1 functions - level-2_solutions

1 I I T AKASH MULTIME-

Functions I I T MATHSSOLUTIONS FOR FUNCTIONS

CLASS WORK-LEVEL-II01. ( 2) | 1|y x x

3 12 1 1 23 2

xy x x

x

02. Solution: Case I: For x<1/2, |2x-1| =1-2x 1-2x = 3[x]+2{x} 1-2x=3(x-{x})+2{x} = 5x-1. Now 0 {x}<1 0 5x-1<1

15 x<

25 [x]=0 x={x}

x=5x-1 x=1/4, which is a solu-tion.Case II: For x1/2,|2x-1|=2x-1 2x-1=3x[x]+2{x} 2x-1=3(x-{x})+2{x}.{x}=x+1Now 0{x}<1 0x+1<1 -1x<0Which is not possible since x1/2.Hence x=1/4 is the only solution.

03. The given equation can be written as

2 24 5 2 3 4 5 2 3x x x x x x

2 4 5 2 3 0x x x (as |a+b| = |a|+ |b| provided ab 0 )

2 3 0x (as 2 4 4 0x x x R )

32

x .

Hence solutions set is 3 ,2

.

04. Let; x=I+f where I integer, f frac-tional part (i.e. 0 f<1)

2 2 25x x

2 2 25I f I f

22 1 25I I

2 2 2 1 25I I I 22 2 24 0I I

2 12 0I I 4 3 0I I

4 3I or I

Here, x I f

So, 4 3x f or x f

Since, 0 1f

4 4x and x

Hence, , 4] [4,x

05. Let ,x I f I integer, f fractional

part (i.e., 0 1f ) [ ] 2 4x x

[ ] 2 4I f I f

2 2 4I I f

2 4I f ,

which is only possible if, 1 02

f or

If 12

f

1 4I 1 4I

So, 3, 5I and 12

f

If 0,f

Then | | 4I

4I and 0f Thus number of solutions are

7 94, ,2 2

x

i.e., 4 solutions.

06. In [x]+2011

1

{ }2011r

x r

we know that {x+r}={x} as r integer

2011

1 2011[ ]

r

xx

......... upto2011times

2011 2011[ ]

x xx

20112011

[ ] xx

[ ] x xx

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Functions I I T MATHS

Thus, 2011

1 2011[ ]

r

x rxx

07. Let f(x) = x + 1 is increasing in [-1, 1], then

f(-1) = 0, f(1) = 2 ( ) 1f x x Let f(x) = x + 1 is decreasing in [-1, 1], then

f(-1) = 2, f(1) = 0 ( ) 1f x x

08. Option (a) : xy y x 2

y 2x f (y)y 1

Option (b): f(1) is not defined

Option (c): 3y f (x) 1

x 1

y decreases for all x except x = 1

Option (d): x 2f (x)x 1

rational function x

09.5( )3

xf xx

, 2 5( )( )

2xfof x

x

4( )( ) ( )fofofof x x f x x 8 12 16 2012( ) ( ) ( ) ...... ( )f x f x f x f x x

and 2009 ( ) ( )f x f x

10. AG OA or AG OA

if AG OA

Let 2aPA x x

Area of 21 ( )(2 )2

APQ x x x

If AG OAArea of APQ area of square ABCD -

Area of PCQ

2 1 2 .2. 22

a a x a x

2 22 2ax x a 11. 2 ( ) cos ( ) ( ) (1)f u f u f u

put u = 0, x in (1)2 (0)cos ( ) ( )f x f x f x

( ) ( ) 2 cos (2)f x f x a x

put 1 (1)2 2

u x in

( ) ( ) 0f x f x

put / 22

u and x is (1)

( ) ( ) 2 sin 2 sin (4)2

f x f x f x b x

from 2, 3, 2 ( ) 2 cos 2 sin ,( ) cos sinf x a x b x

f x a x b x

12. 2( 1) ( ) ( )f x y f x f y ;

put 20 (1) 2 4x y f

put x = 0, y = 1 2(2) (1 2) 9f ;

put 2 21 (3) (2 2) 4 16x y f 2( ) ( 1)f x x

13. If 1 2 1 2( ) ( )2 2

x x f x f xf

then the

function will be convex and the convexity will remain true for any number of variables. Here

ex is convex for x and tan x isconvex

for 02

x

14. 2( ) ( ) ( ) ......(1)f x f x f y f x y Put 0x

2( ) (0)f f y f y R

Range of f codomains of f Rf is onto ( ) 0b R f b

from (1),

2( ) ( ) ( ).....(2), ,

f b f b f y f b yy y R x y

2( ) ( ) ( ) (2)f x y f x f x f y from

( ) ( ) ( )f f x f f x f y replace

( )xby f x

2( )f f x y






22 2( ) ( )f x y f f x y x y

2 2( )f x x ( )f x x 15. The relavant wavy curve of the given function

isf(x) > 0 x (-5, -2) (-1, 3) (7, ) andf(x) < 0 x ( , -5) (-2, -1) (3, 7).

16. We redefine the functions f(x) and g(x) inthe intervals as shown below:

2 4 3, 3f 4, 3 4

4, 4

x x xx x x

x x

2

3, 3g 3 3 4

2 2 4

x xx x x

x x x

2

1, 3f 4 , 3 4.g 3

4 , 42 2

x xx x xx x

x xx x

17.2 0

( )0

x xf x

x x

2( ) 0

( )( ) 0

f x xfof x

f x x

; 4 0

0x xx x

18.2 0

( )0

x is xf x

x is x

2 ( ) ( ) 0( ) ( )

( ) ( ) 0f x if f x

g x fof xf x if f x

4 02 0

x xx x

19. 2 2( )( ) (4 1) 4fof x x ,3 3( )( ) (4 1) 4fofof x x

ingeneral ( ) 4 1 4n ng x x

1( ) 4 1 1ng x x

20. 1 2 0

f x1 0 2

xx x

1 2 01 0 1

1 1 2

xx x

x x

f(|x|) = 1 2 0

1 0 2

x

x x

1 01 2 0

1 0 1

xx x

x x

21. 3 6( ) log 1f x x x

3 6 3 6( ) log 1 log 1 ( )f x x x x x f x

21

6

3( ) 01

xf x Rx

f is odd and increasing

22.

