PRE-RMO MATHEMATICS FUNCTION

121 14641 
 
4
25 64
1. Evaluate : (i) log 3 27 / 729 (ii) log5 (iii) log 2 (iv) log11  3 
625 8  1331 

Solution


log 3 27 / 729 
33
 log 3  log 3 1  0
33
4
25
(ii) log5
625
 4 25 
 log5  
 625 

 25
1/4

 log 5
 25
2

 log 5  51/ 2  4 

 7 7
    log5 5  
 2 2

64
(iii) log 2
8

 log 2 8

 log 2 23/ 2
3

2
121 14641 
(iv) log11  3 
 1331 
 112  112 
 log11  
 11 

 log11 11
3

3
2. Prove that: 3log10 1.5  log10 240  2log10 9  1

Solution: 3log10 1.5  log10 240  2log10 9  1

Taking L.H.S.

log10 1.5   log10 240  log10 9 2

3

 1.5 3 .240 
 log10  
 92 
 
 15 15 15  240 
 log10  
 1000  9  9 
 log10 10  1 R.H.S.

3. Prove that: 3log 4  2 log 6  log 18 

3/2

 log 96 2 
Solution: Taking L.H.S.

3log 4  2 log 6  log 18 

3/2

 log  4   log  6   log 18 

3 2 3/ 2

  4 3  18 3/ 2 
 log  
 6 2

 
 4  4  4   9 3/2   2 3/2 
 log  
 6  6 
 


 log 96 2  R.H.S.

 1 
4. Find x when, (i) x  log 7   (ii) log 0.4 x  4
 49 
Solution:
 1 
x  log 7  
 49 
  log 7 1/2  7 
2
 

  2  2  log7 7 

 4
(ii) log 0.4 x  4

x   0.4 
4

x  0.0256
5. Prove that: (i) logb a  logc b  log d c  logd a

a2 b2 c2
(ii) log  log  log 0
bc ca ab
1 1 1
(iii)   1
log a abc logb abc log c abc
Solution:
logb a  logc b  logd c  logd a
Taking L.H.S.
logb a  logc b  log d c

 logb a  log d b

 log d a = R.H.S.

a2 b2 c2
(ii) Given, log  log  log 0
bc ca ab
Taking L.H.S.
a2 b2 c2
log  log  log
bc ca ab
 a 2 b2 c2 
 log    
 bc ca ab 

 log 1  0 R.H.S.

1 1 1
(iii) Given,   1
log a abc logb abc log c abc
Taking L.H.S.
 1 1 1
 
log a abc logb abc log c abc

log abc a  log abc b  log abc c

logabc  abc   1 R.H.S.

6. Solve for x: log3  log  x  1  log  2 x  7 

Solution:
Given, log3  log  x  1  log  2 x  7 

log  3x  3  log  2 x  7 

3x  3  2 x  7
x4
7. Solve for x: log 2  log  x  2  log  3x  5  log3

Solution:
Given, log 2  log  x  2  log  3x  5  log3

log  2 x  4  log  9 x  15

2 x  4  9 x  15
7 x  19
19
x
7
log10  x  3  1
8. Solve for x: 
log10  x  21
2
2

Solution:
log10  x  3  1
Given, 
log10  x  21
2
2

2 log10  x  3  log10  x 2  21

 x  3  x 2  21
2

x 2  9  6 x  x 2  21
6 x  30
 30
x 5
6
Thus, x = 5.
9. Solve for x: log 4 log3 log 2 x  0
Solution:
Given, log 4 log3 log 2 x  0

log3 log 2 x  40

log 2 x  31

x  23  8

If a 3  b 3  0 and  a  b   0 , prove that : log  a  b  

1
10.  log a  log b  log 3
2
Solution: Given, a 3  b 3  0

a 3
 b 3    a  b   3ab  a  b 
3

or  a  b   3ab  a  b 
3
 a 3
 b3  0 

Taking log both sides, we get

 3  log  a  b    log3  log a  log b  log  a  b 

 2log  a  b   log a  log b  log 3

1
log  a  b    log a  log b  log 3
2
Hence proved
x2  2
11. Find range of : y 
x2  2
x2  2
Solution: Given, y  2
x 2
x2  2
y ;xR
x2  2
yx 2  2 y  x 2  2

