LoginSIGN IN Create Account

Rd Sharma XII Vol 1 2020 Solutions for Class 12 Science Math Chapter 16 Increasing And Decreasing Functions are provided here with simple step-by-step explanations. These solutions for Increasing And Decreasing Functions are extremely popular among Class 12 Science students for Math Increasing And Decreasing Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Science Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 16.10:

Question 1:

Prove that the function f(x) = loge x is increasing on (0, ∞).

Answer:

Let x1, x2 ∈0, ∞ such that x1<x2. Then, x1<x2⇒loge x1<loge x2⇒fx1<fx2∴ x1<x2⇒fx1<fx2, ∀ x1, x2 ∈0, ∞So, fx is increasing on 0, ∞.

Page No 16.10:

Question 2:

Prove that the function f(x) = loga x is increasing on (0, ∞) if a > 1 and decreasing on (0, ∞), if 0 < a < 1.

Answer:

fx=loga xLet x1, x2 ∈0, ∞ such that x1<x2.Case 1: Let a>1Here, x1<x2⇒loga x1<loga x2⇒fx1<fx2∴ x1<x2⇒fx1<fx2, ∀ x1, x2 ∈0, ∞So, fx is increasing on 0, ∞.Case 2: Let 0<a<1Here, x1<x2⇒loga x1>loga x2⇒fx1>fx2∴  x1<x2⇒fx1>fx2, ∀ x1, x2 ∈0, ∞So, fx is decreasing on 0, ∞.

Page No 16.10:

Question 3:

Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.

Answer:

Here,fx=ax+bLet x1, x2 ∈R such that x1<x2. Then, x1<x2⇒a x1<a x2      ∵ a >0⇒a x1+b<a x2+b⇒fx1<fx2∴ x1<x2⇒fx1<fx2, ∀ x1, x2 ∈R So, fx is increasing on R.

Page No 16.10:

Question 4:

Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.

Answer:

fx=ax+bLet x1, x2 ∈R such that x1<x2.Then, x1<x2⇒a x1>a x2       (∵ a <0)⇒a x1+b>a x2+b⇒fx1>fx2Thus, x1<x2⇒fx1>fx2, ∀ x1, x2 ∈RSo, fx is decreasing on R.

Page No 16.10:

Question 5:

Show that f(x) = 1x is a decreasing function on (0, ∞).

Answer:

Here,fx=1xLet x1, x2 ∈0, ∞ such that x1<x2. Then, x1<x2⇒1x1>1x2⇒fx1>fx2∴ x1<x2⇒fx1>fx2, ∀ x1, x2 ∈0, ∞So, fx is decreasing on 0, ∞.

Page No 16.10:

Question 6:

Show that f(x) = 11+x2 decreases in the interval [0, ∞) and increases in the interval (−∞, 0].

Answer:

Here,fx=11+x2Case 1: Let x1, x2 ∈0, ∞ such that x1<x2. Then, x1<x2⇒x12<x22⇒1+x12<1+x22⇒11+x12>11+x22⇒fx1>fx2 ∀ x1, x2 ∈0, ∞So, fx is decreasing on 0, ∞.Case 2: Let x1, x2 ∈(-∞, 0] such that x1<x2. Then, x1<x2⇒x12>x22⇒1+x12>1+x22⇒11+x12<11+x22⇒fx1<fx2⇒fx1<fx2, ∀ x1, x2 ∈(-∞, 0]So, fx is increasing on (-∞, 0].

Page No 16.10:

Question 7:

Show that f(x) = 11+x2 is neither increasing nor decreasing on R.

Answer:

fx=11+x2R can be divided into two intervals 0, ∞ and (-∞, 0].Case1: Let x1, x2 ∈0, ∞ such that x1<x2. Then, x1<x2⇒x12<x22⇒1+x12<1+x22⇒11+x12>11+x22⇒fx1>fx2∀ x1, x2 ∈0, ∞So, fx is decreasing on 0, ∞.Case 2: Let x1, x2 ∈(-∞, 0] such that x1<x2. Then, x1<x2⇒x12>x22⇒1+x12>1+x22⇒11+x12<11+x22⇒fx1<fx2 ∀ x1, x2 ∈(-∞, 0]So, fx is increasing on (-∞, 0].Here,  fx is decreasing on 0, ∞ and increasing on (-∞, 0].Thus, fx is neither increasing nor decreasing on R.

Page No 16.10:

Question 8:

Without using the derivative, show that the function f (x) = | x | is
(a) strictly increasing in (0, ∞)
(b) strictly decreasing in (−∞, 0).

Answer:

Here,fx=x(a)  Let x1, x2 ∈0, ∞ such that x1<x2. Then, x1<x2⇒x1<x2⇒fx1<fx2∴ x1<x2⇒fx1<fx2, ∀ x1, x2 ∈0, ∞So, fx is increasing on 0, ∞.(b)  Let x1, x2 ∈ (-∞, 0]. such that x1<x2. Then, x1<x2⇒x1>x2⇒fx1>fx2∴ x1<x2⇒fx1>fx2, ∀ x1, x2∈ (-∞, 0].So, fx is decreasing on (-∞, 0].

Page No 16.10:

Question 9:

Without using the derivative show that the function f (x) = 7x − 3 is strictly increasing function on R.

Answer:

Here,fx=7x-3Let x1, x2 ∈R such that x1<x2. Then, x1<x2⇒7 x1<7 x2      ∵ 7 >0⇒7 x1-3<7 x2-3⇒fx1<fx2∴ x1<x2⇒fx1<fx2, ∀ x1, x2 ∈RSo, fx is  strictly increasing on R.



Page No 16.33:

Question 1:

Find the intervals in which the following functions are increasing or decreasing.
(i) f(x) = 10 − 6x − 2x2

(ii) f(x) = x2 + 2x − 5

(iii) f(x) = 6 − 9x − x2

(iv) f(x) = 2x3 − 12x2 + 18x + 15

(v) f(x) = 5 + 36x + 3x2 − 2x3

(vi) f(x) = 8 + 36x + 3x2 − 2x3

(vii) f(x) = 5x3 − 15x2 − 120x + 3

(viii) f(x) = x3 − 6x2 − 36x + 2

(ix) f(x) = 2x3 − 15x2 + 36x + 1

(x) f(x) = 2x3 + 9x2 + 12x + 20

(xi) f(x) = 2x3 − 9x2 + 12x − 5

(xii) f(x) = 6 + 12x + 3x2 − 2x3

(xiii) f(x) = 2x3 − 24x + 107

(xiv) f(x) = −2x3 − 9x2 − 12x + 1

(xv) f(x) = (x − 1) (x − 2)2

(xvi) f(x) = x3 − 12x2 + 36x + 17

(xvii) f(x) = 2x3 − 24x + 7

(xviii) fx=310x4-45x3-3x2+365x+11

(xix) f(x) = x4 − 4x

(xx) fx=x44+23x3-52x2-6x+7

(xxi) f(x) = x4 − 4x3 + 4x2 + 15

(xxii) f(x) = 5x32-3x52, x > 0

(xxiii) f(x) = x8 + 6x2

(xxiv) f(x) = x3 − 6x2 + 9x + 15

(xxv) fx=x(x-2)2

(xxvi) fx=3x4-4x3-12x2+5

(xxvii) fx=32x4-4x3-45x2+51

(xxviii) fx=log2+x-2x2+x, x∈R

(xxix) fx=x44-x3-5x2+24x+12

Answer:

When x-ax-b>0 with a<b, x<a or x>b.

When x-ax-b<0 with a<b, a<x<b.

(i) f(x) =10-6x-2x2f'(x)=-6-4xFor f(x) to be increasing, we must have f'(x) >0⇒-6-4x>0⇒-4x>6⇒x<-32⇒x∈-∞,-32So, f(x) is increasing on -∞,-32.For f(x) to be decreasing, we must have f'(x) <0⇒-6-4x<0⇒-4x<6⇒x>-64⇒x>-32⇒x∈-32,∞So, f(x) is decreasing on -32,∞.

iifx=x2+2x-5f'x=2x+2For f(x) to be increasing, we must havef'x>0⇒2x+2>0⇒2x+1>0⇒x+1>0⇒x>-1⇒x∈-1, ∞So, f(x) is increasing on -1, ∞.For f(x) to be decreasing, we must havef'x<0⇒2x+2<0⇒2x+1<0⇒x+1<0⇒x<-1⇒x∈-∞, -1So, f(x) is decreasing on -∞, -1.

iii fx=6-9x-x2f'x=-2x-9For f(x) to be increasing, we must havef'x>0⇒-2x-9>0⇒-2x>9⇒x<-92⇒x∈-∞, -92So, f(x) is increasing on -∞, -92.For f(x) to be decreasing, we must havef'x<0⇒-2x-9<0⇒-2x<9⇒x>-92⇒x∈ -92, ∞So, f(x) is decreasing on  -92, ∞.

iv fx=2x3-12x2+18x+15f'x=6x2-24x+18        =6 x2-4x+3        =6 x-1x-3For f(x) to be increasing, we must havef'x>0⇒6 x-1x-3>0⇒x-1x-3>0           Since 6>0, 6 x-1x-3>0⇒x-1x-3>0⇒x<1 or x>3⇒x∈-∞, 1∪3, ∞So, f(x) is increasing on -∞, 1∪3, ∞.

 


For f(x) to be decreasing, we must havef'x<0⇒6 x-1x-3<0⇒x-1x-3<0      Since 6>0, 6 x-1x-3<0⇒ x-1x-3<0⇒1<x<3 ⇒x∈1, 3So, f(x) is decreasing on 1, 3.




v fx=5+36x+3x2-2x3f'x=36+6x-6x2         =-6 x2-x-6         =-6 x-3x+2For f(x) to be increasing, we must havef'x>0⇒-6 x-3x+2>0     ⇒x-3x+2<0        Since -6<0,-6 x-1x+2>0⇒ x-1x+2<0⇒-2<x<3 ⇒x∈-2, 3So, f(x) is increasing on -2, 3.





For f(x) to be decreasing, we must havef'x<0⇒-6 x-3x+2<0⇒x-3x+2>0    Since -6<0, -6 x-1x+2<0⇒ x-1x+2>0⇒x<-2 or x>3 ⇒x∈-∞, -2∪3, ∞So, f(x) is decreasing on -∞, -2∪3, ∞.


vi fx=8+36x+3x2-2x3f'x=36+6x-6x2       =-6 x2-x-6      =-6 x-3x+2For f(x) to be increasing, we must havef'x>0⇒-6 x-3x+2>0     ⇒x-3x+2<0       Since -6<0, -6 x-3x+2>0 ⇒x-3x+2<0⇒-2<x<3⇒x∈-2, 3So, f(x) is increasing on -2, 3.