2

0 for x 0

f( x) ( x) sin for 1 x 1( x 0)x

x| x| for x 1or x 1

for x = 0= -f(x), f(x) is an odd function.

f '( x) f '(x) , f '( x) f '(x) f(x) is an even function.Hence, option (a) and (d) are correct.

23. 1 1( ) sin sin { } cos cos 1 { } 1f x x x

( ) { } 1 { } 1f x x x

( ) 0 ( ) 0x z f x f x

{ } 1{ }x z x x ,

( ) 1 { } { } 1 ( )f x x x f x and f(x) is a periodic with period 1





Functions I I T MATHS24. x = y = 0 f(0) = 0 or f(0) = 1

If f(0) = 0 thus f(0 + y) + f(0 - y) =2 f(0) f(y) f(y) + f(- y) = 0 f is an odd functionIf y = 0 f(x) = 0, x R If f(0) = 1, then f(0 + y) + f(0 - y) = 2 f(0) f(y)f(y) + f(-y) = 2 f(y) f is an even function

25.2 1( ) 2 ( ).....(1)x f x f g x

x

2 21 14 ( ) 2 2 .....(2)x f x f x gx x

22

1 1( ) ( ) 23

f x g x x gxx

Since ( )g x is odd and 2x is even

( )f x is an odd function.

But given ( )f x is an even function.

( ) 0,f x R

26. f(-x)= 3 2 , for 0 2

2 for 2 4x x x

x x

3 2

2, for 4 2for 2 0

x xx x x

Therefore odd extension of f(x) is as follows :

h(x)= 3 2

2, for 4 2, for 2 0

x xx x x

27.

2

2

1f(x) 1 1 1 1

1

x xx x x

x x

2

2

1 1f(-x) 1 1 1 1 1 1

1 1

x x xx x x x

x x x

= f(x) f(x) is even.

28. 21( ) ( ) ( )2

f x a f x f x x

21 1 1 1( 2 ) ( ) ( ) ( )2 2 2 2

f x a f x f x f x

( )f x is periodic.

29. Option (a):

1, x 0Sgn(x) 0, x 0

1, x 0

xSgn(e ) 1 ; xSgn(e ) is constant function.Hence, it is periodic.Option(b): Period of sin x is 2and period of |sin x| is Period of sin x + |sinx| is LCM {2 , } , 2Option (c): Let f(x) = min {sin x, |x|} = sin x

Option (d): 1 1f (x) x x 2[ x]2 2

1 1 1 1x x x x 2( x { x})2 2 2 2

1 1x x 2{ x}2 2

Hence, f(x) is periodic30. Option (a): Periods of sin x and |sin x| are

2 and respectively..

period of f(x) = LCM of {2 , } 2 Option (b): Periods of sinx, cosx, cosec x are2 hence, period of g(x) = 2Option (c): Period of h(x) = LCM{2 ,2 } 2 Option (d):

1 2 1 2p(x) [x] x x x x x 11

3 3 3 3

1 211 {x} x x3 3

period of p(x) is 1.31. (10 ) (10 )f x f x

(10 (10 )) (10 (10 ))f x f x

(20 ) ( )f x f x

(20 ) (20 ) ( )f x f x f x





Functions I I T MATHS( ) (20 )f x f x

(20 ) ( )f x f x

( ) (10 ( 10))(10 10)(20 ) ( )

f x f xf xf x f x

f is on odd functions( ) (20 ) (20 (20 )) (40 )f x f x f x f x

f is a periodic functions.32. f(x)+f(x+a)+f(x+2a)+......+f(x+na)=k

replacing x=x+af(x+a)+f(x+2a)+.........+f(x+(n+1)a)=ksubstracting we gert,f(x)-f(x+(n+1)a)=0f(x)=f(x+(n+1)a)

T=(n+1)a33. f(x, y) = f(2x + 2y , 2y - 2x)

= f(2(2x + 2y) + 2(2y - 2x), 2(2y - 2x) -2(2x + 2y)f(x, y) = f(8y, -8x) = f(8(-8x), -8 (8y)) =f(-64x, -64y) = f(64(64x), 64(64(y))= f(212x, 212y)f(x, 0) = f(212x, 0)f(2y, 0) = f(212. 2y, 0) = f(212+y, 0) g(y) = g(y + 12)Hence g(x) is periodic and its period is12.

34.3x+2

f

+f(x) = f(x+1)

+1x+2

f

................(i)

Replacing x by 12

x+

f(x+2) + 1x+2

f

=3x+2

f

+f(x+1).........(ii)Adding (i) and (ii), we have f(x+2) +f(x)=2f(x+1) f(x+2) - f(x+1) = f(x+1) - f(x) =c(say) f(x+1) - f(x) = cf(x+2) -f(x+1) = cf(x+3) - f(x+2) = c............f(x+n) - f(x+n-1) = c addimg, f(x+n) -f(x) = nc

since |f(x)| 2 x R. So. |f(x+n) - f(x)| 4. x R, n N Now, if c 0, then we can select suffi-ciently large values on n for which |nc| >4, which is a contradiction. So, c=0.Hence f(x+1) - f(x) = 0 f(x+1) =f(x) f is periodic.

