# Class 12 RD Sharma Solutions – Chapter 17 Increasing and Decreasing Functions – Exercise 17.2 | Set 1

### Question 1. Find the intervals in which the following functions are increasing or decreasing.

### (i) f(x) = 10 – 6x – 2x^{2}

**Solution:**

We are given,

f(x) = 10 – 6x – 2x

^{2}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 – 6 – 4x

f'(x) = – 6 – 4x

For f(x) to be increasing, we must have,

=> f'(x) > 0

=> – 6 – 4x > 0

=> – 4x > 6

=> x < –6/4

=> x < –3/2

=> x ∈ (–∞, –3/2)

For f(x) to be decreasing, we must have,

=> f'(x) < 0

=> – 6 – 4x < 0

=> – 4x < 6

=> x > –6/4

=> x > –3/2

=> x ∈ ( –3/2, ∞)

Thus, f(x) is increasing on the interval x ∈ (–∞, –3/2) and decreasing on the interval x ∈ ( –3/2, ∞).

### (ii) f(x) = x^{2} + 2x – 5

**Solution:**

We are given,

f(x) = x

^{2}+ 2x – 5On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 2x + 2 – 0

f'(x) = 2x + 2

For f(x) to be increasing, we must have,

=> f'(x) > 0

=> 2x + 2 > 0

=> 2x > –2

=> x > –2/2

=> x > –1

=> x ∈ (–1, ∞)

For f(x) to be decreasing, we must have,

=> f'(x) < 0

=> 2x + 2 < 0

=> 2x < –2

=> x < –2/2

=> x < –1

=> x ∈ (–∞, –1)

Thus, f(x) is increasing on the interval x ∈ (–1, ∞) and decreasing on the interval x ∈ ( –∞, –1).

### (iii) f(x) = 6 – 9x – x^{2}

**Solution:**

We are given,

f(x) = 6 – 9x – x

^{2}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 – 9 – 2x

f'(x) = – 9 – 2x

For f(x) to be increasing, we must have,

=> f'(x) > 0

=> –9 – 2x > 0

=> –2x > 9

=> x > –9/2

=> x ∈ (–9/2, ∞)

For f(x) to be decreasing, we must have,

=> f'(x) < 0

=> –9 – 2x < 0

=> –2x < 9

=> x < –9/2

=> x ∈ (–∞, –9/2)

Thus, f(x) is increasing on the interval x ∈ (–9/2, ∞) and decreasing on the interval x ∈ ( –∞, –9/2).

### (iv) f(x) = 2x^{3} – 12x^{2} + 18x + 15

**Solution:**

We are given,

f(x) = 2x

^{3}– 12x^{2 }+ 18x + 15On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{2}– 24x + 18 + 0f'(x) = 6x

^{2}– 24x + 18For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x

^{2}– 24x + 18 = 0=> 6 (x

^{2}– 4x + 3) = 0=> x

^{2}– 4x + 3 = 0=> x

^{2}– 3x – x + 3 = 0=> x (x – 3) – 1 (x – 3) = 0

=> (x – 1) (x – 3) = 0

=> x = 1, 3

Clearly, f'(x) > 0 if x < 1 and x > 3.

Also, f'(x) < 0, if 1 < x < 3.

Thus, f(x) is increasing on the interval x ∈ (–∞, 1)∪ (3, ∞) and decreasing on the interval x ∈ (1, 3).

### (v) f(x) = 5 + 36x + 3x^{2} – 2x^{3}

**Solution:**

We are given,

f(x) = 5 + 36x + 3x

^{2}– 2x^{3}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 + 36 + 6x – 6x

^{2}f'(x) = 36 + 6x – 6x

^{2}For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 36 + 6x – 6x

^{2}= 0=> 6 (– x

^{2}+ x + 6) = 0=> 6 (–x

^{2}+ 3x – 2x + 6) = 0=> –x

^{2}+ 3x – 2x + 6 = 0=> x

^{2}– 3x + 2x – 6 = 0=> (x – 3) (x + 2) = 0

=> x = 3, – 2

Clearly, f’(x) > 0 if –2 < x < 3.

Also f’(x) < 0 if x < –2 and x > 3.

Thus, f(x) is increasing on x ∈ (–2, 3) and f(x) is decreasing on interval x ∈(–∞, –2) ∪ (3, ∞).