 

 

For f(x) to be decreasing, we must havef'x<0⇒-6 x-3x+2<0⇒x-3x+2>0     Since -6<0, -6 x-3x+2<0⇒x-3x+2>0⇒x<-2 or x>3 ⇒x∈-∞, -2∪3, ∞So, f(x) is decreasing on -∞, -2∪3, ∞.



vii fx=5x3-15x2-120x+3f'x=15x2-30x-120        =15 x2-2x-8        =15 x-4x+2For f(x) to be increasing, we must havef'x>0⇒15 x-4x+2>0     ⇒x-4x+2>0        Since 15>0, 15 x-4x+2>0  ⇒x-4x+2>0⇒x<-2 or x>4⇒x∈-∞, -2 ∪ 4, ∞So, f(x) is increasing on x∈-∞, -2 ∪ 4, ∞.



 

For f(x) to be decreasing, we must have,f'x<0⇒15 x-4x+2<0⇒x-4x+2<0         Since 15>0, 15 x-4x+2<0⇒x-4x+2<0⇒-2<x<4⇒x∈-2, 4So, f(x) is decreasing on x∈-2, 4.


viii fx=x3-6x2-36x+2f'x=3x2-12x-36        =3 x2-4x-12        =3 x-6x+2For f(x) to be increasing, we must have,f'x>0⇒3 x-6x+2>0⇒x-6x+2>0     Since 3>0, 3 x-6x+2>0⇒x-6x+2>0⇒x<-2 or x>6⇒x∈-∞, -2 ∪ 6, ∞So, f(x) is increasing on x∈-∞, -2 ∪ 6, ∞.

 

 

For f(x) to be decreasing, we must havef'x<0⇒3 x-6x+2<0⇒x-6x+2<0       Since 3>0, 3 x-6x+2<0⇒x-6x+2<0⇒-2<x<6 ⇒x∈-2, 6So, f(x) is decreasing on x∈-2, 6.


ix fx=2x3-15x2+36x+1f'x=6x2-30x+36        =6 x2-5x+6        =6 x-2x-3For f(x) to be increasing, we must havef'x>0⇒6 x-2x-3>0⇒x-2x-3>0       Since 6>0, 6x-2x-3>0⇒x-2x-3>0⇒x<2 or x>3⇒x∈-∞, 2 ∪ 3, ∞So, f(x) is increasing on x∈-∞, 2 ∪ 3, ∞.





For f(x) to be decreasing, we must havef'x<0⇒6 x-2x-3<0⇒x-2x-3<0      Since 6>0, 6x-2x-3<0⇒x-2x-3<0⇒2<x<3 ⇒x∈2, 3So, f(x) is decreasing on x∈2, 3.


x fx=2x3+9x2+12x+20f'x=6x2+18x+12       =6 x2+3x+2       =6 x+1x+2For f(x) to be increasing, we must havef'x>0⇒6 x+1x+2>0⇒x+1x+2>0       Since 6>0, 6 x+1x+2>0⇒x+1x+2>0⇒x<-2 or x>-1⇒x∈-∞, -2 ∪ -1, ∞So, f(x) is increasing on x∈-∞, -2 ∪ -1, ∞.

 



For f(x) to be decreasing, we must havef'x<0⇒6 x+1x+2<0⇒x+1x+2<0     Since 6>0, 6 x+1x+2<0⇒x+1x+2<0⇒-2<x<-1 ⇒x∈-2, -1So, f(x) is decreasing on x∈-2, -1.



xifx=2x3-9x2+12x-5f'x=6x2-18x+12       =6 x2-3x+2       =6 x-1x-2For f(x) to be increasing, we must havef'x>0⇒6 x-1x-2>0⇒x-1x-2>0       Since 6>0, 6 x-1x-2>0⇒ x-1x-2>0⇒x<1 or x>2⇒x∈-∞, 1 ∪ 2, ∞So, f(x) is increasing on x∈-∞, 1 ∪ 2, ∞.



 

For f(x) to be decreasing, we must havef'x<0⇒6 x-1x-2<0⇒x-1x-2<0        Since 6>0, 6 x-1x-2<0⇒ x-1x-2<0⇒1<x<2⇒x∈1, 2So, f(x) is decreasing on x∈1, 2.


xiifx=6+12x+3x2-2x3f'x=12+6x-6x2        =-6 x2-x-2        =-6 x-2x+1For f(x) to be increasing, we must havef'x>0⇒-6 x-2x+1>0⇒x-2x+1<0          Since -6<0, -6 x-2x+1>0⇒x-2x+1<0  ⇒-1<x<2 ⇒x∈-1, 2So, f(x) is increasing on -1, 2.

 

 

For f(x) to be decreasing, we must havef'x<0⇒-6 x-2x+1<0⇒x-2x+1>0       Since -6<0, -6 x-2x+1<0⇒x-2x+1>0⇒x<-1 or x>2 ⇒x∈-∞, -1∪2, ∞So, f(x) is decreasing on -∞, -1∪2, ∞.


xiii fx=2x3-24x+107f'x=6x2-24=6 x2-4=6 x+2x-2For f(x) to be increasing, we must havef'x>0⇒6 x+2x-2>0⇒x+2x-2>0       Since 6>0, 6 x+2x-2>0⇒x+2x-2>0 ⇒x<-2 or x>2⇒x∈-∞, -2 ∪ 2, ∞So, f(x) is increasing on x∈-∞, -2 ∪ 2, ∞.



 

For f(x) to be decreasing, we must havef'x<0⇒6 x+2x-2<0⇒x+2x-2<0        Since 6>0, 6 x+2x-2<0⇒x+2x-2<0 ⇒-2<x<2 ⇒x∈-2, 2So, f(x) is decreasing on x∈-2, 2.


xiv fx=-2x3-9x2-12x+1f'x=-6x2-18x-12        =-6 x2+3x+2        =-6 x+1x+2For f(x) to be increasing, we must havef'x>0⇒-6 x+1x+2>0⇒x+1x+2<0     Since -6<0, -6 x+1x+2>0⇒x+1x+2<0⇒-2<x<-1 ⇒x∈-2, -1So, f(x) is increasing on -2, -1.




 

For f(x) to be decreasing, we must havef'x<0⇒-6 x+1x+2<0⇒x+1x+2>0        Since -6<0,-6 x+1x+2<0⇒x+1x+2>0  ⇒x<-2 or x>-1 ⇒x∈-∞, -2∪-1, ∞So, f(x) is decreasing on -∞, -2∪-1, ∞.



xv fx=x-1x-22              =x-1x2-4x+4              =x3-5x2+8x-4f'x=3x2-10x+8       =3x2-6x-4x+8       =x-23x-4For f(x) to be increasing, we must havef'x>0⇒x-23x-4>0⇒x<43 or x>2⇒x∈-∞, -43 ∪ 2, ∞So, f(x) is increasing on x∈-∞, 43 ∪ 2, ∞.


For f(x) to be decreasing, we must havef'x<0⇒x-23x-4<0⇒43<x<2 ⇒x∈43, 2So, f(x) is decreasing on x∈43, 2.



xvi fx=x3-12x2+36x+17f'x=3x2-24x+36       =3 x2-8x+12       =3 x-2x-6For f(x) to be increasing, we must havef'x>0⇒3 x-2x-6>0⇒x-2x-6>0       Since 3>0, 3 x-2x-6>0⇒x-2x-6>0⇒x<2 or x>6 ⇒x∈-∞, 2 ∪ 6, ∞So, f(x) is increasing on x∈-∞, 2 ∪ 6, ∞.

 

 

For f(x) to be decreasing, we must havef'x<0⇒3 x-2x-6<0⇒x-2x-6<0      Since 3>0, 3 x-2x-6<0⇒x-2x-6<0⇒2<x<6 ⇒x∈2, 6So, f(x) is decreasing on x∈2, 6.


xvii fx=2x3-24x+7f'x=6x2-24       =6 x2-4       =6 x+2x-2For f(x) to be increasing, we must havef'x>0⇒6 x+2x-2>0⇒x+2x-2>0              Since 6>0, 6 x+2x-2>0⇒x+2x-2>0⇒x<-2 or x>2⇒x∈-∞, -2 ∪ 2, ∞So, f(x) is increasing on x∈-∞, -2 ∪ 2, ∞.



 

For f(x) to be decreasing, we must havef'x<0⇒6 x+2x-2<0⇒x+2x-2<0          Since 6>0, 6 x+2x-2<0⇒x+2x-2<0  ⇒-2<x<2⇒x∈-2, 2So, f(x) is decreasing on x∈-2, 2.



xviii fx=310x4-45x3-3x2+365x+11                  =3x4-8x3-30x2+72x+11010f'x=12x3-24x2-60x+7210        =1210x3-2x2-5x+6       = x-1x2-x-610        =1210x-1x+2x-3Here, 1, 2 and 3 are the critical points.The possible intervals are -∞ -2, -2, 1, 1, 3 and 3, ∞.    For f(x) to be increasing, we must havef'x>0⇒1210x-1x+2x-3>0⇒x-1x+2x-3>0⇒x∈-2, 1∪3, ∞So, f(x) is increasing on x∈-2, 1∪3, ∞.



 
For f(x) to be decreasing, we must havef'x<0⇒1210x-1x+2x-3<0⇒x-1x+2x-3<0⇒x∈-∞ -2∪1, 3 So, f(x) is decreasing on x∈-∞ -2∪1, 3.


xix fx=x4-4xf'x=4x3-4        =4x3-1For f(x) to be increasing, we must havef'x>0⇒4x3-1>0      ⇒x3-1>0⇒x3>1⇒x>1⇒x∈1, ∞So, f(x) is increasing on 1, ∞.


For f(x) to be decreasing, we must havef'x<0⇒4x3-1<0⇒x3-1<0⇒x3<1⇒x<1⇒x∈-∞, 1So, f(x) is decreasing on -∞, 1. 



xxfx=x44+23x3-52x2-6x+7              =3x4+8x3-30x2-72x+8412f'x=12x3+24x2-60x-7212       =x3+2x2-5x-6       = x+1x2+x-6      =x+1x-2x+3Here, -1, 2 and -3 are the critical points.The possible intervals are -∞ -3, -3, -1, -1, 2 and 2, ∞.    For f(x) to be increasing, we must havef'x>0⇒x+1x-2x+3>0⇒x∈ -3, -1∪2, ∞So, f(x) is increasing on x∈-3, -1∪2, ∞.



For f(x) to be decreasing, we must havef'x<0⇒x+1x-2x+3<0⇒x∈-∞ -3∪-1, 2                          From eq. (1)So, f(x) is decreasing on x∈-∞ -3∪-1, 2.





xxi fx=x4-4x3+4x2+15f'x=4x3-12x2+8x       =4x x2-3x+2       =4x x-1x-2Here, 0, 1 and 2 are the critical points. The possible intervals are -∞, 0, 0, 1, 1, 2 and 2, ∞.       ...(1)For f(x) to be increasing, we must havef'x>0⇒4x x-1x-2>0                      Since 4>0, 4x x-1x-2>0⇒x x-1x-2>0  ⇒x x-1x-2>0⇒x∈ 0, 1∪2, ∞                     From eq. (1)So, f(x) is increasing on x∈0, 1∪2, ∞.