35. We have; 1 1 3f x f x f x

Putting x=x+1, we get;

( ) 2 3 1f x f x f x

again putting x=x+1, we get ;

( 1) 2 3 2f x f x f x

adding (i) and (iii), we get,

( 1) 2 1 3 3{ ( ) 2}f x f x f x f x f x

3 3 1f x

1 3 3 1 2 1f x f x f x f x

1 5 3f x f x f x

Adding (iv) and (v), we get;

1 5 0f x f x

1 5f x f x

6f x f x

6 12f x f x

Using (vi) and (vii); we get;

12f x f x

12f x f x

Thus f(x) is periodic with period 12.36. By verifying a, b, c are correct





Functions I I T MATHS37. For the function to be defined, we must have

21 [2 4x ] 1 21 2 4x 2 21 2 4x 2

23 4x 0

38. f (x ) x xee e e f (x) log (e e )

x 1 xe e 0 e e x 1

Let x y xey f (x) log (e e ) e e e

x y yee e e x log (e e )

y 1 ye e 0 e e ; y 1

fR ( ,1)

39. 1 2f (x) sec [1 cos x]

Domain of f(x) is R and 20 cos x 1 21 (1 cos x) 2 2[1 cos x] 1and 2

Then, range of f(x) is sec-1 and sec-1 21 1{sec 1,sec 2}

40. Domain of f = R, Domain g = R - [-1, 0);because is 1 0 1 [ ] 0x thus x Domain of f - g = R - [-1, 0); since

0 (1 [ ]) 0xe x y

0 1 [ ] 0y is x ,0 1 [ ] 0y is x

{0}y R .

41.2 2 2 21 2 1n n n n n

2 20 0n n x n x n

22 20 1 ,2 1n x n x n n

( ) , 0,1,2,3,....f x K K n Since

2 2( ) , [ ,2 1)f x n n x x n n

2 , 0,1,2,..n n x K K n

( ) 0f x if

2 2 1n x n n n x n

[0,2 1)x n 42. by verification.43. cos (sinx) is defined for all real x

logx {x}0Case I, 0<x<1{x}1x (0,1)Case II,x>1 and {x} >0{x}1 but 0< {x}<1not possible (D)

44. [ ] 0,x x x R

[ ]( ) 0,1 [ ]

x xf x x Rx x

45. By verfication

46. f (A) y ;1 y 02

1 f (A) 02

; 2

1 A 02 A 1

From first two relations 2

2

(A 1) 02(A 1)

;

A R And from last two relations we get A < 0

A ( ,0)

47. 1( ) ( ) 2 tan , 0 1f x g x x for x Domains = [0, 1]

10 1 0 tan4

x x ,

10 2 tan2

x

0,2

range 48. As we know,

2 2 2 2 2

4 4a b c d a b c d

.................(i)

(using Tcheby cheff’s Inequality) Where8a b c d e and





Functions I I T MATHS2 2 2 2 2 16a b c d e

Equation(i)reduces to, 2 28 16

4 4e e

2 264 16 4 16e e e

25 16 0e e

5 16 0e e (using number linerule)

1605

e

Thus range of 160,5

e

49. Let 1 2cot 2 ,x x where 0,

2cot 2 ,x x where 0,

2cot 1 1 2 ,x x where

0,

2cot 1 1 ,x where

0,

cot 1 , where 0,

4

Range od f(x) ,4

50. ( ) sin[ ] cos[ ]f x x x ; is defined x R 1 2( ) sec (1 sin )f x x ; is defind x R

becuase 21 sin 1x x R ;

( ) tan(log )f x x/ 2 , ( )x f x is not defind;

9( ) cos cos 28

f x x x

2 1 12cos cos 2 cos8 4

x x x ; is

defind x R 51. (gof )x g{f (x)}

1/ 3g{2x sin x}, (2x sin x) y (say)

2y 2x sin x Range of (gof) x is R.Option (A) is correct for

1 2 1 2x x f (x ) f (x ) Options (C) is correct gof is one - oneOptions B) is correct Range of gof is RHence, both f and g are ontoOption (D) is correct

52. Option (a) is sin(sin-1 x) = x

1

x

y

10

Option (b) x [ 1,1]

1

x

y

0 1 / 2 / 2 1

Options (c):

x

1 , x 0Sgn(x) Sgn(x) x

1ln e x , x 0x

0

y

x






Option (d):

3

3 3

x , x 0x Sgn(x) x x 0

0, x 0

0

y

x

53. 2f (x) [x] [x 1] 3 ; 2[x] 1 [x] 3

([x] 2)([x] 1) ; f (x) 0, then [x] = - 22 x 1

and [x] = 1 1 x 2 ; many one and into

54. Since ( )f x and ( )g x are mirror images of

each other about the line , ( )y a f x and

( )g x are at equal distances from the liney a

Let R ( )f a k ( )g a k ( ) 2 ,h a R

( )h x is a constant function ( )h x ismany one and in to.

55.2 3( )

1xy f xx

13 3( )2 2y xx f x

y x

1 2( ) ( ) 3 0f x f x x x

1 132

x

1( ) ( ) ( )f x x f x f x 56. 2 1cosx y

since the power of x is even it is symmetricabout y - axis and 1 1y

57. f(x) = 2 sin 26

x

1 1 2( ) sin2 6

xf x

since 1 1 [0, 4]2n x

also 1 1sin cos2

x x

1 12 2( ) cos3 2

xf x

58. 20092 2 4 21 1 1 ..... 1 1

1

x x x xRHS

x

2009

2009

2

21 1

( ) 11

xg x x

x

1(2) 2, 42

g g

59. Given f(x) = 2

, 1, 1 4

8 , 4

x xx x

x x

Let f(x) = y 1( )...............(1)x f y

2 2

, 1

, 1 4

, 464 64

y y

x y y

y y

1

2

, 1

( ) , 1 16 From(i)

, 1664

y y

f y y y

y y






Hence 1

2

, 1

( ) , 1 16

, 1664

x x

f x x xx x

60. By verification.61. Here x,y,z are integers and 5 is prime

number.

