This exercise explains the monotonicity of a given function. RD Sharma Solutions are framed by a team of experts who have extensive knowledge about the respective subjects. The PDF of solutions of exercise 17.2 primarily helps students from the exam point of view and by practising regularly, students can obtain worthy results in the Class 12 exams. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 17 Exercise 17.2 is available here. Some of the important topics of this exercise are listed below.

- 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

**RD Sharma Solutions For Class 12 Increasing and Decreasing Functions Exercise 17.2:**Download PDF Here

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**Page No: 17.33**

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

**(i) f (x) = 10 â€“ 6x â€“ 2x ^{2}**

**Solution:**

**(ii) f (x) = x ^{2}Â + 2x â€“ 5**

**Solution:**

**(iii) f (x) = 6 â€“ 9x â€“ x ^{2}**

**Solution:**

**(iv) f(x) = 2x ^{3}Â â€“ 12x^{2}Â + 18x + 15**

**Solution:**

**(v) f (x) = 5 + 36x + 3x ^{2}Â â€“ 2x^{3}**

**Solution:**

Given f (x) = 5 + 36x + 3x^{2}Â â€“ 2x^{3}

â‡’Â

â‡’Â fâ€™(x) = 36 + 6x â€“ 6x^{2}

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

â‡’Â fâ€™(x) = 0

â‡’Â 36 + 6x â€“ 6x^{2}Â = 0

â‡’Â 6(â€“x^{2}Â + x + 6) = 0

â‡’Â 6(â€“x^{2}Â + 3x â€“ 2x + 6) = 0

â‡’Â â€“x^{2}Â + 3x â€“ 2x + 6 = 0

â‡’Â x^{2}Â â€“ 3x + 2x â€“ 6 = 0

â‡’Â (x â€“ 3) (x + 2) = 0

â‡’Â x = 3, â€“ 2

Clearly, fâ€™(x) > 0 if â€“2< x < 3 and fâ€™(x) < 0 if x < â€“2 and x > 3

Thus, f(x) increases on xÂ âˆˆÂ (â€“2, 3) and f(x) is decreasing on interval (â€“âˆž, â€“2)Â âˆªÂ (3, âˆž)

**(vi) f (x) = 8 + 36x + 3x ^{2}Â â€“ 2x^{3}**

**Solution:**

Given f (x) = 8 + 36x + 3x^{2}Â â€“ 2x^{3}

Now differentiating with respect to x

â‡’Â

â‡’Â fâ€™(x) = 36 + 6x â€“ 6x^{2}

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

â‡’Â fâ€™(x) = 0

â‡’Â 36 + 6x â€“ 6x^{2}Â = 0

â‡’Â 6(â€“x^{2}Â + x + 6) = 0

â‡’Â 6(â€“x^{2}Â + 3x â€“ 2x + 6) = 0

â‡’Â â€“x^{2}Â + 3x â€“ 2x + 6 = 0

â‡’Â x^{2}Â â€“ 3x + 2x â€“ 6 = 0

â‡’Â (x â€“ 3) (x + 2) = 0

â‡’Â x = 3, â€“ 2

Clearly, fâ€™(x) > 0 if â€“2 < x < 3 and fâ€™(x) < 0 if x < â€“2 and x > 3

Thus, f(x) increases on xÂ âˆˆÂ (â€“2, 3) and f(x) is decreasing on interval (â€“âˆž, 2)Â âˆªÂ (3, âˆž)

**(vii) f(x) = 5x ^{3}Â â€“ 15x^{2}Â â€“ 120x + 3**

**Solution:**

Given f(x) = 5x^{3}Â â€“ 15x^{2}Â â€“ 120x + 3

Now by differentiating above equation with respect x, we get

â‡’Â

â‡’Â fâ€™(x) = 15x^{2}Â â€“ 30x â€“ 120

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

â‡’Â fâ€™(x) = 0

â‡’Â 15x^{2}Â â€“ 30x â€“ 120 = 0

â‡’Â 15(x^{2}Â â€“ 2x â€“ 8) = 0

â‡’Â 15(x^{2}Â â€“ 4x + 2x â€“ 8) = 0

â‡’Â x^{2}Â â€“ 4x + 2x â€“ 8 = 0

â‡’Â (x â€“ 4) (x + 2) = 0

â‡’Â x = 4, â€“ 2

Clearly, fâ€™(x) > 0 if x < â€“2 and x > 4 and fâ€™(x) < 0 if â€“2 < x < 4

Thus, f(x) increases on (â€“âˆž,â€“2)Â âˆªÂ (4, âˆž) and f(x) is decreasing on interval xÂ âˆˆÂ (â€“2, 4)