For f(x) to be decreasing, we must havef'x<0⇒4x x-1x-2<0                Since 4>0, 4x x-1x-2<0⇒x x-1x-2<0 ⇒x x-1x-2<0⇒x∈ -∞, 0∪1, 2                          From eq. (1)So, f(x) is decreasing on x∈-∞, 0∪1, 2.





xxii fx=5x32-3x52, x>0f'x=152x12-152x32       =152x121-xHere, 0,1 are the roots.The possible intervals are -∞,0, 0, 1 and 1, ∞          ...(1)For f(x) to be increasing, we must havef'x>0⇒152x121-x>0⇒x∈0, 1So, f(x) is increasing on 0, 1.

 



For f(x) to be decreasing, we must havef'x<0⇒152x121-x<0⇒x∈1, ∞So, f(x) is decreasing on 1, ∞. 


xxiii fx=x8+6x2f'x=8x7+12x        =4x 2x6+3For f(x) to be increasing, we must havef'x>0⇒4x 2x6+3>0     Since 2x6+3>0, 4x 2x6+3>0⇒x>0⇒x>0⇒x∈0, ∞So, f(x) is increasing on x∈0, ∞.



For f(x) to be decreasing, we must havef'x<0⇒4x 2x6+3<0⇒x<0                    Since 2x6+3>0, 4x 2x6+3<0⇒x<0 ⇒x∈-∞, 0So, f(x) is decreasing on x∈-∞, 0.



 


xxiv fx=x3-6x2+9x+15f'x=3x2-12x+9        =3 x2-4x+3        =3 x-1x-3For f(x) to be increasing, we must havef'x>0⇒3 x-1x-3>0 ⇒x-1x-3>0                Since 3>0, 3 x-1x-3>0 ⇒x-1x-3>0⇒x<1 or x>3⇒x∈-∞, 1 ∪ 3, ∞So, f(x) is increasing on x∈-∞, 1 ∪ 3, ∞.



 

For f(x) to be decreasing, we must havef'x<0⇒3 x-1x-3<0⇒x-1x-3<0           Since 3>0, 3 x-1x-3<0 ⇒x-1x-3<0 ⇒1<x<3 ⇒x∈1, 3So, f(x) is decreasing on x∈1, 3.


xxv fx=xx-22                =x2-2x2                =x4+4x2-4x3f'x=4x3+8x-12x2        =4x x2-3x+2        =4x x-1x-2Here, 0, 1 and 2 are the critical points.The possible intervals are -∞, 0, 0, 1, 1, 2 and 2, ∞.           For f(x) to be increasing, we must havef'x>0⇒4x x-1x-2>0⇒x-1x-2>0⇒x∈ 0, 1∪2, ∞ So, f(x) is increasing on x∈0, 1∪2, ∞.





For f(x) to be decreasing, we must have f'(x) <0⇒4xx-1x-2<0⇒xx-1x-2<0⇒x∈-∞,0∪1,2So, f(x) is decreasing on x∈-∞,0∪1,2.




xxvi fx=3x4-4x3-12x2+5f'x=12x3-12x2-24x       =12xx2-x-2       =12xx+1x-2Here, x=0, x=-1 and x=2 are the critical points. The possible intervals are -∞, -1, -1, 0, 0, 2 and 2, ∞.       .....(1)For f(x) to be increasing, we must havef'x>0⇒12xx+1x-2>0                      Since, 12>0, 12xx+1x-2>0⇒xx+1x-2>0⇒xx+1x-2>0⇒x∈-1, 0∪2, ∞                     From eq. (1)So, f(x) is increasing on x∈-1, 0∪2, ∞.


For f(x) to be decreasing, we must havef'x<0⇒12xx+1x-2<0                Since 12>0, 12xx+1x-2<0⇒xx+1x-2<0 ⇒x x+1x-2<0⇒x∈ -∞, -1∪0, 2                          From eq. (1)So, f(x) is decreasing on x∈-∞, -1∪0, 2.


xxvii fx=32x4-4x3-45x2+51f'x=6x3-12x2-90x       =6xx2-2x-15       =6xx-5x+3Here, x=-3, x=0 and x=5 are the critical points. The possible intervals are -∞, -3, -3, 0, 0, 5 and 5, ∞.       .....(1)For f(x) to be increasing, we must havef'x>0⇒6xx-5x+3>0                      Since, 6>0, 6xx-5x+3>0⇒xx-5x+3>0⇒xx-5x+3>0⇒x∈-3, 0∪5, ∞                     From eq. (1)So, f(x) is increasing on x∈-3, 0∪5, ∞.


For f(x) to be decreasing, we must havef'x<0⇒6xx-5x+3<0                Since 6>0, 6xx-5x+3<0⇒xx-5x+3<0 ⇒xx-5x+3<0⇒x∈-∞, -3∪0, 5                          From eq. (1)So, f(x) is decreasing on x∈-∞, -3∪0, 5.



xxviii fx=log2+x-2x2+x, x∈Rf'x=12+x-2+x2-2x2+x2       =2+x-4+2x-2x2+x2       =2+x-42+x2       =x-22+x2, x≠-2Here, x=2 is the critical point. The possible intervals are -∞, 2 and 2, ∞.       .....(1)For f(x) to be increasing, we must havef'x>0⇒x-22+x2>0⇒x-2>0, x≠-2⇒x>2⇒x∈2, ∞                     From eq. (1)So, f(x) is increasing on x∈2, ∞.


For f(x) to be decreasing, we must havef'x<0⇒x-22+x2<0⇒x-2<0, x≠-2⇒x<2⇒x∈-∞, 2                         From eq. (1)So, f(x) is decreasing on x∈-∞, 2.


(xxix)
fx=x44-x3-5x2+24x+12f'x=x3-3x2-10x+24For increasingf'x>0⇒x3-3x2-10x+24>0⇒-3<x<2For decreasingf'x<0⇒x3-3x2-10x+24<0⇒2<x<4
Thus, for the increasing function the interval is -3,2∪4,∞ and for the decreasing function -∞,-3∪2,4.



Page No 16.34:

Question 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.

Answer:

Here,fx=x2-6x+9f'x=2x-6For f(x) to be increasing, we must havef'x>0⇒2x-6>0⇒2x>6⇒x>3⇒x∈3, ∞So, f(x) is increasing on 3, ∞.For f(x) to be decreasing, we must havef'x<0⇒2x-6<0⇒2x<6⇒x<3⇒x∈-∞, 3So, f(x) is decreasing on -∞, 3.

Let (x, y) be the coordinates on the given curve where the normal to the curve is parallel to the given line.
Slope of the given line = 1

Slope of tangent = dydxx, y= 2x-6Slope of normal = -1Slope of tangent=-12x-6Now,Slope of normal = Slope of the given line-12x-6=1-1=2x-62x=5x=52Given curve isy=x2-6x+9  =254-15+9  =14x, y=52, 14Hence, the coordinates are 52, 14.

Page No 16.34:

Question 3:

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

Answer:

fx=sin x - cos x, x∈0, 2πf'x=cos x + sin xFor f(x) to be increasin, we must havef'x>0⇒cos x + sin x>0⇒sin x>-cos x⇒tan x>-1⇒x∈0, 3π4∪7π4, 2πSo, f(x) is increasing on 0, 3π4∪7π4, 2π.

For f(x) to be decreasing we must have,f'x<0⇒cos x + sin x<0⇒sin x<-cos x⇒tan x<-1⇒x∈3π4, 7π4So, f(x) is decreasing on 3π4, 7π4.

Page No 16.34:

Question 4:

Show that f(x) = e2x is increasing on R.

Answer:

fx=e2xf'x=2e2xNow, x∈RSince the value of e2x is always positive for any real value of x, e2x>0.⇒2e2x>0⇒f'x>0So, f(x) is increasing on R.

Page No 16.34:

Question 5:

Show that f(x) = e1/x, x ≠ 0 is a decreasing function for all x ≠ 0.

Answer:

fx=e1xf'x=e1xddx1x        =e1x-1x2        =-e1xx2Here, e1x>0 and x2>0, for any real value of x ≠ 0.∴ f'x=-e1xx2<0,∀x∈R, x≠ 0So, f(x) is a decreasing function.

Page No 16.34:

Question 6:

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

Answer:

fx=loga x      =log xlog af'x=1x log aSince 0<a<1 and x>0, f'x=1x log a<0.So, fx is decreasing for all x>0.    

Page No 16.34:

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).

Answer:

Here,fx=sin xDomain of sin x is 0, π.f'x=cos xFor x∈0, π2, cos x>0      ∵ cos x is positive in first quadrant f'x>0So, f(x) is increasing for 0, π2.For x∈π2, π, cos x<0      ∵ cos x is negative in second quadrant So, f(x) is decreasing for π2, π.Since f(x) is increasing on 0, π2 and decreasing on π2, π, fx is neither decreasing nor increasing on 0, π.

Page No 16.34:

Question 8:

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

Answer:

Here,fx=log sin xDomain of log sin x is 0, π.f'x=1sin xcos x       =cot xFor x∈0, π2, cot x>0         ∵ Cot function is positive in first quadrant⇒f'x>0 So, f(x) is increasing on 0, π2.For x∈π2, π, cot x<0      ∵ Cot  function is negative in second quadrant ⇒f'x<0   So, f(x) is decreasing on π2, π.

Page No 16.34:

Question 9:

Show that f(x) = x − sin x is increasing for all x ∈ R.

Answer:

fx=x-sin xf'x=1-cos xFor f(x) to be increasing, we must havef'x>0⇒1-cos x>0⇒f'(x)⩾0 for all x∈R            ∵ Cos x⩽1So, f(x) is increasing for all x∈R.

Page No 16.34:

Question 10:

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

Answer:

fx=x3-15x2+75x-50f'x=3x2-30x+75       =3 x2-10x+25       =3x-52>0, ∀x∈R       ∵ Square of any function is always greater than zero So,  f(x)is an increasing function for all x∈R.

Page No 16.34:

Question 11:

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

Answer:

fx=cos2xf'x=2 cos x -sin x⇒f'x=-sin 2x           ...1Now,0<x<π2⇒0<2x<π ⇒sin 2x>0             ∵ Sine fuction is positive in first and second quadrant ⇒-sin 2x<0⇒f'x<0               From eq. (1)So, f(x) is decreasing on 0, π2.

Page No 16.34:

Question 12:

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

Answer:

fx=sin xf'x=cos x>0 ∀x∈-π2, π2     ∵ Cos function is positive in first and fourth quadrant So, fx is increasing on -π2, π2.

Page No 16.34:

Question 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).

Answer:

Here,fx=cos xDomain of cos  x is -π, π.⇒f'x=-sin xFor x∈-π, 0, sin x<0       ∵ sine function is negative in third and fourth quadrant ⇒- sin x>0⇒f'x>0So, cos x is increasing in -π,0.For x∈0, π), sin x>0        ∵ sine function is positive in first and second quadrant ⇒-sin x<0⇒f'x<0So, f(x) is decreasing on 0, π.Thus, f(x) is neither increasing nor decreasing in -π, π.

Page No 16.34:

Question 14:

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

Answer:

fx=tan xf'x=sec2 xHere,-π2<x<π2⇒sec x>0        ∵ Sec function is positive in first and fourth quadrant ⇒sec2 x>0⇒f'x>0, ∀x∈-π2, π2So, f(x) is increasing on -π2, π2.

