Inequation & Equation Solutions Final

7/29/2019 Inequation & Equation Solutions Final

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Inequation & Equation Solutions

Ex.1(A)

1.1 5

21

xx

x x

5121

x x

x x

2(2x + 1) = 5 1 x x

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

16 x2 + 4 + 16x = 25x +x2

9x2 + 9x 4 = 0

(3x 1) (3x + 4) = 0

1

3x

only satisfies

Ans (b)

2. logx(x + 3)2

x = 16

log2

3 16xx

x

(x + 3)2 = 16

x2 + 6x 7 = 0

(x+ 7) (x 1) = 0

x = 7 , 1

According to definition of logarithm no value satisfies.

Ans : (d)

3. 3 910 10

log ( 1) log ( 1)x x

3 3

10 10

1log ( 1) log ( 1)

2x x

1/23 9

10 10

log ( 1) log ( 1)x x

(x 1) > (x 1)1/2


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(x 1)2 > (x 1)

(x 1)2 (x 1) > 0

(x 1) [x 2) > 0

Using wavy curve method,

x (, 1) u (2, )

Ans : (a)

4. log4((x 1) = log2((x 3)

1

2log2(x 1) log2(x 3) = 0

21

log 0

3

x

x

1

13

x

x

x 1 = (x 3)2

x2 7x + 10 = 0

x = 5, 2

According to the definition of logarithm x = 2 is rejected.

x = 5 is the only solution

Ans : (b)

5. f(n) = log2002n2

Now, N = f(11) + f(13) + f(14)

= log2002(112) + log2002(13)

2 + log2002(14)2

= log2002(112. 132. 142)

= 2 log 2002

Case 1 : x - 4

-(x + 4) (5 x) = 9

-9 = 9 (haha..)

no solution in this region - 1Case II

-4< x 5


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(x + 4) + (5 x) = 9

9 = 9 (always true)

all value satisfies

x > 5 (3) From 1, 2, 3 x [5, ] Ans : (b)

6. Given root are 1, i 5

We know that complex root always exists as a pair.

3rdroot is also present and its-i 5

(x 1) (x + i 5 ) (x - i 5 ) = 0

(x 1) (x2 + 5) = 0 [ i2 = -1)

x3 x2 + 5x 5 = 0

Ans : (d)

7. |x 1| + |5 2x| = |3x 6|

The options are such that, we can do by guess work.

Put x = 0 (simplest)

It stratifies [active method is given in Q.1]

Ans : (d)

8. |x + 4| - |5 x| = 9

Here, x = -4 and x = 5 are the critical point

9. x + y = 1 (1)

x3 + y3 = 19 (2)

squaring (1), x2 + y2 + 2xy = 1

x2 + y2 = 1 2xy ..(3)

For xy,

(2) (x + y)3 3xy (x + y) = 19

1 3xy = 19

xy = -6

(3) x2 + y2 = 13


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ans : (d)

10. 3x + 1 =62log6

3x + 1 =62log6 [ log

xba = log

abx ]

x + 1 = 62log

x = 62log 1

= 6 22 2log log

= 26

log2

= 32log

Ans (a)

11. log111

13

+ log111

14

+ . + log1

1242

= log112

3

+ log113

4

+ .. log11241

242

= log112 3 4 241

. . . ...........3 4 5 242

= log112

242

= log111

121

= log111(121)

= -2 log11

= -2

Ans (b)

12. a = log35 and b = log1725

a =5

1

log 3and b =

5

2

log 17

in a denominator is a faction and in b denominator is greater than 1

a> b

13. 2 log2(x2 + 3x) 0


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2 log2(x2 + 3x)

4 x2 + 3x

x2 + 3x 4 0

(x + 4) (x 1) 0

x [-4, 1] .(1)

According the validity of log function,

x2 + 3x > 0

x(x + 3) > 0

x (-, -3) U (0, ) (2)

From (1) & (2)

x [-4, 3] U [0, 1]

Ans (b)

14. for s1

x = 11log 7 ; y = 7log 11

xy = 1 x =1

y

y in 7 x in 11

=1

11yIn Inye

=

2 7 11y In In

ye

=

7log 11 log7 11In

y

e

=ln7 11ln

ye

= e0

= 1

Ans (T)

For S1 : Take any freetion from given method Ans : F

For S2 : by25(2 + tan2) = 0.5

2 + tan2 = (25)1/2


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tan2 = 3

tan2 = tan21

2

= n 2

15. y =1

2

2

| log |

log

x

x

= 2

2

| 1 log |

log

x

x

= 2

2

| log |

log

x

x

For 0 < x < 1, log2x < 0

y = -1 And for x > 1, log2x > 0

y = -1 for x = 1, its undefined

Ans : (c)

16. |x 1| 3 and |x 1| 1

-3 x 1 3 x 1 - 1 or x 1 1

-2 x 1 3 x 0 or x 2

-2 x 4 ..(4) x (-, 0) U [2, ] . from (1) & (2)

17. if log27 =p

q(a rational)

7 = 2p/q

Right hand side will need K odd

contradictory

Itss andirrational

Ans : (c)

18.

3 5

log2 2 log 24 4

2 ......(1)

x x

x

at log2x = t


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(1) 23 5

1/24 4 2t t

x

23 5

1/24 42log log 2

t t

x

23 5 1

4 4 2t t t

(3t2 + 4t 5) t = 2

3t2 + 4t2 5t 2 = 0

(t 1) (3t2 + 7t + 2) = 0

(t 1) (3t + 1) (t + 2) = 0

t = +1,3

4

, -2

For t = 1, log2x = 1 x = 2

For t =3

4

, log2x =

3

4

x =

1

3 2

Fro t = -2, log2x = -2 x =1

4

19. 2 log10x logx(0.01) = 2log10x logx(102

)

= 2 log10x + 2logx10

= 2[log10x + logx10]

4 [A.M G.M]

Ans : (d)

20. log0.4(x 5) < log0.16(x 5)

Log0.4(x 5) < log(0.4)(x 5)

x 5 > 5x

(x 5)2 > x 5

x2 11x + 30 > 0

x (-, 5) U (6, )

Ans : (d)

21. log16x + log4x + log2x = 7


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1

4log2x +

1

2log2x + log2x = 7

1

1/242log . . 7

x x x

77

4 2x

4

7 472 2 x

x = 16

22. log3(x + 2) (x + 4) + log1/3(x + 2)3

1log 7

2

3

( 2)( 4)log 0( 2)7

x x

x

( 4)

17

x x < 3 ..(1)

From definition of log, x + 2 > 0

x > - 2 ..(2)

23. conceptual

24. 7 7log log 7 7 7

= log7(log77)

= log71

= 0

25. Take x + y + z = loga(bc) + logb(ca) + logc(ab)

= not matching

- 1

Take (1 + x) 1 + (1 + y) 1 + (1 + 3) 1

= (1 + logabc) 1 + logbca) 1 + (1 + logcab) 1

= (logaa + logabc) 1 + (logbb + logbca)

1 + (logcc + logcab) 1

= logabca + logabcb + logabcc

= logabcabc

= 1


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26. 1 + abc = 1 +log12 log 24 log 36

log 24 log36 log 48

= 1 + log4812

= log4848 + log4812

1 + abc = log48(48 12) = log48(24)2

Now, option (c) =log 24 log 36

2log 36 log 48

= 2 log4824

= log48(24)2

27. x = log5(100) y = log72058x = 3 + log58 ..(1) y = 3 + log76 .(2)

from (1) & (2) it obvious

x > y

Ans (a)

28. 271/x + 121/x 2.81/x =

1/ 1/27 8

1 2. 0

12 12

x x

1/ 1/

9 21 2. 0

4 3

x x

21/ 1/3 2

1 2 02 3

x x

Put

1/3

2

x

t

2 32

1 0 2 0 t t tt

(t 1) (t2 + t + 2) = 0

t = 1

1 10

3 3 31

2 2 2

x x

10 . x no solution

x


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29.2

12

4 9

x

x

1

144x2 16x4 + 81 + 72x2

16x4 72x2 + 81 0

(4x2 9)2 0

Its always true.

option (A) is correct.

