# 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,∞).

_{e}x is increasing on (0,∞).

**Solution:**

Let x1, x2 ∈ (0, ∞)

We have, x1<x2

⇒ log

_{e}x_{1 < }log_{e}x_{2 }⇒ f(x

_{1}) < f(x_{2})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 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})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 x

_{1}<x_{2}⇒ log x

_{1}<log x_{2}⇒ ( log x

_{1}/log a) > (log x_{2}/log a) [log a<0]⇒ f(x

_{1}) > f(x_{2})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 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})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 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})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 x

_{1}, x_{2}∈ (0,∞) and x_{1}> x_{2}⇒ 1/x

_{1}< 1/x_{2}⇒ f(x

_{1}) < f(x_{2})Thus, x

_{1}> x_{2}⇒ f(x_{1}) < f(x_{2})Therefore, f(x) is decreasing function.

### Question 6: Show that f(x) = 1/(1+x^{2}) decreases in the interval [0, ∞] and increases in the interval [-∞,0].

**Solution:**

We have,

f(x) = 1/1+ x

^{2 }Case 1:

when x ∈ [0, ∞]

Let x

_{1}, x_{2}∈ [0,∞] and x_{1}> x_{2}⇒ x

_{1}^{2}> x_{2}^{2 }⇒ 1+x

_{1}^{2}< 1+x_{2}^{2}⇒ 1/(1+ x

_{1}^{2 })> 1/(1+ x_{2}^{2 })⇒ f(x1) < f(x2)

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

Case 2:

when x ∈ [-∞, 0]

Let x

_{1}> x_{2}⇒ x

_{1}^{2}< x_{2}^{2 }[-2>-3 ⇒ 4<9]⇒ 1+x

_{1}^{2}< 1+x_{2}^{2}⇒ 1/(1+ x

_{1}^{2})> 1/(1+ x_{2}^{2})⇒ f(x

_{1}) > f(x_{2})Therefore, f(x) is increasing in [-∞,0].

### Question 7: Show that f(x) = 1/(1+x^{2}) is neither increasing nor decreasing on R.

**Solution:**

We have,

(x) = 1/1+ x

^{2}R can be divided into two intervals [0, ∞] and [-∞,0]

Case 1:

when x ∈ [0, ∞]

Let x

_{1 }> x_{2}⇒ x

_{1}^{2}> x_{2}^{2}⇒ 1+x

_{1}^{2}< 1+x_{2}^{2}⇒ 1/(1+ x

_{1}^{2})> 1/(1+ x_{2}^{2})⇒ f(x

_{1}) < f(x_{2})Therefore, f(x) is decreasing in [0, ∞].

Case 2:

when x ∈ [-∞, 0]

Let x

_{1}> x_{2}⇒ x

_{1}^{2}< x_{2}^{2}[-2>-3 ⇒ 4<9]⇒ 1+x

_{1}^{2}< 1+x_{2}^{2}⇒ 1/(1+ x

_{1}^{2})> 1/(1+ x_{2}^{2})⇒ f(x

_{1}) > f(x_{2})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 x

_{1}, x_{2}∈ [0,∞] and x_{1}> x_{2}⇒ f(x

_{1}) > f(x_{2})Thus, f(x) is strictly increasing in [0,∞].

(ii). Let x

_{1}, x_{2}∈ [-∞, 0] and x_{1}> x_{2}⇒ -x

_{1}<-x_{2}⇒ f(x

_{1}) < f(x_{2})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 x

_{1}, x_{2}∈ R and x_{1}>x_{2}⇒ 7x

_{1}> 7x_{2}⇒ 7x

_{1 }– 3 > 7x_{2}– 3⇒ f(x

_{1}) > f(x_{2})Thus, f(x) is strictly increasing on R