2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)

8/13/2019 2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)

1/150

Preface Page No

1. Functions

Exercise 01 - 28

2 LimitsExercise 29 - 53

3 Continuity and derivabilityExercise 54 - 78

4 Method of differentiationExercise 79 - 100

5 Application of derivativesExercise 101 - 126

6 Solution of triangleExercise 127 - 149

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MATHEMATICSCLASS : XII

CONTENT


2/150

SOLUTIONS (XII) # 1

FUNCTIONS

EXERCISE # 1PART - I

Section (A) :

A3. (i) f(x) =1x

3x5x2

3

!

"!

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

3x5x3

"!

"!

Division by zero is undefined # x $ 1# Domain x %R {1, 1} & x %(', 1) ((1, 1) ((1, ')

(ii) f(x) =x

xsin 1!

For sin1x, x %[1, 1]and division by zero is undefined x $0 # Domain x %[1, 0) ((0, 1]

(iii) f(x) =|x|x

1

"

for function to be defined x + |x| > 0 for x > 0, x + |x| = 2x > 0for x )0, x + |x| = 0 # Domain is x %(0, ')

(iv) f(x) = ex + sin x

Domain x %R as there is no restriction for exponent of e.

(v) f(x) = )x1(log

1

10 !+ 2x"

1 x > 0 and x + 2 *0 and 1 x $1& x %(', 1) {0} and x *2 & x %[2, 0) ((0, 1)

(vi) f(x) = x21! + 3 sin1 +

,

-./

0 !2

1x3

1 2x *0 and 1 )2

1x3 !)1 & x )

2

1and

3

1)x )1

Taking intersection

11/21/3x# Domain x % 1

2

345

6!

2

1,

3

1

(vii) f(x) = xsin1

2!

2x

1

!

1 )x )1 and x > 2 & x %7

(viii) f(x) = logx ++

,

-../

0+,

-./

0! 2/1x1

log2

In case of composite function in log.We start with outer log.

x > 0, x $1 and

2

1x

1

!> 1 & x %(0, 1) ((1, ') and 0 < x

2

1< 1

& x %(0, ') {1} and2

1< x 0 & x %(0, 1]

Range 0 < sin1x )2

8

'< !n (sin1x) )!n +,

-./

082

Inequality doesn't change as !n is increasing function(iv) f(x) = 2 3x 5x2

Domain x %RMethod 1y = 5x23x + 2opening downward parabola

D/4a2

0

Range y % 12

3./

0 !'!

a4

D,

& y % 12

3./

0'! 20

49,

Method 25x2+ 3x + (y 2) = 0

D *0 & 9 20 (y 2) *0 & 20y 49 )0 & y )20

49

(v) f(x) = 3 |sin x| 4|cos x|f(x) is a periodic function with period 8. So analysis is limited in [0, 8]

fmax

= 3.1 4.0 = + 3 at x =2

8, |sin x| = 1, |cos x| = 0

fmin

= 3.0 4.1 = 1 at x = 0, |sin x| = 0, |cos x| = 1 # Range y %[4, 3]

(vi) f(x) = xtan1

xsin

2" + xcot1

xcos

2"f(x) = sin x |cos x| + cos x |sin x|periodic period = 28

f(x) =

9

9999

:

99999

;

= 0

e12

x

1.e

1x1

2

1x

1

?+,-.

/0

+,

-./

0

increasing function

Hence one - one

(vi) f(x) = )xcos(4

x3 28

8even function

Hence many - one

(vii) f(x) = sin1x cos1x = 2 sin1x 2

8monotonically increasing.

C5. (i) f(x) = sin (x2+1)& f(x) = f (x) = even function

(ii) f(x) = x + x2

& f(

x) = x2

x

$f (x) or

f(x) Neither even nor odd function(iii) f(x) = x x3 & f(x) = x + x3= f(x) odd function

(iv) f(x) = x ++,

-../

0

"1a

1ax

x

& f(x) = x ++,

-../

0

"1a

1ax

x

& f(x) = x ++,

-../

0

"1a

1ax

x

= f(x) even function

(v) f(x) = log (x + 1x2 " ) & f(x) = log (x + 1x2 " )

f(x) + f(x) = log

12

3

45

6 """" )1xx)(1xx( 22 = log [(x2 + 1) x2] = 0 hence odd function

(vi) f(x) = sin x + cos x & f(x) = sin x + cos x $f(x) or f(x)Neither even nor odd.