30. 11

| 1 .( 1)

xx

x xx x

( 1)

.( 1)

x

xx

0

2( 1)x

x 0

1

x 0

x > 0

x (0, ) U {- 1} Ans (b)

31.1

72 1 13 . 13 3

x

0

721

13 . 1

3

x

72 1

3 1

x

713 1 x

71 03 3 x

71 0 x

71x

2(71)x

0 x < (71)2


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32. 5x + 2

2 3x

- 169 > 0

5x + 12x > 132

We know that 52 + 122 = 132

for x > 2,

5x + 12x > 132

Ans (2, )

Ex.1(B)

1. 184x 3 = 3 4

54 2x

184x 3 = 3 4

2 9 3 2

x

184x 3 = 3 4

18. 18x

184x 3 =

3 43

218

x

4x 3 = 3

3 42

x

x = 6

2. Use rule of divisibility

3. x2 6x 15x 15|- 5 = 0

Case 1 Case 2

5x 15 0 5x 15 < 0

x 3 (i) x < 3 ... (i)

a. x2 6x (5x 15) 5 = 0 x2 6x + (5x 15) 15 = 0

x2 11x + 10 = 0 x2 x 20 = 0

(x 10) (x 1) = 0 (x 5) (x + 4) = 0

x = 10, 1 ..(ii) x = 5, -4 (ii)

From (1) & (2) from (i) & (ii)

x = 10 x = -4


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l = 1 & m = 1

4. Conceptual

5. 11

1log ( 0.5)

log ( 0.5)

x

x

xx

2

1log 0.5 1 x x

logx + 1(x 0.5) = 1

Logx + 1(x 0.5) = 1 logx + 1(x 0.5) = -1

1

12

x x 1 1

2 1

x

x

1

12

21

12 2

x

x x

1

12

23

02 2

x

x

1 5

2 22

x

3, 1

2

xx

But according to the definition of log

X + 1 > 0 and1

02

x

x > 1 and1

2

x

x = 1 is the only solution

6. log[11/5 + (32)1/5 + (243)1/5]

= log[1 + 2 + 3]

= log 6

option (c)

Take option (D) = 1

5(log 1 + log32 + log 243)


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=1

5log (32 243)

=1

5log(65) = log6

option (D) is also correct.

7. x3 2x + 3 = 0

L + + = 0 .. (1)

= - 2 .. (2)

= -3 .. (3)

At y =1

Y =1

{ + + = 0}

=1

is the root of x3 2x + 3 = 0

33

2 3 01 1

y y

y y

3

3

23 0

11

y y

yy

- y3 + 2y (y + 1)2 + 3(y + 1)3 = 0

-y3 + 2y3 + 4y2 + 2y + 3y3 + 3 + 9y2 + 9y = 0

4y3 + 13y2 + 11y + 3 = 0

Transformed equation is

4x3 + 13x2 + 11x + 3 = 0

It root are , ,1 1 1

13

1 4

Similarly check the other options.


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8. S1 :|x| + 1 = 0

Case 1 Case 2

Of x > 0 of x < 0

Then x2 + x + 1 = 0 x2 x + 1 = 0

D < 0 D < 0

no real solution.

For S2: x2 5|x| + 6 = 0

|x|2 5 |x| + 6 = 0

(|x| - 3) (|x| - 2) = 0

|x| = 3 or |x| = 2

x = 3 or x = 2

For S3: x2 - |x| - 2 = 0

|x|2 - |x| - 2 = 0

|x| = 2 or |x| = -1

x = 2 (Rejected).

9. 51 log 0.04 xx

51 log 25 xx

51 log 25 5log log 5 x

x

(1 - log5x) log5x = -2

Put log5x = t

(1 t) = -2

t t2 + 2 = 0 t2 t 2 = 0

(t 2) (t + 1) = 0

t = 2 or -1

Log5x = 5 log5x = -1

x = 25 x =1

5


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10. 102/x + 251/x = 1/17

504

x

1 1

1/17100 75 50

4

xx x

11100 75 17

50 50 4

x

x

1 11 172

2 4

x x

Put1

2 x t

t +1 17

4 t

t

4t2 17t + 4 = 0

(4t 1)(t 4) = 0

t =1

4or 4

122 2 x

122 2x

1 2 x

1 2x

x =1

2

x =

1

2

11. Case1 x2 + 4x + 3 0 (x + 3) (x + 1) 0

x (-, -3) U (-1, ) .. (1)

G.E (x2

+ 4x + 3) + 2x + 5 = 0

x2 + 6x + 8 = 0 (x + 4) (x + 2) = 0

x = -4, -2

x = -4

Case2 x2 + 4x + 3 < 0 x (-3, -1)

G.E -(x2 + 4x + 3) + 2x + 5 = 0

-x2 - 2x + 2 = 0

x2 + 2x 2 = 0


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x =2 4 8

2

= 1 3

But x 1 3 is rejected not bounding

x 1 3

12.1

2log0.1x 2

1

2log101x and 1

2

log 2x

1

2log10

1

x

and 101

log 2

x

1/21 10x

and1

100x

1

10x and

1

100x

13. log2(32x + 2 + 7) = log24 + log2(3

x 1 + 1)

2 2

1

3 7log 2 04 3 1

x

x

2 2

1

3 71

4 3 1

x

x

32x.9 + 7 =3

4 43

x

Put 3x = t

9t2 + 7 =4

43

t

27t2 + 21 = 4t + 12

27t2 4t + 9 = 0

D < 0

no soln.


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14. loga1a2 a1 = 4

1

14

log 1 2

aa a

loga1(a1 a2) =1

4

1 + loga1a2 =1

4

loga1a2 =3

4

Now,

1 2 1 2 1 2

1/3

1

1/2

2

1 1log log 1 log 2

3 2

a a a a a a

aa a

a

1 1 2

1 1 143 2 log

a a a

2 1

4 1 1.

3 2 1 log

a a

4 1 1

43 2 13

4 3

3 2

17

6

< 3

PASSAGE 1

(i), (ii), (iii) Very trivial just observation.

