x 4 x 30

x 3

x 4 0

x , 4

But for this case x < 0

So we get

x , 0

take union of (1) & (2)

x ,3

22. [C]

2

2

x 3x 13

x x 1

3

2

x 3x 13 3x x 1

Case 1 :2

2

x 3x 13

x x 1


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as 2x x 1 is always greater than zero

so 2 2x 3x 1 3 x x 1

2

4x 2 0

x R .(1)

Case 2 :2

2

x 3x 13

x x 1

2 2x 3x 1 3x 3x 3

22x 6x 4 0

2x 3x 2 0

(x + 1) (x + 2) > 0

x , 2 1, .(2)

Take intersection of (1) & (2)

x , 2 1,

23. [D]

x 3 x

1x 2

x 3 x

1 0x 2

x 3 x x 2

0x 1

x 3 2 0x 2

Case 1 : x 3 x 3

x 3 2

0x 2

2 1


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x 1

0x 2

x , 2 1,

but for this case x 3

so we get x 3, 2 1,

case 2 : x + 3 < 0

x


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So, x , 1 2, .(1)

2

x 2 21

x 3

2

x 2 41

x 9

2

5 x 20

9 x

2

2

5x 9x 180

9x

25x 6 x 3

0x

06

5 3

So, 6

x , 0 0,35

(2)

By taking (1) (2)

6

x , 1 2,35

25.

4 1 x 2 x 0

Here 1 x 0

2x > 0

x 1 .(i)

4 1 x 0

4 1 x


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16 1 x

x 15 ..(ii)

So, (i) (ii)

x 15,1 ..(iii) feasible region

Take intersection of (iv) & (iii)

x 2, 1 (1)

Now

4 1 x 2 x

Take square both side

4 1 x 2 x

2 x 1 x

here 2 x 0 x 2 (iv)

so, 2 x 1 x

take square2

4 x 4x 1 x

2x 5x 3 0

5 13 5 13

x x2 2

5 13 13 5

x , ,2 2

(2)

(1) (2)

13 5

x , 12

26. [A]


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2 x

4 x 0x

Here2

4 x 0

2x 4

x 2, 2 .(i)

Case 1 :x > 0

2 x

4 x 0x

x 0

24 x 1 0

24 x 1 true for x R

Case 2 : x < 0

2 x4 x 0

x

24 x 1 0

24 x 1

Take square 24 x 1 2x 3

x 3, 3

but x < 0

so x 3, 0

so case (1) case (2)

x R .(ii)

But our feasible region is x 2, 2

So greatest integral x = 2


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27. [C]

3

x 2 x0

4 x

here

34 x 0

3 3x 4 x , 4 (1)

as3

4 x is greater than zero

x 2 x 0

x 2 x

2 2(x 2) x

x 2 x x 2 x 0

2 2x 2 0

x 1 .(2)

So (1) (2)

3x 1, 4

28. [A]

4x 1

log 0x 2

x 1

1x 2

x 1 x 2

0x 1

1

0x 2

1

0x 2


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x , 2

29. [A]

3x ax b 0

0

a

b

3 3 3 2 2 3

2 32

2 33

3 3b

22 2 2 2

3 3 3

2 2 2

3b 3b

2a 2a

30. [A]

3 2f x x 2x k

if (x1) is divisor of f (x)

then f (1) = 0

f (1) = 1 + 2 + k = 0

k =3

31. [D]

3 24x 20x 23x 6 0

Lets take root are , ,


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20

2 54

..(1)

2 23

4

2 6

4

2 6

5 24

2 3 6

5 2 04

3 2

8 20 6 0

Which gives1

2

So1

2 52

6

So roots are1 1

6, ,2 2

32. [D]

3 22x 5x 3x 1 0

1

2

5

2

5

1 1 1 25

1

2

33. [A]

3 2x 4x x 6 0


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Roots are 2 , 3 ,

So, 2 3 4

5 4 ..(1)

2 3 6

2 1 (2)

by (1) & (2)

2 4 5 1

3 25 4 1 0

By this we get 1

So roots are 2, 3,1

34. [B]

3 24x 12x 11x k 0

sum of roots 12

a d a a d 34

a = 1

as a is root of equation so it will satisfy it.