### (vi) f(x) = 8 + 36x + 3x^{2} – 2x^{3}

**Solution:**

We are given,

f(x) = 8 + 36x + 3x

^{2}– 2x^{3}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 + 36 + 6x – 6x

^{2}f'(x) = 36 + 6x – 6x

^{2}For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 36 + 6x – 6x

^{2}= 0=> 6 (– x

^{2}+ x + 6) = 0=> 6 (–x

^{2}+ 3x – 2x + 6) = 0=> –x

^{2}+ 3x – 2x + 6 = 0=> x

^{2}– 3x + 2x – 6 = 0=> (x – 3) (x + 2) = 0

=> x = 3, –2

Clearly, f’(x) > 0 if –2 < x < 3.

Also f’(x) < 0 if x < –2 and x > 3.

Thus, f(x) is increasing on x ∈ (–2, 3) and f(x) is decreasing on interval x ∈(–∞, –2) ∪ (3, ∞).

### (vii) f(x) = 5x^{3} – 15x^{2} – 120x + 3

**Solution:**

We are given,

f(x) = 5x

^{3}– 15x^{2}– 120x + 3On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 15x

^{2 }– 30x – 120 + 0f'(x) = 15x

^{2}– 30x – 120For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 15x

^{2}– 30x – 120 = 0=> 15(x

^{2}– 2x – 8) = 0=> 15(x

^{2}– 4x + 2x – 8) = 0=> x

^{2}– 4x + 2x – 8 = 0=> (x – 4) (x + 2) = 0

=> x = 4, –2

Clearly, f’(x) > 0 if x < –2 and x > 4.

Also f’(x) < 0 if –2 < x < 4.

Thus, f(x) is increasing on x ∈ (–∞,–2) ∪ (4, ∞) and f(x) is decreasing on interval x ∈ (–2, 4).

### (viii) f(x) = x^{3} – 6x^{2 }– 36x + 2

**Solution:**

We are given,

f(x) = x

^{3}– 6x^{2}– 36x + 2On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x

^{2}– 12x – 36 + 0f'(x) = 3x

^{2}– 12x – 36For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 3x

^{2}– 12x – 36 = 0=> 3(x

^{2}– 4x – 12) = 0=> 3(x

^{2}– 6x + 2x – 12) = 0=> x

^{2}– 6x + 2x – 12 = 0=> (x – 6) (x + 2) = 0

=> x = 6, –2

Clearly, f’(x) > 0 if x < –2 and x > 6.

Also f’(x) < 0 if –2< x < 6

Thus, f(x) is increasing on x ∈(–∞,–2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (–2, 6).

### (ix) f(x) = 2x^{3} – 15x^{2} + 36x + 1

**Solution:**

We are given,

f(x) = 2x

^{3}– 15x^{2}+ 36x + 1On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{2 }– 30x + 36 + 0f'(x) = 6x

^{2}– 30x + 36For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x

^{2}– 30x + 36 = 0=> 6 (x

^{2}– 5x + 6) = 0=> 6(x

^{2}– 3x – 2x + 6) = 0=> x

^{2}– 3x – 2x + 6 = 0=> (x – 3) (x – 2) = 0

=> x = 3, 2

Clearly, f’(x) > 0 if x < 2 and x > 3.

Also f’(x) < 0 if 2 < x < 3.

Thus, f(x) is increasing on x ∈ (–∞, 2) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (2, 3).

### (x) f(x) = 2x^{3} + 9x^{2} + 12x + 1

**Solution:**

We are given,

f(x) = 2x

^{3}+ 9x^{2}+ 12x + 1On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{2}+ 18x + 12 + 0f'(x) = 6x

^{2}+ 18x + 12For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x

^{2}+ 18x + 12 = 0=> 6 (x

^{2}+ 3x + 2) = 0=> 6(x

^{2}+ 2x + x + 2) = 0=> x

^{2}+ 2x + x + 2 = 0=> (x + 2) (x + 1) = 0

=> x = –1, –2

Clearly, f’(x) > 0 if –2 < x < –1.

Also f’(x) < 0 if x < –1 and x > –2.

Thus, f(x) is increasing on x ∈ (–2,–1) and f(x) is decreasing on interval x ∈(–∞, –2) ∪ (–2, ∞).