PASSAGE 2

N = 727

log 7log 9007 900 900 30 N

A = 2 2 2 2log 4 log 4 log 2 log 32 3 4 4

= 4 + 9 + 4 9

A = 8


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D = (log549) (log7125)

=2 log 7 3log 5

log5 log 7

D = 6

(i) logAD = a

Log86 = a 1

3log26 = a log26 = 3a

Now, log612 = log6(6 2) = 1 + log62

= 1 +1

3a=

3 1

3

aa

(ii) Here, m = n = 3, N = 30

Take option (A) log303 < log330 = log330

log303 < log330

Its true

(iii)530

810

log 30 8 6 3 3 log 2

= log550 log52

PASSAGE 3

y = 4 14x2 - 91

(i) B1co3 of mod max.valur of y = 4 for |4x2 9| = 0 & there is no limit for any bigger we can subject

from 4.

y (-, 4]

(ii) -3 = 4 |4x2 9|

|4x2 9| = 7

4x2 9 = 7

4x2 = 9 7

4x2

= 16 4x2

= 2

x = 2 x =1

2


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(iii) y = |Z| + 4 |Z| = y 4

Z = (y 4)

Z = y 4 Z = 4 y

Z = -|4x2 9| Z = |4x2 9|

Z takes all-ve values. Z akes all the value

Infinite (x, Z)

But for every x there are two values of Z

Two ordered points.

PASSAGE 4

(i) 4 x > 0 .. (1) x< 4

x> 0 . (2)

x + 1 . (3)

2 + x > 0 .. (4) x> - 2

For answer (1) (2) (3) (4)

x (0, 4) {,}

(ii) 03

x

x.. (1)

log 2 03

x

x. (2) 1

3

x

x

Froans (1) (2) 3 03 x

x (-, -3)

(iii)4

01

x

x (1)

x 1 0 . (2) x 1

(1) (2) x [1, )

-3 0

-4 -1


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Assertion & Reasons

(1) Conceptual

(2) log3(x2 + 1) + |log4y

2| = 0

log3(x2 + 1) + |log24| = 0

Both terms are positive on L + s.

x2 + 1 and log24 has to be O.

But x2 + 1 > O

students (1) False

(3) 10 110

log 13 12 log 14 13

10 10 1log 13 12 log 14 13

10 10log 13 12 log 14 13

13 12 14 13

Its true

statement 1 is true

And statement 2 is true theory

option (A)

Matrix-Match Type

(1) (A)

21 1 2

05

x x x

x

x [-1, 2] (5, )

(B)

| | 50

2

x x

x

5

02

x

x[ |x| 0]

-1 1 2 5

2 5


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(C) at Q = log1010 = 1

Q = log1099 < 2 Q [1, 2)

P = log109999 < 4 p [3, 4)

P Q [1, 2) [3, 4)

(D)

21 1 20

5

x x x

x

x (-, -1] [1, 5)

(2) (A) 2 23 7 7log 5 8log (5 4log 7 | | k

log39 = |k|

|k| = 2 k = 2

(B) 2

0.5 1

2

log 4 log 4 | 2 | 2

(C) logx(x2 1) = 0

x> 0 .. (1)

x 1 . (2)

x2 1 > 0 (3)

and x2 1 = x0

x2 = 2

x = 2 . (4)

For Ans (1) (2) (3) (4)

x = 2

(D) 9 77 x

21 2

11

x

x

=2

1 42 4

11 x

x

-1 1 2 5


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=1 4

9 77 411 9 77

= 4 9 771

9 77 411 4

= 1

9 77 9 77 411

= 2

EXRECISE -1 (c)

Subjective solutions

1.2

2

X 5x 4

x 4

= 1

2

2

X 5x 4

x 4

= 1

2

2

X 5x 4

x 4

= 12

2

X 5x 4

x 4

= -1

X2- 5x + 4 = x2- 4 X2 5 x + 4 = - x2 + 4

5x = 8 2x2- 5 x = 0

X= 85

x (2x 5) = 0

x = 0, 52

2. 2x 3 = x 3 - (I)

Validity conditions

2x -3 0 x 32

- (1)

x 3 ,2

.

x + 3 0 x -3 - (2)

verify (I), we set 2x 3 = x + 3

x = 6


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X = 6 E3

,2

Ans. x = 6

3. 26 4x x = x + 4

6 4x x2 = (x + 4)

6 4x x2 = x2 + 16 + 8x

2x2 + 12 x + 10 = 0

x2+ 6x + 5 = 0

(x + 5) (x + 1) = 0

x = -5, -1

Validity conditions (i) 6 - 4x x2 0

x2 + 4x -6 0

X 2 10 X 2 10 0

X E 2 10 , 2 10 - (1)

(ii) x + 4 0 x -4 - (2)From (1) & (2)

XE 4, 2 10

X = -1 only E 4, 2 10

Ans. x = 1

4. |log x log x2

| = -x

LHS is positive

RHS has to be positive

x< 0

But x cannot be negative in log x

No values satisfies

5. 4 log 2 7 + log23. Log 3

5. Log 52 + 6

2 log 2(74) + log3 log 5 log 2 + 6


24/67

= 74 + 1 + 6

= 2408

6. x2 - |5x + 8| > 0

Case 1 Case 2

% 5x + 8 0 % 5x + 8 < 0 x 0

Now < 0

X2 (5x + 8) > 0

X2 5x - 8 > 0 No real root

5 57 5 57

x x2 2

> 0 XER

x5 57 5 57

, u ,

2 2

- (ii)

From (i) & (ii)

Ans : x5 57 5 57

, u ,2 2

(7)2

|x 3|2. 0

x 5x 6

Validity x 3, 2

2

|x 3|2. 0

x 5x 6

|x 3|

2 0x 3 x 2

Case 1 Case 2


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x-3 0 x-3 < 0 x < 3 (i)

x 3 (i) G.1

2 0x 2

x 32 0

x 3 x 2

12 o

x 2

12 o

x 2

2x 30

x 2

1 2x 40

x 2

3x ,2

2

-(ii)

2x 50

x 2

From (i) & (ii)

3x ,2

2

From (i) & (ii)

No sol.

For final answer :

(Case 1 ) u (Case 2) 3

x ,2

2

8. |x-2| |x+4|

|x-2|2 |x+4|2

x +1 0

x -1

9. |2x-4| < x-1

Case 1 Case 2

2x-4 0 2x-4 < 0 x 0

x< 3 _ (ii) x > 53

- (ii)

From (i) & (ii) From (i) & (ii)


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X 2,3 -(1) X (5, 3) (2)

From (1) & (2) & (Case1) U (Case 2) (7)X 5

,33

10. 2|x+1| > x+4

Case 1 Case 2

X+1 0 x +1 < 0 x x+4

G.E 2 (x+1) > x+4 3x 0 x< - 2 (ii)

x> 2 (ii) (i) (ii) gives

(i) (ii) gives

x 2, -(1) X , 2 -(2)

(Case1) U (Case 2)

x , 2 U (Case 2)

11. |x + 2| - |x 1| < x -3

2

Case I : of x - 2 -(i)

Then G.E -(x + 2) + (x-1) < x -3

2

x >3

2 - (ii)

(i) (ii) x _(1)

Case (1) : -2 < x < 1 -(i)

G.E. x + 2 + (x-1) < x -3

2


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x 9

2(ii)

(i) (ii) x9

,2

-(3)

For final answer ; (Case I) U (Case II) U ( Case III)

x9

,2

12.| 2|

02

x

x

|x-2| is always positive

G.E.1

02x

x> 2

13.2

14x

2

14x

2 > |x-4|

Case 1 Case 2

x-4 o x-4 < 0 x -2

x< 6 (ii) x > 2 _(ii)