(vii) f(x) = (x21) |x| & f(x) = f(x) even function.

(viii) f(x) =9:

9;

0, Jx %R # f(x) = g(t) = t2+ t + 1, t > 0

g(t) =

2

2

1t +

,

-./

0" +

4

3

+,

-./

0"

2

1t >

2

1&

2

2

1t +

,

-./

0" >

4

1&

2

2

1t +

,

-./

0" +

4

3> 1 Range is (1, ')

Section (B) :

B2.* (A) f(x) =)x(secn 1e! = sec1x, x %(', 1] ((1, ')

g(x) = sec1x, x %(', 1] ( [1, ')non-identical functions

(B) f(x) = tan (tan1x) = x, x %R g(x) = cot (cot1x) = x, x %Ridentical functions

(C) f(x) = sgn (x) =9

:

9;

0, Jx %R and 7x2+ 2x + 10 > 0 Jx %R" a = 2 > 0 and " a = 7 and D = 4 280 < 0

D = 1 40 = 39 < 0# f(x) > 0 Jx %R

Also f(x) never tends to 'as 7x2+ 2x + 10 has no real roots, Range $Codomain so into function.

C3. f(x) = x3+ x2+ 3x + sin x, x %Rf>(x) = 3x2+ 2x + 3 + cos x

" 3x2+ 2x + 3 *12

32as a = 3 > 0 and D < 0

1 )cos x )1 so f>(x) > 0 Jx %R

'Gxlim f(x) = + ' !'Gx

lim f(x) = '

Hence f(x) is one-one and onto function (as f(x) is continuous function)

C6. f(g(x1)) = f(g(x

2)) & g(x

1) = g(x

2)

as f is one - one function & x1= x

2as g is one - one function

hence f(g(x1)) = f(g(x

2))

& x1= x

2 & f(g(x)) is one - one function

Section (D) :

D2. f(x) = sin L Mx]a[ . Period = ]a[28

= 8

[a] = 4 & a %[4, 5)

D3. f(x) = x + a [x + b] + sin 8x + cos 28x + sin (38x) + cos (48x) + ........ + sin (2n 1)8+ cos (2px)

f(x) = {x + b} + a

b + sin (8x) + cos (28x) + sin (38x) + cos (48x) + .... + sin (2n

1) + cos (2n8x)

Period of f(x) = L.C.M (1, 2,3

2,

4

2, .........,

1n2

2

!,

n2

2) = 2

# period of f(x) = 2since f(1 + x) $f(x) , hence fundamental period is 2

D7.* (A) f(x) = cos (cos1 x)= x, x %[1, 1] odd function

(B) f (x + 8) = cos (sin (x +8)) + cos (cos (x + 8))f (x + 8) = cos (sin x) + cos (cos x) = f(x)

f +,

-./

0 8"

2x = cos ++

,

-..

/

0+,

-./

0 8"

2xsin + cos ++

,

-../

0+,

-./

0 8"

2xcos

= cos (cos x) + cos (sin x) = f(x)

fundamental period =2

8

(C) f(x) = cos (3 sin x), x %[1, 1]3 sin1 ) 3 sin x ) 3 sin 1& cos (3 sin 1) ) cos (3 sin x) ) 1 # Range is [cos (3 sin1), 1]

Section (E) :

E1.1

y= xx

xx

ee

ee!

!

"

!

By compnendo and dividendo

y1

y1

!"

=2

e2 x& x = !n ++

,

-../

0

!"

y1

y1# f1(x) = !n +

,

-./

0!"

x1

x1


10/150

SOLUTIONS (XII) # 9

E7. f(1) = 1 = 2 1f(n + 1) = 2f(n) + 1 # f(2) = 2f(1) + 1 = 2. 1 + 1 = 3 = 221

f(3) = 7 = 231f(4) = 15 = 241

Similarly f(n) = 2n1

E8. Method 1 : (usual but lengthy)

x2

f(x) + f(1

x) = 2x

x4

.....(1)replace x by (1 x) in equation (1)(1 x)2f(1 x)+ f(x) = 2 (1x) (1 x)4 .....(2)eliminate f(1 x) by equation (1) and (2)we getf(x) = 1 x2