### (xi) f(x) = 2x^{3} – 9x^{2} + 12x – 5

**Solution:**

We are given,

f(x) = 2x

^{3}– 9x^{2}+ 12x – 5On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{2}– 18x + 12 – 0f'(x) = 6x

^{2}– 18x + 12For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x

^{2}– 18x + 12 = 0=> 6 (x

^{2}– 3x + 2) = 0=> 6(x

^{2}– 2x – x + 2) = 0=> x

^{2}– 2x – x + 2 = 0=> (x – 2) (x – 1) = 0

=> x = 1, 2

Clearly, f’(x) > 0 if x < 1 and x > 2.

Also f’(x) < 0 if 1 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (1, 2).

### (xii) f(x) = 6 + 12x + 3x^{2} – 2x^{3}

**Solution:**

We are given,

f(x) = 6 + 12x + 3x

^{2}– 2x^{3}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 + 12 + 6x – 6x

^{2}f'(x) = 12 + 6x – 6x

^{2}For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 12 + 6x – 6x

^{2}= 0=> 6 (–x

^{2}+ x + 2) = 0=> x

^{2}– x – 2 = 0=> x

^{2}– 2x + x – 2 = 0=> (x – 2) (x + 1) = 0

=> x = 2, –1

Clearly, f’(x) > 0 if –1 < x < 2.

Also f’(x) < 0 if x < –1 and x > 2.

Thus, f(x) is increasing on x ∈ (–1, 2) and f(x) is decreasing on interval x ∈(–∞, –1) ∪ (2, ∞).

### (xiii) f(x) = 2x^{3} – 24x + 107

**Solution:**

We are given,

f(x) = 2x

^{3}– 24x + 107On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{2}– 24 + 0f'(x) = 6x

^{2 }– 24For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x

^{2}– 24 = 0=> 6x

^{2}= 24=> x

^{2}= 4=> x = 2, –2

Clearly, f’(x) > 0 if x < –2 and x > 2.

Also f’(x) < 0 if –2 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, –2) ∪ (2, ∞), and f(x) is decreasing on interval x ∈ (–2, 2).

### (xiv) f(x) = –2x^{3} – 9x^{2} – 12x + 1

**Solution:**

We are given,

f(x) = –2x

^{3}– 9x^{2}– 12x + 1On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = –6x

^{2}– 18x – 12 + 0f'(x) = –6x

^{2 }– 18x – 12For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> –6x

^{2}– 18x – 12 = 0=> 6 (–x

^{2}– 3x – 2) = 0=> x

^{2}+ 3x + 2 = 0=> x

^{2}+ 2x + x + 2 = 0=> (x + 2) (x + 1) = 0

=> x = –2, –1

Clearly, f’(x) > 0 if x < –1 and x > –2.

Also, f’(x) < 0 if –2 < x < –1.

Thus, f(x) is increasing on x ∈ (–2, –1) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (–1, ∞).

### (xv) f(x) = (x – 1) (x – 2)^{2}

**Solution:**

We are given,

f(x) = (x – 1) (x – 2)

^{2}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = (x – 2)

^{2}+ 2 (x – 1) (x – 2)f'(x) = (x – 2) (x – 2 + 2x – 2)

f'(x) = (x – 2) (3x – 4)

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> (x – 2) (3x – 4) = 0

=> x = 2, 4/3

Clearly, f’(x) > 0 if x < 4/3 and x > 2.

Also, f’(x) < 0 if 4/3 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, 4/3) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (4/3, 2).

### (xvi) f(x) = x^{3} – 12x^{2} + 36x + 17

**Solution:**

We are given,

f(x) = x

^{3}– 12x^{2}+ 36x + 17On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x

^{2}– 24x + 36 + 0f'(x) = 3x

^{2}– 24x + 36For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 3x

^{2}– 24x + 36 = 0=> 3 (x

^{2}– 8x + 12) = 0=> x

^{2}– 8x + 12 = 0=> x

^{2}– 6x – 2x + 12 = 0=> (x – 6) (x – 2) = 0

=> x = 6, 2

Clearly, f’(x) > 0 if x < 2 and x > 6.

Also, f’(x) < 0 if 2 < x < 6.

Thus, f(x) is increasing on x ∈ (–∞, 2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (2, 6).