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(i) (ii) (i) (ii)

X (4, 6) -(1) X (2, 4) -(2)

(Case1) U (Case 2)

(2, 4) U (4, 6)

From validity x 4

Ans : X (2, 4) U (4, 6)

14.2 1

21

x

x

2 1 21

x

x

or 2 1 21

x

x

2 1

2 01

x

x

2 12 0

1

x

x

2 1 2 2

01

x x

x

10

1x

4 3

0

1

x

x

x> 1 (2)

(1) U (2)

X3

,14

U (1, )

15.1 1

3 2x

2 1 03x

Case 1 Case 2

X 0 (i) x < 0 (i)

G.E 23x

-1 < 0 G.E 23x

-1 < 0


29/67

5

03

x

x

21 0

3x

5

03

x

x

50

3

x

x

x (- , 3) U (5, ) -(ii) x (- , 5) U (-3, ) -(ii)

(i) (ii) (i) (ii)

x (0 , 3) U (5, ) -(1) x (- , -5) U (-3, 0) -(2)

(1) U (2)

x (- , -5) U (-3, 3) U (5, )

Exercise -2 A

1. x3 + 2x2 3 x -1 =0

Transforming x =1

ywe get.

Y3 + 3y2 2y -1 = 0 (1)

1

,1

,1

y

are the roots of the above eqn.

1

+

1

+

1

y= -3,

12

,

11

y

2 2221 1 1 1 1 1 1

2y y

= 9 + 4 = 13

1

,

1

,

1

ysatisfy eqn. (1)

3 2

1 3 21 0

3 2

1 1 13 2 1 0

2

31 3(13) 2( 3) 3 0

3

142


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4 3 2

1 1 1 13 2 0

S + 3 (-42) 2 (13) (-3) = 0

2. Let the root be diminished by h

Let y = -h

= y + h

2 (y + h)3= 9 (y + h)2 -6 = 0

Given that the second term is missing

2 (3h) 9 = 0

H =

3

2

3. Let y = y

Y = y -

Y = p - . = p y

The transformed eqn. is

(p y)3 p (p y)2 q (p y) r = 0

q3 3p2y + 3py2 y3 q3 + 2p2y py2 + pq - qy r = 0

Y3 2py2 + (p2 + q) y + (r pq) = 0

4. Let y =11 3 1y

y

4

y

Transformed eqn

3 2

64 32 203 0

y y y

3y3 20y2 + 32y 64 = 0

5. f(x ) = x3 3x2 + 4

f(x) has a repeated root, f (x) and f1(x) have a common root

f1(x) = 3x2 6x = 0 x = 0, 2


31/67

Clearly x = 2 satisfies

6. x = 3 2

X2 = 3 + 2 - 2 6

(x2 5) = - 2 6

(x2 5)2 = 24

X4 10 x2 + 1 = 0

7. 1 1 4 1x x x

Squaring on both sides we get.

(x+1) + (x-1) -

2 21 4 1x x

1-2x = 2 2 1x

1-4x + 4x2 = 4x2 4

8.2 9

3log 2log1x x

x

Clearly, x >0 ; x 1

2 9

3log 2log 7x

x

Put log 3x = t

2t -4

t=7

2t2 7t -4 = 0

2t2 8t + t -4 = 0

t =1

2

, 4

Two solutions

9.3

2 22

xx

Observe that x < 2

3 - 2

2 2 2x x


32/67

2

2 2 2 3 0x x

2 1x

2-x > 1

X < 1

10. Let y = r1 + 2

R1 = y 2

(y-2) 2 (y-2)2 + 4 (y-2) + 5074 = 0

The above eqn. has r1 + 2, r2 + 2 and r3 + 2 as the roots.

(r1 + 2) (r2 + 2) (r3 + 2) = product of roots = -5050

11. 2log 2x = log (x2+ 75),

Clearly x > 0 and 4x2 = x2 + 75 x2 = 25

x = 5 only (x = -5) rejected)

12.

28 15

2

x x

x

|x-| =1

Notice that x 2.

Now obvious solution is |x-3| =1

x = 4, (x = 2 rejected)

And also2

8 152

x xx

= 0

X = 5, 3

But x = 3 rejected Base becomes

So, final answer is x = 4, 5

13. log2x (x2- 5x + 6) > 1

x is prime x > 1


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x2 5x + 6 > 2x

X2 -7x + 6 > 0

(x-1) (x-6) > 0

x (- , 1) (6, )

14. Gives y2

7 12x xy

= 1 and x + y = 6

(6 - x) (x + 4) (x + 3) = 1

This is true when x = 5 ( Base is 1)

X = -3, -4 (Exponent gero)

X = 7 ( -ve power ever)

But x > 0 x = 5 ; x = 7

15. xy + 3y2 x + 4y -7 = 0 -(1)

2 x y + y2 2x 2y + 1 = 0 -(2)

(2) 2x (y 1) + (y 1)2 = 0

( y 1) (2x + y -1) = 0

When y = 1 eqn. 91) is always true XER

Options b, c, d

When 2x + y -1 = 0 ;eqn (1) x = 2 ; y = -3 , (a)

16. 2| 2| 3 9x y

Now L.H.S. 3

` and R.H.S. 3

Equality exists only when L.H.S. L. H. S. = R. H. S. = 3

2| 2| 3 9x y = 3

x = -2 ; y = 0

17. log10 (-x) = 10log | |x

Clearly x < 0

Given eqn log 10 (-x) = 10log ( )x


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t = t

t = 0 or 1

log10 (-x) = 0 or 1

X = -1 or -10

18. Given logx2 . log2x

2 = log 4x2

base> 0, 1 x > 0 ; x 1,1 1

,2 4

Given eqn2 4

2 2 2

1 1

log .log logx x x

22 2

1 1

2 loglog . 1 logxx x

Put log2x = t

t = 2

log2x = 2

X = 2 22 ,2

19.2

3 3

1 1

log (2 1) log (2 1)x

x

Observe that x > 0. And R.H.S. is always +ve

, Both L.H.S and R.H.S. +ve.

log3 (22x-1) < log3 (2

x+1)

22x 1 < 2x + 1

(2x + 1) (2x 2) < 0

2x (-1, 2)

x (- , 1) -(1)

But for both to be +ve.