**(viii) f(x) = x ^{3}Â â€“ 6x^{2}Â â€“ 36x + 2**

**Solution:**

Given f (x) = x^{3}Â â€“ 6x^{2}Â â€“ 36x + 2

â‡’Â

â‡’Â fâ€™(x) = 3x^{2}Â â€“ 12x â€“ 36

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

â‡’Â fâ€™(x) = 0

â‡’Â 3x^{2}Â â€“ 12x â€“ 36 = 0

â‡’Â 3(x^{2}Â â€“ 4x â€“ 12) = 0

â‡’Â 3(x^{2}Â â€“ 6x + 2x â€“ 12) = 0

â‡’Â x^{2}Â â€“ 6x + 2x â€“ 12 = 0

â‡’Â (x â€“ 6) (x + 2) = 0

â‡’Â x = 6, â€“ 2

Clearly, fâ€™(x) > 0 if x < â€“2 and x > 6 and fâ€™(x) < 0 if â€“2< x < 6

Thus, f(x) increases on (â€“âˆž,â€“2)Â âˆªÂ (6, âˆž) and f(x) is decreasing on interval xÂ âˆˆÂ (â€“2, 6)

**(ix) f(x) = 2x ^{3}Â â€“ 15x^{2}Â + 36x + 1**

**Solution:**

Given f (x) = 2x^{3}Â â€“ 15x^{2}Â + 36x + 1

Now by differentiating above equation with respect x, we get

â‡’Â

â‡’Â fâ€™(x) = 6x^{2}Â â€“ 30x + 36

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

â‡’Â fâ€™(x) = 0

â‡’Â 6x^{2}Â â€“ 30x + 36 = 0

â‡’Â 6 (x^{2}Â â€“ 5x + 6) = 0

â‡’Â 3(x^{2}Â â€“ 3x â€“ 2x + 6) = 0

â‡’Â x^{2}Â â€“ 3x â€“ 2x + 6 = 0

â‡’Â (x â€“ 3) (x â€“ 2) = 0

â‡’Â x = 3, 2

Clearly, fâ€™(x) > 0 if x < 2 and x > 3 and fâ€™(x) < 0 if 2 < x < 3

Thus, f(x) increases on (â€“âˆž, 2)Â âˆªÂ (3, âˆž) and f(x) is decreasing on interval xÂ âˆˆÂ (2, 3)

**(x) f (x) = 2x ^{3}Â + 9x^{2}Â + 12x + 20**

**Solution:**

Given f (x) = 2x^{3}Â + 9x^{2}Â + 12x + 20

Differentiating above equation we get

â‡’Â

â‡’Â fâ€™(x) = 6x^{2}Â + 18x + 12

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

â‡’Â fâ€™(x) = 0

â‡’Â 6x^{2}Â + 18x + 12 = 0

â‡’Â 6(x^{2}Â + 3x + 2) = 0

â‡’Â 6(x^{2}Â + 2x + x + 2) = 0

â‡’Â x^{2}Â + 2x + x + 2 = 0

â‡’Â (x + 2) (x + 1) = 0

â‡’Â x = â€“1, â€“2

Clearly, fâ€™(x) > 0 if â€“2 < x < â€“1 and fâ€™(x) < 0 if x < â€“1 and x > â€“2

Thus, f(x) increases on xÂ âˆˆÂ (â€“2,â€“1) and f(x) is decreasing on interval (â€“âˆž, â€“2)Â âˆªÂ (â€“2, âˆž)

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

**Solution:**

Given f(x) = x^{2}Â â€“ 6x + 9

â‡’Â

â‡’Â fâ€™(x) = 2x â€“ 6

â‡’Â fâ€™(x) = 2(x â€“ 3)