### (xvii) f(x) = 2x^{3} – 24x + 7

**Solution:**

We are given,

f(x) = 2x

^{3}– 24x + 7On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{2}– 24 + 0f'(x) = 6x

^{2}– 24For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x

^{2}– 24 = 0=> 6x

^{2}= 24=> x

^{2}= 4=> x = 2, –2

Clearly, f’(x) > 0 if x < –2 and x > 2.

Also f’(x) < 0 if –2 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, –2) ∪ (2, ∞), and f(x) is decreasing on interval x ∈ (–2, 2).

### (xviii) f(x) = 3x^{4}/10 – 4x^{3}/5 -3x^{2} + 36x/5 + 11

**Solution:**

We are given,

f(x) = 3x

^{4}/10 – 4x^{3}/5 -3x^{2}+ 36x/5 + 11On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{3}/5 – 12x^{2}/5 -3(2x) + 36/5f'(x) = 6/5[(x – 1)(x + 2)(x – 3)]

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6/5[(x – 1)(x + 2)(x – 3)] = 0

=> x = 1, –2, 3

Clearly, f’(x) > 0 if –2 < x < 1 and if x > 3

Also f’(x) < 0 if 1 < x < 3.

Thus, f(x) is increasing on x ∈ (3, ∞) and f(x) is decreasing on interval x ∈ (1, 3).

### (xix) f(x) = x^{4} – 4x

**Solution:**

We are given,

f(x) = x

^{4}– 4xOn differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 4x

^{3}– 4For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 4x

^{3}– 4 = 0=> 4 (x

^{3}– 1) = 0=> x

^{3}– 1 = 0=> x

^{3}= 1=> x = 1

Clearly, f’(x) > 0 if x > 1.

Also f’(x) < 0 if x < 1.

Thus, f(x) is increasing on x ∈ (1, ∞), and f(x) is decreasing on interval x ∈ (–∞, 1).

### (xx) f(x) = x^{4}/4 + 2/3x^{3} – 5/2x^{2} – 6x + 7

**Solution:**

We have,

f(x) = x

^{4}/4 + 2/3x^{3}– 5/2x^{2}– 6x + 7On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 4x

^{3}/4 + 6x^{2}/3 – 10x/2 – 6 + 0f'(x) = x

^{3}+ 2x^{2}– 5x – 6For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> x

^{3}+ 2x^{2}– 5x – 6 = 0=> (x + 1) (x + 3) (x – 2) = 0

=> x = –1, –3, 2

Clearly f'(x) > 0 if –3 < x < –1 and x > 2.

Also f'(x) < 0 if x < –3 and –1 < x < 2.

Thus, f(x) is increasing on x ∈ (–3, –1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, –3) ∪ (–1, 2).

### (xxi) f(x) = x^{4 }– 4x^{3} + 4x^{2} + 15

**Solution:**

We have,

f(x) = x

^{4}– 4x^{3}+ 4x^{2}+ 15On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 4x

^{3}– 12x^{2}+ 8x + 0f'(x) = 4x

^{3}– 12x^{2}+ 8xFor f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 4x

^{3}– 12x^{2}+ 8x = 0=> 4x (x

^{2}– 3x + 2) = 0=> 4x (x – 2) (x – 1) = 0

=> x = 0, 1, 2

Clearly f'(x) > 0 if 0 < x < 1 and x > 2.

Also f'(x) < 0 if x < 0 and 1 < x < 2.

Thus, f(x) is increasing on x ∈ (0, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0) ∪ (1, 2).

### (xxii) f(x) = , x > 0

**Solution:**

We have,

f(x) =

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> = 0

=> = 0

=> x

^{1/2}(1 – x) = 0=> x = 0, 1

Clearly f'(x) > 0 if 0 < x < 1.

Also f'(x) < 0 if x > 0.

Thus, f(x) is increasing on x ∈ (0, 1) and f(x) is decreasing on interval x ∈ (1, ∞).

### (xxiii) f(x) = x^{8} + 6x^{2}

**Solution:**

We have,

f(x) = x

^{8}+ 6x^{2}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 8x

^{7}+ 12xFor f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 8x

^{7}+ 12x = 0=> 4x (2x

^{6}+ 3) = 0=> x = 0

Clearly f'(x) > 0 if x > 0.

Also f'(x) < 0 if x < 0.

Thus, f(x) is increasing on x ∈ (0, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0).