35/67

Log3(22x -1) > 0 22x> 2 x >

1

2

20.2

2 1 30

log log logx ax a x

a a a

Put log ax = t

2 1 20

1 2t t t

2 (1 + 1) (2 + t) + (2 + t) + 2t (1 + t) = 0

2t2 + 6t + 4 + 2t + t2 + 2t + 2t2 = 0

5t2 + 9t + 4 = 0

(5t + 4) (t + 1) = 0

t = -1;4

5

Two solutions

21.135 5

3 3

3 3

15 405

log

log log

log

= (log3135). (log3

15) (log35) (log 3

405)

= (3+ log35) (1 + log35) (log35) (4+ log35)

= 3

22. Let N = 12300

Log 10 N = 300 (log 1012)

= 300 (2 log 102 + log 10

3)

= 300 (0.602 + 0.4771)

Log10N = 323.73

N has 324 digits

23.( 3)

log ( 2 4 3)1

110

x

x x

log(x-3)(x2-4x +3) 0

1 + log (x+3)(x-1) 0

log(x-3)(x-1) -1


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Given x is integer

x-11

3xwhich is not possible for on

24. log102 = 0.3010 log210 =

1

0.3010

1+ log25=

1

0.3010 log2

5 =1

10.301

=699

301

Log 52 =

301

699

Log 564 = 6 log 52= 6 x 301 602699 233

25. log2 x + 3 (6x2 + 23 x + 21 ) = 4 log 3x-7 (4x

2 + 12 x + 9)

1 + log 2x+3 (3x + 7) = 4 2 log 3x + 7 (2x + 3)

Put t = log 2x+3 (3x + 7)

1 + t = 4 -2

t

Solving t = 1; t = 2

When t = 1 log 2x+3 3x + 7 = 1 x = -4

t = 2 log 2x+3 3x + 7 = 2 (3x + 7) = (2x + 3)2

x = -2,1

1

26. 2 log10 3 3 = 3 k log 102

2 log3

23

10= 3 k log 10

2

2 log3

23

2 . Log. 102 = 3 k log 10

2

2

23 102 10

logloglog 3

22 3k

2 log3

23

2. = 3 k

3

23 = 3 k


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k = 32

27. 2( 1)

2log

5x

x

( 1)

2log

2 5x

x

(x 1)2> x + 5 x2 3 x 4 > 0

(x-4) (x + 1) > 0 x > 4 or x < -1 (reject)

28. x = log 10005

= 3 log 510 = 3 + 3 log 5

2

= 3 + 3 log 58

Y = log 7 2058 = log 7 (73 . 6) = 3 + log 7

6

Clearly log 58> 1 x > 4

And log 76 < 1 y < 4

x > y

29. Given eqn. 2 log xa + log ax

a + 3 log a2xa = 0

2 1 3

0log 1 log 2 log

x x x

a a a

Put log ax = t

2 1 3

01 2t t t

2 (2 + 3t + t2) + 2t + t2 + 3t + 3t2 = 0

6t2 + 11t + 4 = 0

30. Let 1x t x = t2 + 1 (Remember t 0)

2 21 3 4 1 8 6 1t t t t

|t-2| + |t-3| = 1

2 t 3

2 1 3x

5 x 10


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31. The given equation can be written as

2x+1 21

4x

= 2 |x-3| 24 1x

= 4. 2 |x-3| 21

4x

2x-1 = 2 |x-3|( XEI 2 14

x

is cancelled.)

X-1 = (x-3) X=2 No ve integral value.

32. Clearly, x, y are roots of the eqn.

Log2t + log t

2 =10

3

Log2t= 3 ;

1

3

t = 8 ;1

32

x = 8, y =1

32

x + y = 8 +1

32

33. The required a satisfies the inequality

2a2 2 (2a + 1) a + a (a + 1) < 0

a2 + a > 0

a (- , -1) (0, )

34. When x 0 2x + 2x 2 2

2x 2 2 x 1

2

When x < 0 2x + 2-x 2 2

2

22 2 2 1 0x x

2 2 1 2 2 1 0x x


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2 2 1 2 2 1x xor

- < x log 2 2 1 only 0x

x 2 1,log 2 1 ,2

35. Let y =1

1

By C & D 1

1

y

y

=1

Or =1

1

y

y

Satisfies the eqn. x3 x- 1 = 0

3- -1 = 0

3

1 11 0

1 1

y y

y y

(y 1)3 (y 1) (y + 1)2 (y + 1)3 = 0

y3

+ 7y2

y + 1 = 0

36. Given , , are the roots of

X3 + 2x2 + 3x + 2 = 0.

2; 3; 2

Let y = 3

= 23 3

Y =2

3

= ?

37. 1 4 2 3x x x x

Squaring on both sides we get

2 21 4 1 4 2 3 2 3x x x x x x x x

Squaring again 4 = 6 No solution.


40/67

38. xx-y = y x+y and 1x y

2

1x

y

22

11

2

1y

yy

yy

y

2y 2 2

2 1,y

y y

y

y =1 or 2y -2 2

2 1y

y y y3 =3

39. log2x = log4

y + log4(4-x)

Log2x = log4

y(4-x)

X2 = y (4-x) _(1)

Given eqn. log3 (x+y) = log 3x log3

y

x + y =x

y-(2)

From (1)and (2)

X2 = y (4-x) and x =2

1

y

y-(3)

4 2

24

11

y yy

yy

3

24 4

1

yy y

y

y3 = 4 4y y2 -4y + 4y2 + y3

3y2 -8y + 4 = 0

Y =

40. 2 x+1 = y2 + 4 and 2 x-1 y

Y2 = 2x+1 4

2

4

y= 2 x-1 -1 y 1

Y2 4y + 4 0

y =2


41/67

41. Cos 2x >3

4

Cos2 x > Cos26

Thus, for3

2 2x

x5 7

, ,6 6 6 6

42. x8 x5 + x2 x + 1 > 0

For x > 1 x8 x5>0 ; x2 x > 0 thus always true

0 < x < 1 -x5 + x2>0 ; -x + 1 > 0

For x< 0 ; always true.

43. x2 + y2 xy x y + 1 0

X2 x (y + 1) + (y2 y + 1) 0

D = (y + 1)2 4 (y2 y + 1) = -3 (y 1)2 0

Q.E in x is always > 0

44. x2 + (a. log (1 a2) ) x + (1 a2 ) = 0

Clearly 1 a2 has to be greater than gero for log to be defined.

And Product of roots be less than gero for roots to be of opposite .

Both cant happen simultaneously.

Hence, .

45. Given xy (x + y) = 0

x = 0 OR y = 0 OR x + y = 0

When x = 0 |y| = k y = k

When y= 0 |x| = k x = k

When x + y = 0 |x| + | -x | =2

k x = 2

k

Y = 2

k


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Solutions of (x, y) are ( k, 0) , (0, k) , (2

k,

2

k,

46. x1, x2 are the roots of the eqn x2 -2m x + m = c

x1 + x2 = 2m x1 x2 = m

x12

+ x22

(x1 + x2)2

2 x1 x2 = 4m2

2m

Also x13 + x2

3 = (x1 + x2) (x12 + x2

2 x1 x2) = 2m (4m2 3m)

Given x12 + x2

2 = x13 + x2

3

4m2 2m = 2m (4m2 3m)

m = 0 OR 2m 1 = 4m2 3m

4m2 5m + 1 = 0

Sum of roots =5

4

47. Let , , , be the roots.

Given 4

, , , = 1

We know that 44

And equality exists when

Hence, = 1 is the only possibility

Now a = 6

B = 4

48. |x2 + 3x| = 2 x2

Clearly, 2 x2 has to be positive

x (- 2 , 2 )

Case (i) 2 0x

-x2 -3 x = 2 x2

x = 23


43/67

Case (ii) 0 < x < 2

x2 + 3 x = 2 x2

2x2 + 3 x 2 = 0

x =1

2

, -2 (reject).

x = 2 1,3 2

49. ax + by = 1 and px2 + qy2 = 1 have only one solution

px2 + q2

11

ax

b

should have D = 0

(pb2

+ qa2

)x2

2aqx + (q b2

) = 0

D = 0

4a2q2 = =4 (pb2 + qa2) (q b2)

Pq2 on both sides

2 2 2 2

1a b a b

p q p q

On, Simplifying2 2

1a b

p q

50. Given 4 32x y

y x

5

2

x y

y x -(1)

And

Also log3 (x y) = 1 log3 (x + y)

x2 y2 = 3 -(2)

And from (1) we get 2x

y OR

1

2

Case (i) 2x

y

x = 2y and x2

y

2

= 3

y2 = 1 y = 1

solutions (x, y) (2, 1) ; (-2, -1) (reject) as x + y > c


44/67

Case (ii)1

2

x

y

x =y

zthen from (2) we get not possible.

