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.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 17 |

Exercise | 17.1 |

Category | RD Sharma Solutions |

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

**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

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