. . 5z x yy z xx y z

xyz

1 1 1. . 5z x yy z xx y z is only possible if

Case I:1 1 15, 1. 1

z x yy z xx y z

5, 1 1 2 2 1z zx y y y and z

, , 5, 2,1x y z

Case II: 1 1 11, 5, 1z x yy z xx y z

, , 1,5, 2x y z

Case III: 1 1 11, 1, 5z x yy z xx y z

, , 2,1,5x y z

62. Note: These type of questions where f(x)are either maximum or minimum shouldbe solved graphically for better repre-sentation.Let 1 x xf

and 22 x xf

Now from graph for 1 x xf and

22 ,x xf Here, neglecting the graph,

i.e., below point of intersection since, wewant to find the maximum of two func-tions 1 xf and 2 xf

2 , 0 1, 0 1

x x or xx

x xf

EXERCISE - II_SOLUTIONSPassage - I:

g(x) 1, g(x) 1f (x)

2g(x) 1, 1 g(x) 2

2 2

2 2

x 1, x 1, 1 x 2x 2 1, x 2 1, 2 x 3

f (g(x))2x 1, 1 x 2, 1 x 22(x 2) 1, 1 x 2 2, 2 x 3

2

2

x 1, 1 x 1f (g(x))

2x 1 1 x 2

1. Ans.C

Hence the domain of f(x) is 1, 2 2. Ans. C

For 1 x 1 , we have

2 2x 0,1 x 1 1,2

For 1 x 2 we have2 2x (1, 2] 2x 1 (3,5)

Hence the range is 1,2 3,53. Ans.B

For2 2f (g(x)) 2 x 1 2 and 2x 1 2

1x 1 or x2

x 1 only hence 2 roots2y 4y 1 0

y ( 2 3 y ( 2 3) 0

y , 2 3 2 3, Passage - II:

Given f(1) =8 and f(2) = 32 putting x=1and y= 1f(1+1) -a = f(1) + bf(2) = a + f(1) +b = a+b+8 or, 32 =a+b+8 a + b = 24..........(i) putting x = 2, y = -1f(1)+2a = f(2) + b or, 8 + 2a = 32 + b2a - b = 24..........(ii)from (i) & (ii)a = 16, b = 8 (i-C)





Functions I I T MATHSEXERCISE - III_SOLUTIONS

1. a)

1 1 1 121( ) cos sin cos sinf x x x x xx

11 2sin2 2

x

1 3 1 1,2 2 4 2 2 4

b a

1b a

b)

21 1

21( ) 2sin cos

4f x x x

21 1

21 2sin sin

4 2x x

221 1

21 2 sin sin

4x x

221 1

22 sin sin

2 8x x

2 21

22 sin

4 16x

2 2

22 5

2 4 16 4b

2

22 1

8 4a

1b a .

c) The functions 1 1cos sinx and x decrease as x varies from -1 to 1

33

31 9( 1)

2 8b f

3

31 1(1)

8 8a f

54

b a

d)

1 13

21 1 1 1

1 cos sin

cos sin 3sin cos

x x

x x x x

21 1

21 sin sin

4 22x x

221 1

21 33 sin sin

2 42x x

221 1

23 sin sin

2 122x x

21

23 sin

4 482x

2 2

23 9 7

16 48 82b

2 2

23 1

16 48 82a

34

b a

4.






5.1 ( )( ) ( ( )) log1 ( )

g xfog x f g xg x

32 3

2 3

1 3 3 1log log 3 ( )1 3 3 1

x x x x f xx x x x

and 1 (log ) (1) 11

eg of g e ge

6. The function f(x) = sin(2x + 1) decreases from1 to 0 on [ 3 / 4 1/ 2 / 2 1/ 2] os f isonto but one-one. Since

( / 2, 1/ 2) 0 ( 1/ 2)f f so f is notone-one on [ 3 / 4 1/ 2, 1/ 2] but is onto.On [ / 4 1/ 2,3 / 4 1/ 2] f decreasesfrom 1 to -1 and is continuous. Hence f isone-one onto as well.

7. (A) Inx = x

1e and x>0

It is possible only for one value of x i.e.at x=1(A-q)(B) x2log 3x 16x (x+3)3=16x+3= 4x=-7, 1but x -7, 1 because base should not benegative or equal to 1. B-p(C) 2tanx= / 2 -xtanx = / 4 -x/2 no. of roots of equation = 3 (C-q)

(D) period of sin 3 t=2 23 3

period of sin 4 t=2 24 4

period of sin 3 t + sin 4 t=L.C.M of 2, 2 2 2H.C.F of 3,4 1

(D-r)

EXERCISE - IV_INTEGER TYPEQUESTION SOLUTIONS

1. 1 1( ) ;f f x x f fx x

for

0x Least Value of 1xx

is 2.

2. The given equation

( ) 1 ( ) 1 ( ) 1f x f y f xy

If ( ) ( ) 1,g x f x then

( ) ( ) ( ) ( ) , ( ) 1n ng x g y g xy g x x f x x

(2) 7 2 1 7 3nf n 3( ) 1, ( 2) 9f x x f

3. ( 2) 5 ( 1) 6 ( ) 0f x f x f x

2 5 6 ( ) 0E E f x

Solving 2 5 6 0,E E the roots are 2and 3

( ) .3 .2x xf x A B

(0) 0f B A

(1) 1 1f A ( ) 3 2x xf x

2008 20082008 3 2f

Which is divisible by 3 2 5 .