Method 2 :Since R.H.S. is polynomial of 4thdegree and also by options consider f(x) = ax2+ bx + cx2f(x) + f(1 x) = 2x x4

& x2(ax2 + bx + c) + a (1 x)2+ b (1 x) + c = 2x x4

by comparing coefficientsa = 1b = 0

c = 1# f(x) = x2+ 1


1. (i) f(x) = xx 2.223 !!!3 2x2 .2x*0 or (2x)23.2x+ 2 )0

or (2x1) (2x2) )0 & 2x%[1, 2]& x % [0, 1]

(ii) f(x) = 2

x11 !!1 2x1! *0 & 2x1! ) 1 & 0 )1 x

2 )1 & x % [1, 1](iii) f(x) = (x2+ x + 1)3/2

D : x %R

(iv) f(x) =2x

2x

"!

+x1

x1

"!

2x

2x

"!

* 0 andx1

x1

"!

*0

x %(', 2) ([2, ') and x %(1, 1]D : 7

(v) f(x) = xtanxtan 2!

tan x tan2x *0 or 0 )tan x )1 or x % $N%n

12

345

6 8"88

4n,n

(vi) f(x) =2

xsin2

1

sin2

x$0 or x $2n8

(vii) f(x) = ++,

-../

0 !4

xx5log

2

4/1

4

xx5 2!)1 and 5x x2> 0 or x %(0, 1] ([4, 5)


11/150

SOLUTIONS (XII) # 10

(viii) f(x) = log10

(1 log10

(x25x + 16))1 log

10(x25x + 16) > 0 or x25x + 6 < 0

or x %(2, 3)

2. (i) f(x) = 1 |x 2||x 2| %[0, ') # f(x) %(', 1]

(ii) f(x) =5x

1

!D : x %(5, ')R : f(x) %(0, ')

(iii) f(x) =x3cos2

1

!

range of cos 3x is [1, 1] cos 3x [1, 1] # f(x) % 12

345

61,

3

1

(iv) f(x) =4x8x

2x2 !!

"= y

x + 2 = yx28yx 4y or yx2x (8y + 1) (4y + 2) = 0

for x to be real D *0(8y + 1)2+ 4y (4y + 2) *064y2+ 16y + 1 + 16y2+ 8y *0

80y2+ 24y + 1 *0 or y % 12

3./

0!'!

4

1, ( +

,

-45

6'! ,

20

1

(v) f(x) =4x2x

4x2x2

2

""

"!= y

x22x + 4 = yx2+ 2xy + 4yx2(1 y) 2x(1 + y) + 4(1 y) = 0D *0

4(1 + y)216(1 y)2*0 or y % 1234563,

31

(vi) f(x) = 3 sin2

2

x16

!8

D : x % 12

345

6 88!

4,

4

22

x16

!8

% 12

345

6 84

,0 # f(x) % 12

345

6

2

3,0

(vii) f(x) = x32x2+ 5 = (x21)2+ 4R : [4, ')

(viii) f(x) = x312x , x %[3, 1] = x (x212)f>(x) = 3x212 = 0 or x = 2 R : [11, 16]

(ix) f(x) = sin2x + cos4x= sin2x + 1 + sin4x 2 sin2x= sin4x sin2x + 1

= +,

-./

0!

2

1xsin2 +

4

3R : 1

2

345

61,

4

3.

7. (i) f(x) = 6x

7x

)2(

)12( "

neither even non add

(ii) f(x) =xsinx

9xxsec 2 !"= f(x) even

(iii) f(x) = f(x) odd


12/150


(iv) f(x) =9:

9;

0

period G28; 8,3

28

Period of f(x) = L.C.M. +,

-./

0 888

3

2,,2 = 28

(v) f(x) = +,

-./

0"""""+

,

-./

0""""

n531n42 2

xtan...

2

xtan

2

xtan

2

xtan

2

xsin....

2

xsin

2

xsinxsin

Period of f(x) = L.C.M. of 8888888 n3n53 2.....2,2,2,......2,2,2

= 2n8

(vi) f(x) =x3cosxcos

x3sinxsin

""

period of f(x) = L.C.M. +,

-./

0 88

88

3

2,2,

3

2.2 = 28

For fundamental period

f(x + 8) =)3x3cos()xcos(

)3x3sin()x(sin

8""8"8""8"

=x3cosxcos

x3sinxsin# Fundamental period = 8

11. f(x) = 2x1

x1

"

!

