RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 (Updated for 2024)

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Exercise 17.2 Page No: 17.33

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

(i) f (x) = 10 – 6x – 2x²

Solution:

(ii) f (x) = x² + 2x – 5

Solution:

(iii) f (x) = 6 – 9x – x²

Solution:

(iv) f(x) = 2x³ – 12x² + 18x + 15

Solution:

(v) f (x) = 5 + 36x + 3x² – 2x³

Solution:

Given f (x) = 5 + 36x + 3x² – 2x³

⇒

⇒ f’(x) = 36 + 6x – 6x²

For f(x), now we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x² = 0

⇒ 6(–x² + x + 6) = 0

⇒ 6(–x² + 3x – 2x + 6) = 0

⇒ –x² + 3x – 2x + 6 = 0

⇒ x² – 3x + 2x – 6 = 0

⇒ (x – 3) (x + 2) = 0

⇒ x = 3, – 2

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

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

(vi) f (x) = 8 + 36x + 3x² – 2x³

Solution:

Given f (x) = 8 + 36x + 3x² – 2x³

Now differentiating with respect to x

⇒

⇒ f’(x) = 36 + 6x – 6x²

For f(x), we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x² = 0

⇒ 6(–x² + x + 6) = 0

⇒ 6(–x² + 3x – 2x + 6) = 0

⇒ –x² + 3x – 2x + 6 = 0

⇒ x² – 3x + 2x – 6 = 0

⇒ (x – 3) (x + 2) = 0

⇒ x = 3, – 2

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

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

(vii) f(x) = 5x³ – 15x² – 120x + 3

Solution:

Given f(x) = 5x³ – 15x² – 120x + 3

Now by differentiating the above equation with respect x, we get

⇒

⇒ f’(x) = 15x² – 30x – 120

For f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 15x² – 30x – 120 = 0

⇒ 15(x² – 2x – 8) = 0

⇒ 15(x² – 4x + 2x – 8) = 0

⇒ x² – 4x + 2x – 8 = 0

⇒ (x – 4) (x + 2) = 0

⇒ x = 4, – 2

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

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

(viii) f(x) = x³ – 6x² – 36x + 2

Solution:

Given f (x) = x³ – 6x² – 36x + 2

⇒

⇒ f’(x) = 3x² – 12x – 36

For f(x), we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 3x² – 12x – 36 = 0

⇒ 3(x² – 4x – 12) = 0

⇒ 3(x² – 6x + 2x – 12) = 0

⇒ x² – 6x + 2x – 12 = 0

⇒ (x – 6) (x + 2) = 0

⇒ x = 6, – 2

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

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

(ix) f(x) = 2x³ – 15x² + 36x + 1

Solution:

Given f (x) = 2x³ – 15x² + 36x + 1

Now by differentiating the above equation with respect x, we get

⇒

⇒ f’(x) = 6x² – 30x + 36

For f(x), we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 30x + 36 = 0

⇒ 6 (x² – 5x + 6) = 0

⇒ 6(x² – 3x – 2x + 6) = 0

⇒ x² – 3x – 2x + 6 = 0

⇒ (x – 3) (x – 2) = 0

⇒ x = 3, 2

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

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

(x) f (x) = 2x³ + 9x² + 12x + 20

Solution:

Given f (x) = 2x³ + 9x² + 12x + 20

Differentiating the above equation we get

⇒

⇒ f’(x) = 6x² + 18x + 12

For f(x), we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² + 18x + 12 = 0

⇒ 6(x² + 3x + 2) = 0

⇒ 6(x² + 2x + x + 2) = 0

⇒ x² + 2x + x + 2 = 0

⇒ (x + 2) (x + 1) = 0

⇒ x = –1, –2

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

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

2. Determine the values of x for which the function f(x) = x² – 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x² – 6x + 9 where the normal is parallel to the line y = x + 5.

Solution:

Given f(x) = x² – 6x + 9

⇒

⇒ f’(x) = 2x – 6

⇒ f’(x) = 2(x – 3)

For f(x), let us find critical point, we must have

⇒ f’(x) = 0

⇒ 2(x – 3) = 0

⇒ (x – 3) = 0

⇒ x = 3

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

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

Now, let us find the coordinates of the point

Equation of curve is f(x) = x² – 6x + 9

The slope of this curve is given by

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

Solution:

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

Solution:

Given f (x) = e^2x

⇒

⇒ f’(x) = 2e^2x

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

⇒ f’(x) > 0

⇒ 2e^2x > 0

⇒ e^2x > 0

Since the value of e lies between 2 and 3

So, whatever be the power of e (that is x in domain R) will be greater than zero.

Thus f(x) is increasing on interval R

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

Solution:

6. Show that f(x) = log_a x, 0 < a < 1 is a decreasing function for all x > 0.

Solution:

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:

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

Solution:

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

Solution:

Given f (x) = x – sin x

⇒

⇒ f’(x) = 1 – cos x

Now, as given x ϵ R

⇒ –1 < cos x < 1

⇒ –1 > cos x > 0

⇒ f’(x) > 0

Hence, the condition for f(x) to be increasing

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

10. Show that f(x) = x³ – 15x² + 75x – 50 is an increasing function for all x ϵ R.

Solution:

Given f(x) = x³ – 15x² + 75x – 50

⇒

⇒ f’(x) = 3x² – 30x + 75

⇒ f’(x) = 3(x² – 10x + 25)

⇒ f’(x) = 3(x – 5)²

Now, as given x ϵ R

⇒ (x – 5)² > 0

⇒ 3(x – 5)² > 0

⇒ f’(x) > 0

Hence, the condition for f(x) to be increasing

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

11. Show that f(x) = cos² x is a decreasing function on (0, π/2).

Solution:

Given f (x) = cos² x

⇒

⇒ f’(x) = 2 cos x (–sin x)

⇒ f’(x) = –2 sin (x) cos (x)

⇒ f’(x) = –sin2x

Now, as given x belongs to (0, π/2).