51. a> 1

Log x + ax 2 log a

x

log 2log

log( ) log

x x

x a a

1 2

log 0log( ) log

xx a a

log 2log( )

log 0log( ).log

a x ax

x a a

x> 0 ; a > 1

x + a > 1 and also a > 1

2

loglog log 0

( )

ax

x a

log x 0 x 1

52. log3(x + 2)> log x + 2

81

log3(x + 2)>

2

3

4

logx

But x < - 1 x + 2 < 1 log3(x + 2)< 0

But log3(x + 2)= t. we get t =

4

t

t (-2, 0) log3(x + 2)> -2

53. 4

4

1 1

1 log 3log

2

x x

x

One can easily evaluate the possible values of x (-3, -2) (-1, ) to make log defined.

For x (-3, -2)


45/67

1

12

x

x

L.H.S. is +ve

And x + 3 < 1 R.H.S is ve

, x (-3, -2).

Now when , x (-1, -).

L.H.S is ve and R.H.S +ve

Always true.

54. We know min value of in4

D

a

X2 + 2x (y 3) + (3y 2 + 6y) has a min value of

2 24( 3) 4(3 6 )4 4

y y yD

a

= 2y2 + 12y 9

This also attains its min at4

D

a

= -2.

55. Given x1 + x2 x3x4 = 1 (1)

x1 + 2x2 + 3x3 x4 = 2 -(2)

3x1 + 5x2 + 5x3 3x4 = 6 -(3)

Eqn. (1) + 2 x (eqn (2)) gives us

3x1 + 5x2 + 5x3 3x4 = 5 which is a

56. p = a log10b = b log10

a

CONCEPTUAL

P

Exercise 2 (B)

1. (X2 -1) 2 2 0X X -(1)

Feasible region : x2 x -2 0

(x -2) (x + 1) 0

( -, -1] (2, )


46/67

Now, (1) x2 1 0 x( -, -1] (1, )

, x( -, -1] (2, )

2.2

2 15 170

10

x x

x

-(1)

Feasible region : 2x2

+ 15 17 0

x 17

, 1,2

Now , (1) 10 x > 0 x < 10

, x 17

, 1,102

3.2

8 2 6 3x x x

Feasible region : 8 +2 x x2 0 x2 2x -8 0

(x -4) (x +2) 0

x [-2, 4]

When R.H.S is ve i.e., 6 -3 x 0 x 2 -(1)

It is always true

When R.H.S is +ve i.e., x 2

Squaring on both sides, we get

8 + 2x x2> 36 36x + 9x2

10x2 38x + 28 < 0

5x2 19x + 14 < 0

5x2 5x 14 x + 14 < 0

(5x 14) (x 1) < 0

x14

1,5

, x (1, 2] - (2)

Final answer (1) (2)

x (1, 4]

4. 3 15 6x x

Clearly x -3 is the feasible region


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Squaring x + 3 + x 15 + 2 3 15 36x x

2 3 15 18 2x x x

2

3 15 9x x x

+ 18x + 45 < 81 18 x +

36 x < 36

X < 1

Ans : [-3, 1)

5. 3x2 + 15 + 22 5 1 2x x

Put x2 + 5x + 1 = t (Remember t = 0 is the feasible region)

3 (t -1) + 2 2t

2 5 3t t -(1)

4t = 25 + 9t2 30 t

9t2 34 t + 25 = 0

t = 1 ;25

9

(Reject, doesnt satisfy (1) )

6. 12 5x 2 = 23

24 3x

23

2 5 2 3 322 .3 2 .3x x

2 10x-4 . 3 5x-2 =2

29 3

9 3 22 .3

x

x

5x -2 =2

9 32

x

3x2 + 10 x 13 = 0

(3x + 13) (x -1) = 0

X =13

,13

7. (x2 3x 3) |x + 1| = 1

Case (I) Base = 1 x2 3x -3 = 1 x = -1; 4

Case (II) Exponent x = -1


48/67

Ans : {-1, 4}

8. xy = yx and x2 = y3 x = y 32

, 3

2

23

2332

3 9

2 4

1

yy

y yy

y y y

y

9

14

y or y

9. log(x + 1)(x3 9x + 8) . log (x 1)

(x + 1)= 3

. log(x 1)(x3 9x + 8) = 3

x3 9x + 8 = (x -1)3 = x3 3x2 + 3x - 1

3x2 12x + 9 = 0

x2 4x + 3 = 0

x = 3 , 1 (Reject, because base 0 )

, x = 3

10. |x + 3|

28 15

2 1

x x

x

Clearly, x 2

(Case(i) Base = 1 |x + 3| = 1 x = 4, 2 (reject)

(Case) (ii) Exponent = 0

x2 8 x + 15 = 0 x = 5 ,3 (reject)

(Because base 0)

, {4, 5}

11. log3xx = log 9x x

Clearly, x > 0 ; x 1 1

,3 9

Given eqnWhen x 1} 3 9

x x

1 1

log 1 log 1

No solution

When x = 1 ; 0 = 0 Which is true.

x = 1


49/67

12.x 5

10 10log log25 5 4.x

x

1 0 102log log5 5 4.5x

x10 1 0

2log log

5 - 4. 5 5 0x

x

10log5 5, 1 (Reject)

10

log 1x

x = 10

13. 25x 12. 2x -625 16

. 0100 100

x

25x 12.2x -625 2

. 0100 25

x

x

(25x)2 12. (2x) (25x) -

25

2 04

x

225 25

4. 48. 25. 02 2

x x

25 25

2 25 2. 1 02 2

x x

25 25

2 2

x

x = 1

14.3

log 2 24x x

Feasible 1 x >

3

8

Case (i) 0< x< 1 2x 23

4x

X2 2x +3

04

x1 3

, ,2 2

, x3 1

,8 2

-(1)


50/67

Case (ii) x > 1

2x 23

4x

x1 3

,2 2

, x3

1,2

-(2)

Ans : (1) (2) 3 1

,8 2

3

1,2

15. x 2

1

10 10log 3log 1000

x x

Put 10log x = t

xt2 3t + 1 > 1000 (1)

Case (i) 0 < x < 1

Eqn. (1) t2 3t + 1 < log x 1000 = 3 log x10

t2 3t + 1 < 3t

t3 3t2 + t -3 >0 ( , t < 0)

(t2 + 1) (t 3) < 0 t > 3 log 10 x > 3 x> 100 x

Case(ii) eqn. (1) t2 3t + 1 >3

t t > 3

16. |x |2

x x 2 < 1 (Clearly x 0)

Case (i) 0< |x| 0

(x 2) (x + 1) > 0

x (- , -1) (2, ) -(2)

(1) (2) Case (ii) |x| > 1 -(3)

x2 x 2 < 0

x (-1, 2) -(4)