x2  y  1  2  2 y
 2  2y
x2 
1 y

2  2y
x
1 y
Now, x is defined if
2  2y
0
1 y

y   1,1

x2  x  1
12. Find range of : y 
x2  x  1
x2  x  1
Solution: Given, y 
x2  x  1
x2  x  1
y 2 ;xR
x  x 1
yx2  yx  y  x2  x  1

x2  y  1  x  y  1   y  1  0

D0

 y  1  4  y  1  0
2 2

3 y 1 y  3  0
1
 y3
3

13. Find range of : y  9  x 2

Solution:

Given, y  9  x 2

y  9  x2

as 9  x  0,9
2

 y  9  x 2   0, 3
14. Check Surjectivity:
f : 0,2   1,26 , defined as f  x    3sin x  4 cos x 2  1
Solution:
f  x    3sin x  4 cos x   1
2

As  25  3sin x  4cos x  25
5  3sin x  4 cos x  5
Thus f  x min  0  1  1
f  x  max   5   1  26
2

Thus f  x 1,26
Hence f(x) is onto
15. f : R  S , defined as f  x   [sin x] is an onto function then S must be, (where [.] is GIF)
(a) [−1, 1] (b) {−1, 0, 1} (c) {−sin1, 0 , sin1} (d) none
Solution:
f  x   [sin x]
For all real values of ‘x’
Sin x gives values between −1 and 1
Thus all integers between them will be range of [sin x]

Thus S 1,0,1

16. 
Find the range of log3 x 2  9 . 
(a)  0,   (b)  2,   (c)  3,   (d) 9,  

Solution:
We know that the range of x 2  9 is 9,   .

 
Thus, range of log3 x 2  9 will be  log 3 9, log 3   , that is  2,   .

17. Find the range of the function y  x 2  7 .

(a) y   7,   (b) y   7,   (c) y    7,   (d) y   7,  

Solution:
We know that x 2  7   7,   .
 Thus, x 2  7   7,  
3
18. Find the range of the function y  .
5  4cos x
1  1 1  1  1 1 
(a) y   , 1 (b) y   ,  (c) y   , 1 (d) y   , 
5  5 4 9  9 4
Solution:
3
Given y 
5  4cos x
We know that  1  cos x  1
 4  4cos x  4
 5  4  5  4cos x  5  4
 1  5  4cos x  9
1 1
1 
5  4cos x 9
1 3
  1
9 5  4cos x
1 
 y   , 1
9 

19. 
Find the range of y  log16 x 2  x  1 . 
3   3 
(a) y   ,   (b) y  8,   (c) y   2 4 ,   (d) y  1,   3
4   
Answer: (b)
Solution:

Given y  log16  x 2  x  1
3 
We know that x 2  x  1   ,  
4 
 3 
 log16  x 2  x  1  16 4 , 16  , that is, log16  x 2  x  1  8,  
 
 y  8,  

20. The function f :    defined as f  x   x 2  x  12 is _____.

(a) one-one (b) many-one (c) onto (d) into

Answer: (b), (d)
Solution:
The given function is f :    defined as f  x   x 2  x  12 .

Consider,
f  x1   f  x2 
 x12  x1  12  x2 2  x2  12
 x12  x2 2  x1  x2  0
  x1  x2  x1  x2  1  0
 x1  x2 or x1  x2  1
 f is many-one.
Consider,
f  x   x 2  x  12
2 2
1 1 1
 x  2  x           12
2

2 2 2

2
 1  49
x  
 2 4
 49 
 f  x    ,  
 4 
 49 
Thus, Range    ,   , but Codomain   .
 4 
Since range and codomain are not equal, the given function is into.
Thus, the given function is a many-one into function.