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Page No 17.10:

Question 1:

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

Answer:

Let x1, x2 0,  such that x1<x2. Then, x1<x2loge x1<loge x2fx1<fx2∴ x1<x2fx1<fx2,  x1, x2 0, So, fx is increasing on 0, .

Page No 17.10:

Question 2:

Prove that the function f(x) = logax 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<x2loga x1<loga x2fx1<fx2 x1<x2fx1<fx2,  x1, x2 0, So, fx is increasing on 0, .Case 2: Let 0<a<1Here, x1<x2loga x1>loga x2fx1>fx2  x1<x2fx1>fx2,  x1, x2 0, So, fx is decreasing on 0, .

Page No 17.10:

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<x2a x1<a x2      ∵ a >0a x1+b<a x2+bfx1<fx2 x1<x2fx1<fx2,  x1, x2 R So, fx is increasing on R.

Page No 17.10:

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<x2a x1>a x2       (∵ a <0)a x1+b>a x2+bfx1>fx2Thus, x1<x2fx1>fx2,  x1, x2 RSo, fx is decreasing on R.

Page No 17.10:

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<x21x1>1x2fx1>fx2∴ x1<x2fx1>fx2,  x1, x2 0, So, fx is decreasing on 0, .

Page No 17.10:

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<x2x12<x221+x12<1+x2211+x12>11+x22fx1>fx2  x1, x2 0, So, fx is decreasing on 0, .Case 2: Let x1, x2 (-, 0] such that x1<x2. Then, x1<x2x12>x221+x12>1+x2211+x12<11+x22fx1<fx2fx1<fx2,  x1, x2 (-, 0]So, fx is increasing on (-, 0].

Page No 17.10:

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<x2x12<x221+x12<1+x2211+x12>11+x22fx1>fx2 x1, x2 0, So, fx is decreasing on 0, .Case 2: Let x1, x2 (-, 0] such that x1<x2. Then, x1<x2x12>x221+x12>1+x2211+x12<11+x22fx1<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 17.10:

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<x2x1<x2fx1<fx2∴ x1<x2fx1<fx2,  x1, x2 0, So, fx is increasing on 0, .(b)  Let x1, x2  (-, 0]. such that x1<x2. Then, x1<x2x1>x2fx1>fx2∴ x1<x2fx1>fx2,  x1, x2 (-, 0].So, fx is decreasing on (-, 0].

Page No 17.10:

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<x27 x1<7 x2      ∵ 7 >07 x1-3<7 x2-3fx1<fx2∴ x1<x2fx1<fx2,  x1, x2 RSo, fx is  strictly increasing on R.



Page No 17.32:

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) = 5x32-3x52x > 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, xR

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>6x<-32x-,-32So, f(x) is increasing on -,-32.For f(x) to be decreasing, we must have f'(x) <0-6-4x<0-4x<6x>-64x>-32x-32,So, f(x) is decreasing on -32,.

iifx=x2+2x-5f'x=2x+2For f(x) to be increasing, we must havef'x>02x+2>02x+1>0x+1>0x>-1x-1, So, f(xis increasing on -1, .For f(x) to be decreasing, we must havef'x<02x+2<02x+1<0x+1<0x<-1x-, -1So, f(xis 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>9x<-92x-, -92So, f(xis increasing on -, -92.For f(x) to be decreasing, we must havef'x<0-2x-9<0-2x<9x>-92x -92, So, f(xis 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>06 x-1x-3>0x-1x-3>0           Since 6>0, 6 x-1x-3>0x-1x-3>0x<1 or x>3x-, 13, So, f(xis increasing on -, 13, .

 


For f(x) to be decreasing, we must havef'x<06 x-1x-3<0x-1x-3<0      Since 6>0, 6 x-1x-3<0 x-1x-3<01<x<3 x1, 3So, f(xis 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(xis increasing on -2, 3.





For f(x) to be decreasing, we must havef'x<0-6 x-3x+2<0x-3x+2>0    Since -6<0, -6 x-1x+2<0 x-1x+2>0x<-2 or x>3 x-, -23, So, f(xis decreasing on -, -23, .