& f>(x) = 0 at x = 1 2

for x % B A21,12 ""! f is bijective function hence f is invertible.


13/150


2x1

x1

"

!= y

or x2y + x + (y 1) = 0

or x =y2

)1y(y411 !!O!=

y2

1y4y41 2 "!O!

f1(x) =9:

9; 3x > 2

8or x % +

,

-./

0 8! 0,

6

Domain :+

,

-./

0 8! 0,

6 P(

1, 0) K +,

-

./

0 8! 0,

6


14/150


3. f(x) = sin1 ++,

-../

0 "2/3

3

x2

x1+ )xsin(sin + log

(3{x} + 1)(x2+ 1)

Domain : 3{x} + 1 $1 or 0 & x Q N

and 1 ) 2/3

3

x2

x1")1

2x

3/2

)1 + x3

)2x3/2

1 + x3+ 2x3/2*0(1 + x3/2)2*0 & x %R1 + x32x3/2)0 or (1 x3/2)2)0

or 1 x3/2= 0 or x = 1Hence domain x% 7

6. f(x) = (sin1x + cos1x)33 sin1x cos1x (sin1x + cos1x)

=8

383 sin1x +

,

-./

0!

8 ! xcos2

1

2

8=

8

38

4

3 28sin 1x + 3

2

8(sin1x)2

=

8

38+

2

38

1

1

2

3

4

4

5

6 8"

8! !!

16xsin

2)x(sin

2121

32

3 38=

32

38+

2

38

21

4xsin +

,

-.

/

0 8!!

maximum value of f(x) at x = 1 & fmaximum

=32

38+

2

38

16

9 38=

8

7 38

8. Here (2 log2(16 sin2x + 1) > 0

& 0 < 16 sin2x + 1 < 4 & 0 ) sin2x 0 & 2log

2*

2log (2 log

2(16 sin2x + 1)) > '

& 2 *y > '

Hence range is y %('R 2]

11. (A) f(x) = e1/2 !n x= x , D : x > 0

g(x) = x , D : x *0

(B) tan1(tan x) = x D : x $ (2n +1)2

8

cot 1(cot x) = x D : x $ n8(C) f(x) = cos2x + sin4x = cos2x + (1 cos2x)2 = 1 cos2x + cos4x = sin2x + cos4x

g(x) = sin2x + cos4x

(D) f(x) =x

|x|, D : x $0

g(x) = sgn (x), D : x %R

12. f(6{x}25{x} + 1) & f((3{x} 1) (2 {x} 1))

(3{x} 1) (2{x} 1) ) 0 or {x}% 12

345

62

1,

3

1# x %$

N%n

12

345

6""

2

1n,

3

1n

13. f(x) = cot1x R+ G +,

-./

0 82

,0

g(x) = 2x x2 R GRf(g(x)) = cot1(2x x2), where x %(0, 1]

hence f(g(x)) % +,

-45

6 882

,4


15/150


20. f +,

-./

0"

3

1x = 1

2

345

6"

3

1x + 1

2

345

6"

3

2x + [x + 1] 3 +

,

-./

0"

3

1x + 15

= 12

345

6"

3

1x + 1

2

345

6"

3

2x + [x] 3x + 15 = f(x) # fundamental period is 1/3

21. f(x) = |x 1| f : R+GRg(x) = ex, g : [1, ') GRfog(x) = f[g(x)] = |ex1|D : [1, ')R : [0, ')

25. f(x) =}x{

])x[sin(8= 0 , x QN

(A) By graph fundamental period is one(B) f(x) = 0 = f(x) # even function(C) Range y %{0}

(D) y = sgn++

,

-

.

.

/

0

}x{

}x{sgn 1, x QN

y = sgn (1) 1 & y = 1 1y = 0, x QN Identical to f(x)

28. f(x) = sin x + tan x + sgn (x26x + 10)f(x) = sinx + tan x + sgn ((x 3)2+1)f(x) = sin x + tan x + 1period = L.C.M. (28, 8) = 28fundamental period = 28

29. f : N GI

f(n) =9:

9;

0 &3 < x < 3 ...(2)Hence from (1) & (2)we get 2 )x < 3# Domain = [2, 3).