4. 0 0 9899 101x x x to

1 101 19799 101x x x to

2 202 296...99 101x x x to

49 494999 101x x x

The desired number is

21 3 5 ... 99 1 50 1 2499

5. Since ,4x x x x x x becomes

4 4 3 5x x x x x x






0 0x x 513

x x

102 ,3

x x not a solution.

51 ,3

x x not a solution.

The solutions are 0 and 5/3.6. Putting x = f(y) =0 then f(0) = f(0) +0 +

f(0) - 1f(0)=1

Putting x = f(y)We get f(0) = f(x) + x2+f(x)-1 f(x)=1- x2/2

f(16) =1 -256 1 128 127

2

|f(16)|=127.7. f(1)+2f(2) + 3f(3)+.........+n(f(n)=n(n+1)

f(n)f(1) + 2(2) + 3(3) +..........(n+1) f(n+1) =(n+1) (n+2) f(n+1) n f(n)=n+1)f(n+1)i.e., 2f(2) = 3f(3)=.............= x(f(n) i.e.,

f(n) = 1

2n2126 f (1063) = 2.

8. f(x-2) + f(x) = 2f(x-1) ...................(i)

f(x) + f(x+2) = 2f(x+1) ...................(ii)(i) & (ii) f(x-2) + f(x+2) + 2f(x) =

2 (f(x-1) + f(x+1) f(x-2) + f(x+2) + 2f(x) = 2f(x) f(x-2) + f(x+2) = 0f(x) = f(x+8)

17

r=0f 2+8r 7 18 126.

9. f2(x). 31-x =x1+x

f

.................(i)

replace x by 1-x1+x

f21-x1+x

f(x) = 31-x

1+x

..........(i)

by using (i) and (ii)

f3(x) = x6 31+x

1-x

f(x) = x3 1+x1-x

f(-2) = 83 [f(-2)] = 2

|[f(-2)]| = 2.

10. Replacing ‘x’ by 1-x1+x we have

f(x). 1x

f

= f(x) + 1x

f

f(x) = xn+1

f(x) = x3 + 1 f(3) = 28

Now 10 10

3 2

n=1 n=1f(n)-1 n 55 3025

10

n=1

1 f(n)+1 5.605

SOLUTIONS FOR ADDITIONAL EXERCISE

1. &xy e y ax b

0, 0b a one real root Two curves on touch each other at x = log a

loga a b a one reapted real root

loga a b a two distinct real roots

loga a b a No real root

2. '( ). ( ) cos 2f x f x x

'( ). ( ) cos 2f x f x dx x dx ( ) (sin2 ) ( ) (sin2 ) 1f x x c f x x

( )f x is periodic with period . Clearly

2 is also a period.

3. 2x (a b y)x ab cy 0 ;20 (a b y) 4(ab cy) 0

2 2y 2(a b 2c) (a b) 0 y ;2 20 (a b 2c) (a b) 0

(2a 2c)(2b 2c) 0 ;c (a,b) if a b





Functions I I T MATHSand c (b,a) if b a ; i. e. a < c < b or a >c > b

4. We havef(x) [1 sinx] [1 sin2x] [1 sin3x] ..... [1 sin nx]

1 [sinx] 1 [sin2x] 1 [sin3x] ..... 1 [sinnx]

n [sin x] [sin2x] [sin3x] ...... [sin nx]

x (0, ) , sin x 0 is n + 1 (if n is odd)

5.

( ) ( ) ( )( ) ( ) ( ) 0, , , .( ) ( ) ( )

f g ff g f andf g f

6. 2( ) 1f x x x 21 3

2 4y x

;

11 3 ( )2 4

x y f y

1 1 3( ) ( )2 4

f x x x

since,the graphs of the original and inversefunction can intersect only on

2, 1y x y x x x 2( 1) 0 1x x

7. f(x) = x4 - 2x2 + 3 = 2 2( 1) 2x

function is even in x, so it is symmetric about y-axisf(x) = x4 - 2x2 + 3 = (x2 - 1)2 + 2function is even in x, so it is symmetric about y-axisf(x) attains minimum value at 1x f(x) decreases in

] , 1[ ]0, [

[ 1, 0] [1, [

similarly and increases

in and

Domain of ( ) ] , [f x Range of f(x) = [2, [

8. 22( )

1xf x

x

f(x) is odd in x, so graph is symmetric aboutoriginThe minimum and maximum of the function isrespectively -1 and 1.Domain of ( ) ] , [f x Range of f(x) = [-1, 1]

9.. 219 9 2 0x x 1 92

x ; 4 0x

4,x also g(x) and ( )h x I x = 4, 5, 8 and also for these values

( ) ( )g x h x10. Putting value of x = 4,5,8 in f(x)

Range = {36, 105}

11. 1

( ) (1 [sin ]n

k

f x k x






12. a)

b)

c)

d)

no maximum, no minimum

13. a) 1 2

'2 2

cos cos' ,cos1 sin

sin sin( )sin1 cos

x xf xxx

x xf xxx

'3 2

'4 2

cos cos( ) ,cos1 sin

sin sin( ) ,3 ,sin 21 cos

x xf xxx

x xf xxx

' ' ' '1 2 3 4(3) 1, (3) 1, 3 1, 3 1f f f f

b)

'1

' ' '2 3 4

34 , , 4 1,2

4 1, 4 1, 4 1

f

f f f

c)

' '1 2

' '3 4

35 , 2 , 5 1, 5 1,2

5 1, 5 1

f f

f f

d)

' '1 2

' '3 4

57 2 , , 7 1, 7 1,2

7 1, 7 1

f f

f f

14. a) ( ) ( ) (0)f x ax b a x f x b x f

ax a ax b b x ax b

2 1 0 1,

0 ( )

a x ab a

b f x x

b) 2 2( ) ( )f x x f x x

( ) , , ,f x x x x x

c) 2

0''( ) 2 '(2) (2) (0xf x dx f f f

2 3 1 1 4 by integration by parts

d) 1 ''( ) 1f x . Integrating from 0 to x.

