advanced inequations exercise - 2(a)

8/10/2019 Advanced Inequations Exercise - 2(a)

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2

3

2 2

3 3 3

2y 3

x

2x

y 3

49. [B]

2ax bx c 0 ; (a > 0)

f (1) = a + b + c > 0

50.

x 1 x 4 x 2 x 3

x 1 x 4 x 2 x 3 2 x 2 x 3 2 x 1 x 4

(x + 1) (x + 4) = (x + 2) (x + 3)

4 = 5 contradiction no solution.

51.

x R 1

2

2

x x 7y

x x 7

2x 1 y x 1 y 7 1 y 0

for x R; D 0

21 y 4 1 y 7 1 y 0

y 1 27y 29 0

1


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29y ,1

27

for y = 1, no value of x possible.

So y = 1 is not possible.

52. [C]

x y x y 1x y ; yx

x y12x y 1x

x

x yx y

2

3x = y

3x x 1 231

x3

x = 1 = y is also possible solution.

So total 2 numbers

53. [D]

2 4 4log x log y log 4 x

2 4log x log y 4 x

2 21

log x log y 4 x2

2x y 4 x ..( 1)

3 3x

log x y logy


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xx y

y

2yx

1 y ..(2)

By equation (2) if a < x < 4 then it is not possible for y > 1 element (1) is incorrect

By equation (1) & (2)

22 2y yy 4

1 y 1 y

2y 2,

3

for y = 2 x = 4 not possible as x > 0

for2 4

y x3 3

unique solution

54. [B]

x 1 22 y 4 ; x 12 y

2x y 4

2 2 ..(1)

x2 2y ..(2)

By (1) & (2)

2y 42y

2

2y 4 4y 0

2y 2 0

y = 2


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55. [A]

2 32cos2

2 3cos x4

3cosx

2 or

3cosx

2

5 7x , ,

6 6 6 6

56.

8 5 2x x x x 1 0

57.

2 2x y xy x y 1 0

2x y x 1 y 1 0

58. [D]

2 2 2

x a log 1 a x a 1 0

for domain 21 a 0 a 1,1 ..(1)

f (0) < 0 for opposite sign roots.

2a 1 < 0

2a 1 0

a , 1 1, ...(2)

By (1) & (2)

a

59.

3y

2

3y

2


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x y k ; k > 0

(x + y) xy = 0

60. [D]

2x 2mx m 0

1 2x x 2m

1 2x x m

3 3 2 21 2 1 2x x x x

3 21 2 1 2 1 2 1 2 1 2x x 3x x x x x x 2x x

3 28m 3m 2m 4m 2m

22m 4m 5m 1 0

2m [4m 1] (m 1) = 0

1m 0, ,1

4

1 5sum 0 14 4

61. [C]

4 3 2x 4x ax bx 1 0

All roots positive

4

14

1

A.M. = G. M.

It means all roots are equal.


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1

So, a 6, b 3

62.

2 2x 3x x 2 0

Case 1 : x 3,0

2 2x 3 x 2 0

5 = 0

No solution.

Case 2 :

x , 3 0,

2 2x 3x x 2 0

1x , 2

2

x = 2 not valid.

63. [A]

ax + by = 1 (1)

2 2 px qy 1 .(2)

Only one solution then line is tangent to the ellipse

So, 2 2 2 2c a m b

2 2

1 1 a 1 b p b q

2 2a b1

p q

64.


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x yy x4 32 .(1)

2 23log x y 1

2 2x y 3 ..(2)

By eq. (1)

x y 5y x 24 4

x y 5y x 2

x 12,y 2

for x = 2y ; 2 24y y 3

y 1

(2 , 1) , ( 2, 1)

1x y

2 not possible

Also, ( 2, 1) not possible , as not in domain.

65. [C]

a > 1,

2x a alog x log x

x a alog x 2 log x

x > 0 ; x + a > ( a > 1)Case 1 :

x 0,1

It is clear that alog x is more negative than

3log x

2log x

1


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a xlog x .

So given inequality not true in this case.

Case 2 :

x 1,

As it is clear by graph , given inequality is true for this case.

66.

f (4 + x) = f (4 x)

f (x) is symmetric about x = 4

x = 4 is the mid-point of x & x

8

So sum of all roots = 8 + 8 + 8 + 8 = 32

67. [C]

Given x < 1

3 x 2log (x 2) log 81 so, x + 2 > 0 ; x > 2 ..(1)

381

1log (x 2)

log (x 2)

33

4log (x 2)

log (x 2)

13

3

log (x 2) 40

log (x 2)

3 33

log (x 2) 2 log (x 2) 10

log (x 2)

(2)

2 x 1

0 < x + 2 < 1

x 4

02


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30 log (x 2)

3log (x 2) negative

So by (2) 3log x 2 2 0

2(x 2) 3

1x 2

9

17x

9

So,17

x , 19

68. [C]

44

1 1x 1 log x 3logx 2

Domain is x 3, 2 1,

Case 1 : x 3, 2

0 < x + 3 < 1

4x 1

log negativex 2

4log x 3 negative

But inequality not true for this case.

Case 2 : x 1,

4x 1

log negativex 1

4log x 3 positive


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So inequality is true.

69.

2 2x 2xy 3y 6x 6y f x, y

f 2x 2y 6 0

y ..(1)

f 2x 6y 6 0

y ..(2)

By (1) & (2) x = 6, y = 3

So, 2 2minf 6 2 6 3 3 3 6 6 6 3

= 27

advanced inequations exercise - 2(a)

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