8. 3xx7 P !

! is defined if

7 x *0, x 3 *0 and 7 x *x 3&3 )x )5 and x %N # x = 3, 4, 5

# f(3) = 3337 P !

! = 04P = 1

f(4) = 3447 P !

! = 13P = 3

f(5) =35

57 P!

! =2

2P = 2

Hence range = {1, 2, 3}.


21/150


9. f(x) = tan1 +,

-./

0

! 2x1

x2= 2 tan1x for x %(1, 1)

If x %(1, 1) &tan1x % +,

-./

0 88!

4,

4

&2 tan1

x %+

,

-./

0 88!

2,

2

Clearly, range of f(x) = +,

-./

0 88!

2,

2

for f to be onto, co-domain = range # Codomain of function = B = +,

-./

0 88!

2,

2.

10. f(2a x) = f(a (x a)) = f(1) f(x a) f(a a) f(a + x a)= f(1) f(x a) f(0) f(x)= f(x) [" x = 0, y = 0, f(0) = f2(0) f2(1) &f2(1) = 0 &f(1) = 0]&f(2a x) = f(x).

11. f(x) is defined if 1 ) 2x 1 )1 and cos x > 0

or 0 )2

x)2 and

2

8< x x &x < 0 #x %(', 0) Ans.

14. f(x) = (x 1)2+ 1, x *1f : [1, ') G[1, ') is a bijective function & y = (x 1)2+ 1 & (x 1)2= y 1

& x = 1 1y & f 1(y) = 1 1y

& f 1(x) = 1 + 1x {#x *1}

so statement-2 is correctNow f(x) = f 1(x) & f(x) = x & (x 1)2+ 1 = x& x23x + 2 = 0 & x = 1, 2so statement-1 is correct

ADVANCE LEVEL PROBLEMPART-I

1. f(x) = +,-

./0

"!

!" x3

1x2loglog

2

2

4x

For domain :

2

4xlog " +

,

-./

0"!x3

1x2log2 )0

Case-I

0 1 or x > 2 ..........A

2

4xlog " +

,

-./

0"

!x3

1x2log2 ) 0 & 0 < log2 x3

1x2

"!

) 1 & 1 < x31x2

"!

) 2

& x %(4, ' ) ..........(ii) # (i) ((ii) Domain x %(4, 3) ((4, ')

2. f(x) = (x12x9+ x4x + 1)1/2

Dr : x12x9 + x4x + 1 > 0For x )0 it is obvious that for f(x) to be defined Dr > 0.For x *1, (x12x9) + (x4x) + 1 is positiveSince x12*x9, x4*x.For 0 < x < 1, Dr = x12+ (x4x9) + (1 x) > 0

Since x4> x9, x < 1.Hence Dr > 0 for all x %RDomain is x %R

3. f(x) = |x|x

|x|x

ee

ee

"

! !

if x * 0, f(x) = x

xx

e2

ee !!=

2

1 2x )e(2

1=

2

1 +

+,

-../

0!

2x )e(

11 ; f(x) % +

,

-45

62

1,0 ........(i)

f(x) % +,

-45

62

1,0

if x < 0, f(x) = xx

xx

ee

ee!"

!= 0 .........(ii) # range of f(x) is (i) ((ii) = +

,

-45

62

1,0

4. f(a) = aa2 2 ! for domain of f(x)

2a2a *0 & a(2a 1) *0 # a %(', 0] ( +,

-45

6',

2

1

Let g(x) Kx2+ (a + 1)x + (a 1) = 0(i) D *0(a + 1)24(a 1) *0 & a %R ...(i)

(ii) 2 < A2

B< 1 & 2 <

2

)1a( " < 1 & a % L M3,3! ....(ii)

(iii) g(2) > 0 & 4 2(a + 1) + (a 1) > 0 & a < 1(iv) g(1) > 0 & 1 + a + 1 + a 1 > 0 & a > 1/2Now (i) P (ii) P (iii) P (iv) we get

5. f(x) = sin1 12

345

6"

2

1x2 + cos1 1

2

345

6!

2

1x2

Domain : 1 ) 12

345

6!

2

1x2 ) 1 & x % +

+,

-../

0!