51/67

(3) (4) x (1, 2)17. log2

2

x 2 + 2x > 2+ (x + 1) . log2 (2 x)

(2 ) .2 22log 2 2 1 log (2 )x

x x x

(1 x)2log (2 ) 2(1 )x x

(1 x) 2 02log (2 )x

1 x> 0 and2

log (2 ) 2x

OR

1 x< 0 and2

log (2 ) 2x

x< -2 OR x> 1

Remember that x < 2 is the feasible region,

, x (- , -2) ( 1, 2)

18. It log 5 (x2 + 1) log 5 (px

2 + 4x +p)

5x2 + 5 px2 + 4x + p

(5 P) x2 -4 + (5 P) 0 -x

D 0 and (5 P) 0 -(1)

16 _ 4 (5 P)2 0

p (- , 3] [7, ) -(2)

16 - p2< 0 -(3)

19. log1

2 11

1x

x

x

Feasible region

X >0 ; x 1

-log2 1

11

x

x

x

2 10

1

x

x

log2 1

11

x

x

x

x

1, 1,2

0 < x < 1 x > 1

2 11

x xx

2 11

x xx

2x 1 x2 x (, x < 1) 2x 1 x2 x


52/67

X2 3x + 1 0 (1) X2 3x + 1 0

From Feasible region and (1)

x3 5 1

,2 2

OR , x

3 51,

2

20. log3 (16x 2 (12x) ) 2x + 1

16x 2 (12x) 32x . 3

2

4 42. 3 0

3 3

x x

21. log4x log x

43

2

Clearly x >0 ; x 1

Put log 4x = t

t -1 3

2t

2

1 30

2

t

t

22 3 2

0t t

t

(2 1)( 2)

0t t

t

t 1

, 0,22

log4x 1

, 0,2

2

4 4

3 1 03 3

x x

43

3

x

x4

3

log x

Now Feasible region : 16x 2 . 12x> 0

4x 23x > 0


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4

23

x

X > 24

3

log

, x 2 3

4 4

3 3

log ,log

22. 4 15 4 15 8x x

Put 4 15x

t

t +1

8t

2 8 1 0t t

8 64 4

4 152

t

4 15 4 15x

OR1

4 15

x = 2 OR -2

23. Put (5 + 26

) x2-3 = t

Given eqn t +1

10t

t2 10t + 1 = 0

t = 5 2 6

2

32 2

2

15 6 5 6

5 6

x

or

x2 3 = 1 OR -1

24. Feasible region x + 2 > 0 x > -2

And x + 2 1 x -1

Put log5 x + 2 = t

Given eqn 1 +

2

t = t

t + 2 = t2


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t2 t 2 = 0

t = -1 , 2

log5 (x+2) = -1, 2

9

,235

x

25.2 1

8

1 1.log ( 2) log 3 5

6 3x x

(Clearly x > 2 and x >5

3 x >2 )

2

1 1.log ( 2)

6 3x

2

1.log (3 5)

6x

21 1log ( 2) 3 56 3

x x

2log ( 2) 3 5 2x x

3x2 11x + 6 = 0

x = 3;2

3(Reject)

26. log4 log2 x + log2 log4 x = 2

2 2 2 21 1

log log log log 22 2

x x

2 2 2 21

log log log log 1 22

x x

2 2log log 2x

2log 4x

x = 16

27. 2

2 3log 49

27 5 6 2 .x

x

2

3log 2

2

2 5 67 49

2

x x

2

32 5 6

7 72

x x

2

2 5 63

2

x x


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x =3

2

, 4

28. 3x + 4x = 5x 3 4

15 5

x x

L.H.S> continuously decreases as the value of x increases.

Hence, only one solution by Trial and error

X = 2

29. Put 5 24x

= t

t +1

10t

t = 5 24

x 1

5 24 5 24 OR5 24

x = 2 OR -2

30. log2x+3(2x + 3) (3x + 7) = 4 log 3x + 7

(2x + 3)2

1 + log (2x + 3)(3x + 7) = 4 2. log3x + 7

(2x + 3)

Put log (2x + 3)(3x + 7) = t

1 + t = 4 -2

t

t = 1, 2

log(2x + 3)(3x + 7) = 1 OR 2

3x + 7 = 2x + 3 OR 3x + 7 = 4x2 + 12 x + 9

(Reject values for which base < 0 or base = 1)

x = -4 OR x = -2

(Reject) (Reject)


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Exercise 3

1. |x2-2x| + y = 1 and x2 + |y| = 1

y = x2 -1 OR 1 x2

Case (i) y 0

|x2-2x| + y = 1 and x2 + y = 1

|x2-2x| = x2

x = 0 OR |x-2| = x

x = 0 OR x = 1

y = 1 OR y = 0

Case (ii) y < 0 |y| = -y

|x2-2x| + y = 1 and x2 y = 1

|x2- 2x| + x2 = 2

X2 2x + x2 = 2 2x x2 + x2 = 2

X2 x 1 = 0 x = 1 y = 0

X =1 5

2

(Reject)

But given x N x = 1 ;y = 0

2. , , are the root of 2x3 + x2 7 = - 0

1 7; 0;

2 2

Consider

1 1 1 3

1

0 32

= -3

3. Sum of integers div. by 3 = 3 + 6 + 9 + .. + 99

= 3(1 + 2 + 3 + + 33)


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= 3. 33 34

2= 1683

Sum if int. div by 5 = 5 + 10 + 15 + .. + 100

= 5(1 + 2 + 3 + . + 20) = 1050

Sum of int. div by 15 = (1 + 2 + . + 6) 315

Sum of iunt.Div by 3or 5 = 1683 + 1050 315 = 2418

4. x4 3x3 x + 3 < 0

x3(x 3) 1 (x 3) < 0

(x3 1) (x 3) < 0

(x 1) (x 3) (x2 + x + 1) < 0

x (1, 3)

Integers satisfying are x = 2

5. (i) y 2x + 1 = 0 y - |x| - 1 = 0

Y = |x| + 1

|x| + 1 2x + 1 = 0

|x| = 2x 2

If x > 0 x = 2x 2 x = 2

x< 0 -x = 2x 2 x =2

3(Reject x < 0)

x = 2 ; y = 3

(ii) x + 2y 6 = 0 ; |x 3| = y

x + 2|x 3| = 6

If x 3 x + 2x 6 = 6

x = 4

x< 3 x + 2(3 x) = 6

x = 0

x = 4 ; y = 1 and x = 0 ; y = 3

(iii) xy + x + y = 11 ; xy (x + y) = 30

xy +30

xy= 11


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xy = 6 or 5

x + y = 5 or 6 respectively

x = 2 ; y = 3 or x = 3 ; y = 2

?

