RD Sharma Class 12 Solutions - Increasing and Decreasing Functions | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


17. Increasing and Decreasing Functions
Exercise 17.1
1. Question
Prove that the function f(x) = log
e
 x is increasing on (0, 8).
Answer
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
2. Question
Prove that the function f(x) = log
a
 x is increasing on (0, 8) if a > 1 and decresing on (0, 8), if 0 < a < 1.
Answer
case I
When a > 1
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
case II
When 0 < a < 1
f(x) = log
a
 x 
when a<1 log a < 0
let x
1
<x
2
? log x
1
 < log x
2
?  [ ]
? f(x
1
) > f(x
2
)
So, f(x) is decreasing in (0, )
3. Question
Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.
Answer
we have,
f(x) = ax + b, a > 0
Page 2


17. Increasing and Decreasing Functions
Exercise 17.1
1. Question
Prove that the function f(x) = log
e
 x is increasing on (0, 8).
Answer
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
2. Question
Prove that the function f(x) = log
a
 x is increasing on (0, 8) if a > 1 and decresing on (0, 8), if 0 < a < 1.
Answer
case I
When a > 1
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
case II
When 0 < a < 1
f(x) = log
a
 x 
when a<1 log a < 0
let x
1
<x
2
? log x
1
 < log x
2
?  [ ]
? f(x
1
) > f(x
2
)
So, f(x) is decreasing in (0, )
3. Question
Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.
Answer
we have,
f(x) = ax + b, a > 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 > ax
2
 for some a > 0
? ax
1
 + b> ax
2
 + b for some b
? f(x
1
) > f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) > f(x
2
)
So, f(x) is increasing function of R
4. Question
Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.
Answer
we have,
f(x) = ax + b, a < 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 < ax
2
 for some a > 0
? ax
1
 + b< ax
2
 + b for some b
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function of R
5. Question
Show that  is a decreasing function on (0, 8).
Answer
we have
let x
1
,x
2
 (0, ) We have, x
1
 > x
2
? 
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function
6. Question
Show that  decreases in the interval [0, 8) and increases in the interval (-8, 0].
Answer
We have,
Page 3


17. Increasing and Decreasing Functions
Exercise 17.1
1. Question
Prove that the function f(x) = log
e
 x is increasing on (0, 8).
Answer
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
2. Question
Prove that the function f(x) = log
a
 x is increasing on (0, 8) if a > 1 and decresing on (0, 8), if 0 < a < 1.
Answer
case I
When a > 1
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
case II
When 0 < a < 1
f(x) = log
a
 x 
when a<1 log a < 0
let x
1
<x
2
? log x
1
 < log x
2
?  [ ]
? f(x
1
) > f(x
2
)
So, f(x) is decreasing in (0, )
3. Question
Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.
Answer
we have,
f(x) = ax + b, a > 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 > ax
2
 for some a > 0
? ax
1
 + b> ax
2
 + b for some b
? f(x
1
) > f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) > f(x
2
)
So, f(x) is increasing function of R
4. Question
Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.
Answer
we have,
f(x) = ax + b, a < 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 < ax
2
 for some a > 0
? ax
1
 + b< ax
2
 + b for some b
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function of R
5. Question
Show that  is a decreasing function on (0, 8).
Answer
we have
let x
1
,x
2
 (0, ) We have, x
1
 > x
2
? 
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function
6. Question
Show that  decreases in the interval [0, 8) and increases in the interval (-8, 0].
Answer
We have,
Case 1
When x [0, )
Let ,  (0, ] and 
? 
? 
? 
? f(x
1
)< f(x
2
)
 f(x) is decreasing on[0,8).
Case 2
When x (- ,0]
Let > 
? 
? 
? 
? 
 f(x) is increasing on(-8,0].
Thus, f(x) is neither increasing nor decreasing on R.
7. Question
Show that  is neither increasing nor decreasing on R.
Answer
We have,
Case 1
When x [0, )
Let > 
? 
? 
? 
f(x
1
) < f(x
2
)
?  f(x) is decreasing on[0,8).
Case 2
When x (- ,0]
Page 4


17. Increasing and Decreasing Functions
Exercise 17.1
1. Question
Prove that the function f(x) = log
e
 x is increasing on (0, 8).
Answer
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
2. Question
Prove that the function f(x) = log
a
 x is increasing on (0, 8) if a > 1 and decresing on (0, 8), if 0 < a < 1.
Answer
case I
When a > 1
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
case II
When 0 < a < 1
f(x) = log
a
 x 
when a<1 log a < 0
let x
1
<x
2
? log x
1
 < log x
2
?  [ ]
? f(x
1
) > f(x
2
)
So, f(x) is decreasing in (0, )
3. Question
Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.
Answer
we have,
f(x) = ax + b, a > 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 > ax
2
 for some a > 0
? ax
1
 + b> ax
2
 + b for some b
? f(x
1
) > f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) > f(x
2
)
So, f(x) is increasing function of R
4. Question
Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.
Answer
we have,
f(x) = ax + b, a < 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 < ax
2
 for some a > 0
? ax
1
 + b< ax
2
 + b for some b
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function of R
5. Question
Show that  is a decreasing function on (0, 8).
Answer
we have
let x
1
,x
2
 (0, ) We have, x
1
 > x
2
? 
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function
6. Question
Show that  decreases in the interval [0, 8) and increases in the interval (-8, 0].
Answer
We have,
Case 1
When x [0, )
Let ,  (0, ] and 
? 
? 
? 
? f(x
1
)< f(x
2
)
 f(x) is decreasing on[0,8).
Case 2
When x (- ,0]
Let > 
? 
? 
? 
? 
 f(x) is increasing on(-8,0].
Thus, f(x) is neither increasing nor decreasing on R.
7. Question
Show that  is neither increasing nor decreasing on R.
Answer
We have,
Case 1
When x [0, )
Let > 
? 
? 
? 
f(x
1
) < f(x
2
)
?  f(x) is decreasing on[0,8).
Case 2
When x (- ,0]
Let > 
? 
? 
? 
? 
 f(x) is increasing on(-8,0].
Thus, f(x) is neither increasing nor decreasing on R.
8. Question
Without using the derivative, show that the function f(x) = | x | is
A. strictly increasing in (0, 8)
B. strictly decreasing in (-8, 0).
Answer
We have,
f(x) = |x| = 
(a)Let ,  (0, ) and 
? 
So, f(x) is increasing in (0, )
(b) Let ,  (-8, 0)and 
? 
? 
 f(x) is strictly decreasing on(-8, 0).
9. Question
Without using the derivative show that the function f(x) = 7x - 3 is strictly increasing function on R.
Answer
Given,
f(x) = 7x – 3
Lets ,  R and 
? 7 > 7
? 7 > 7
? 
 f(x) is strictly increasing on R.
Exercise 17.2
1 A. Question
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 10 – 6x – 2x
2
Page 5