2

5,

2

5

and 1 ) 123

45

6 "2

1x2 ) 1 & x % +

+,

-../

0!2

3,

2

3


23/150


& domain is x % ++,

-../

0!

2

3,

2

3or x2% +

,

-45

62

3,0

if (i) x2% +,

-45

62

1,0 , then f(x) = 8

if (ii) x2 % +,

-45

61,2

1, then f(x) = 8

if (iii) x2% +,

-45

62

3,1 , then f(x) = 8 & range = {8}

6. +,

-./

02

1f = 1

2

345

62

1+ 1

2

345

6"

2000

1

2

1+ 1

2

345

6"

2000

2

2

1+....... 1

2

345

6"

2000

1999

2

1= 1000

7. f(x) = ax2+ bx + cf(0) = c &c %Nf(1) = a + b + c &(a + b + c) %N & (a + b) %N

8. f(x) = 9:

9; 12

f(x) =1112]x[1]x[

1

!!"! & f(x) = )12]x([21

!

Now for f(x) to be defined [x] $12 & x Q [12, 13) but x > 12x Q (12, 13)Case-## 1 )x )12

f(x) =11]x[121]x[

1

!!""! =

9:

9; 0, x + 0.5 $ 1 & x %(0.5, ' ) & x $ 0.5 .....(A)

&3x4x4

3x2x2

2

!!

!"> 0 or

)1x2)(3x2(

)1x)(3x(

"!!"

> 0

or x %( ' , 3) ( +,

-./

0! 1,

2

1 ( +

,

-./

0',

2

3.....(B)

(A) P(B) # Domain of f(x) : x % +

,

-.

/

0! 1,

2

1( +

,

-.

/

0',

2

3

S

TU

:

; 0 or x % +,

-./

0! 6,

6

1

& Domain +,-

456 8(1

23.

/0 8! 6,

3

5

3,

6

1


25/150


(v) f(x) =

123

456 !

21x

5!

21 xsin3

!+

1x

!)1x7(

"

"

12

345

6 !2

1x $ 0 & x Q [1, 3) & x %[1, 1]

& x + 1 > 0 & x %(1,

' ) & 7x + 1 %w

# DomainSTU

:;(x) =1x2

1

!

x3

1

!= 0 or f>(x) = 0 at x =

5

7

f> ++

,

-../

0 !

5

7> 0 & f> +

+,

-../

0 "

5

7< 0 & maxima at x =

5

7Range : B A10,2

(ii) For domain (i) [x] > 0 and [x] $1 so [x] *2, so x %[2, ')

for range if x %[2, '), thenx

|x|= 1 so f(x) = cos10 =

2

8

Range of f(x) =STU

:; 0 & x %(2n8,(2n+1)8)here [x 1] > 0 & [x 1] $ 1 & x %[3, ' ]

Domain x %[3, 8) ( L M12

345

68"8

'

C)1n2(,n2

1n$ .

For range sin x %(0, 1] and [x 1] %[2, ') so range %(', 0]

(vi) f(x) = tan1 |x|2]x[]x[ !"!" + 2x

1

Domain : (i) [x] + [x] *0 & x %N (ii) 2 | x | *0 & |x| )2 & x %[2, 2] (iii) x $0

For domain (i) P (ii) P (iii) Domain : {

2,

1, 1, 2}

Range :STU

:; 0If x %N, x > 0

& x : {1, 2, 3, 4, .......... } .......(A)If x Q N

[x] 1 > 0

& [x] > 1 & x %[2, ') .......(B)A( [

# x %{1} ( [2, ')

7. Case # y = x x < 1x = y y < 1f1(x) = x x < 1

Case ## y = x2

1 )x )4x2= y 1 )y )16

x = y 1 )y )16

f1(x) = x 1 )x )16

Case ### y = x8 x > 4

x =64

y2y > 16

f1(x) =64

x2x > 16

8. Let x = y = 1f(x) + f(y) + f(xy) = 2 + f(x) . f(y)3f (1) = 2 + (f(1))2 & f(1) = 1, 2. But given thatf(1) $1 so f(1) = 2

Now put y =x

1

f(x) + f +,

-./

0x

1+ f(1) = 2 + f(x) . f +

,

-./

0x

1& f(x) + f +

,

-./

0x

1= f(x) . f +

,

-./

0x

1

so f(x) = xn + 1Now f(4) = 17 & (4)n+ 1 = 17 &n = 2

f(x) = +(x)