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<3x-2, 3So, f(xis increasing on -2, 3.

 

 

For f(x) to be decreasing, we must havef'x<0-6 x-3x+2<0x-3x+2>0     Since -6<0, -6 x-3x+2<0x-3x+2>0x<-2 or x>3 x-, -23, So, f(xis decreasing on -, -23, .



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>015 x-4x+2>0     x-4x+2>0        Since 15>0, 15 x-4x+2>0  x-4x+2>0x<-2 or x>4x-, -2  4, So, f(xis increasing on x-, -2  4, .



 

For f(x) to be decreasing, we must have,f'x<015 x-4x+2<0x-4x+2<0         Since 15>0, 15 x-4x+2<0x-4x+2<0-2<x<4x-2, 4So, f(xis 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>03 x-6x+2>0x-6x+2>0     Since 3>0, 3 x-6x+2>0x-6x+2>0x<-2 or x>6x-, -2  6, So, f(xis increasing on x-, -2  6, .

 

 

For f(x) to be decreasing, we must havef'x<03 x-6x+2<0x-6x+2<0       Since 3>0, 3 x-6x+2<0x-6x+2<0-2<x<6 x-2, 6So, f(xis 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>06 x-2x-3>0x-2x-3>0       Since 6>0, 6x-2x-3>0x-2x-3>0x<2 or x>3x-, 2  3, So, f(xis increasing on x-, 2  3, .





For f(x) to be decreasing, we must havef'x<06 x-2x-3<0x-2x-3<0      Since 6>0, 6x-2x-3<0x-2x-3<02<x<3 x2, 3So, f(xis decreasing on x2, 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>06 x+1x+2>0x+1x+2>0       Since 6>0, 6 x+1x+2>0x+1x+2>0x<-2 or x>-1x-, -2  -1, So, f(xis increasing on x-, -2  -1, .

 



For f(x) to be decreasing, we must havef'x<06 x+1x+2<0x+1x+2<0     Since 6>0, 6 x+1x+2<0x+1x+2<0-2<x<-1 x-2, -1So, f(xis 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>06 x-1x-2>0x-1x-2>0       Since 6>0, 6 x-1x-2>0 x-1x-2>0x<1 or x>2x-, 1  2, So, f(xis increasing on x-, 1  2, .



 

For f(x) to be decreasing, we must havef'x<06 x-1x-2<0x-1x-2<0        Since 6>0, 6 x-1x-2<0 x-1x-2<01<x<2x1, 2So, f(xis decreasing on x1, 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>0x-2x+1<0          Since -6<0, -6 x-2x+1>0x-2x+1<0  -1<x<2 x-1, 2So, f(xis increasing on -1, 2.

 

 

For f(x) to be decreasing, we must havef'x<0-6 x-2x+1<0x-2x+1>0       Since -6<0, -6 x-2x+1<0x-2x+1>0x<-1 or x>2 x-, -12, So, f(xis decreasing on -, -12, .


xiii fx=2x3-24x+107f'x=6x2-24=6 x2-4=6 x+2x-2For f(x) to be increasing, we must havef'x>06 x+2x-2>0x+2x-2>0       Since 6>0, 6 x+2x-2>0x+2x-2>0 x<-2 or x>2x-, -2  2, So, f(xis increasing on x-, -2  2, .



 

For f(x) to be decreasing, we must havef'x<06 x+2x-2<0x+2x-2<0        Since 6>0, 6 x+2x-2<0x+2x-2<0 -2<x<2 x-2, 2So, f(xis 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>0x+1x+2<0     Since -6<0, -6 x+1x+2>0x+1x+2<0-2<x<-1 x-2, -1So, f(xis increasing on -2, -1.




 

For f(x) to be decreasing, we must havef'x<0-6 x+1x+2<0x+1x+2>0        Since -6<0,-6 x+1x+2<0x+1x+2>0  x<-2 or x>-1 x-, -2-1, So, f(xis 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>0x-23x-4>0x<43 or x>2x-, -43  2, So, f(xis increasing on x-, 43  2, .