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

â‡’Â fâ€™(x) = 0

â‡’Â 2(x â€“ 3) = 0

â‡’Â (x â€“ 3) = 0

â‡’Â x = 3

Clearly, fâ€™(x) > 0 if x > 3 and fâ€™(x) < 0 if x < 3

Thus, f(x) increases on (3, âˆž) and f(x) is decreasing on interval xÂ âˆˆÂ (â€“âˆž, 3)

Now,Â let us find coordinates of point

EquationÂ of curve is f(x) = x^{2}Â â€“ 6x + 9

Slope of this curve is given by

**3. Find the intervals in which f(x) = sin x â€“ cos x, where 0 < x < 2Ï€ is increasing or decreasing.**

**Solution:**

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

**Solution:**

Given f (x) = e^{2x}

â‡’Â

â‡’Â fâ€™(x) = 2e^{2x}

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

â‡’Â fâ€™(x) > 0

â‡’Â 2e^{2x}Â > 0

â‡’Â e^{2x}Â > 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 â‰ 0 is a decreasing function for all x â‰ 0.**

**Solution:**

**6. Show that f(x) = log _{a}Â x, 0 < a < 1 is a decreasing function for all x > 0.**

**Solution:**

**7. Show that f(x) = sin x is increasing on (0, Ï€/2) and decreasing on (Ï€/2, Ï€) and neither increasing nor decreasing in (0, Ï€).**

**Solution:**

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

**Solution:**

**9. Show that f(x) = x â€“ sin x is increasing for all xÂ ÏµÂ R.**

**Solution:**

Given f (x) = x â€“ sin x

â‡’Â

â‡’Â fâ€™(x) = 1 â€“ cos x

Now, as given xÂ ÏµÂ R

â‡’Â â€“1 < cos x < 1

â‡’Â â€“1 > cos x > 0

â‡’Â fâ€™(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval xÂ âˆˆÂ R

**10. Show that f(x) = x ^{3}Â â€“ 15x^{2}Â + 75x â€“ 50 is an increasing function for all xÂ ÏµÂ R.**

**Solution:**

Given f(x) = x^{3}Â â€“ 15x^{2}Â + 75x â€“ 50

â‡’Â

â‡’Â fâ€™(x) = 3x^{2}Â â€“ 30x + 75

â‡’Â fâ€™(x) = 3(x^{2}Â â€“ 10x + 25)

â‡’Â fâ€™(x) = 3(x â€“ 5)^{2}

Now, as given xÂ ÏµÂ R

â‡’Â (x â€“ 5)^{2}Â > 0

â‡’Â 3(x â€“ 5)^{2}Â > 0

â‡’Â fâ€™(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval xÂ âˆˆÂ R

**11. Show that f(x) = cos ^{2}Â x is a decreasing function on (0, Ï€/2).**

**Solution:**

Given f (x) = cos^{2}Â x

â‡’Â

â‡’Â fâ€™(x) = 2 cos x (â€“sin x)

â‡’Â fâ€™(x) = â€“2 sin (x) cos (x)

â‡’Â fâ€™(x) = â€“sin2x

Now, as given x belongs to (0, Ï€/2).

â‡’Â 2xÂ âˆˆÂ (0, Ï€)

â‡’Â Sin (2x)> 0

â‡’Â â€“Sin (2x) < 0

â‡’Â fâ€™(x) < 0

Hence, condition for f(x) to be decreasing

Thus f(x) is decreasing on intervalÂ (0, Ï€/2).

Hence proved

**12. Show that f(x) = sin x is an increasing function on (â€“Ï€/2, Ï€/2).**

**Solution:**

Given f (x) = sin x

â‡’Â

â‡’Â fâ€™(x) = cos x

Now, as given x âˆˆ (â€“Ï€/2, Ï€/2).

That is 4^{th}Â quadrant, where

â‡’Â Cos x> 0

â‡’Â fâ€™(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on intervalÂ (â€“Ï€/2, Ï€/2).

