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Class 12 RD Sharma Solutions – Chapter 17 Increasing and Decreasing Functions – Exercise 17.1
  • Last Updated : 03 Jan, 2021

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

⇒ 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+ x

Case 1:

when x ∈ [0, ∞]

Let x1, x2 ∈  [0,∞] and x1 > x2

⇒  x12 > x22  

⇒  1+x12 < 1+x22

⇒ 1/(1+ x12 )> 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 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

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