Page No 16.34:

Question 15:

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

Answer:

fx=tan-1sin x+cos xf'x=11+sin x+cos x2cosx-sinx       =11+1+2 sin x cos xcosx-sinx       =cosx-sinx2+sin 2xHere,π4<x<π2⇒π2<2x<π⇒sin 2x>0⇒2+sin 2x>0              ...1Also,π4<x<π2cos x<sin x⇒cos x-sin x<0         ...2f'x=cos x-sin x2+sin 2x<0, ∀x∈π4, π2       From eqs. (1) and (2)So, fx is decreasing on π4, π2.

Page No 16.34:

Question 16:

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

Answer:

fx=sin 2x+π4f'x=2 cos 2x+π4Here,3π8<x<5π8⇒3π4<2x<5π4⇒π<2x+π4<3π2⇒cos 2x+π4<0        ∵ Cos function is negative in third quadrant⇒2 cos 2x+π4<0⇒f'x<0, ∀x∈3π8, 5π8So, fx is decreasing on 3π8, 5π8.

Page No 16.34:

Question 17:

Show that the function f(x) = cot-l(sinx + cosx) is decreasing on 0,π4 and increasing on π4,π2.

Answer:

We have,fx=cot-1sinx+cosx⇒f'x=-11+sinx+cosx2×cosx-sinx=sinx-cosx1+sin2x+cos2x+2sinxcosx=sinx-cosx1+1+2sinxcosx=sinx-cosx2+2sinxcosx=12×sinx-cosx1+sinxcosxFor fx to be decreasing, we must havef'x<0⇒12×sinx-cosx1+sinxcosx<0⇒sinx-cosx1+sinxcosx<0⇒sinx-cosx<0         In first quadrant⇒sinx<cosx⇒tanx<1⇒0<x<π4So, fx is decreasing on 0,π4.For fx to be increasing, we must havef'x>0⇒12×sinx-cosx1+sinxcosx>0⇒sinx-cosx1+sinxcosx>0⇒sinx-cosx>0         In first quadrant⇒sinx>cosx⇒tanx>1⇒π4<x<π2So, fx is increasing on π4,π2.

Page No 16.34:

Question 18:

Show that f(x) = (x − 1) ex + 1 is an increasing function for all x > 0.

Answer:

fx=x-1 ex+1f'x=x-1 ex+ex       =xex-ex+ex       =xexGiven: x>0 We know,  ex>0⇒ xex>0⇒f'x>0, ∀x>0So, f(x) is increasing on for all x>0.

Page No 16.34:

Question 19:

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

Answer:

fx=x2-x+1f'x=2x-1For f(x) to be increasing, we must havef'x>0⇒2x-1>0⇒2x>1⇒x>12⇒x∈12, 1So, f(x) is increasing on 12, 1.For f(x) to be decreasing, we must havef'x<0⇒2x-1<0⇒2x<1⇒x<12⇒x∈0, 12So, f(x) is decreasing on 0, 12.Since f(x) is increasing on 12, 1 and decreasing on 0, 12, fx is neither increasing nor decreasing on (0, 1).

Page No 16.34:

Question 20:

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

Answer:

fx=x9+4x7+11f'x=9x8+28x6≥0, ∀x∈R     ∵ x8, x6≥0, for ∀x∈RSo, f(x) is increasing on R.



Page No 16.35:

Question 21:

Prove that the function f(x) = x3 − 6x2 + 12x − 18 is increasing on R.

Answer:

fx=x3-6x2+12x-18f'x=3x2-12x+12       =3x2-4x+4       =3x-22≥0, ∀x∈R         ∵ 3>0 & x-22≥0So, f(x) is increasing on R.

Page No 16.35:

Question 22:

State when a function f(x) is said to be increasing on an interval [a, b]. Test whether the function f(x) = x2 − 6x + 3 is increasing on the interval [4, 6].

Answer:

A function f(x) is said to be increasing on a, b if it is increasing at x=a and x=b.Here,fx=x2-6x+3f'x=2x-6⇒f'x=2x-3Now, f'4=24-3                  = 2∴ f'4>0  So, f(x) is increasing on x=4 &, f'6=26-3                  =6∴ f'6>0   So, f(x) is increasing on x=6 Hence,  fx is increasing on [4, 6].

Page No 16.35:

Question 23:

Show that f(x) = sin x − cos x is an increasing function on (−π/4, π/4).

Answer:

fx=sin x-cos xf'x=cos x+sin x       =cos x1+sin xcos x       =cos x1+cot xHere,-π4<x<π4⇒cos x>0          ... 1Also,-π4<x<π4⇒-1<cot x<1⇒0<1+cot x<2⇒1+cot x>0      ... 2cos x1+cot x>0, ∀x∈-π4, π4        From eqs. (1) and (2) ⇒f'x>0, ∀x∈-π4, π4So, fx is increasing on -π4, π4.

Page No 16.35:

Question 24:

Show that f(x) = tan−1 x − x is a decreasing function on R.

Answer:

fx=tan-1x-xf'x=11+x2-1       =1-1-x21+x2      =-x21+x2We know,x2≥0, 1+x2>0, ∀ x∈R∴ -x21+x2<0, ∀ x∈R⇒f'x<0, ∀ x∈RSo, fx is decreasing on R.

Page No 16.35:

Question 25:

Determine whether f(x) = −x/2 + sin x is increasing or decreasing on (−π/3, π/3).

Answer:

fx=-x2+sin xf'x=-12+cos xHere, -π3<x<π3⇒cos x>12⇒-12+cos x>0⇒f'x>0, ∀x∈-π3, π3So, fx is increasing on -π3, π3.

Page No 16.35:

Question 26:

Find the intervals in which f(x) = log (1 + x) − x1+x is increasing or decreasing.

Answer:

fx=log 1+x-x1+xDomain of fx is -1, ∞.f'x=11+x-1+x-x1+x2       =11+x-11+x2       =x1+x2For f(x) to be increasing, we must havef'x>0⇒x1+x2>0⇒x>0             ∵ 1+x2>0, Domain: -1, ∞⇒x∈0, ∞So, f(x) is increasing on 0, ∞.For f(x) to be decreasing, we must havef'x<0⇒x1+x2<0⇒x<0              ∵1+x2>0, Domain: -1, ∞⇒x∈-1, 0So, f(x) is decreasing on -1, 0.

Page No 16.35:

Question 27:

Find the intervals in which f(x) = (x + 2) e−x is increasing or decreasing.

Answer:

 fx=x+2 e-xf'x=-e-xx+2+e-x        =-xe-x-2e-x+e-x        =-xe-x-e-x        =e-x-x-1For f(x) to be increasing, we must havef'x>0⇒e-x-x-1>0⇒-x-1>0          ∵ e-x>0, ∀x∈R⇒-x>1⇒x<-1⇒x∈-∞, -1So, f(x) is increasing on -∞, -1.For f(x) to be decreasing, we must havef'x<0⇒e-x-x-1<0⇒-x-1<0     ∵ e-x>0, ∀x∈R⇒-x<1⇒x>-1⇒x∈-1, ∞So, f(x) is decreasing on -1, ∞.

Page No 16.35:

Question 28:

Show that the function f given by f(x) = 10x is increasing for all x.

Answer:

fx=10xf'x=10x log 10>0, ∀x∈RSo, f(x) is increasing for all x∈R. 

Page No 16.35:

Question 29:

Prove that the function f given by f(x) = x − [x] is increasing in (0, 1).

Answer:

fx=x-xLet x1, x2 ∈0, 1 such that x1<x2. Then,x1=x2= 0               ...(1)Now, x1<x2⇒x1-x1<x2-x2         From eq. (1)⇒fx1<fx2∴ x1<x2⇒fx1<fx2, ∀ x1, x2 ∈0, 1So, fx is increasing on 0, 1.

Page No 16.35:

Question 30:

Prove that the following functions are increasing on R.
(i) f(x)=3x5 + 40x3 + 240x
(ii) fx=4x3-18x2+27x-27

Answer:

(i)

fx=3x5+40x3+240xf'x=15x4+120x2+240       =15 x4+8x2+16       =15 x2+42>0, ∀x∈R        ∵ 15>0 and x2+42>0

So, f(x) is increasing on R.

(ii) fx=4x3-18x2+27x-27

f'x=12x2-36x+27⇒f'x=34x2-12x+9⇒f'x=32x-32>0, ∀x∈R

So, f(x) is increasing on R.

Page No 16.35:

Question 31:

Prove that the function f given by f(x) = log cos x is strictly increasing on (−π/2, 0) and strictly decreasing on (0, π/2).

Answer:

fx=log cos xf'x=1cos x-sin x       =-tan xNow, x∈-π2, 0⇒tan x<0⇒-tan x>0 ⇒f'(x)>0So, f(x) is strictly increasing on -π2, 0.Now, x∈0,π2⇒tan x>0⇒-tan x<0 ⇒f'(x)<0So, f(x) is strictly decreasing on 0, π2.

Page No 16.35:

Question 32:

Prove that the function f given by f(x) = x3 − 3x2 + 4x is strictly increasing on R.

Answer:

fx=x3-3x2+4xf'x=3x2-6x+4        =3x2-2x+4        =3x2-2x+1-3+4        =3x-12+1>0, ∀x∈RHence, f(x) is strictly increasing on R.

Page No 16.35:

Question 33:

Prove that the function f(x) = cos x is:
(i) strictly decreasing in (0, π)
(ii) strictly increasing in (π, 2π)
(iii) neither increasing nor decreasing in (0, 2π)

Answer:

fx=cos xf'x=-sin xi Here,0<x<π⇒sin x>0      ∵ Sine function is positive in first and second quadrant⇒-sin x<0⇒f'x<0, ∀x∈0, πSo, f(x) is strictly decreasing on 0, π.ii Here,π<x<2π⇒sin x<0     ∵ Sine function is negative in third and fourth quadrant⇒-sin x>0⇒f'x>0, ∀x∈π, 2πSo, f(x) is strictly increasing on π, 2π.iii From eqs. (1) and (2), we get f(x) is strictly decreasing on 0, π and is strictly increasing on π, 2π.So, fx is neither increasing nor decreasing on 0, 2π.

Page No 16.35:

Question 34:

Show that f(x) = x2 − x sin x is an increasing function on (0, π/2).

Answer:

fx=x2-x sin xf'x=2x-x cos x-sin xHere,0<x<π2⇒0<sin x<1 and 0<cos x<1⇒2x-x cos x-sin x>0⇒f'x>0, ∀x∈0, π2So, fx is increasing on 0, π2.

Page No 16.35:

Question 35:

Find the value(s) of a for which f(x) = x3 − ax is an increasing function on R.

Answer:

fx=x3-axf'x=3x2-aGiven: fx is increasing on R.⇒f'x≥0  ∀ x∈R⇒3x2-a≥0  ∀ x∈R⇒a≤3x2∀ x∈RThe least value of 3x2 is 0.∴ a≤0

Page No 16.35:

Question 36:

Find the values of b for which the function f(x) = sin x − bx + c is a decreasing function on R.