### (xxiv) f(x) = x^{3} – 6x^{2} + 9x + 15

**Solution:**

We are given,

f(x) = x

^{3}– 6x^{2}+ 9x + 15On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x

^{2}– 12x + 9 + 0f'(x) = 3x

^{2}– 12x + 9For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 3x

^{2}– 12x + 9 = 0=> 3 (x

^{2}– 4x + 3) = 0=> x

^{2}– 4x + 3 = 0=> x

^{2}– 3x – x + 3 = 0=> (x – 3) (x – 1) = 0

=> x = 3, 1

Clearly f'(x) > 0 if x < 1 and x > 3.

Also f'(x) < 0 if 1 < x < 3.

Thus, f(x) is increasing on x ∈ (–∞, 1) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (1, 3).

### (xxv) f(x) = [x(x – 2)]^{2}

**Solution:**

We are given,

f(x) = [x(x – 2)]

^{2}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

f'(x) = 2 (x

^{2}– 2x) (2x – 2)f'(x) = 4x (x – 2) (x – 1)

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 4x (x – 2) (x – 1) = 0

=> x = 0, 1, 2

Clearly f'(x) > 0 if 0 < x < 1 and x > 2.

Also f'(x) < 0 if x < 0 and 1< x < 2.

Thus, f(x) is increasing on x ∈ (0, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0) ∪ (1, 2).

### (xxvi) f(x) = 3x^{4} – 4x^{3} – 12x^{2} + 5

**Solution:**

We are given,

f(x) = 3x

^{4}– 4x^{3}– 12x^{2 }+ 5On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 12x

^{3}– 12x^{2}– 24xFor f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 12x

^{3}– 12x^{2}– 24x = 0=> 12x (x

^{2}– x – 2) = 0=> 12x (x + 1) (x – 2) = 0

=> x = 0, –1, 2

Clearly f'(x) > 0 if –1 < x < 0 and x > 2.

Also f'(x) < 0 if x < –1 and 0< x < 2.

Thus, f(x) is increasing on x ∈ (–1, 0) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, –1) ∪ (0, 2).

### (xxvii) f(x) = 3x^{4}/2 – 4x^{3} – 45x^{2} + 51

**Solution:**

We have,

f(x) = 3x

^{4}/2 – 4x^{3}– 45x^{2}+ 51On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x

^{3}– 12x^{2}– 90xFor f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x

^{3}– 12x^{2}– 90x = 0=> 6x (x

^{2}– 2x – 15) = 0=> 6x (x + 3) (x – 5) = 0

=> x = 0, –3, 5

Clearly f'(x) > 0 if –3 < x < 0 and x > 5.

Also f'(x) < 0 if x < –3 and 0< x < 5.

Thus, f(x) is increasing on x ∈ (–3, 0) ∪ (5, ∞) and f(x) is decreasing on interval x ∈ (–∞, –3) ∪ (0, 5).

### (xxvii) f(x) =

**Solution:**

We have,

f(x) =

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

f'(x) =

f'(x) =

f'(x) =

f'(x) =

Clearly f'(x) > 0 if x > 2.

Also f'(x) < 0 if x < 2

Thus, f(x) is increasing on x ∈ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 2).

### Question 2. Determine the values of x for which the function f(x) = x^{2} – 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x^{2} – 6x + 9 where the normal is parallel to the line y = x + 5.

**Solution:**

Given f(x) = x

^{2}– 6x + 9On differentiating both sides with respect to x, we get

=> f’(x) = 2x – 6

=> f’(x) = 2(x – 3)

For f(x), we need to find the critical point, so we get,

=> f’(x) = 0

=> 2(x – 3) = 0

=> (x – 3) = 0

=> x = 3

Clearly, f’(x) > 0 if x > 3.

Also f’(x) < 0 if x < 3.

Thus, f(x) is increasing on (3, ∞) and f(x) is decreasing on interval x ∈ (–∞, 3).

Equation of the given curve is f(x) = x

^{2}– 6x + 9.Slope of this curve is given by,

=> m

_{1}= dy/dx=> m

_{1}= 2x – 6And slope of the line is y = x + 5

Slope of this curve is given by,

=> m

_{2}= dy/dx=> m

_{2}= 1Now according to the question,

=> m

_{1}m_{2}= –1=> 2x – 6 = –1

=> 2x = 5

=> x = 5/2

Putting x = 5/2 in the curve y = x

^{2}– 6x + 9, we get,=> y = (5/2)

^{2}– 6 (5/2) + 9=> y = 25/4 – 15 + 9

=> y = 1/4

Therefore, the required coordinates are (5/2, 1/4).

