# Class 12 RD Sharma Solutions – Chapter 17 Increasing and Decreasing Functions – Exercise 17.1

### Question 1: Prove that the function f(x) = loge x is increasing on (0,âˆž).

Solution:

Let x1, x2 âˆˆ (0, âˆž)

We have, x1<x2

â‡’ loge x1 < loge x

â‡’ f(x1) < f(x2)

Therefore, f(x) is increasing in (0, âˆž).

### Question 2: Prove that the function f(x) = loga (x) is increasing on (0,âˆž) if a>1 and decreasing on (0,âˆž) if 0<a<1.

Solution:

Case 1:

When a>1

Let x1, x2 âˆˆ (0, âˆž)

We have, x1<x2

â‡’ loge x1 < loge x2

â‡’ f(x1) < f(x2)

Therefore, f(x) is increasing in (0, âˆž).

Case 2:

When 0<a<1

f(x) = loga x = logx/loga

When a<1 â‡’ log a< 0

let x1<x2

â‡’ log x1<log x2

â‡’ ( log x1/log a) > (log x2/log a)                                [log a<0]

â‡’   f(x1) > f(x2)

Therefore, f(x) is decreasing in (0, âˆž).

### Question 3: Prove that f(x) = ax + b,  where a, b are constants and a>0 is an increasing function on R.

Solution:

We have,

f(x) = ax + b, a > 0

Let x1, x2 âˆˆ R and x1 >x2

â‡’ ax1 > ax2 for some a>0

â‡’ ax1 + b > ax2 + b for some b

â‡’ f(x1) > f(x2)

Hence, x1 > x2  â‡’   f(x1) > f(x2)

Therefore, f(x) is increasing function of  R.

### Question 4: Prove that f(x) = ax + b, where a, b are constants and a<0 is a decreasing function on R.

Solution:

We have,

f(x) = ax + b, a < 0

Let x1, x2 âˆˆ R and x1 >x2

â‡’ ax1 < ax2 for some a>0

â‡’ ax1 + b <ax2 + b for some b

â‡’ f(x1) <f(x2)

Hence, x1 > x2  â‡’   f(x1) <f(x2)

Therefore, f(x) is decreasing function of  R.

### Question 5: Show that f(x) = 1/x is a decreasing function on (0,âˆž).

We have,

f(x) = 1/x

Let x1, x2 âˆˆ  (0,âˆž) and x1 > x2

â‡’  1/x1 < 1/x2

â‡’ f(x1) < f(x2)

Thus, x1 > x2 â‡’ f(x1) < f(x2)

Therefore, f(x) is decreasing function.

### Question 6: Show that f(x) = 1/(1+x2) decreases in the interval [0, âˆž] and increases in the interval [-âˆž,0].

Solution:

We have,

f(x) = 1/1+ x

Case 1:

when x âˆˆ [0, âˆž]

Let x1, x2 âˆˆ  [0,âˆž] and x1 > x2

â‡’  x12 > x22

â‡’  1+x12 < 1+x22

â‡’ 1/(1+ x12 )> 1/(1+ x2 )

â‡’ f(x1) < f(x2)

Therefore, f(x) is decreasing in [0, âˆž].

Case 2:

when x âˆˆ [-âˆž, 0]

Let x1 > x2

â‡’  x12 < x2                           [-2>-3 â‡’   4<9]

â‡’  1+x12 < 1+x22

â‡’ 1/(1+ x12)> 1/(1+ x22  )

â‡’ f(x1) > f(x2)

Therefore, f(x) is increasing in [-âˆž,0].

### Question 7: Show that f(x) = 1/(1+x2) is neither increasing nor decreasing on R.

Solution:

We have,

(x) = 1/1+ x2

R can be divided into two intervals [0, âˆž] and [-âˆž,0]

Case 1:

when x âˆˆ [0, âˆž]

Let x1 > x2

â‡’  x12 > x22

â‡’  1+x12 < 1+x22

â‡’ 1/(1+ x12 )> 1/(1+ x22  )

â‡’ f(x1) < f(x2)

Therefore, f(x) is decreasing in [0, âˆž].

Case 2:

when x âˆˆ [-âˆž, 0]

Let x1 > x2

â‡’  x12 < x22                            [-2>-3 â‡’   4<9]

â‡’  1+x12 < 1+x22

â‡’ 1/(1+ x12)> 1/(1+ x22 )

â‡’ f(x1) > f(x2)

Therefore, f(x) is increasing in [-âˆž,0].

Here, f(x) is decreasing in [0, âˆž] and f(x) is increasing in [-âˆž,0].

Thus, f(x) neither increases nor decreases on R.

### (i) strictly increasing in (0,âˆž)                      (ii) strictly decreasing in (-âˆž,0)

Solution:

(i). Let x1, x2 âˆˆ  [0,âˆž] and x1 > x2

â‡’  f(x1) > f(x2)

Thus, f(x) is strictly increasing in [0,âˆž].

(ii). Let x1, x2 âˆˆ  [-âˆž, 0] and x1 > x2

â‡’  -x1<-x2

â‡’  f(x1) < f(x2)

Thus, f(x) is strictly decreasing in [-âˆž,0].

### Question 9: Without using the derivative show that the function f(x) = 7x – 3 is strictly increasing function on R.

Solution:

f(x) = 7x-3

Let x1, x2 âˆˆ R and x1 >x2

â‡’  7x1 > 7x2

â‡’  7x1 – 3 > 7x2 – 3

â‡’  f(x1) > f(x2)

Thus, f(x) is strictly increasing on R

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Previous
Next