Answer:

fx=sin x-bx+cf'x=cos x-bGiven: fx is decreasing on R.f'x<0, ∀x∈R⇒cos x-b<0, ∀x∈R⇒cos x-b<0,∀x∈R ⇒cos x<b, ∀x∈R⇒ b⩾1            ∵ -1⩽cos x⩽1    

Page No 16.35:

Question 37:

Show that f(x) = x + cos x − a is an increasing function on R for all values of a.

Answer:

fx=x+cos x-af'x=1-sin xWe know,sin x≤1, ∀x∈R⇒-sin x≥-1, ∀x∈R⇒1-sin x≥0, ∀x∈R⇒f'x≥0, ∀x∈RHence, fx is increasing on R for all values of a.

Page No 16.35:

Question 38:

Let f defined on [0, 1] be twice differentiable such that | f"(x) | ≤ 1 for all x ∈ [0, 1]. If f(0) = f(1), then show that | f'(x) | < 1 for all x ∈ [ 0, 1].

Answer:

If a function is continuous and differentiable and f(0) = f(1) in given domain x ∈ [0, 1],
then by Rolle's Theorem;
f'(x) = 0 for some x ∈ [0, 1]
Given: |f"(x)| ≤ 1
On integrating both sides we get,
|f'(x)| ≤ x
Now, within interval x ∈ [0, 1]
We get, |f' (x)| < 1.

Page No 16.35:

Question 39:

Find the intervals in which f(x) is increasing or decreasing:

(i) f(x) = x|x|, x ∈R

(ii) f(x) = sinx + |sinx|, 0 < x ≤2π

(iii) f(x) = sinx(1 + cosx), 0 < x < π2
                                                                                                                                                                                                     [CBSE 2014]

Answer:

i fx=xx, x∈RCase I: When x≥0fx=xx=xx=x2⇒f'x=2x≥0 ∀x≥0So, fx is increasing for x≥0.Case II:  When x<0fx=xx=x-x=-x2⇒f'x=-2x≥0 ∀x<0So, fx is increasing for x<0.Hence, fx is increasing for x∈R.ii fx=sinx+sinx, 0<x≤2πCase I: When x∈0,πfx=sinx+sinx=2sinx⇒f'x=2cosxAs, cosx>0 for x∈0,π2 and cosx<0 for x∈π2,πSo, f'x>0 for x∈0,π2 and f'x<0 for x∈π2,π∴ fx is increaing on 0,π2 and fx is decreasing on π2,π.Case II: When x∈π,2πfx=sinx-sinx=0⇒f'x=0So, fx is neither increaing nor decreasing on π,2π.iii fx=sinx1+cosx,0<x<π2⇒fx=sinx+sinxcosx⇒f'x=cosx+sinx-sinx+cosxcosx⇒f'x=cosx-sin2x+cos2x⇒f'x=cosx+cos2x-1+cos2x⇒f'x=2cos2x+cosx-1⇒f'x=2cos2x+2cosx-cosx-1⇒f'x=2cosxcosx+1-1cosx+1⇒f'x=2cosx-1cosx+1For fx to be increasing, we must havef'x>0⇒2cosx-1cosx+1>0This is only possible when2cosx-1>0 and cosx+1>0⇒2cosx-1>0 and cosx+1>0⇒cosx>12 and cosx>-1⇒x∈0,π3 and x∈0,π2So, x∈0,π3∴ fx is increasing on 0,π3.For fx to be decreasing, we must havef'x<0⇒2cosx-1cosx+1<0This is only possible when2cosx-1<0 and cosx+1>0⇒2cosx-1<0 and cosx+1>0⇒cosx<12 and cosx>-1⇒x∈π3,π2 and x∈0,π2So, x∈π3,π2∴ fx is decreasing on π3,π2.



Page No 16.39:

Question 1:

The interval of increase of the function f(x) = x − ex + tan (2π/7) is
(a) (0, ∞)
(b) (−∞, 0)
(c) (1, ∞)
(d) (−∞, 1)

Answer:

(b) (−∞, 0)

fx=x-ex+tan2π7f'x=1-exFor f(x) to be increasing, we must havef'x>0⇒1-ex>0⇒ex<1⇒x<0⇒x∈-∞, 0So, f(x) is increasing on -∞, 0.

Page No 16.39:

Question 2:

The function f(x) = cot−1 x + x increases in the interval
(a) (1, ∞)
(b) (−1, ∞)
(c) (−∞, ∞)
(d) (0, ∞)

Answer:

(c) (−∞, ∞)

fx=cot-1x+xf'x=-11+x2+1     =-1+1+x21+x2        =x21+x2≥0, ∀x∈RSo, f(x) is increasing on -∞, ∞.

Page No 16.39:

Question 3:

The function f(x) = xx decreases on the interval
(a) (0, e)
(b) (0, 1)
(c) (0, 1/e)
(d) none of these

Answer:

(c) (0, 1/e)

Given: fx=xxApplying log with base e on both sides, we getlog fx=x loge xf'xfx=1+loge xf'x=fx1+loge x=xx1+loge xFor f(x) to be decreasing, we must havef'x<0⇒xx1+loge x<0Here, logaritmic function is defined for positive values of x.⇒xx>0⇒1+loge x<0      Since xx>0, xx1+loge x<0⇒1+loge x<0  ⇒loge x<-1⇒x<e-1                  ∵ logax<N⇒x<aN for a>1 Here, e>1⇒logex<-1⇒x<e-1⇒x∈0, e-1So, f(x) is decreasing on 0, 1e.

Page No 16.39:

Question 4:

The function f(x) = 2 log (x − 2) − x2 + 4x + 1 increases on the interval
(a) (1, 2)
(b) (2, 3)
(c) (1, 3)
(d) (2, 4)

Answer:

(b) (2, 3)

Given: fx=2 log x-2-x2+4x+1Domain of fx is 2, ∞.f'x=2x-2-2x+4       =2-2x2+4x+4x-8x-2       =-2x2+8x-6x-2       =-2 x2-4x+3x-2For f(x) to be increasing, we must havef'x>0⇒-2 x2-4x+3x-2>0⇒x2-4x+3+<0            ∵ x-2>0 & -2<0⇒x-1x-3<0⇒1<x<3⇒x∈1, 3Also, the domain of fx is 2, ∞.⇒x∈1, 3∩2, ∞⇒x∈2, 3



Page No 16.40:

Question 5:

If the function f(x) = 2x2 − kx + 5 is increasing on [1, 2], then k lies in the interval
(a) (−∞, 4)
(b) (4, ∞)
(c) (−∞, 8)
(d) (8, ∞)

Answer:

(a) (−∞, 4)

fx=2x2-kx+5f'x=4x-kFor f(x) to be increasing, we must havef'x>0⇒4x-k>0⇒k<4xSince x∈1, 2, 4x∈4, 8.So, the minimum value of 4x is 4.Since k<4x, k<4.⇒k∈-∞, 4

Page No 16.40:

Question 6:

Let f(x) = x3 + ax2 + bx + 5 sin2x be an increasing function on the set R. Then, a and b satisfy
(a) a2 − 3b − 15 > 0
(b) a2 − 3b + 15 > 0
(c) a2 − 3b + 15 < 0
(d) a > 0 and b > 0

Answer:

(c) a2 − 3b + 15 < 0


fx=x3+ax2+bx+5sin2xf'x=3x2+2ax+b+5 sin 2xGiven: fx is increasing on R.⇒f'x>0, ∀x∈R⇒3x2+2ax+b+5 sin 2x>0, ∀x∈R Since this quadratic function is >0, its discriminant is <0.⇒2a2-43b+5 sin 2x<0⇒4a2-12b-60 sin 2x<0⇒a2-3b-15 sin 2x<0We know that the minimum value of sin 2x is -1.∴ a2-3b-15<0 

Page No 16.40:

Question 7:

The function fx=logex3+x6+1 is of the following types:
(a) even and increasing
(b) odd and increasing
(c) even and decreasing
(d) odd and decreasing

Answer:

(b) odd and increasing

f(x) =logex3+x6+1⇒f(-x)=loge-x3+x6+1               =loge-x3+x6+1x3+x6+1x3+x6+1               =logex6+1-x6x3+x6+1               =loge1x3+x6+1               =-logex3+x6+1               =-f(x) Hence, f(-x)=-f(x)Therefore, it is an odd function.

f(x)=logex3+x6+1ddxf(x)=1x3+x6+1×3x2+12x6+1×6x5                =1x3+x6+1×6x2x6+1+6x52x6+1                =1x3+x6+1×6x2x6+1+x32x6+1                =6x22x6+1>0Therefore, given function is an increasing function.

Page No 16.40:

Question 8:

If the function f(x) = 2 tan x + (2a + 1) loge | sec x | + (a − 2) x is increasing on R, then
(a) a ∈ (1/2, ∞)
(b) a ∈ (−1/2, 1/2)
(c) a = 1/2
(d) a ∈ R

Answer:

f(x)= 2 tan x+2a+1logesec x+a-2 xWhen sec x>0⇒sec x=sec xddxfx=2sec2x+2a+11sec x×sec x tan x+a-2                  =2sec2x+2a+1tan x+a-2 For f(x) to be  increasing, 2sec2x+2a+1tan x+a-2⩾0⇒2+2 tan2x+2a+1tan x+a-2⩾0⇒2 tan2x+2a+1tan x+a⩾0Its discriminant ⩽0           ∵ax2+bx+c⩾0⇒b2-4ac⩽0⇒2a+12-4.2.a⩽0⇒4a2-4a+1⩽0⇒2a-12⩽02a-12<0 cannot be possible .∴ 2a-12=0⇒a=12When sec x<0⇒sec x=-sec xddxfx=2sec2x+2a+11-sec x×sec x tan x+a-2                 =2sec2x-2a+1tan x+a-2 For f(x) to be  increasing, 2sec2x-2a+1tan x+a-2⩾0⇒2+2 tan2x-2a+1tan x+a-2⩾0⇒2 tan2x-2a+1tan x+a⩾0    Its discriminant ⩽0          ∵ax2+bx+c⩾0⇒b2-4ac⩽0⇒-2a+12-4.2.a⩽0⇒4a2-4a+1⩽0⇒2a-12⩽02a-12<0 cannot be possible .∴ 2a-12=0⇒a=12

Page No 16.40:

Question 9:

Let fx=tan-1gx, where g (x) is monotonically increasing for 0 < x < π2. Then, f(x) is
(a) increasing on (0, π/2)
(b) decreasing on (0, π/2)
(c) increasing on (0, π/4) and decreasing on (π/4, π/2)
(d) none of these

Answer:

(a) increasing on (0, π/2)
 
Given: gx is increasing on 0, π2. Then,x1<x2, ∀ x1, x2∈0, π2⇒gx1<gx2Taking tan-1 on both the sides, we gettan-1gx1<tan-1gx2⇒fx1<fx2, ∀ x1, x2∈0, π2So, fx is increasing on 0, π2.

