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

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

Solution:

Let x1, x2 ∈ (0, ∞)

We have, x1<x2

⇒ log_e x_{1 <} log_e x₂ 

⇒ f(x₁) < f(x₂)

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

Question 2: Prove that the function f(x) = log_a (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

⇒ log_e x₁ < log_e x₂

⇒ f(x₁) < f(x₂)

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

Case 2:

When 0<a<1

f(x) = log_a x = log_x/log_a

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

⇒ 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 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 ∈ R and x₁ >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+ x² 

Case 1:

when x ∈ [0, ∞]

Let x₁, x₂ ∈  [0,∞] and x₁ > 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].

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 x₁ > 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.

Question 8: Without using the derivative, show that the function f(x) = |x| is,

(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

⇒  7x₁ – 3 > 7x₂ – 3

⇒  f(x₁) > f(x₂)

Thus, f(x) is strictly increasing on R