15. 2( ) log ( 1af x x x is defined for all x,

since 2 1 | |x x and is odd. It increasesfor all value of x and has an inverse. Solving

2 2log 1 , 1yay x x a x x

so 2 1.ya x x Adding we get

1 ( ).2

y yx a a

Thus

1 ( ) 2 21( ) ( ) 0 0. ( ) 12

x x f xf x a a for x a x x

which increases on [0, ) .






16. 2

22

1'( )1

xf xx

which changes sign at

1x and so not one-one lim

2 01x

xx

and range is 10, .2

f

is not onto.

b) 2

'( ) 1 02xf x x f is one-

one, and onto.

c) 2 2'( ) 3 2 0

3f x x x

'( )f x changes sign f is not one-one,

but onto.

d) 2'( ) 3 1 0f x x f is one-oneand onto.

17. 2'( ) 0, 0bf x a if a bx

( )f x increases for 0x and for 0x ( )f x is onto '( ) 0f x if

0, 0, ( )a b f x is onto.

2'( ) 0. ( )bf x a f xx

has both

maximum and minimum at /x b a for

0, 0a b or 0, 0a b .

( )f x is neither one-one nor onto.SOLUTIONS FOR HOME WORK SHEET - I

1. If 0, 0x y y x

0, 0 | | | | 0 , .x y y y x x x y 2. f(x) = cos 9x + cos 10x

By verfication3. (C) |x-1| + |x-2| + |x-3| 6

Consider f(x) = |x-1| +|x-2| + |x-3|

=

6 3 , 14 , 1 2

, 2 33 6 3

x xx x

x xx x

graph of f(x) shows f(x)6 for x 0 orx4.

4. We are given that 3[x]+1=2([x]-3)+5 [x] = -2 y 3(-2) +1 = -5. Hence [x+y] = [x] + y = -2-5 = -7.

5. 2 2 23 3 3{ ( ) }4 4 4

a x g x

2 2{ ( )} 1x g x ; 2( ) 1g x x

6. 2( ) .f x Ax Bx C 2 ( ) .Ax Ax A B x C

( 1)2 ( ) .2

x xA A B x C

Hence (D) is the correct answer.

7.

1 11 1 xxf f (x)1x 1 x1x

Option (b) is correct.

and

1 11 1 xxf 1x 1 x1x

1 x x 11 x x 1

1f (x)





Functions I I T MATHS8. x+2y=2y=p

x-2y=q

solving we get x=p+q

2

y=p-q2

2 2p -qf(p,q)8

f(x,y)=2 2x -y8

9. ( ) ( )f cx d g ax d

acx ad b acx bc d ad b bc d

( ) , ( )f d ad b g b bc d

( ) ( )f d g b

10. y=1+αxy-1x=

f-1(x)=x-1α =f(x)= 1+αx

y-1x=

=1+αx 21=x- α+α xEquating the coeffient of x

2α =1&α=-1α= 1α= 1 .

11. f(x) = [x], [.] denotes the greatest integerfunction. and g(x) = |x|Now, (fog)x f (g(x)) f (| x |) [| x |]

and (gof )x g(f (x)) g([x]) | [x] | Optiona (a):

5 5 5(gof fog) (gof ) (fog)3 3 3

5| 2 | 2 1 13

Option (b): (f 2g)( 1) f ( 1) 2g( 1)

[ 1] 2 | 1| ; 1 2 1 Option (c):

5 5 5(gof fog) (gof ) (fog)3 3 3

5 53 3

5|1| 1 1 03

Option (d):(f 2g)(1) 2g(1) [1] 2[1] 1 2 3

12.2

2

1( )1

x

x

ef xe

( ) ( )f x f x f is odd

21

2 2

4.( ) 0,( 1)

x

x

ef x x Re

is increasing and odd

13. 2 21f(x) 2cos x 2cos x 2cosxcos x

2 3 3

1 2f(x) (1 cos2x) 1 cos 2x cos 2x cos

2 3 3 3

1 3 2cos2x cos 2x cos 2x2 2 3 3

1 3 32cos 2x cos cos 2x2 2 3 3 3 4

14. Option (a):

x x

x x

a 1 1 af ( x) ( x) ( x)a 1 1 a

x

x

a 1x f (x)a 1

Option (b): 2 2

1g '(x) ,(x a )

is even, g(x)

is oddOption (c):

2 23 3h( x) (1 x) (1 x) h(x) Option (d): If x is rational, then - x is alsorational and if x is irrational, then -x is alsoirrational

0, if x is rationalp(x)

1, if x is irrational






15. f(-x)

cos( ) cos1 112 2

x xx x

x xas x n I, so as - 1x

=

cos f(x) f(x) is an odd function.12

xx

16. By defination of odd fucntion

17,18,19It is easy to see that

1 1 1 11 1 2 2 3 3 4 5, , , .f f f f f f f f

For 11 4 1 4 1 4.f o F f F f o f f o f

Thus

1 4( ) (1/(1 )) 1/(1 ) ( ).F x f x x f x For, G o

13 6 6 3 6 3. ( )f f G f o f f f SoG x

6 5(1 ) (1 ) /(1 ) 1 1/ ( ).f x x x x x f x

For , 14 5 4 5 5 5 .f o H f H f o f f o f

Therefore,

5 5 5 5

4

( ) ( ( )) (( 1)) (( 1) / )( 1) / 1 1 ( ).

