Advanced Inequations Exercise - 1(a)

8/10/2019 Advanced Inequations Exercise - 1(a)

1/22


2/22

3

log9log 9

log3

= 2

5. [D]

95 71 4

log 36log 3 log 981 27 3

23 3log 36log 5 3 4981 3 3 log 7

3 33

l 13 log 36 4 log 7log 5

4 2 23 3 3

3

4 2233 3log 36log 5 log 73 3 3

3

4 225 36 7

625 36 6 49 890

6. [C]

k 5 xlog x log k log 5 ; given k 1 , k > 0

log x log k log 5log k log 5 log x

5 xlog x log 5

x = 5 is the only possible solution

7. [A]

5 alog a log x 2

log a log x

2log5 log a

5log x 2

2x 5


3/22

= 125

8. [C]

2 2 4 2A log log log 256 2log 2

12

4

2 2 42

log log log 4 log 2

2 2 21

log log 4 2 log 21

2

2log 2 4 1 4

5

9. [D]

10log x y (given)

32

1000 10log x 2 log x

101

2 log x3

bk aa1

by formula log log bk

2 y3

11. [A]

12

log sin x 0 ; x 0, 4

as base1

2lies between 0 to 1 to satisfy given inequality, 0 < sin x < 1

x 0, 2 ,3

As we can see in this interval

we get3 9 11

, , ,4 4 4 4

as integral

1y

2

4

34

2

9

4

114

3 4 x

y

0


4/22

multiples of4

12. [B]

2

1

2log x 6x 12 2

1 22log x 6x 12 2

221 log x 6x 12 2

22log x 6x 12 2

22 2log x 6x 12 log 4 0

2

2

x 6x 12log 0

4

2x 6x 12

0 14

case1

2x 6x 12

0 4

2x 6x 12 0

x R ..(1) as discriminant if quadratic expression 2x 6x 12 is less than zero.

Descriminant 2

D 6 4 12 1

D =12

case

2

2x 6x 12

44

2x 6x 12 4

2x 6x 8 0

1


5/22

x 4 x 2 0

x 2, 4 .(ii)

By taking intersection of (i) & (ii) we get

x 2, 4

13. [B]

2log (x 1)2 x 5 here x1 > 0 ; x > 1 (i)

1

2(2 )

log (x 1)

2 x 5

22 log x 12 x 5 bka1 b

or by formula log logk a

2

2log x 12 x 5

2(x 1) x 5

2x 1 2x x 5

2x 3x 4 0

(x4) (x + 1) > 0

So we get x , 1 4, (ii)

By taking intersection of (i) & (ii)

14. [C]

210log (x 2 x 2) 0

As base is greater than 1 so to hold the inequality true

20 x 2x 2 1

So, 2 20 x 2x 2 and x 2x 2 1

Case1

42

4

1

x 4,

1

1 3 1 3


6/22

2x 2x 2 0

x 1 3 x 1 3 0

So, x ,1 3 1 3, .(i)

Case2

2x 2x 2 1

2x 2x 3 0

x 3 x 1 0

So we get x 1, 3 .(ii)

By taking intersection of (i) & (ii) we get,

x 1,1 3 1 3, 3

15. [A]

0.2x 2

log 1x

As base of log is less than 1 so hold the inequality true

x 2

0.2x

x 2

0.2 0x

x 2 0.2x

0x

0.8x 2

0x

So, 5

x , 0,2

16. [C]

1 3

5

2

0


7/22

m

a amlog n log n ma a n

17. [C]

nmn ma a

Take log both side

nmn m

a alog a log a

nmn m

n 1m n

18.