⇒ 2x ∈ (0,
π)

⇒ Sin (2x)> 0

⇒ –Sin (2x) < 0

⇒ f’(x) < 0

Hence, the condition for f(x) to be decreasing

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

Hence proved

12. Show that f(x) = sin x is an increasing function on (–π/2, π/2).

Solution:

Given f (x) = sin x

⇒

⇒ f’(x) = cos x

Now, as given x ∈ (–π/2, π/2).

That is 4^th quadrant, where

⇒ Cos x> 0

⇒ f’(x) > 0

Hence, the condition for f(x) to be increasing

Thus f(x) is increasing on interval (–π/2, π/2).

13. Show that f(x) = cos x is a decreasing function on (0, π), increasing in (–π, 0) and neither increasing nor decreasing in (–π, π).

Solution:

Given f(x) = cos x

⇒

⇒ f’(x) = –sin x

Taking different regions from 0 to 2π

Let x ∈ (0, π).

⇒ Sin(x) > 0

⇒ –sin x < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in (0, π)

Let x ∈ (–π, o).

⇒ Sin (x) < 0

⇒ –sin x > 0

⇒ f’(x) > 0

Thus f(x) is increasing in (–π, 0).

Therefore, from the above condition, we find that

⇒ f (x) is decreasing in (0, π) and increasing in (–π, 0).

Hence, condition for f(x) neither increasing nor decreasing in (–π, π)

14. Show that f(x) = tan x is an increasing function on (–π/2, π/2).

Solution:

Given f (x) = tan x

⇒

⇒ f’(x) = sec²x

Now, as given

x ∈ (–π/2, π/2).

That is 4^th quadrant, where

⇒ sec²x > 0

⇒ f’(x) > 0

Hence, the condition for f(x) to be increasing

Thus f(x) is increasing on interval (–π/2, π/2).

15. Show that f(x) = tan^–1 (sin x + cos x) is a decreasing function on the interval (π/4, π /2).

Solution:

16. Show that the function f (x) = sin (2x + π/4) is decreasing on (3π/8, 5π/8).

Solution:

Thus f (x) is decreasing on the interval (3π/8, 5π/8).

17. Show that the function f(x) = cot^–1 (sin x + cos x) is decreasing on (0, π/4) and increasing on (π/4, π/2).

Solution:

Given f(x) = cot^–1 (sin x + cos x)

18. Show that f(x) = (x – 1) e^x + 1 is an increasing function for all x > 0.

Solution:

Given f (x) = (x – 1) e^x + 1

Now differentiating the given equation with respect to x, we get

⇒

⇒ f’(x) = e^x + (x – 1) e^x

⇒ f’(x) = e^x(1+ x – 1)

⇒ f’(x) = x e^x

As given x > 0

⇒ e^x > 0

⇒ x e^x > 0

⇒ f’(x) > 0

Hence, the condition for f(x) to be increasing

Thus f(x) is increasing on interval x > 0

19. Show that the function x² – x + 1 is neither increasing nor decreasing on (0, 1).

Solution:

Given f(x) = x² – x + 1

Now by differentiating the given equation with respect to x, we get

⇒

⇒ f’(x) = 2x – 1

Taking different regions from (0, 1)

Let x ∈ (0, ½)

⇒ 2x – 1 < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in (0, ½)

Let x ∈ (½, 1)

⇒ 2x – 1 > 0

⇒ f’(x) > 0

Thus f(x) is increasing in (½, 1)

Therefore, from the above condition, we find that

⇒ f (x) is decreasing in (0, ½)  and increasing in (½, 1)

Hence, the condition for f(x) is neither increasing nor decreasing in (0, 1)

20. Show that f(x) = x⁹ + 4x⁷ + 11 is an increasing function for all x ϵ R.

Solution:

Given f (x) = x⁹ + 4x⁷ + 11

Now by differentiating the above equation with respect to x, we get

⇒

⇒ f’(x) = 9x⁸ + 28x⁶

⇒ f’(x) = x⁶(9x² + 28)

As given x ϵ R

⇒ x⁶ > 0 and 9x² + 28 > 0

⇒ x⁶ (9x² + 28) > 0

⇒ f’(x) > 0

Hence, the condition for f(x) to be increasing

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

RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2: Important Topics From The Chapter

Some of the essential topics of RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2 are listed below.

• The solution of rational algebraic inequations
• Strictly increasing functions
• Strictly decreasing functions
• Monotonic functions
• Monotonically increasing function
• Monotonically decreasing functions
• Necessary and sufficient conditions for monotonicity
• Finding the intervals in which a function is increasing or decreasing
• Proving the monotonicity of a function on a given interval
• Finding the interval in which a function is increasing or decreasing

We have included all the information regarding CBSE RD Sharma Solutions Class 12 Maths Chapter 17 Exercise 17.2. If you have any queries feel free to ask in the comment section. 

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