17. Increasing and Decreasing Functions
Exercise 17.1
1. Question
Prove that the function f(x) = log
e
 x is increasing on (0, 8).
Answer
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
2. Question
Prove that the function f(x) = log
a
 x is increasing on (0, 8) if a > 1 and decresing on (0, 8), if 0 < a < 1.
Answer
case I
When a > 1
let x
1
,x
2
 (0, )
We have, x
1
<x
2
? log
e
 x
1
 < log
e
 x
2
? f(x
1
) < f(x
2
)
So, f(x) is increasing in (0, )
case II
When 0 < a < 1
f(x) = log
a
 x 
when a<1 log a < 0
let x
1
<x
2
? log x
1
 < log x
2
?  [ ]
? f(x
1
) > f(x
2
)
So, f(x) is decreasing in (0, )
3. Question
Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.
Answer
we have,
f(x) = ax + b, a > 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 > ax
2
 for some a > 0
? ax
1
 + b> ax
2
 + b for some b
? f(x
1
) > f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) > f(x
2
)
So, f(x) is increasing function of R
4. Question
Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.
Answer
we have,
f(x) = ax + b, a < 0
let x
1
,x
2
 R and x
1
 > x
2
? ax
1
 < ax
2
 for some a > 0
? ax
1
 + b< ax
2
 + b for some b
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function of R
5. Question
Show that  is a decreasing function on (0, 8).
Answer
we have
let x
1
,x
2
 (0, ) We have, x
1
 > x
2
? 
? f(x
1
) < f(x
2
)
Hence, x
1
 > x
2
? f(x
1
) < f(x
2
)
So, f(x) is decreasing function
6. Question
Show that  decreases in the interval [0, 8) and increases in the interval (-8, 0].
Answer
We have,
Case 1
When x [0, )
Let ,  (0, ] and 
? 
? 
? 
? f(x
1
)< f(x
2
)
 f(x) is decreasing on[0,8).
Case 2
When x (- ,0]
Let > 
? 
? 
? 
? 
 f(x) is increasing on(-8,0].
Thus, f(x) is neither increasing nor decreasing on R.
7. Question
Show that  is neither increasing nor decreasing on R.
Answer
We have,
Case 1
When x [0, )
Let > 
? 
? 
? 
f(x
1
) < f(x
2
)
?  f(x) is decreasing on[0,8).
Case 2
When x (- ,0]
Let > 
? 
? 
? 
? 
 f(x) is increasing on(-8,0].
Thus, f(x) is neither increasing nor decreasing on R.
8. Question
Without using the derivative, show that the function f(x) = | x | is
A. strictly increasing in (0, 8)
B. strictly decreasing in (-8, 0).
Answer
We have,
f(x) = |x| = 
(a)Let ,  (0, ) and 
? 
So, f(x) is increasing in (0, )
(b) Let ,  (-8, 0)and 
? 
? 
 f(x) is strictly decreasing on(-8, 0).
9. Question
Without using the derivative show that the function f(x) = 7x - 3 is strictly increasing function on R.
Answer
Given,
f(x) = 7x – 3
Lets ,  R and 
? 7 > 7
? 7 > 7
? 
 f(x) is strictly increasing on R.
Exercise 17.2
1 A. Question
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 10 – 6x – 2x
2
Answer
Given:- Function f(x) = 10 – 6x – 2x
2
Theorem:- Let f be a differentiable real function defined on an open interval (a, b).
(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)
Algorithm:-
(i) Obtain the function and put it equal to f(x)
(ii) Find f’(x)
(iii) Put f’(x) > 0 and solve this inequation.
For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.
Here we have,
f(x) = 10 – 6x – 2x
2
? 
? f’(x) = –6 – 4x
For f(x) to be increasing, we must have
? f’(x) > 0
? –6 –4x > 0
? –4x > 6
? 
? 
? 
Thus f(x) is increasing on the interval 
Again, For f(x) to be increasing, we must have
f’(x) < 0
? –6 –4x < 0
? –4x < 6
? 
? 
? 
Thus f(x) is decreasing on interval 
1 B. Question
Find the intervals in which the following functions are increasing or decreasing.
f(x) = x
2
 + 2x – 5
Answer