341 2 2

5 3 93 3 1

x 16x x4

35 42 2 3 2

2 3 13 92 2 2x 4 x 4 x

5 4 4 33

23 3 9 22x 4 x 4 x

5 2 4 3

22 3 3 2x 4

17

364 x

19. [D]

mn 1 2n n

mm 1 2m

2 2 21

2 2

nm m 2n n

mn m 2m

2 21

2 2

1

n 1m n


8/22

nm m 2n n mn n 2m2 1

2n m2 1

2nm = 0

20. [D]

3 xx x 3x x x here x 0

1

13

x1

1x 3x x

4

3

4x

x 3x x

Take log both side

4

3x x

4x log x x log x

3

4

3 4

x x2

1

3 4

x x3

1

3 4

x3

3

4x

3

64

27

21. [C]

if1 1

3 3x 2 2

m = 2n


9/22

then

31 1 1 1

3 3 3 3 32x 6x 2 2 2 6 2 2

21 1 1 1

3 3 3 32 2 2 2 2 3

1 1 2 2

3 3 3 32 2 2 2 2 2 3

1 1 2 2

3 3 3 32 2 2 2 2 1

=

3 31 1

3 32 2 2

3 3 2 2

by a b a b a b ab

12 2 2

= 3

22. [A]

xx x

x x x (here x 0)

3

2

x3

x 2x x

3

2

3x

x 2x x

take log both sides

3

2x x

3x log x x log x

2

3

2 3

x x2

1

2 3

x2


10/22

9

x4

23. [C]

x 2x 1

5 5 0.2 26

x 2

x 1 15 5 265

x 1 1 x 25 5 26

x 1 3 x5 5 26

at x = 1, 3 above equation satisfy

24. [D]

1

x 2x

2

4 2

4 2

1 1x x 2

x x

22

1x 2 2

x

222 2 2

= 362 = 24

25. [C]

1 1 1b c abc ca ab

c a b

x x x

x x x

1 1 1 1 1 1

c b a c b ax x x

1 1 1 1 1 1

c b a c b ax

0x 1


11/22

26. [C]

m n mna a a

m n mna a

m + n = mn ..(1)

then m (n2) + n (m2) = ?

2mn2m2n

2 (m + n)2 (m + n)

27. [B]

log (x + 1) + log (x1) = log 3

log (x + 1) (x1) = log 3

2x 1 3

2x 4

x 2

But x + 1 > 0 & x1 > 0

x >1 & x > 1

so, x 1, is our feasible region.

Only x = 2 lies in the feasible region.

28. [D]

f (x) 5 x 3

f (x) 5 (x 3) ; x 3,

= 5 + (x3) ; x ,3

f (x) = 8x ; x 3,

= 2 + x ; x ,3


12/22

So, greatest value of function occur at x = 3

So f (3) = 83 = 5

29. [B]

m 3 2m 2n m 3 n n 1

m 1 n 3 m

2 3 5 6

6 10 15

m 3 2m 2 m n 3 n 1 n 1

m 1 m 1 n 3 n 3 m m

2 3 5 2 3

2 3 2 5 3 5

m 3 n 1 m 1 n 3 2m n n 1 m 1 m m n 3 n 3 m2 3 5

0 0 02 3 5

1

30. [A]

2 1

3 3x x 2

Take cube both side

1 2 12

2 3 3 32x x 3x x x x 8

2x x 3x 2 8

2x 7x 8 0

(x + 8) (x1) = 0

x = 1 ,8

31. [D]

if a + b + c = 0

2 2 2 3 3 3a b c a b c

bc ca ab abc


13/22

3 33 3a b 3ab a b c c 3ab c c

abc abc

3abc

3

abc

32. [C]

x x 12 2 4

x

x 22 4

2

x x2 2 2 8

x

2 8

x = 3

so,x 3

x 3 27

33. [B]

2 2

2 2

log x log y logx log y

log x log y log x log y

log x log y 2 log x log y

2 log x log y log x log y

1

34. [B]

210log 2x 7x 16 1

2 12x 7x 16 10

22x 7x 6 0

22x 4x 3x 6 0

(2x + 3) (x + 2) = 0


14/22

3

x , 22

35. [C]

a > 0, b > 0, c > 0

a b c 1

log a b c logabc

a b c

a b clog

abc

a 1 b 1 c 1log a b c

36. [B]

10 10 10log log log x 0

010 10log log x 10 1

110log x 10 10

0x 10

37. [C]

x 2 2x 4

25 125

x 2 2x 4

2 35 5

2x 4 6x 125 5

6 x 12 2 x 45 1

4x 8 o5 1 5

So, 4x8 = 0

x = 2


15/22

38. [C]

x 1 x 13

1 x x 6

here

x0

1 x

;

x0

x 1

so, x 0, 1 is the feasible region for the equation.

x 1 x 1361 x x

1 13

6x 1 x

Taking square both side

36

x 1 x169

2 36

x x 0169

9 4

x x 013 13

9 4

x ,13 13

Here values lies in the feasible region.