Page No 16.40:

Question 10:

Let f(x) = x3 − 6x2 + 15x + 3. Then,
(a) f(x) > 0 for all x ∈ R
(b) f(x) > f(x + 1) for all x ∈ R
(c) f(x) is invertible
(d) none of these

Answer:

(c) f(x) is invertible
f(x) =x3 − 6x2 + 15x + 3
f'(x) =3x2-12x+15         =3x2-4x+5         =3x2-4x+4+1         =3x-22+13>0Therefore, f(x) is strictly increasing function.⇒ f-1(x) exists.Hence, f(x) is an invertible function.

Page No 16.40:

Question 11:

The function f(x) = x2 e−x is monotonic increasing when
(a) x ∈ R − [0, 2]
(b) 0 < x < 2
(c) 2 < x < ∞
(d) x < 0

Answer:

(b) 0 < x < 2

fx=x2e-xf'x=2xe-x-x2e-x       =e-x x2-xFor f(x) to be monotonic increasing, we must havef'x>0⇒e-x x2-x>0       ∵ e-x>0⇒ x2-x>0⇒ xx-2<0⇒0<x<2

Page No 16.40:

Question 12:

Function f(x) = cos x − 2 λ x is monotonic decreasing when
(a) λ > 1/2
(b) λ < 1/2
(c) λ < 2
(d) λ > 2

Answer:

(a) λ > 1/2

fx=cos x-2 λ xf'x=-sin x-2 λ For f(x) to be decreasing, we must havef'x<0⇒-sin x-2 λ <0⇒sin x+2 λ>0 ⇒2 λ>-sin xWe know that the maximum value of -sin x is 1.⇒2 λ>1⇒λ>12

Page No 16.40:

Question 13:

In the interval (1, 2), function f(x) = 2 | x − 1 | + 3 | x − 2 | is
(a) monotonically increasing
(b) monotonically decreasing
(c) not monotonic
(d) constant

Answer:

(b) monotonically decreasing

If 1<x<2, then x>1 and x<2.⇒x-1>0 and x-2<0⇒x-1=x-1 and x-2=-x-2Now,fx=2 x-1+3 x-2=2x-1-3x-2=2x-2-3x+6=-x+4f'x=-1<0, ∀x∈1, 2So, fx is monotonically decreasing.

Page No 16.40:

Question 14:

Function f(x) = x3 − 27x + 5 is monotonically increasing when
(a) x < −3
(b) | x | > 3
(c) x ≤ −3
(d) | x | ≥ 3

Answer:

(d) | x | ≥ 3

fx=x3-27x+5f'x=3x2-27        =3 x2-9For f(x) to be increasing, we must havef'x>0⇒3 x2-9>0⇒x2-9>0       Since 3>0, 3 x2-9>0⇒x2-9>0|⇒x+3x-3>0⇒x<-3 or x>3⇒x>3

Page No 16.40:

Question 15:

Function f(x) = 2x3 − 9x2 + 12x + 29 is monotonically decreasing when
(a) x < 2
(b) x > 2
(c) x > 3
(d) 1 < x < 2

Answer:

(d) 1 < x < 2

fx=2x3-9x2+12x+29f'x=6x2-18x+12        =6 x2-3x+2        =6x-1x-2For f(x) to be decreasing, we must havef'x<0⇒6x-1x-2<0         ⇒x-1x-2<0            Since 6>0, 6x-1x-2<0⇒x-1x-2<0⇒1<x<2So, f(x) is decreasing for 1<x<2.

Page No 16.40:

Question 16:

If the function f(x) = kx3 − 9x2 + 9x + 3 is monotonically increasing in every interval, then
(a) k < 3
(b) k ≤ 3
(c) k > 3
(d) k ≥ 3

Answer:


(c) k > 3

fx=kx3-9x2+9x+3f'x=3kx2-18x+9       =3 kx2-6x+3Given: f(x) is monotonically increasing in every interval. ⇒f'x>0⇒3 kx2-6x+3>0⇒kx2-6x+3>0⇒k>0 and -62-4k3<0       ∵ ax2+bx+c>0⇒a>0 and Disc<0 ⇒k>0 and -62-4k3<0⇒k>0 and 36-12k<0⇒k>0 and 12k>36⇒k>0 and k>3⇒k>3

Page No 16.40:

Question 17:

f(x) = 2x − tan−1 x − log x+x2+1 is monotonically increasing when
(a) x > 0
(b) x < 0
(c) x ∈ R
(d) x ∈ R − {0}

Answer:

(c) x ∈ R

Given: fx=2x-tan-1x-log x+x2+1f'x=2-11+x2-1x+x2+11+12x2+1.2x       =2-11+x2-1x+x2+11+xx2+1       =2-11+x2-1x+x2+1x+x2+1x2+1       =2-11+x2-1x2+1       =2+2x2-1-x2+11+x2       =1+2x2-x2+11+x2For f(x) to be monotonically increasing, f'x>0.⇒1+2x2-x2+11+x2>0         ⇒1+2x2-x2+1>0          ∵  1+x2>0⇒1+2x2>x2+1⇒1+2x22>x2+1⇒1+4x4+4x2>x2+1⇒4x4+3x2>0Thus, f(x) is monotonically increasing for x∈R.

Page No 16.40:

Question 18:

Function f(x) = | x | − | x − 1 | is monotonically increasing when
(a) x < 0
(b) x > 1
(c) x < 1
(d) 0 < x < 1

Answer:

(d) 0 < x < 1

fx=x-x-1Case 1: Let x<0       If x<0 , then x=-x⇒ x-1=-x-1Now, fx=x-x-1      =-x--x+1      =-x+x-1      =-1f'x=0So, fx is not monotonically increasing when x < 0.Case 2: Let 0<x<1Here,x=x⇒ x-1=-x-1Now, fx=x-x-1      =x+x-1     =2x-1f'x=2>0So, fx is monotonically increasing when 0<x<1.Case 3: Let x>1 If x>0, then x=x⇒ x-1=x-1Now, fx=x-x-1      =x-x+1      =1f'x=0So, fx is not monotonicallyincreasing when x >1.Thus, fx is monotonically increasing when 0<x<1.

Page No 16.40:

Question 19:

Every invertible function is
(a) monotonic function
(b) constant function
(c) identity function
(d) not necessarily monotonic function

Answer:

(a) monotonic function

We know that "every invertible function is a monotonic function".

Page No 16.40:

Question 20:

In the interval (1, 2), function f(x) = 2 | x − 1 | + 3 | x − 2 | is
(a) increasing
(b) decreasing
(c) constant
(d) none of these

Answer:

(b) decreasing

Given: fx=2x-1+3x-2If 1<x<2, then x-1=x-1.⇒x-2=-x-2Now, fx=2x-1+3x-2      =2 x-1+3 -x+2      =2x-2-3x+6      =-x+4f'x=-1<0So, fx is decreasing when 1<x<2.



Page No 16.41:

Question 21:

If the function f(x) = cos |x| − 2ax + b increases along the entire number scale, then

(a) a = b

(b) a=12b

(c) a≤-12

(d) a>-32

Answer:

(c) a≤-12

Given: fx=cos x-2ax+bNow, x=x , x≥0               -x, x<0and cos x=cosx ,         x≥0               cos-x=cosx,   x<0∴cosx=cos x      ,∀x∈R∴ fx=cos x-2ax+b⇒f'x=-sin x-2aIt is given that f(x) is increasing.⇒f'x≥0⇒-sin x-2a≥0⇒sin x+2a≤0⇒2a ≤-sin xThe least value of -sin x is -1.⇒2a≤-1⇒a≤-12

Page No 16.41:

Question 22:

The function fx=x1+x is
(a) strictly increasing
(b) strictly decreasing
(c) neither increasing nor decreasing
(d) none of these

Answer:

(a)  strictly increasing

fx=x1+xCase 1: When x>0, x=xfx=x1+x      =x1+x⇒f'x=1+x1-x11+x2            =11+x2>0, ∀x∈RSo, fx is strictly increasing when x > 0.Case 2: When x<0, x=-xfx=x1+x      =x1-x⇒f'x=1-x1-x-11-x2        =11-x2>0, ∀x∈RSo, fx is strictly increasing when x < 0.Thus, fx is strictly increasing on R.
 

Page No 16.41:

Question 23:

The function fx=λ sin x+2 cos xsin x+cos x is increasing, if
(a) λ < 1
(b) λ > 1
(c) λ < 2
(d) λ > 2

Answer:

(d) λ > 2

fx=λ sin x+2 cos xsin x+cos xf'x=sin x+cos xλ cos x-2 sin x+λ sin x+2 cos xcos x-sin xsin x+cos x2=λcos x sin x+λcos2x-2 sin2x-2 sinx cos x-λsinx cos x-2cos2x+λ sin2x+2 cosx sinx sin x+cos x2=-2 sin2x+cos2x+λ sin2x+cos2xsin x+cos x2=-2 +λ sin x+cos x2For f(x) to be increasing, we must havef'x>0⇒-2 +λ sin x+cos x2>0     ⇒λ -2>0         ∵sin x+cos x2>0⇒λ >2

Page No 16.41:

Question 24:

Function f(x) = ax is increasing on R, if
(a) a > 0
(b) a < 0
(c) 0 < a < 1
(d) a > 1

Answer:

(d) a > 1

fx=axf'x=ax log aGiven: f(x) is increasing on R.⇒f'x>0⇒ax log a>0⇒ax>0       Logarithmic function is defined for positive values of aWe know,ax log a>0It can be possible when ax>0 and log a>0 or  ax<0 and log a<0  Not possible, logarithmic function is defined for positive values of a⇒log a>0⇒a>1So, f(x) is increasing when a > 1.

Page No 16.41:

Question 25:

Function f(x) = loga x is increasing on R, if
(a) 0 < a < 1
(b) a > 1
(c) a < 1
(d) a > 0

Answer:

(b) a > 1

fx=loga x=log xlog af'x=1x log aGiven: f(x) is increasing on R.⇒f'x>0, ∀x∈R⇒1x log a>0, ∀x∈R⇒a>1

Page No 16.41:

Question 26:

Let Ď•(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, Ď• (x)
(a) increases on [0, a]
(b) decreases on [0, a]
(c) increases on [−a, 0]
(d) decreases on [a, 2a]

Answer:

Given: Ď•(x) = f(x) + f(2a − x)

Differentiating above equation with respect to x we get,

Ď•'(x) = f'(x) − f(2a − x)        .....(1)

Since, f''(x) > 0, f'(x) is an increasing function.

Now,

when
x∈0, a
x≤2a-xf'x≤f2a-x        .....2

Considering equation (1) and (2) we get,

Ď•'(x) ≤ 0

⇒ Ď•'(x) is decreasing in [0, a]



 

Page No 16.41:

Question 27:

If the function f(x) = x2 − kx + 5 is increasing on [2, 4], then
(a) k ∈ (2, ∞)
(b) k ∈ (−∞, 2)
(c) k ∈ (4, ∞)
(d) k ∈ (−∞, 4).

