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Revision Sheet

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0% found this document useful (0 votes)
39 views3 pages

Revision Sheet

Uploaded by

Khushi Jain
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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TM

REVISION PRACTICE SHEET MATHS


Path to success KOTA (RAJASTHAN )

Solving quadratic and rational inequalities. (Method of intervals)

Type-1 : Quadratic inequality involving non-repeated linear factors.


(1) 3x2 – 7x + 6 < 0
(2) (x2 – x – 6)(x2 + 6x)  0 )
5 2 x3
(3) Solve f ' (x)  g ' (x) where f (x) = 5 – 3x + x – , g (x) = 3x – 7.
2 3

Type-2 : Quadratic inequality involving Repeated linear factos.


(1) (x + 1)(x – 3)(x – 2)2  0.
(2) x(x + 6)(x + 2)2(x – 3) > 0
(3) (x – 1)2(x + 1)3(x – 4) < 0
x3 (2x  3)2 ( x  4)6
(4) Number of positive integral solution of 0
( x  3)3 (3x  8) 4
(A) only one (B) 2 (C) 3 (D) 4

f ( x)
Type-3 : Quadratic / algebraic inequality of the type of . (Rational inequality)
g( x )

2x  3 x 2  5x  12
(1) >0 (2) >3
3x  7 x 2  4x  5

x 2  5x  6 ( x  1)2 (x  1)3
(3) <0 (4) <0
x2  x  1 x 4 ( x  2)
x 1 x  5 2( x  4) 1
(5)  (6) 
x 1 x 1 ( x  1)(x  7) x  2

x 2  6x  7 x 2  4x  4
(7) <0 (8) >0
| x  4| 2x 2  x  1

1 /2
TM
REVISION PRACTICE SHEET MATHS
Path to success KOTA (RAJASTHAN )

Type-4 : Double inequality and biquadratic inequality.


3x 2  7 x  8
(1) 1< 2
x2  1

(2) (x2 + 3x + 1) (x2 + 3x – 3)  0


( 2 x  3)
(3) (x2 + 3x)(2x + 3) – 16  0.
( x 2  3x )

MISCELLANEOUS EQUATIONS INEQUATIONS AND


LOGARITHMIC INEQUALITIES :
A. LINEAR EQUATION / INEQUATIONS INVOLVING MODULUS :

1. |x–3|+2|x+1|=4
3
2. |x+2|–|x–1|<x 
2
3. Find the least +ve integer satisfying | x + 1 | + | x – 4 | > 7.
2 x  1 3x  1
4. Greater integer satisfying – >1
3 2

B. QUADRATIC EQUATION / INEQUATION INVOLVINGMODULUS &


EXPONENTIAL :
5x  16
1. | x2 + 4x + 2 | =
3
2. ( | x – 1 | – 3) (| x + 2 | – 5) < 0
3. | x – 5 | > | x2 – 5x + 9 |
4. 2| x + 2 | – | 2x + 1 – 1 | = 2x + 1 + 1

2 /2
TM
REVISION PRACTICE SHEET MATHS
Path to success KOTA (RAJASTHAN )

x 2  5x  4
5. 1
x2  4

x 2  3x  1
6. <3
x2  x  1

C. LOGARITHMIC EQUATION :
4x  6
1. log 1 0
x
5
2. (a) log2x+3(x2) < log2x + 3(2x + 3)
(b) logx+3(x2 – x) < 1
 2x  6 
3. log 7   >0
 2x  1 
4. log3 | 3 – 4x | > 2
5. log0.2 (x2 – x – 2 ) > log0.2 (– x2 + 2x + 3)
3x  6
log 1 log 2
x 2 2
6. (0.3) 3 >1


 x 2  x 
7. log0.5  log 6  <0
 x  4 

| x 2  4x | 3
8. log3 2 0
x | x 5|

 4x  3  1
9. log 22   >–
 4  3x  2
10. 2 log32 x  3log3 x  8 2 log32 x  3 log3 x  6  3

3 /2

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