### Question 3. Find the intervals in which f(x) = sin x – cos x, where 0 < x < 2π is increasing or decreasing.

**Solution:**

We have,

f(x) = sin x – cos x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(sin x – cos x)

f'(x) = cos x + sin x

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> cos x + sin x = 0

=> 1 + tan x = 0

=> tan x = –1

=> x = 3π/4 , 7π/4

Clearly f'(x) > 0 if 0 < x < 3π/4 and 7π/4 < x < 2π.

Also f'(x) < 0 if 3π/4 < x < 7π/4.

Thus, f(x) is increasing on x ∈ (0, 3π/4) ∪ (7π/4, 2π) and f(x) is decreasing on interval x ∈ (3π/4, 7π/4).

### Question 4. Show that f(x) = e^{2x} is increasing on R.

**Solution:**

We have,

=> f(x) = e

^{2x}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 2e

^{2x}For f(x) to be increasing, we must have

=> f’(x) > 0

=> 2e

^{2x}> 0=> e

^{2x}> 0Now we know, the value of e lies between 2 and 3. Therefore, f(x) will be always greater than zero.

Thus, f(x) is increasing on interval R.

Hence proved.

### Question 5. Show that f(x) = e^{1/x}, x ≠ 0 is a decreasing function for all x ≠ 0.

**Solution:**

We have,

=> f(x) = e

^{1/x}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = -e

^{x}/x^{2}As x ∈ R, we have,

=> e

^{x}> 0Also, we get,

=> 1/x

^{2}> 0This means, e

^{x}/x^{2}> 0=> -e

^{x}/x^{2}< 0Thus, f(x) is a decreasing function for all x ≠ 0.

Hence proved.

### Question 6. Show that f(x) = log_{a} x, 0 < a < 1 is a decreasing function for all x > 0.

**Solution:**

We have,

=> f(x) = log

_{a}xOn differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 1/xloga

As we are given 0 < a < 1,

=> log a < 0

And for x > 0, 1/x > 0

Therefore, f'(x) is,

=> 1/xloga < 0

=> f'(x) < 0

Thus, f(x) is a decreasing function for all x > 0.

Hence proved.

### Question 7. Show that f(x) = sin x is increasing on (0, π/2) and decreasing on (π/2, π) and neither increasing nor decreasing in (0, π).

**Solution:**

We have,

f(x) = sin x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = cos x

Now for 0 < x < π/2,

=> cos x > 0

=> f'(x) > 0

And for π/2 < x < π,

=> cos x < 0

=> f'(x) < 0

Thus, f(x) is increasing on x ∈ (0, π/2) and f(x) is decreasing on interval x ∈ (π/2, π).

Hence f(x) is neither increasing nor decreasing in (0, π).

Hence proved.

### Question 8. Show that f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).

**Solution:**

We have,

f(x) = log sin x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = (1/sinx)cosx

f'(x) = cot x

Now for 0 < x < π/2,

=> cot x > 0

=> f'(x) > 0

And for π/2 < x < π,

=> cos x < 0

=> f'(x) < 0

Thus, f(x) is increasing on x ∈ (0, π/2) and f(x) is decreasing on interval x ∈ (π/2, π).

Hence proved.

### Question 9. Show that f(x) = x – sin x is increasing for all x ∈ R.

**Solution:**

We have,

f(x) = x – sin x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 1 – cos x

Now, we are given x ∈ R, we get

=> –1 < cos x < 1

=> –1 > cos x > 0

=> f’(x) > 0

Thus, f(x) is increasing on interval x ∈ R.

Hence proved.

### Question 10. Show that f(x) = x^{3} – 15x^{2} + 75x – 50 is an increasing function for all x ∈ R.

**Solution:**

We have,

f(x) = x

^{3}– 15x^{2}+ 75x – 50On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x

^{2}– 30x + 75 – 0f'(x) = 3x

^{2}– 30x + 75f’(x) = 3(x

^{2}– 10x + 25)f’(x) = 3(x – 5)

^{2}Now, as we are given x ϵ R, we get

=> (x – 5)

^{2}> 0=> 3(x – 5)

^{2}> 0=> f’(x) > 0

Thus, f(x) is increasing on interval x ∈ R.

Hence proved.

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