For f(x) to be decreasing, we must havef'x<0x-23x-4<043<x<2 x43, 2So, f(xis decreasing on x43, 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>03 x-2x-6>0x-2x-6>0       Since 3>0, 3 x-2x-6>0x-2x-6>0x<2 or x>6 x-, 2  6, So, f(xis increasing on x-, 2  6, .

 

 

For f(x) to be decreasing, we must havef'x<03 x-2x-6<0x-2x-6<0      Since 3>0, 3 x-2x-6<0x-2x-6<02<x<6 x2, 6So, f(xis decreasing on x2, 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>06 x+2x-2>0x+2x-2>0              Since 6>0, 6 x+2x-2>0x+2x-2>0x<-2 or x>2x-, -2  2, So, f(xis increasing on x-, -2  2, .



 

For f(x) to be decreasing, we must havef'x<06 x+2x-2<0x+2x-2<0          Since 6>0, 6 x+2x-2<0x+2x-2<0  -2<x<2x-2, 2So, f(xis 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, 11, 3 and 3, .    For f(x) to be increasing, we must havef'x>01210x-1x+2x-3>0x-1x+2x-3>0x-2, 13, So, f(xis increasing on x-2, 13, .



 
For f(x) to be decreasing, we must havef'x<01210x-1x+2x-3<0x-1x+2x-3<0x- -21, 3 So, f(xis decreasing on x- -21, 3.


xix fx=x4-4xf'x=4x3-4        =4x3-1For f(x) to be increasing, we must havef'x>04x3-1>0      x3-1>0x3>1x>1x1, So, f(xis increasing on 1, .


For f(x) to be decreasing, we must havef'x<04x3-1<0x3-1<0x3<1x<1x-, 1So, f(xis 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>0x+1x-2x+3>0x -3, -12, So, f(xis increasing on x-3, -12, .



For f(x) to be decreasing, we must havef'x<0x+1x-2x+3<0x- -3-1, 2                          From eq. (1)So, f(xis 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 -, 00, 11, 2 and 2, .       ...(1)For f(x) to be increasing, we must havef'x>04x x-1x-2>0                      Since 4>0, 4x x-1x-2>0x x-1x-2>0  x x-1x-2>0x 0, 12,                      From eq. (1)So, f(xis increasing on x0, 12, .



For f(x) to be decreasing, we must havef'x<04x x-1x-2<0                Since 4>0, 4x x-1x-2<0x x-1x-2<0 x x-1x-2<0x -, 01, 2                          From eq. (1)So, f(xis decreasing on x-, 01, 2.





xxii fx=5x32-3x52, x>0f'x=152x12-152x32       =152x121-xHere, 0,1 are the roots.The possible intervals are -,00, 1 and 1,           ...(1)For f(x) to be increasing, we must havef'x>0152x121-x>0x0, 1So, f(xis increasing on 0, 1.

 



For f(x) to be decreasing, we must havef'x<0152x121-x<0x1, So, f(xis decreasing on 1, 


xxiii fx=x8+6x2f'x=8x7+12x        =4x 2x6+3For f(x) to be increasing, we must havef'x>04x 2x6+3>0     Since 2x6+3>0, 4x 2x6+3>0x>0x>0x0, So, f(xis increasing on x0, .



For f(x) to be decreasing, we must havef'x<04x 2x6+3<0x<0                    Since 2x6+3>0, 4x 2x6+3<0x<0 x-, 0So, f(xis 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>03 x-1x-3>0 x-1x-3>0                Since 3>0, 3 x-1x-3>0 x-1x-3>0x<1 or x>3x-, 1  3, So, f(xis increasing on x-, 1  3, .



 

For f(x) to be decreasing, we must havef'x<03 x-1x-3<0x-1x-3<0           Since 3>0, 3 x-1x-3<0 x-1x-3<0 1<x<3 x1, 3So, f(xis decreasing on x1, 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 -, 00, 11, 2 and 2, .           For f(x) to be increasing, we must havef'x>04x x-1x-2>0x-1x-2>0x 0, 12,  So, f(xis increasing on x0, 12, .





For f(x) to be decreasing, we must have f'(x) <04xx-1x-2<0xx-1x-2<0x-,01,2So, f(x) is decreasing on x-,01,2.