Answer:


(d) k ∈ (−∞, 4)

fx=x2-kx+5f'x=2x-kGiven: f(x) is increasing on[2, 4].⇒ f'x>0⇒2x-k>0⇒k<2x∵  x∈2, 4, maximum value of k is 4, k < 4.∴ k∈-∞, 4

Page No 16.41:

Question 28:

The function f(x) = −x/2 + sin x defined on [−π/3, π/3] is
(a) increasing
(b) decreasing
(c) constant
(d) none of these

Answer:

f(x) =-x2+sin x defined on -π3,π3∴ f'(x)=-12+cos x  ⇒ f'(x)⩾0   ∀x∈-π3,π3∵ for x∈-π3,π3  , cos x⩾12

Hence, the given function is increasing .

Page No 16.41:

Question 29:

If the function f(x) = x3 − 9kx2 + 27x + 30 is increasing on R, then
(a) −1 ≤ k < 1
(b) k < −1 or k > 1
(c) 0 < k < 1
(d) −1 < k < 0

Answer:

(a)

fx=x3-9kx2+27x+30f'x=3x2-18kx+27        =3 x2-6kx+9Given: f(x) is increasing on R.⇒f'x>0  for all x∈R⇒3 x2-6kx+9>0 for all x∈R⇒x2-6kx+9>0  for all x∈R⇒-6k2-419<0      ∵ ax2+bx+c>0 for all x∈R⇒a>0 and Disc<0⇒36k2-36<0⇒k2-1<0⇒k+1k-1<0It can be possible when k+1<0 and k-1>0.⇒k<-1 and k>1       (Not possible)or k+1>0 and k-1<0⇒k>-1 and k<1⇒-1<k<1Disclaimer: (a) part should be −1< k < 1 instead of −1 ≤ k < 1.

Page No 16.41:

Question 30:

The function f(x) = x9 + 3x7 + 64 is increasing on
(a) R
(b) (−∞, 0)
(c) (0, ∞)
(d) R0

Answer:

(a) R

fx=x9+3x7+64f'x=9x8+21x6>0, ∀x∈RSo, f(x) is increasing on R.

Page No 16.41:

Question 31:

The interval on which the function f(x) = 2x3 + 9x2 + 12x - 1 is decreasing is 
(a) [-1, ∞)            (b) [-2, -1]             (c) (∞, -2]               (d) [-1, 1]

Answer:


fx=2x3+9x2+12x-1

⇒f'x=6x2+18x+12

⇒f'x=6x2+3x+2

⇒f'x=6x+2x+1

For f(x) to be decreasing,

f'x<0

⇒6x+2x+1<0

⇒x--2x--1<0

⇒-2<x<-1         [For a < b, if (x − a)(x − b) < 0 ⇒ a < x < b]

Thus, the interval on which the given function f(x) is decreasing is [−2, −1].

Hence, the correct answer is option (b).

Page No 16.41:

Question 32:

y = x(x-3)2 decrease for the values of x given by
(a) 1 < x < 3                (b) x < 0                (c) x > 0                    (d) 0 < x < 32

Answer:


y = x(x − 3)2  

Differentiating both sides with respect to x, we get

dydx=x×2x-3+x-32×1

⇒dydx=x-32x+x-3

⇒dydx=x-33x-3

⇒dydx=3x-1x-3

For y to be decreasing,

dydx<0

⇒3x-1x-3<0

⇒x-1x-3<0

⇒1<x<3                  [For a < b, if (x − a)(x − b) < 0 ⇒ a < x < b]

Thus, y decreases for 1 < x < 3.

Hence, the correct answer is option (a).

Page No 16.41:

Question 33:

The function f(x) = 4sin3x - 6sin2x + 12sin x + 100 is strictly
(a) increasing in  π, 3π2                         (b) decreasing in π2,π
(c) decreasing in -π2,π2                      (d) decreasing in 0,π2

Answer:


The given function is f(x) = 4sin3x − 6sin2x + 12sinx + 100.

f(x) = 4sin3x − 6sin2x + 12sinx + 100

Differentiating both sides with respect to x, we get

f'x=4×3sin2x×cosx-6×2sinx×cosx+12cosx

f'x=12sin2xcosx-12sinxcosx+12cosx

⇒f'x=12cosxsin2x-sinx+1

⇒f'x=12cosxsinx-122+34

Now,

sinx-122+34>0 ∀ x ∈ R

When x∈-π2,π2, cosx ≥ 0

∴f'x≥0

So, f(x) is increasing in -π2,π2.

When x∈0,π2, cosx ≥ 0

∴f'x≥0

So, f(x) is increasing in 0,π2.

When x∈π, 3π2, cosx ≤ 0

∴f'x≤0

So, f(x) is decreasing in π, 3π2.

When x∈π2,π, cosx < 0

∴f'x<0

So, f(x) is strictly decreasing in π2,π.

Thus, the function f(x) = 4sin3x − 6sin2x + 12sin x + 100 is strictly decreasing in π2,π.

Hence, the correct answer is option (b).

Page No 16.41:

Question 34:

Which of the following functions is decreasing in 0,π2?
(a) sin 2x          (b) tan x              (c) cos x            (d) cos 3x

Answer:


Let f(x) = sin2x

∴f'x=2cos2x

0<x<π2    (Given)

⇒0<2x<π

Now, cos2x > 0 when 0<2x<π2 and cos2x < 0 when π2<2x<π.

⇒f'x>0 when 0<x<π4 and f'x<0 when π4<x<π2

⇒ f(x) is increasing when 0<x<π4 and f(x) is decreasing when π4<x<π2

Thus, f(x) = sin2x is both increasing and decreasing in the interval 0,π2.


Let g(x) = tanx

∴g'x=sec2x

Now, sec2x > 0 when 0<x<π2

⇒g'x>0 when 0<x<π2

 ⇒ g(x) = tanx is increasing when 0<x<π2


Let h(x) = cosx

∴h'x=-sinx

Now, sinx > 0 when 0<x<π2

⇒h'x<0 when 0<x<π2

 ⇒ h(x) = cosx is decreasing when 0<x<π2


Let p(x) = cos3x

∴p'x=-3sin3x

0<x<π2    (Given)

⇒0<3x<3π2

Now, sin3x > 0 when 0<3x<π and sin3x < 0 when π<3x<3π2.

⇒p'x<0 when 0<x<π3 and p'x>0 when π3<x<π2

⇒ p(x) is decreasing when 0<x<π3 and p(x) is increasing when π3<x<π2

⇒ p(x) = cos3x is both increasing and decreasing in the interval 0,π2.


Thus, the function cosx is decreasing in 0,π2.

Hence, the correct answer is option (c).



Page No 16.42:

Question 35:

The function f(x) = tan x - x
(a) always increases                    (b) always decreases
(c) never increases                      (d) sometimes increases sometime decreases

Answer:


The given function is fx=tanx-x.

fx=tanx-x

Differentiating both sides with respect to x, we get

f'x=sec2x-1

We know

secx∈(-∞,-1]∪[1,∞)

⇒sec2x∈[1,∞)

Or sec2x ≥ 1 for all real values of x

∴f'x≥0 ∀x∈R

⇒ f(x) is increasing for all x ∈ R

Thus, the function f(x) always increases for all real values of x.

Hence, the correct answer is option (a).

Page No 16.42:

Question 1:

The values of 'a' for which the function f(x) = sin x − ax + b increases on R are _______________.

Answer:


The given function is f(x) = sinx − ax + b.

f(x) = sinx − ax + b

Differentiating both sides with respect to x, we get

f'x=cosx-a

It is given that, f(x) increases on R.

∴f'x≥0 ∀ x ∈ R

⇒cosx-a≥0 ∀ x ∈ R

⇒a≤cosx ∀ x ∈ R

⇒a≤-1             -1≤cosx≤1

∴ a ∈ (−∞, −1]

Thus, the values of 'a' for which the function f(x) = sin x − ax + b increases on R are (−∞, −1].


The values of 'a' for which the function f(x) = sin x − ax + b increases on R are ___(−∞, −1]___.

Page No 16.42:

Question 2:

The function f(x) = 2x2-1x4, x > 0, decreases in the interval ________________.

Answer:


The given function is fx=2x2-1x4, x>0.

fx=2x2-1x4

⇒fx=2x2-1x4

Differentiating both sides with respect to x, we get

f'x=-4x3+4x5

⇒f'x=-4x2-1x5

⇒f'x=-4x+1x-1x5

For f(x) to be decreasing,

f'x<0

⇒-4x+1x-1x5<0

⇒x+1x-1>0            x5>0 for x>0

⇒x--1x-1>0

⇒x∈-∞,-1∪1,∞          [For a < b, if (x − a)(x − b) > 0 ⇒ x ∈ (−∞, a) ∪ (b, ∞)]

But, x > 0       (Given)

∴ x ∈ (1, ∞)

Thus, the given function f(x) decreases in the interval (1, ∞).


The function f(x) = 2x2-1x4, x > 0, decreases in the interval ____(1, ∞)____.

Page No 16.42:

Question 3:

The function g(x) = x + 1x,x≠0 decreases in the closed interval ____________________.

Answer:


The given function is gx=x+1x, x≠0.

gx=x+1x

Differentiating both sides with respect to x, we get

g'x=1-1x2

For g(x) to be decreasing,

g'x<0

⇒1-1x2<0

⇒x2-1x2<0

⇒x2-1<0       x2>0 ∀x∈R-0

⇒x+1x-1<0

⇒x--1x-1<0                 [For a < b, if (x − a)(x − b) < 0 ⇒ a < x < b]

⇒-1<x<1

∴ x ∈ (−1, 1)

Thus, the given function g(x) decreases in the interval (−1, 1).


The function g(x) = x + 1x,x≠0 decreases in the closed interval _____[−1, 1]_____.

Page No 16.42:

Question 4:

The largest open interval in which the function f(x) = 11+x2 decreases is _______________.

Answer:


The given function is fx=11+x2.

fx=11+x2

Differentiating both sides with respect to x, we get

f'x=-2x1+x22

For f(x) to be decreasing,

f'x<0

⇒-2x1+x22<0

⇒-2x<0          1+x22>0 ∀x∈R

⇒x>0

∴ x ∈ (0, ∞)

Thus, the largest open interval in which the given function f(x) decreases is (0, ∞).


The largest open interval in which the function f(x) = 11+x2 decreases is ____(0, ∞)____.

Page No 16.42:

Question 5:

The set of values of x for which f(x) = tan x - x is increasing is _______________.

Answer:


The given function is fx=tanx-x.

fx=tanx-x

Differentiating both sides with respect to x, we get

f'x=sec2x-1

For f(x) to be increasing,

f'x>0

⇒sec2x-1>0

⇒sec2x>1

We know

secx∈(-∞,-1]∪[1,∞)

⇒sec2x∈[1,∞)

Or sec2x ≥ 1 for all real values of x

∴f'x>0 ∀x∈R 

Thus, the set of values of x for which f(x) is increasing is the set of all real numbers i.e. R.


The set of values of x for which f(x) = tan x − x is increasing is ___the set of all real numbers i.e. R___.