( 1) / 1

H x f f x f x f x xx x f x

x x x

For 1 13 4 2 3 4 2.J f o f o f f o f o f Thus

3 4 3 3( ) (1/ ) (1/(1 1/ )) ( / 1)J x f o f x f x f x x

41 / 1 1/1 ( ).x x x f x 20,21,22

Putting x = 1 = y = in f(x + y) = f(x) + f(y), we

obtain f(2) = 2f(1). Assuming that f(k) = kf(1).

f(k+1) = f(k) + f(1) +f(1) = (k + 1) f(1). Thus

by induction, f(n) = nf(1) for all .n N If n = -

m, m N then -f(m)=(-m)f(1) but f(n+(-n)=f(n)+f(-n) and f(0)=0 so f(-n) = -f(n). Thusf(n)=nf(1) for all .n I

3 33 ( ) (4 3 ) (4 3 ) 3 ( ) 0f x f x x f x x f x

( 1,1), ( 1) ( 2)x f x or f x may’t be

defined. The equation in (c) is also not mean-

ingful.

LHS is meaningful but R.H.S = f(1) is notdefined.

1 2 1 21 2

1 2 1 2

1 1 (1 )(1 )( ) ( ) log log log1 1 (1 )(1 )

x x x xf x f xx x x x

1 2

1 2 1 2 1 2

1 21 2 1 2

1 2

11 1log log1 1

1

x xx x x x x x

x xx x x xx x

1 2

1 21x xf

x x

23. a)

( ) ,1 1 1

x

x x xx x xef x f x

e e e

( ) , ( ) ( )f x f x f x f x

( )f x is neither even nor odd.

b) 2 1( ) 12 21 1

x

x xx x ef x

e e

/2 2

2 22

xx

x xx e e odd odd even

e e

c) 2 21 1 1 1( )

2

x x x xf x

x

2 1 1x even oddx odd

d) 24 2 2ln 1 ln 1x x x x

4 2 2

2 22

1 1ln ln11

x x x xx xx x






( )f x f x f is odd

24.

1 5( )

2 3x x

f xx x

The sign of

( )f x is that of

1 2 3 5x x x x

( ) 0 ( 1,1)f x in

1( ) 1 1

2 3x

f xx x

since

1 0

2 3x

x x

in 1,1

a) ( ) 0, ( ) 1,0 ( ) 1f x f x f x b) ( ) 0 ( ) 1 (1,2)f x f x in

c) ( ) 0 ( ) 1 (3,5)f x f x in

d)

1( ) 0, ( ) 1 12 3xf x f x

x x

since 1 0

2 3x

x x

in (5, )

25. ( ) 1 0xf x has roots 1,2,3, ......,9

( ) 1 1 2 .... 9xf x A x x x

Setting 0x , we find 19!

A Setting

10,x we find

10 (10) 1 1f since 19!

A

2 1(10)10 5

f

26. The equation

( ) 1 ( ) 1 1f x f y f xy

If ( ) ( ) 1,g x f x then

( ) ( ) ( )g x g y g xy

( ) , ( ) 1 ,n ng x x f x x

(2) 5 1 2 5 2nf n

2( ) 1 , (4) 17f x x f

27. 31/3 1/31 1, 1f x x f x f

3( ) 3 1 1g f f f 3(5) 3 4 4 63g

28. Setting 1y in the given relation,

2 ( ) ( ) (1) ( ) 2x xf x f x f f x

( ) 2xf x 9

2 9 10

1

( ) 2 2 ... 2 2 2 1022n

f n

29. Since

1 1 1 1x x x x

13

x

1 0 1 1 1 1x x x x x 0 1 1 1 1 1x x x x x

11 1 1 13

x x x x x .

1 1,1, 1,3 3

x does not satisfy the given

equation. No solution.30. The equation implies

1( ) 1 1 1f x fx

If

( ) ( ) 1g x f x

1( ) 1then g x gx

( ) , ( ) 1 , ( 2) 33n ng x x f x x f






1 2 33 5n n 5( ) 1 , (1) 0f x x f

SOLUTIONS FOR HOME WORK SHEET - II1. If f(x) = sinx + cosax is periodic

period of sin 2 ; period of

2cos| |

xa

f(x) is periodic | |a must be rational2. period of x [x]e is 1

and period of 2cosax isa

period of f(x) = LCM of 21,a

ie, period of f(x) is possible only when a isirrational of type Option (a): If a Then, period of f(x) = LCM of {1, 2} = 2Option (b): if a 2 Then, period of f(x) = LCM if {1, 1} = 1Option (c): if a 3

Then, period of f(x) = LCM of 21,3

2 2 2LCM of , 22 3 1

3. Period of 1sin tan 22 2

rr r

x x is

LCM of 2 3{2 ,2 , 2 ,......2 } 2n nis 4. f(x)+f(x+4)=f(x+2)+f(x+6)........(i)

Replacing x by x+2 in equation (i), we getf(x+2)+f(x+6)=f(x+4)+f(x+8)......(ii)Adding (i) & (ii) f(x)=f(x+8)f(5)=f(13)=10f(13)=f(21)=10.........f(805)=10

100

r=1f(5+8r)=f(13)+f(21)+......+f(805)

=10+10+..........+100 times.=10 100 1000 .