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, 00, 2 and 2, .       .....(1)For f(x) to be increasing, we must havef'x>012xx+1x-2>0                      Since, 12>0, 12xx+1x-2>0xx+1x-2>0xx+1x-2>0x-1, 02,                      From eq. (1)So, f(xis increasing on x-1, 02, .


For f(x) to be decreasing, we must havef'x<012xx+1x-2<0                Since 12>0, 12xx+1x-2<0xx+1x-2<0 x x+1x-2<0x -, -10, 2                          From eq. (1)So, f(xis decreasing on x-, -10, 2.


xxvii fx=32x4-4x3-45x2+51f'x=6x3-12x2-90x       =6xx2-2x-15       =6xx-5x+3Here, x=-3x=0 and x=5 are the critical points. The possible intervals are -, -3-3, 00, 5 and 5, .       .....(1)For f(x) to be increasing, we must havef'x>06xx-5x+3>0                      Since, 6>0, 6xx-5x+3>0xx-5x+3>0xx-5x+3>0x-3, 05,                      From eq. (1)So, f(xis increasing on x-3, 05, .


For f(x) to be decreasing, we must havef'x<06xx-5x+3<0                Since 6>0, 6xx-5x+3<0xx-5x+3<0 xx-5x+3<0x-, -30, 5                          From eq. (1)So, f(xis decreasing on x-, -30, 5.



xxviii fx=log2+x-2x2+x, xRf'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>0x-22+x2>0x-2>0, x-2x>2x2,                      From eq. (1)So, f(xis increasing on x2, .


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



Page No 17.33:

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>02x-6>02x>6x>3x3, So, f(xis increasing on 3, .For f(x) to be decreasing, we must havef'x<02x-6<02x<6x<3x-, 3So, f(xis 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, y2x-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 17.33:

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, x0, 2πf'x=cos x + sin xFor f(x) to be increasin, we must havef'x>0cos x + sin x>0sin x>-cos xtan x>-1x0, 3π47π4, 2πSo, f(xis increasing on 0, 3π47π4, 2π.

For f(x) to be decreasing we must have,f'x<0cos x + sin x<0sin x<-cos xtan x<-1x3π4, 7π4So, f(xis decreasing on 3π4, 7π4.

Page No 17.33:

Question 4:

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

Answer:

fx=e2xf'x=2e2xNow, xRSince the value of e2x is always positive for any real value of x, e2x>0.2e2x>0f'x>0So, f(xis increasing on R.

Page No 17.33:

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,xR, x≠ 0So, f(x) is a decreasing function.

Page No 17.33:

Question 6:

Show that f(x) = logax, 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 17.33:

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 x0, π2, cos x>0       cos x is positive in first quadrant f'x>0So, f(xis increasing for 0, π2.For xπ2, π, cos x<0       cos x is negative in second quadrant So, f(xis decreasing for π2, π.Since f(xis increasing on 0, π2 and decreasing on π2, π, fx is neither decreasing nor increasing on 0, π.

Page No 17.33:

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 x0, π2, cot x>0          Cot function is positive in first quadrantf'x>0 So, f(xis increasing on 0, π2.For xπ2, π, cot x<0       Cot  function is negative in second quadrant f'x<0   So, f(xis decreasing on π2, π.

Page No 17.33:

Question 9:

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

Answer:

fx=x-sin xf'x=1-cos xFor f(x) to be increasing, we must havef'x>01-cos x>0f'(x)0 for all xR             Cos x1So, f(x) is increasing for all xR.

Page No 17.33:

Question 10:

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

Answer:

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

Page No 17.33:

Question 11:

Show that f(x) = cos2x is a decreasing function on (0, π/2).

Answer:

fx=cos2xf'x=2 cos x -sin xf'x=-sin 2x           ...1Now,0<x<π20<2x<π sin 2x>0              Sine fuction is positive in first and second quadrant -sin 2x<0f'x<0               From eq. (1)So, f(xis decreasing on 0, π2.

