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

Here we provide RD Sharma Class 12 Ex 17.1 Solutions Chapter 17 Increasing and Decreasing Functions for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 17.1 Solutions Chapter 17 Increasing and Decreasing Functions book pdf download. Now you will get step-by-step solutions to each question.

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

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

Solution:

Let x1, x2 ∈ (0, ∞)

We have, x1<x2

⇒ loge x1 < loge x

⇒ f(x1) < f(x2)

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

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

Solution:

Case 1:

When a>1

Let x1, x2 ∈ (0, ∞)

We have, x1<x2

⇒ loge x1 < loge x2

⇒ f(x1) < f(x2)

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

Case 2:

When 0<a<1

f(x) = loga x = logx/loga

When a<1 ⇒ log a< 0

let x1<x2

⇒ log x1<log x2

⇒ ( log x1/log a) > (log x2/log a)                                [log a<0]

⇒   f(x1) > f(x2)

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

### 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 ∈ R and x1 >x2

⇒ ax> axfor some a>0

⇒ ax1 + b > ax2 + b for some b

⇒ f(x1) > f(x2)

Hence, x1 > x2  ⇒   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, x∈ R and x>x2

⇒ ax1 < ax2 for some a>0

⇒ ax1 + b <ax2 + b for some b

⇒ f(x1) <f(x2)

Hence, x1 > x2  ⇒   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,∞).

We have,

f(x) = 1/x

Let x1, x2 ∈  (0,∞) and x1 > x2

⇒  1/x1 < 1/x2

⇒ f(x1) < f(x2)

Thus, x1 > x2 ⇒ f(x1) < f(x2)

Therefore, f(x) is decreasing function.

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

Solution:

We have,

f(x) = 1/1+ x

Case 1:

when x ∈ [0, ∞]

Let x1, x2 ∈  [0,∞] and x1 > x2

⇒  x12 > x22

⇒  1+x12 < 1+x22

⇒ 1/(1+ x1)> 1/(1+ x2 )

⇒ f(x1) < f(x2)

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

Case 2:

when x ∈ [-∞, 0]

Let x1 > x2

⇒  x12 < x2                           [-2>-3 ⇒   4<9]

⇒  1+x12 < 1+x22

⇒ 1/(1+ x12)> 1/(1+ x22  )

⇒ f(x1) > f(x2)

Therefore, f(x) is increasing in [-∞,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, ∞] and [-∞,0]

Case 1:

when x ∈ [0, ∞]

Let x> x2

⇒  x12 > x22

⇒  1+x12 < 1+x22

⇒ 1/(1+ x12 )> 1/(1+ x22  )

⇒ f(x1) < f(x2)

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

Case 2:

when x ∈ [-∞, 0]

Let x1 > x2

⇒  x12 < x22                            [-2>-3 ⇒   4<9]

⇒  1+x12 < 1+x22

⇒ 1/(1+ x12)> 1/(1+ x22 )

⇒ f(x1) > f(x2)

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

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

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

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

Solution:

(i). Let x1, x2 ∈  [0,∞] and x1 > x2

⇒  f(x1) > f(x2)

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

(ii). Let x1, x2 ∈  [-∞, 0] and x1 > x2

⇒  -x1<-x2

⇒  f(x1) < f(x2)

Thus, f(x) is strictly decreasing in [-∞,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 ∈ R and x1 >x2

⇒  7x1 > 7x2

⇒  7x– 3 > 7x2 – 3

⇒  f(x1) > f(x2)

Thus, f(x) is strictly increasing on R

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