(iv) y + x 1 = 0 ; |y| = x + 1

|y| = (1 y) + 1

|y| = 2 y

If Y 0 y = 2 y y = 1

Y < 0 -y = 2 y

y = 1 ; x = 0

(v) |x 1| + y = 0 and 2x y = 1

|x 1| + 2x = 1

Case (i) x 1 3x = 2 x = 2/3 (Reject)

Case (ii) x < 1 x = 0

x = 0 ; y = = 1

(vii)put x = 2y + 3 in x2 + y2 x y = 6 2xy

(2y + 3)2 + y2 (2y + 3) y = 6 = 2 (2y +3)y

y = 0 ;5

3

x = ;1

3

respectively

(vi) 2y2 + xy x2 = 0

(2y x) ( y + x) = 0

x y or x = -y

Case (i) x = 2y

x2

xy y2

+ 3x + 7y + 3 = 0

4y2 2y2 y2 + 6y + 7y + 3 = 0

y2 + 13y + 3 = 0


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y I

Case (ii) x + y = 0

x2 xy y2 + 3x + 7y + 3 = 0

Solving we get x = 1 ; y = -1 or x = 3 ; y = -3

6. |x 1| + |x 2| + |x + 1| + |x + 2| = 4k

If x 2 we get

x 1 + x 2 + x + 1 + x + 2 = 4k

4x = 4k.

for k = {2, 3, 4, 5} ; x = 2, 3, 4, 5 are the respectively integral solutions

Observe that |x- 2| + |x + 2| |2 x + x + 2| = 4

|x - | + |xx + 1| |1 x + x + 1| = 2

L.H.S. 6 No solution for k = 1

7. 3

1 2 2a

k

= < -k + 3 a< 1

3 3log log 20 3 1k

3

10 1

log 2k

0 < -k + log23 < 1

-log23 < - k < 1 log23

log23 1 < k < log23

x = 1

.8. (i) 3x-1 + 3 1-x 2.

x

x

3 32

3 3

Possible only when

x3

3 = 1

x = 1


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(ii) 2 cos2 x. sin 2x

2= 5t + 5 t

L.H.S. 2 and R.H.S. 2

, 2cos2 x. sin2x

2= 2 cos2 x = 1 ; sin2

x

2= 1

9. (i) 2x 0 ; 1

x = 5

(ii) log3 (log2x 9) = 2 + log3 (1 -4 log x

4)

log3 (log2x 9) = log3 4x9 1 4 log

log2x 9 = 9 72 log x

2

Put log2x= t t 9 = 9 -

72

t

t2 18t + 72 = 0

t = 6 ; 12 log2x= 6 ; 12

x = 26 , 212

(iii) 2log2(sinx) + log2(4cot2x) = log23

sin2x. 4cot2x = 3

cos2x =3

4

sin2x =1

4

sin2x =1

2(sinx> 0) (feasible region)

n2x = ( 1)6

nx n

(iv) 1 1 13 3 3

5 5 1log cos log cos 2 log

6 6 9x x

2 5 1cos36 9

x

21

cos4

x

2 2cos cos3

x

3x n

(sinx> 0)


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(v) 3 3 38

3 3 log 2 log 9. 2.6 log 33

x

xx x

8

33 .2 3 9. 2.6 .33

x

xx x

8 827 . 27. 6.62 3

x x

x xx

8.81x = 27 16x + 6.36x

2

9 98. 6. 27. 0

4 4

x x

6 309 9 25

; (Re )4 16 4 16

x

jected

x = 1

11. log 2 2 x q x

x 2 = x + a)2

X2 + (2a 1)X + (a2 + 2) = 0

Case (i) D = O (anique sol)

4a2 4ad1 4(a2d2) = 0 a =7

4

Case (ii) x2 + (2a 1)x + (a2 + 2) = 0

D 0 and f(-a) < 0

A 0 & 3 x > 0

y = 1 x (1, 3)

13. (b c) (y + z) a = b c (1)

(c + a) (z + x) by = c a (2)

(a + b) (x + y) cz = a b (3)

Adding (1), (2) and (3)

(a + b + c) (x + y + z) = 0

x + y + z = 0 (4)

Substituting (4) in (1), (2) and (3) we get

x = ; ;

c b a c b a

x y za b c a b c a b c

14. x + y + z = 4 and x2 + y2 + z2 = 6

y + z = 4 x and y2 + z2 = 6 x2

Now,

(y + z)2 = (4 x)2

2 22

22 16 8

6

y zyz x x

x

2yz = 10+ 2x2 8x

We know that


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(y z)2 0 y2 + z2 2y7

6 x2 10 + 2x2 8x

3x2 8x + 4 0

(3x 2) (x 2) 0

x 2 ,23

15. x2 + 2xy + 3y2 + 2x + 6y 3 0

x2 + 2x(y + 1) + 3(y + 1)2 0

Observe that

D = 4(y + 1)2 12(y + 1)2

= -8(y + 1)2 0

Given Expression 0 yR

16. (i) x + y + z = 13 ; x2 y2 + z2 = 91 (1)

From (1) x + z = 13 y (x + z)2 = (13 y)2

x2 + z2 + 2(xz) = 169 26y + y2

xz = y2 ; x2 + z2 = 91 y2

91 y2 + 2y2 = 169 26y + y2

y = 3

when y = 3 x + z = 10 x = 1 ; y = 9 or

(ii) x2 xy + y2 = 21. Y2 2xy + 15 = 0 (2)

Eg. (2) 2x =2 15 15

y

y

y y

(3)

(iii) in(1) 2

15 152

y y y

y y4y2= 84

On minifying 22

2253 84 y

y

(y2 3) (y2 25) = 0

y = 3 ; 5

Now when we know from (3)

Y = 5 x = 4


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Y = -5 x = -4

Y = 3 x = 3 3

Y = 3 x = 3 3

solution are (4, 5), (-4, -5), ( 3 3 , 3 ) and (3 3 , 3 )

17. x 2 x 1 x 2 x 1 2

2 2

x 1 1 x 1 1 2

| x 1 1| | x 1 1| 2

x 1 1| x 1 1| 2

Case (i) x 1 1 x 2

. 2 x 1 2 x = 2

Case (ii) x 1 1 x < 2

(1) OR 2 = 2 which is always true.

Ans : Feasible region : x 1

, Ans : [1, 2]

18. x + y + z = 2 = x2 + y2 + z2

Y + z = 2 x

Y2 + z2 + 2yz = 4 4x + x2

2 x2 + 2yz = 4 - 4x + x2

2yz - = 2 4x + 2x2

Now , (y z)2 0 y2 + z2 2yz

2 x2 2 4x + 2x2

3x2 4 x 0

x4

3x 03

x4

0, 3

Last value = 0 ; greatest value =4

3


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19. xy + 3y2 x + 4y 7 = 0 and 2xy + y 2 2x 2y + 1 = 0

(y 1) (x + 3y + 7) = 0 and (Y 1) (2X + Y 1) = 0

, when y = 1 ; XER (Clearly)

Other solutions

X+3y + 7 = 0 and

2x + y 1 = 0

x = 2 ; y = -3

20. xy + x + y = 23 (x+1) (y+1) = 24 -(1)

Xz + z1 x = 41 (x + 1) (Z + 1) = 42 -(2)

YZ + Y + Z = 27 (Y + 1) (Z + 1) 28 -(3)

(1), (2), (3).

(X + 1)2 (Y + 1)2 (Z + 1)2 = 24.42.28 = 72. 42. 62

(X + 1) (Y + 1) (Z + 1) = 168 -(4)

eqn

4

2 z + 1 = 7 z = 6, -8

similarly,

4

2;

4

3 y = 3, -5

and z = 5, -8 resp.

Solutions are (5, 3, 6) and ( -7, -5,)

Inequation & Equation Solutions Final

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