Page No 17.33:

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 17.33:

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>0f'x>0So, cos x is increasing in -π,0.For x0, π), sin x>0         sine function is positive in first and second quadrant -sin x<0f'x<0So, f(x) is decreasing on 0, π.Thus, f(x) is neither increasing nor decreasing in -π, π.

Page No 17.33:

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<π2sec x>0         Sec function is positive in first and fourth quadrant sec2 x>0f'x>0, x-π2, π2So, f(xis increasing on -π2, π2.



Page No 17.34:

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<π2π2<2x<πsin 2x>02+sin 2x>0              ...1Also,π4<x<π2cos x<sin xcos x-sin x<0         ...2f'x=cos x-sin x2+sin 2x<0, xπ4, π2       From eqs. (1) and (2)So, fx is decreasing on π4, π2.

Page No 17.34:

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<5π83π4<2x<5π4π<2x+π4<3π2cos 2x+π4<0        ∵ Cos function is negative in third quadrant2 cos 2x+π4<0f'x<0, x3π8, 5π8So, fx is decreasing on 3π8, 5π8.

Page No 17.34:

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+cosxf'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<012×sinx-cosx1+sinxcosx<0sinx-cosx1+sinxcosx<0sinx-cosx<0         In first quadrantsinx<cosxtanx<10<x<π4So, fx is decreasing on 0,π4.For fx to be increasing, we must havef'x>012×sinx-cosx1+sinxcosx>0sinx-cosx1+sinxcosx>0sinx-cosx>0         In first quadrantsinx>cosxtanx>1π4<x<π2So, fx is increasing on π4,π2.

Page No 17.34:

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>0f'x>0, x>0So, f(xis increasing on for all x>0.

Page No 17.34:

Question 19:

Show that the function x2x + 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>02x-1>02x>1x>12x12, 1So, f(xis increasing on 12, 1.For f(x) to be decreasing, we must havef'x<02x-1<02x<1x<12x0, 12So, f(xis decreasing on 0, 12.Since f(x) is increasing on 12, 1 and decreasing on 0, 12fx is neither increasing nor decreasing on (0, 1).

Page No 17.34:

Question 20:

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

Answer:

fx=x9+4x7+11f'x=9x8+28x60, xR     ∵ x8, x6≥0, for ∀xRSo, f(xis increasing on R.

Page No 17.34:

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-220, xR          3>0 & x-220So, f(xis increasing on R.

Page No 17.34:

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-6f'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 17.34:

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<π4cos x>0          ... 1Also,-π4<x<π4-1<cot x<10<1+cot x<21+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 17.34:

Question 24:

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

Answer:

fx=tan-1x-xf'x=11+x2-1       =1-1-x21+x2      =-x21+x2We know,x20, 1+x2>0, ∀ xR -x21+x2<0, ∀ xRf'x<0, ∀ xRSo, fx is decreasing on R.

Page No 17.34:

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<π3cos x>12-12+cos x>0f'x>0, x-π3, π3So, fx is increasing on -π3, π3.

Page No 17.34:

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>0x1+x2>0x>0             ∵ 1+x2>0, Domain: -1, x0, So, f(xis increasing on 0, .For f(x) to be decreasing, we must havef'x<0x1+x2<0x<0              1+x2>0, Domain: -1, x-1, 0So, f(xis decreasing on -1, 0.

Page No 17.34:

Question 27:

Find the intervals in which f(x) = (x + 2) ex 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>0e-x-x-1>0-x-1>0          ∵ e-x>0, xR-x>1x<-1x-, -1So, f(xis increasing on -, -1.For f(x) to be decreasing, we must havef'x<0e-x-x-1<0-x-1<0     ∵ e-x>0, xR-x<1x>-1x-1, So, f(xis decreasing on -1, .

Page No 17.34:

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, xRSo, f(xis increasing for all xR. 

Page No 17.34:

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<x2x1-x1<x2-x2         From eq. (1)fx1<fx2∴ x1<x2fx1<fx2,  x1, x2 0, 1So, fx is increasing on 0, 1.

Page No 17.34:

Question 30:

Prove that the function f(x) = 3x5 + 40x3 + 240x is increasing on R.