2

+ 1. # f(5) = 52

+ 1 = 26

9. Put x = 1, y = 1(f(1))2= f(1) + 6 & f(1) = 3, 2f(1) = 3 [Since f(x) > 0]Put y = 1 in given relationf(x) f(1) = f(x) + 2(x + 2)2f(x) = 2x + 4f(x) = x + 2

10. | f(2k) f(2 i)| = | f(2k) f(2k 1) + f(2k1) f(2k2)......... f(2i+1) f(2 i)|

)| f(2k) f(2k1)| + | f(2k1) f(2k2)| + ...........| f(2 i+1) f(2 i)|Consider | f(2k1 + 2k1)| f(2 k1)| )1So | f(2k) f(2i)| )1 + 1 + ........(k i) term


29/150


I |)2(f)2(f| ik ) IC

k

1i

)ik(

)2

)1k(k Hence proved.

11. (x + 1)2f +,-.

/0

" 1x1 = f(x + 1)

f +,

-./

0" 1x1

= 2)1x(

)x(f1

"

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

Also f +,

-./

0" 1x

1= f +

,

-./

0" 1x

x1 = 1+ f +

,

-./

0"1x

x

= 1 f +,

-./

0" 1xx

= 1 f +,-.

/0 "

x1x

2

1xx +

,-.

/0

"

= 2

2

)1x(

x1

"f +

,

-./

0"

x

11 = 2

2

)1x(

x1

" +

,

-./

0"

2x

)x(f1 .........(ii)

from equation (i) and (ii), we can say that

(i) = (ii) & 2)1x(

)x(f1

"

" = 2

2

)1x(

)x(fx1

"

"

& 1+ f(x) = 1 + 2x f(x) # f(x) = x

12. (i) Put x = y = 1 in given relation, we get f(f(1)) = f(1)

(ii) Now put x = 1, y = f(1) in given relation, we get f(f(1)) =)1(f

)1(f= 1

from (i) and (ii)

# f(f(1)) = 1 # f(1) = 1

Put x = 1, f(f(y)) = y

)1(f & f(f(y)) = y

1now substitute y = f(x)

f(f(f(x))) =)x(f

1& +

,

-./

0x

1f =

)x(f

1

13. "

f(x, y) = f(2x + 2y, 2y

2x) .......(i)= f(2(2x + 2y) + 2 (2y 2x), 2(2y 2x) 2(2x + 2y)) from (i)

& f(x, y) = f(8y, 8x) .......(ii)= f(8(8x), 8(8y)) (using (ii))

& f(x, y) = f(64x, 64y) .......(iii)= f((64) (64 x), (64) (64y)) (using (iii))

& f(x, y) = f(212x, 212y)& f(x, 0) = f(212x, 0)Replace x by 2y

# f(2y, 0) = f(212. 2y, 0)& f(2y, 0) = f(212+y, 0)& f(2x, 0) = f(212+x, 0)& g(x) = g(12 + x) ["given g(x) = f(2x, 0)]Hence g(x) is periodic function with period 12.


30/150


LIMITS

EXERCISE # 1

PART - ISection (A) :

A-6. (i) "G0x im! x]x[ = ++

,-..

/0" valueve 0 GNot an indeterminate form

(ii)!'Gx

im! 1x2 " x C'+ '= ' GNot an indeterminate form

(iii)

2x

im8

G

! (tan x)tan2x= (') form GYes

(iv) "G1xim! Y ZL M nx

1

x !

)h1(n

1

0h}h1{im "

G" !!

)h1(n

1

0h}h{im "

G

!! = (0'form) = 0 GNot an indeterminate form

SECTION (B) :

B-2. (i)0x

imG!

x5cos1

x4cos1

!!

+,

-./

0form

0

0

=0x

imG!

2

x5sin2

x2sin2

2

2

=0x

imG! 2

2

22

2

x52

x5

sin25

x2

x2sin2

++++

,

-

.

.

..

/

0

+,-.

/0

+,

-./

0

=25

16

(ii)6

x

im8

G

!

6x

xcosxsin3

8!

!+

,

-./

0form

0

0

using L' Hospital rule

=6

x

im8

G

!