**13. Show that f(x) = cos x is a decreasing function on (0, Ï€), increasing in (â€“Ï€, 0) and neither increasing nor decreasing in (â€“Ï€, Ï€).**

**Solution:**

Given f(x) = cos x

â‡’Â

â‡’Â fâ€™(x) = â€“sin x

Taking different region from 0 to 2Ï€

LetÂ x âˆˆ (0, Ï€).

â‡’Â Sin(x) > 0

â‡’Â â€“sin x < 0

â‡’Â fâ€™(x) < 0

Thus f(x) is decreasing inÂ (0, Ï€)

Let x âˆˆ (â€“Ï€, o).

â‡’Â Sin (x) < 0

â‡’Â â€“sin x > 0

â‡’Â fâ€™(x) > 0

Thus f(x) is increasing inÂ (â€“Ï€, 0).

Therefore, from above condition we find that

â‡’Â f (x) is decreasing in (0, Ï€) and increasing inÂ (â€“Ï€, 0).

Hence, condition for f(x) neither increasing nor decreasing in (â€“Ï€, Ï€)

**14. Show that f(x) = tan x is an increasing function on (â€“Ï€/2, Ï€/2).**

**Solution:**

Given f (x) = tan x

â‡’Â

â‡’Â fâ€™(x) = sec^{2}x

Now, as given

x âˆˆ (â€“Ï€/2, Ï€/2).

That is 4^{th}Â quadrant, where

â‡’Â sec^{2}x > 0

â‡’Â fâ€™(x) > 0

Hence, Condition for f(x) to be increasing

Thus f(x) is increasing on intervalÂ (â€“Ï€/2, Ï€/2).

**15. Show that f(x) = tan ^{â€“1}Â (sin x + cos x) is a decreasing function on the interval (Ï€/4, Ï€ /2).**

**Solution:**

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

**Solution:**

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

**17. Show that the function f(x) = cot ^{â€“1}Â (sin x + cos x) is decreasing on (0, Ï€/4) and increasing on (Ï€/4, Ï€/2).**

**Solution:**

Given f(x) = cot^{â€“1}Â (sin x + cos x)

**18. Show that f(x) = (x â€“ 1) e ^{x}Â + 1 is an increasing function for all x > 0.**

**Solution:**

Given f (x) = (x â€“ 1) e^{x}Â + 1

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

â‡’Â

â‡’Â fâ€™(x) = e^{x}Â + (x â€“ 1) e^{x}

â‡’Â fâ€™(x) = e^{x}(1+ x â€“ 1)

â‡’Â fâ€™(x) = x e^{x}

As given x > 0

â‡’Â e^{x}Â > 0

â‡’Â x e^{x}Â > 0

â‡’Â fâ€™(x) > 0

Hence, condition for f(x) to be increasing

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

**19. Show that the function x ^{2}Â â€“ x + 1 is neither increasing nor decreasing on (0, 1).**

**Solution:**

Given f(x) = x^{2}Â â€“ x + 1

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

â‡’Â

â‡’Â fâ€™(x) = 2x â€“ 1

Taking different region from (0, 1)

Let x âˆˆ (0, Â½)

â‡’Â 2x â€“ 1 < 0

â‡’Â fâ€™(x) < 0

Thus f(x) is decreasing in (0, Â½)

Let x âˆˆ (Â½, 1)

â‡’Â 2x â€“ 1 > 0

â‡’Â fâ€™(x) > 0

Thus f(x) is increasing in (Â½, 1)

Therefore, from above condition we find that

â‡’Â f (x) is decreasing inÂ (0, Â½) Â and increasing inÂ (Â½, 1)

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

**20. Show that f(x) = x ^{9}Â + 4x^{7}Â + 11 is an increasing function for all xÂ ÏµÂ R.**

**Solution:**

Given f (x) = x^{9}Â + 4x^{7}Â + 11

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

â‡’Â

â‡’Â fâ€™(x) = 9x^{8}Â + 28x^{6}

â‡’Â fâ€™(x) = x^{6}(9x^{2}Â + 28)

As given xÂ ÏµÂ R

â‡’Â x^{6}Â > 0 and 9x^{2}Â + 28 > 0

â‡’Â x^{6 }(9x^{2}Â + 28) > 0

â‡’Â fâ€™(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval xÂ âˆˆÂ R