Answer:

fx=3x5+40x3+240xf'x=15x4+120x2+240       =15 x4+8x2+16       =15 x2+42>0, xR         15>0 and x2+42>0So, f(xis increasing on R.

Page No 17.34:

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, 0tan x<0-tan x>0 f'(x)>0So, f(xis strictly increasing on -π2, 0.Now, x0,π2tan x>0-tan x<0 f'(x)<0So, f(xis strictly decreasing on 0, π2.

Page No 17.34:

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, xRHencef(xis strictly increasing on R.

Page No 17.34:

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<πsin x>0       Sine function is positive in first and second quadrant-sin x<0f'x<0, x0, πSo, f(xis strictly decreasing on 0, π.ii Here,π<x<2πsin x<0      Sine function is negative in third and fourth quadrant-sin x>0f'x>0, xπ, 2πSo, f(xis strictly increasing on π, 2π.iii From eqs. (1) and (2), we get f(xis strictly decreasing on 0, π and is strictly increasing on π, 2π.So, fx is neither increasing nor decreasing on 0, 2π.

Page No 17.34:

Question 34:

Show that f(x) = x2x 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<π20<sin x<1 and 0<cos x<12x-x cos x-sin x>0f'x>0, x0, π2So, fx is increasing on 0, π2.

Page No 17.34:

Question 35:

Find the value(s) of a for which f(x) = x3ax is an increasing function on R.

Answer:

fx=x3-axf'x=3x2-aGiven: fx is increasing on R.f'x0   xR3x2-a0   xRa3x2 xRThe least value of 3x2 is 0.∴ a0

Page No 17.34:

Question 36:

Find the values of b for which the function f(x) = sin xbx + 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, xRcos x-b<0, xRcos x-b<0,xR cos x<b, xR b1             -1cos x1    

Page No 17.34:

Question 37:

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

Answer:

fx=x+cos x-af'x=1-sin xWe know,sin x1, xR-sin x-1, xR1-sin x0, xRf'x0, xRHencefx is increasing on R for all values of a.

Page No 17.34:

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 17.34:

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, xRCase I: When x0fx=xx=xx=x2f'x=2x0 x0So, fx is increasing for x0.Case II:  When x<0fx=xx=x-x=-x2f'x=-2x0 x<0So, fx is increasing for x<0.Hence, fx is increasing for xR.ii fx=sinx+sinx, 0<x2πCase I: When x0,πfx=sinx+sinx=2sinxf'x=2cosxAs, cosx>0 for x0,π2 and cosx<0 for xπ2,πSo, f'x>0 for x0,π2 and f'x<0 for xπ2,π fx is increaing on 0,π2 and fx is decreasing on π2,π.Case II: When xπ,2πfx=sinx-sinx=0f'x=0So, fx is neither increaing nor decreasing on π,2π.iii fx=sinx1+cosx,0<x<π2fx=sinx+sinxcosxf'x=cosx+sinx-sinx+cosxcosxf'x=cosx-sin2x+cos2xf'x=cosx+cos2x-1+cos2xf'x=2cos2x+cosx-1f'x=2cos2x+2cosx-cosx-1f'x=2cosxcosx+1-1cosx+1f'x=2cosx-1cosx+1For fx to be increasing, we must havef'x>02cosx-1cosx+1>0This is only possible when2cosx-1>0 and cosx+1>02cosx-1>0 and cosx+1>0cosx>12 and cosx>-1x0,π3 and x0,π2So, x0,π3 fx is increasing on 0,π3.For fx to be decreasing, we must havef'x<02cosx-1cosx+1<0This is only possible when2cosx-1<0 and cosx+1>02cosx-1<0 and cosx+1>0cosx<12 and cosx>-1xπ3,π2 and x0,π2So, xπ3,π2 fx is decreasing on π3,π2.



Page No 17.39:

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>0ax log a>0Logarithmic function is defined for positive values of a.a>0ax>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>0a>1So, f(xis increasing when > 1.