3 24 a 12a 11a k 0

412 + 11 + k = 0

k3

35. [B]

3 2x 11x 37x 35 0

If root is 3 2 , second root has to be 3 2

Lets take third root


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So, 3 2 3 2 11

5

36. [B]

As x + 1 is factor so x =1 will satisfy the expression.

4 3 2

1 p 3 1 3p 5 1 2p q 1 12 0

p = 2

37. [D]

3x ix 1 i 0

roots will be 1 + i, 1i,

1 + i + 1i + =0

2

So equation with roots (1i) &2 is

2x 1 i 2 x 1 i 2 0

2x 1 i x 2 1 i 0

38.

4 3 22x 9x 8x 9x 2 0

39. [A]

3x 1 0

roots will be , ,

0

0

1

1 3 3


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1 1 1 1 1 1 1 1

1 1 1

= 3

1 1 1 1 1

1 2

So equation with 1 , 1 , 1 as roots

3 2x 3x 3x 2 0

40. [B]

3x 3x 2 0

a + b + c = 0

ab + bc + ca = 3

abc =2

3 3 3 2 2 2a b c 3abc a b c a b c ab bc ca = 0

3 3 3a b c 3 2 =6

41. [A]

3x 1 0

0

0

1

2 2 2 2 2

2 2 2 0

2

2 2 2 4 4 4 2 2 2 2 2 22


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24 4 4

0 2 2

4 4 4 2 0 2 1

4 4 40 2 0 2 0

4 4 4 0

42. [A]

3 2x 3x 2x 4 0

3

2

4

3 2x 2 2 2 x 2 x 2 2 2 0

3 2x 6x 8x 32 0

43. [A]

As , , satisfy cubic equation 3 2x 5x 5x 3 0

So, , , are roots of this equation

5

44. [C]

1

xx 2 15

25

x 2 2x

15

5

2

x 2 x5 5 1

2

x 20x5 5

1


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So2

x 2 0x

2x 2x 2

0x

Numerator is always > 0

So1

0x

x 0,

45.

2x 5x 60x 7

x 3 x 2

0x 7

As 7 x 7

So we can cross multiply 7 x

(x3) (x2) < 0

x 2,3

46. [D]

2x 1

2x 1

2x 1

2

x 1

or2x 1

2

x 1

2x 1 2x 2

0x 1

or

2x 1 2x 20

x 1

4x 3

0x 1

or

10

x 1

32


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3

x ,14

or x > 1

So, 3

x , 14

47. [A]

1 1

x 3 2

Case 1 : x 0

1 1

0x 3 2

5 x0

2 x 3

x 50

x 3

x ,3 5,

but x 0

so, x 0,3 5, ..(1)

Case 2 : x < 0

1 1

0x 3 2

1 1

0x 3 2

5 x

0x 3

x , 5 3,


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but x < 0

so, x , 5 3, 0 ..(2)

so (1) (2)

x , 5 3,

so least positive integer value = 1

48. [C]

2x 1 x x 2 0

2x x 2 0 & x 1 0

x 2 x 1 0 & x 1

x , 1 2, & x 1,

So, x 2,

49. [B]

2 2x 1 x x 2 0

2x 1 0 & 2x x 2 0

x , 1 1, ..(1) & x 2 x 1 0

x , 1 2, .(2)

Take (1) (2)

So,

x , 1 2,

Least positive integer = 2

50. [C]

x x 149 7 98 0


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2x x7 7.7 98 0

x7 a ; 2a 7a 78 0

(a + 14) (a7) < 0 xa 14 7 14 0 for x R

x7 7

So, x < 1

advanced inequations exercise - 1(b)

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