So,

39. [D]

3y 1 y 1 ..(1)

3y 1 0 & y 1 0

1

y & y 0

3

y 0, is our feasible region.

By eq.(1), taking square of both side,

3y + 1 = y1

0 1

9 4x ,

13 13


16/22

2y =2

y =1 ; which does not lie in feasible range of y.

so no solution of y.

40. [D]

x 7 4 3 ; x 4 4 3 3

22

x 2 2 2 3 3

2

x 2 3

x 2 3 (1)

1 1

x 2 3

1 1 2 3 2 3

4 3x 2 3 2 3

1

2 3x

(2)

1 1x 2 3x 2 3

2 3 2 3

= 4

41. [A]

3x 3x 9 0

Let the roots are , ,

So, 0

2 0 (1)

3


17/22

2 2 3 (2)

2 q

2q (3)

By equations (2), (1)

2 2 2 3

23 3

2 1

1 .(4)

2q

2 2

3q 2

q 2

42. [D]

16 4 2log x log x log x 14

2 2 21 1

log x log x log x 144 2

bk aa1

by log log bk

27

log x 144

2log x 8

8

x 2 256

43. [C]

1

4 3x2


18/22

Case 1 :4

4 3x 0 x3

(i)

So1

4 3x2

7

3x2

7

x6

..(ii)

By intersection of (i) & (ii)7 4

x ,6 3

.(A)

Case 2 : 43x < 0 4x 3 ..(iii)

So 1

4 3x2

1

4 3x2

9

3x2

3

x2

..(iv)

taking intersection of (iii) & (iv)

4 3

x ,3 2

(B)

So union of A & B is the solution of the given inequality7 3

x ,

6 2

(C)

44. [A]

2x x 6 0

Case 1 : x 0


19/22

2x x 6 0

x = 3,2

but x 0 so x = 3 is the only root.

Case 2:x < 0

2x x 6 0

2x x 6 0

x = 3,2

But x < 0 so x =3 is the solution.

So multiplication = 3 (3) =9

45. [C]

x 1

0x 2

So, x 2 or x 1

x 2 holds true for x R

Now, x 1

x , 1 1,

46. [A]

x 1 x 5 6

This is special case as (x1)(x + 5) =6

So the given expression will hold true if x 1 x 5 0

x 5,1

12

15


20/22

47. [D]

212

log x 1 0

As base of log is less than 1.

So, 212

log x 1 0

20 x 1 1

21 x 2

2 2x 1 and x 2

x , 1 1, & x 2, 2

x 2, 1 1, 2

48. [D]

2 x 2 y10

log x log 2 log y log 23

2 x10

log x log 2

3

22

1 10log x

log x 3

Lets take 2log x a

So,1 10

aa 3

2 10

a a 1 03

9 1

a a 03 3

9 1

a ,3 3


21/22

29 1 1

log x , 3,3 3 3

1

3 3x 2 , 2

1

3x 8, 2

similarly,

1

3y 8, 2

if x y then1

3x y 8 2

49. [A]

2

3

2

log x 3x 2 2

2

2

3 3

2 2

3log x 3x 2 log

2

2

3

2

x 3x 2log 0

3

4

base is less than 1 so

2x 3x 2

0 13

4

2 2 3

0 x 3x 2 & x 3x 24

2 5

x 2 x 1 0 & x 3x 04

2 5 1 5

x ,1 2, ......(1) & x x x 02 2 4

1


22/22

&5 1

x x 02 1

&1 5

x ,2 2

..(2)

taking intersection of (i) & (ii)

1 5

x ,1 2,2 2

50. [A]

2x x 2a a ;0 9 1

2x

2a x 1

a

2x x 2

a 1

2x x 2 0

(x 2)(x 1) 0

x 1, 2

(0, 1)

y

x

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Advanced Inequations Exercise - 1(a)

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