Page No 17.39:

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,xRaxlog a<0,xRHere, logaritmic function is not defined for negative values of a.ax>0  axlog a<0 can be possible when log a<0,xR.  0<a<1

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Question 3:

Write the set of values of 'a' for which f(x) = logax 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<x2loga x1<loga x2fx1<fx2∴ x1<x2fx1<fx2,  x1, x2 0, So, fx is increasing on 0, .Case 2: Let 0<a<1Here, x1<x2loga x1>loga x2fx1>fx2 x1<x2fx1>fx2,  x1, x2 0, Thus, for a>1, f(x) is increasing in its domain.

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Question 4:

Write the set of values of 'a' for which f(x) = logax 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<x2loga x1<loga x2fx1<fx2∴  x1<x2fx1<fx2,  x1, x2 0, So, fx is increasing on 0, .Case 2: Let 0<a<1Here, x1<x2loga x1>loga x2fx1>fx2∴ x1<x2fx1>fx2,  x1, x2 0, So, fx is decreasing on 0, Thus, for 0<a<1fx is decreasing in its domain.

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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>0a 1+cos x>0         ... 1We know, -1cos x1, xR01+cos x2, xR


 a>0          From eq. 1a0, 

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Question 6:

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

Answer:

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

Page No 17.39:

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<0b1-sin x<0         ... 1We know,sin x11-sin x0b<0                    Since 1-sin x0, b1-sin x<0b<0b-,0

Page No 17.39:

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>01-sin x+a>0sin x<1+aWe know that the maximum value of sin is 1.1+a>1a>0a0, 

Page No 17.39:

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>0k-cos x>0cos x<kWe know that the maximum value of cos x is 1.Since cos x < k, the minimum value of k is 1.⇒ k1, 

Page No 17.39:

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<x2gx1>gx2Applying tan-1 on both sides, we gettan-1gx1>tan-1gx2fx1>fx2Thus, x1<x2 fx1>fx2So, fx is decreasing on R.

Page No 17.39:

Question 11:

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

Answer:

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

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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, x0, π2f'x=cos x-sin xFor f(x) to be increasing, we must havef'x>0cos x-sin x>0sin x<cos xsin xcos x<1tan x<1x [0, π4)

Page No 17.39:

Question 13:

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

Answer:

fx=tan x-xf'x=sec2x-1         =tan2 x0, x0,2πSo, f(xis increasing in its domain.

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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, xRa2-sin x>0, xRa2>sin x, xRWe know that the maximum value of sin is 1.Since a2>sin x, a2 is always greater than 1.a2>1a2-1>0a+1a-1>0a(-, -1)(1, )

Page No 17.39:

Question 1:

The interval of increase of the function f(x) = xex + 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>01-ex>0ex<1x<0x-, 0So, f(xis increasing on -, 0.

Page No 17.39:

Question 2:

The function f(x) = cot−1x + 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+x20, xRSo, f(xis increasing on -, .

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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<0xx1+loge x<0Here, logaritmic function is defined for positive values of x.xx>01+loge x<0      Since xx>0, xx1+loge x<01+loge x<0  loge x<-1x<e-1                   logax<Nx<aN for a>1 Here, e>1logex<-1x<e-1x0, e-1So, f(x) is decreasing on 0, 1e.

Page No 17.39:

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>0x2-4x+3+<0             x-2>0 & -2<0x-1x-3<01<x<3x1, 3Also, the domain of fx is 2, .x1, 32, x2, 3

Page No 17.39:

Question 5:

If the function f(x) = 2x2kx + 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>04x-k>0k<4xSince x1, 2, 4x4, 8.So, the minimum value of 4x is 4.Since k<4x, k<4.k-, 4

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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, xR3x2+2ax+b+5 sin 2x>0, xR Since this quadratic function is >0, its discriminant is <0.2a2-43b+5 sin 2x<04a2-12b-60 sin 2x<0a2-3b-15 sin 2x<0We know that the minimum value of sin 2x is -1.∴ a2-3b-15<0 

Page No 17.39:

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+1f(-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 17.40:

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) aR

Answer:

f(x)= 2 tan x+2a+1logesec x+a-2 xWhen sec x>0sec 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-202+2 tan2x+2a+1tan x+a-202 tan2x+2a+1tan x+a0Its discriminant 0           ax2+bx+c0b2-4ac02a+12-4.2.a04a2-4a+102a-1202a-12<0 cannot be possible . 2a-12=0a=12When sec x<0sec 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-202+2 tan2x-2a+1tan x+a-202 tan2x-2a+1tan x+a0    Its discriminant 0          ax2+bx+c0b2-4ac0-2a+12-4.2.a04a2-4a+102a-1202a-12<0 cannot be possible . 2a-12=0a=12

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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, x20, π2gx1<gx2Taking tan-1 on both the sides, we gettan-1gx1<tan-1gx2fx1<fx2, ∀ x1, x20, π2So, fx is increasing on 0, π2.