5. As the given function is symmetricalabout the lines, x = a and x = b, we havef(a-x) =f(a+x) .......................(1)and f(b-x)=f(b+cx) for all x R .Replacing a-x by y in (1), we getf(y)=f(2a-y)=f(b-(y-2a+b))=f(b+y-2a+b)=f(y+2(ba)) f(y)=f(y+2(b-a)) forall y R . Thus f(x) is periodic.

6. f g f g 1 2D D D D D

or 1 1 2D (D / D )7. For f(x) to be defined sin x 0 and sin

x 1 2n x (2n 1) , n I

For n 0, 1,1

0 x , 2 x , 2 x 3 8. Here, f(x) is defined by [-3, 2]

[ 3, 2].x (i.e. the only value of x we can substitutelie between [-3, 2]).For g(x) =f{|[x]|} to be defined, we musthave

3 | [ ] | 2x

0 | [ ] | 2 [ | | 0x as x for allx]

2 |[ ]| 2 [ | | ]x as x a a x a

2 3x [by definition of greatest integer func-tion].Hence, domain g(x) [ 2,3[ [ 2,3).or

9. ( ) 4 2 1xf x x

221 322 2

xy

;

221 322 2

xy

; 34

y






10. 3 2

122 1 2 0

12 3 2 ( 1)2

xx

x x x x x x

1 1( , 1) ,0 ,2 2

x

1 1,3 ,0 .2 2

P and

11. (a,c). The range of |sin x| = [0, 1]A = range of ln|sin x| = ( ,0]

The range of ln | | ( , )x

B = range of sin | | [ 1,1]x

( ,0] [ 1,1] ( ,1]A B

( ,0) [ 1,1] [ 1,0]A B

12. 2( ) 4 5f x x x ; (1) ( 5) 0f f

f is not one - one

1( ) , (0, )f x x xx

1 5(2)2 2

f f

f is not one-one

13. Option (a): f gD R {0},D R {0}

Also, f gR R ; then f (x) g(x)

Option (b): f gD R {0}, D R , f gD D

Option (c): f gD R {0}, D (0, ) ,

f gD D

Option (d): For domain off (x), (x 1)(x 2) 0

fD ( ,1) (2, ) and

gD (2, ) (3, ) (3, ) . f gD D

14. 2

sin([ ] )( ) 0,1

xf x x Rx x

f is constant function; f is

many one and into functions

15. y = x2+(k-1)x+9=2 21 19

2 2k kx

For entire graph to be above x-axis, we

should have 219 0

2k

2 2k-35<0 (k-7)(k+5)<0k

i.e., -5<k<7

16. f(x)=αx , x 1x+1

fof(x))=x 1

1

xx xx

x

2

1 1x xx

2 21 1 0.......(i)x x

1 0 and 21 0 [as true 1x

eq. (i) is an identity] 1. 17. 18. 19.

We have (x) = x + (x) ( )x x

1( ) 0 ( )( )

x x f xx x

is

not defined for any xIf x I then (x) = [x], so f(x) is not defined

If x I then (x) - [x] = 1, so f(x) = 1

Let x I then (x) = [x] so equation yields x= 0Let x I then ( ) [ ] 1x x and theequation becomes

2 2[ ] 2[ ] 1 [ ] 2 2{ }

11 { }2

x x x x x

or x

1 ,2

x n n I

solution set is





Functions I I T MATHS{0} [ 1/ 2, ]x n n I

20. 21. 22.by finding f 1 and its domain

23. a) 21'( ) 0

1f x

x

one - one

( ) 1f x as x not onto

b) 21( ) 1 , ( )f x f xx

increases both

for 0x and 0x (1) ( 1) 0f f

( )f x is onto but not one-one

c) 21'( ) 1f xx

changes sign

not one-one ( )f x is not onto since

( ) 2 ( ) 2f x and f x

d) '( ) 2 cos 0f x x one - one

( )f x as x onto.

24. a)

2

2 21 1 1 31 1

1 412

x xx x x x

x

The range is 7 31, ,4 4

b a

b) 1

3 2sin 2x

range is

1 4,15 5

b a

c) 2

2 2 21 1 11 12 2 1 7

2 4

x xx x x x

x

The range is 3 4,1 ,7 7

b a

d) Domain is ,2

Range is 1

Since cos 1, sin 0x x

0b a

25. ( ) cosln cos ln lnf xy xy x y

cosln cos ln lnx xf x yy y

( ) xf xy fy

cos ln ln cos ln lnx y x y

2cosln .cosln 2x y f x f y 26. f(2a-x) = f(x) f(2a+x) = -f(x) f si

odd f(x+4a) = f(x) f is periodicwith period 4a f(1+4r) = f(r)

Now r=0

f 1 8r

1 8 f 1 7 /81 f 1

8f 1 7

27. If f3(x) - 3f2(x)+3f(x)-1 = x6

(f(x)-1)3 = x6

f(x) -1 = x2

f(x) = x2 + 128. |f(x) + g(x)| |f(x)| + |g(x)|...............(i)

but given |f(x) + g(x)| |f(x) +|g(x)|........(ii)from (i) & (ii) |f(x) + g(x)| = |f(x)| +|g(x)|is possible only when f(x) g(x) 0 butgiven that f(x) g(x) <0

f(x) = 0 g(x) 0 2011

r=1f(r) 0






29. Perid of f(x)=sin 3x is k =2

3

3.k

30. 21 ..........( )2 ( ) 3 x ix

f x f

Replacing x with 1/x

2

1 1 ...........( )3 ( ) 2 iix x

f x f

Solving (i), (ii) 4

2

2 35

( )xx

f x

74

(2) kf

4 7.k




1 functions - level-2_solutions

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