Page No 16.42:

Question 6:

The set of values of  'a' for which the function f(x) = sin x - cos x - ax + b decreases for all the real values of x, is ___________.

Answer:


The given function is f(x) = sin x − cos x − ax + b.

f(x) = sin x − cos x − ax + b

Differentiating both sides with respect to x, we get

f'x=cosx--sinx-a

⇒f'x=cosx+sinx-a

⇒f'x=2cosx×12+sinx×12-a

⇒f'x=2cosxcosπ4+sinxsinπ4-a

⇒f'x=2cosx-π4-a

For f(x) to be decreasing for all x,

f'x≤0

⇒2cosx-π4-a≤0

⇒a≥2cosx-π4

⇒a≥2              -1≤cosx-π4≤1

∴a∈[2,∞)

Thus, the set of values of  'a' for which the given function f(x) decreases for all the real values of x is [2,∞).


The set of values of  'a' for which the function f(x) = sin x − cos x − ax + b decreases for all the real values of x, is       [2,∞)      .

Page No 16.42:

Question 7:

The set of values of  'a' for which the function f(x) = ax + b is strictly increasing for all real x, is _______________.

Answer:


The given function is f(x) = ax + b.

f(x) = ax + b

Differentiating both sides with respect to x, we get

f'x=a

For f(x) to be strictly increasing for all real x,

f'x>0

⇒a>0

∴a∈0,∞

Thus, the set of values of 'a' for which the function f(x) = ax + b is strictly increasing for all real x is (0, ∞).


The set of values of  'a' for which the function f(x) = ax + b is strictly increasing for all real x, is ____(0, ∞)____.

Page No 16.42:

Question 8:

If kπ is the length of the largest interval in which the function f(x) = 3sin x - 4sin3x is increasing, then k = _________________.

Answer:


The given function is f(x) = 3sin x − 4sin3x.

f(x) = 3sin x − 4sin3x = sin3x

Differentiating both sides with respect to x, we get

f'x=3cos3x

For f(x) to be increasing,

f'x≥0

⇒3cos3x≥0

⇒cos3x≥0

⇒-π2≤3x≤π2

⇒-π6≤x≤π6

∴ Length of the largest interval in which the given function f(x) is increasing = π6--π6=2π6=π3

It is given that, the length of the largest interval in which the function f(x) = 3sin x − 4sin3x is increasing is kπ.

∴kπ=π3

⇒k=13

Thus, the value of k is 13.


If kπ is the length of the largest interval in which the function f(x) = 3sin x − 4sin3x is increasing, then k =       13      .

Page No 16.42:

Question 9:

The set of values of λ for which the function f(x) = λsinx+6cosx2sinx+3cosx is strictly increasing, is ___________________.

Answer:


The given function is fx=λsinx+6cosx2sinx+3cosx.

fx=λsinx+6cosx2sinx+3cosx

Differentiating both sides with respect to x, we get

f'x=2sinx+3cosx×λcosx-6sinx-λsinx+6cosx×2cosx-3sinx2sinx+3cosx2

⇒f'x=2λsinxcosx-12sin2x+3λcos2x-18sinxcosx-2λsinxcosx+3λsin2x-12cos2x+18sinxcosx2sinx+3cosx2

⇒f'x=3λsin2x+cos2x-12sin2x+cos2x2sinx+3cosx2

⇒f'x=3λ-122sinx+3cosx2         [sin2x + cos2x = 1]

For f(x) to be strictly increasing, 

f'x>0

⇒3λ-122sinx+3cosx2>0

⇒3λ-12>0             2sinx+3cosx2>0

⇒3λ>12

⇒λ>4

Thus, the set of values of λ for which the function f(x) is strictly increasing is (4, ∞).


The set of values of λ for which the function f(x) = λsinx+6cosx2sinx+3cosx is strictly increasing, is ____(4, ∞)____.

Page No 16.42:

Question 10:

The largest interval in which f(x) = x1/x is strictly increasing is ______________.

Answer:


The given function is fx=x1x.

For f(x) to be defined x > 0.

fx=x1x

⇒logfx=logx1x

⇒logfx=logxx               logab=bloga

Differentiating both sides with respect to x, we get

1fx×f'x=x×1x-logx×1x2

⇒f'x=x1x1-logxx2

For f(x) to be strictly increasing function,

f'x>0

⇒x1x1-logxx2>0

⇒1-logx>0               For x>0, x1x>0 and x2>0

⇒logx<1

⇒logx<loge

⇒x<e

⇒x∈0,e         (x > 0)

Thus, the largest interval in which f(x) = x1/x is strictly increasing is (0, e).


The largest interval in which f(x) = x1/x is strictly increasing is ____(0, e)____.

Page No 16.42:

Question 1:

What are the values of 'a' for which f(x) = ax is increasing on R?

Answer:

fx=axf'x=ax log aGiven: f(x) is increasing on R.⇒f'x>0⇒ax log a>0Logarithmic function is defined for positive values of a.⇒a>0⇒ax>0We know, ax log a>0It can be possible when ax>0 and log a>0 or ax<0 and log a<0.  ⇒log a>0⇒a>1So, f(x) is increasing when a > 1.

Page No 16.42:

Question 2:

What are the values of 'a' for which f(x) = ax is decreasing on R?

Answer:

fx=axf'x=axlog aGiven:  fx is decreasing on R.⇒f'x<0,∀x∈R⇒axlog a<0,∀x∈RHere, logaritmic function is not defined for negative values of a.⇒ax>0 ∴ axlog a<0 can be possible when log a<0,∀x∈R.  ⇒0<a<1

Page No 16.42:

Question 3:

Write the set of values of 'a' for which f(x) = loga x is increasing in its domain.

Answer:

fx=loga xLet x1, x2 ∈0, ∞ such that x1<x2.Since given function is logarithmic, either a>1 or 0<a<1.Case 1: Let a>1Here, x1<x2⇒loga x1<loga x2⇒fx1<fx2∴ x1<x2⇒fx1<fx2, ∀ x1, x2 ∈0, ∞So, fx is increasing on 0, ∞.Case 2: Let 0<a<1Here, x1<x2⇒loga x1>loga x2⇒fx1>fx2∴ x1<x2⇒fx1>fx2, ∀ x1, x2 ∈0, ∞Thus, for a>1, f(x) is increasing in its domain.



Page No 16.43:

Question 4:

Write the set of values of 'a' for which f(x) = loga x is decreasing in its domain.

Answer:

Given: fx=loga xDomain of the given function is 0, ∞.Let x1, x2 ∈0, ∞ such that x1<x2.Since the given function is logarithmic, either a>1 or 0<a<1.Case 1: Let a>1Here,x1<x2⇒loga x1<loga x2⇒fx1<fx2∴  x1<x2⇒fx1<fx2, ∀ x1, x2 ∈0, ∞So, fx is increasing on 0, ∞.Case 2: Let 0<a<1Here, x1<x2⇒loga x1>loga x2⇒fx1>fx2∴ x1<x2⇒fx1>fx2, ∀ x1, x2 ∈0, ∞So, fx is decreasing on 0, ∞Thus, for 0<a<1, fx is decreasing in its domain.

Page No 16.43:

Question 5:

Find 'a' for which f(x) = a (x + sin x) + a is increasing on R.

Answer:

fx=a x+sin x+af'x=a 1+cos xFor f(x) to be increasing, we must havef'x>0⇒a 1+cos x>0         ... 1We know, -1≤cos x≤1, ∀x∈R⇒0≤1+cos x≤2, ∀x∈R


∴ a>0          From eq. 1⇒a∈0, ∞

Page No 16.43:

Question 6:

Find the values of 'a' for which the function f(x) = sin x − ax + 4 is increasing function on R.

Answer:

fx=sin x-ax+4f'x=cosx-aGiven: f(x) is increasing on R.⇒f'x>0⇒cosx-a>0⇒cosx>a We know, cosx≥-1, ∀x∈R ⇒a<-1⇒a∈-∞, -1

Page No 16.43:

Question 7:

Find the set of values of 'b' for which f(x) = b (x + cos x) + 4 is decreasing on R.

Answer:

fx=bx+cos x+4f'x=b1-sin xGiven: f(x) is decreasing on R.⇒f'x<0⇒b1-sin x<0         ... 1We know,sin x≤1⇒1-sin x≥0⇒b<0                    Since 1-sin x≥0, b1-sin x<0⇒b<0⇒b∈-∞,0

Page No 16.43:

Question 8:

Find the set of values of 'a' for which f(x) = x + cos x + ax + b is increasing on R.

Answer:

fx=x+cos x+ax+bf'x=1-sin x+aFor f(x) to be increasing, we must havef'x>0⇒1-sin x+a>0⇒sin x<1+aWe know that the maximum value of sin x is 1.⇒1+a>1⇒a>0⇒a∈0, ∞

Page No 16.43:

Question 9:

Write the set of values of k for which f(x) = kx − sin x is increasing on R.

Answer:

fx=kx-sin xf'x=k-cos xFor f(x) to be increasing, we must havef'x>0⇒k-cos x>0⇒cos x<kWe know that the maximum value of cos x is 1.Since cos x < k, the minimum value of k is 1.⇒ k∈1, ∞

Page No 16.43:

Question 10:

If g (x) is a decreasing function on R and f(x) = tan−1 [g (x)]. State whether f(x) is increasing or decreasing on R.

Answer:

Given: gx is decreasing on R.⇒x1<x2⇒gx1>gx2Applying tan-1 on both sides, we get⇒tan-1gx1>tan-1gx2⇒fx1>fx2Thus, x1<x2 ⇒fx1>fx2So, fx is decreasing on R.

Page No 16.43:

Question 11:

Write the set of values of a for which the function f(x) = ax + b is decreasing for all x ∈ R.

Answer:

fx=ax+bf'x=aFor f(x) to be decreasing, we must havef'x<0⇒a<0⇒a∈-∞, 0

Page No 16.43:

Question 12:

Write the interval in which f(x) = sin x + cos x, x ∈ [0, π/2] is increasing.

Answer:

fx=sin x+cos x, x∈0, π2f'x=cos x-sin xFor f(x) to be increasing, we must havef'x>0⇒cos x-sin x>0⇒sin x<cos x⇒sin xcos x<1⇒tan x<1⇒x∈ [0, π4)

Page No 16.43:

Question 13:

State whether f(x) = tan x − x is increasing or decreasing its domain.

Answer:

fx=tan x-xf'x=sec2x-1         =tan2 x⩾0, ∀x∈0,2πSo, f(x) is increasing in its domain.

Page No 16.43:

Question 14:

Write the set of values of a for which f(x) = cos x + a2 x + b is strictly increasing on R.

Answer:

fx=cos x+a2x+bf'x=a2-sinxGiven: f(x) is strictly increasing on R.⇒f'x>0, ∀x∈R⇒a2-sin x>0, ∀x∈R⇒a2>sin x, ∀x∈RWe know that the maximum value of sin x is 1.Since a2>sin x, a2 is always greater than 1.⇒a2>1⇒a2-1>0⇒a+1a-1>0⇒a∈(-∞, -1)∪(1, ∞)



View NCERT Solutions for all chapters of Class 12