Page No 17.40:

Question 10:

Let f(x) = x3 − 6x2 + 15x + 3. Then,
(a) f(x) > 0 for all xR
(b) f(x) > f(x + 1) for all xR
(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.

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Question 11:

The function f(x) = x2ex is monotonic increasing when
(a) xR − [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>0e-x x2-x>0        e-x>0 x2-x>0 xx-2<00<x<2

Page No 17.40:

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 <0sin x+2 λ>0 2 λ>-sin xWe know that the maximum value of -sin is 1.2 λ>1λ>12

Page No 17.40:

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<0x-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, x1, 2So, fx is monotonically decreasing.

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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>03 x2-9>0x2-9>0       Since 3>0, 3 x2-9>0x2-9>0|x+3x-3>0x<-3 or x>3x>3

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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<06x-1x-2<0         x-1x-2<0            Since 6>0, 6x-1x-2<0x-1x-2<01<x<2So, f(xis decreasing for 1<x<2.

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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>03 kx2-6x+3>0kx2-6x+3>0k>0 and -62-4k3<0        ax2+bx+c>0a>0 and Disc<0 k>0 and -62-4k3<0k>0 and 36-12k<0k>0 and 12k>36k>0 and k>3k>3

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Question 17:

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

Answer:

(c) xR

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>01+2x2>x2+11+2x22>x2+11+4x4+4x2>x2+14x4+3x2>0Thus, f(x) is monotonically increasing for xR.

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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 < 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-1Nowfx=x-x-1      =x-x+1      =1f'x=0Sofx is not monotonicallyincreasing when x >1.Thus, fx is monotonically increasing when 0<x<1.

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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".

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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.

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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 , x0               -x, x<0and cos x=cosx ,         x0               cos-x=cosx,   x<0cosx=cos x      ,xR fx=cos x-2ax+bf'x=-sin x-2aIt is given that f(x) is increasing.f'x0-sin x-2a0sin x+2a02a -sin xThe least value of -sin x is -1.2a-1a-12

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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+xf'x=1+x1-x11+x2            =11+x2>0, xRSo, fx is strictly increasing when > 0.Case 2: When x<0, x=-xfx=x1+x      =x1-xf'x=1-x1-x-11-x2        =11-x2>0, xRSo, fx is strictly increasing when x < 0.Thus, fx is strictly increasing on R.
 



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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>0λ >2

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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>0ax log a>0ax>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 alog a>0a>1So, f(x) is increasing when > 1.

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Question 25:

Function f(x) = logax 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, xR1x log a>0, xRa>1

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Question 26:

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

Answer:

Given: Ï•(x) = f(x) + f(2ax)

Differentiating above equation with respect to x we get,

Ï•'(x) = f'(x) − f(2ax)        .....(1)

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

Now,

when
x0, a
x2a-xf'xf2a-x        .....2


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

Ï•'(x) ≤ 0

⇒ Ï•'(x) is decreasing in [0, a]



 

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Question 27:

If the function f(x) = x2kx + 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>02x-k>0k<2x∵  x2, 4, maximum value of is 4, < 4. k-, 4

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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 x12

Hence, the given function is increasing .

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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 xR3 x2-6kx+9>0 for all xRx2-6kx+9>0  for all xR-6k2-419<0       ax2+bx+c>0 for all xRa>0 and Disc<036k2-36<0k2-1<0k+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<0k>-1 and k<1-1<k<1Disclaimer: (a) part should be 1< k < 1 instead of 1  k < 1.

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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, xRSo, f(xis increasing on R.



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