1

xsinxcos3 "=

2

1

2

3" =2

(iii)0x

imG!

xsinx3

x2x3tan2!

!

0ximG!

x

xsin.xsin3

2x3

x3tan.3

=1.03

2)1(3=

3

23=

3

1

(iv)0x

imG!

x

asina)xasin()xa( 22 !"" +,

-./

0form

0

0

using L' Hospital rule

=0x

imG!

1)xacos()xa()xasin()xa(2 2 """""

= 2a sina + a2cos a


31/150


B-4.ax

imG!

ax

)2a()2x( 25

2

5

!"!"

R.H.L. =ax

)2a()2x(im

2

5

2

5

ax

"""G

!

=aha

)2a()h2a(im

2

5

2

5

0h """"

G! =

h

12a

h1)2a(

im

2

5

2

5

0h

111

2

3

444

5

6

+,

-./

0"

""

G!

= h

1.....)2a(

h.

!2

2

3

2

5

2a

h.

2

51)2a(

im

2

22

5

0h

1111

2

3

4444

5

6

""

+,

-./

0+,

-./

0

""

""

G! =

2

52

3

)2a( "

L.H.L. =h

)2a()h2a(im

2

5

2

5

0h

""G!

=

L M

h

1....2a

h

!2

2

3

2

5

2a

h

2

512a

im

2

2

5

0h

1111

2

3

4444

5

6

"+,

-./

0"

+,

-./

0+,

-./

0

"+,

-./

0"

"

G! =

2

3

)2a(2

5"

L.H. L. = R.H.L.

So)ax(

)2a()2x(im

2

5

2

5

ax

""G

! =2

5 23)2a( "

SECTION (C) :

C-1. (i)'Gx

im! +,

-./

0"""

222 x

x....

x

2

x

1 =

'Gxim!

L M2x

x.....21 """

='Gx

im!2x2

)1x(x "=

'Gxim!

2

1 1

2

345

6"

x

11 =

2

1

(ii) 'Gx im! L M L MZxcos1xcos !"

= 'Gxim! 2sin +

+,

-../

0 ""2

1xx. sin +

+,

-../

0 "2

1xx

= 'Gxim! 2 sin +

+,

-../

0 ""2

1xx. sin +

+,

-../

0

"" )1xx(.2

1xx

=2 'Gxim! sin +

+,

-../

0 ""2

1xx. 'Gx

im! sin112

3

445

6

"" )1xx(.2

1

= 2x (oscillating 1 to 1) 0= 0


32/150


(iii)'Gx

im! Mxx8x2 "!= ('+ ') = '

(iv)5 374 56

3 423

n 1n3n2n6n

1n1n2nim

""!""

"""!'G

! =

745

7

62

3

43

4

32

3

n

n

1

n

31n

n

2

n

61n

n

11n

n

1

n

21n

im

""""

"""

'G!

=

7410

1

6

46

1

3

n

n

1

n

31n

n

2

n

61

n

11n

n

1

n

21

im

""""

"""

'G! =

01

01"= 1

(v) 'Gxim! L M +

+

,

-

.

.

/

0" 3

2

3

2

)1x(1x

= 'Gx im!

L M L M

L M112

3

445

6""""

11

2

3

44

5

6""""

11

2

3

44

5

6"

3

4

3

2

3

2

3

4

3

4

3

2

3

2

3

4

3

23

2

)1x()1x(1x)1x(

)1x()1x(1x)1x(1x)1x(

= 'Gxim!

L M11

2

3

44

5

6""""

"

3

4

3

2

3

2

3

4

22

)1x()1x(1x)1x(

)1x()1x(= 'Gx

im!

111

2

3

444

5

6

+,

-./

0"

"+,

-./

0"

""3

4

3

2

3

4

1x

1x

1x

1x1)1x(

x4

= 'Gxim!

11

1

2

3

44

4

5

6

+,

-./

0

""+

,

-./

0

""""

3

4

3

2

3

1

1x

1x

1x

1x1)1x)(1x(

x4=

]111[)()01(

4

""\'\" = 0

SECTION (D) :

D-1. (i)2x

imG!

x)2x7(

)2x15()2x(

4

1

5

1

2

1

!"

"!"

Let x = 2 + h

= 0himG

!

)h2()h716(

)h1532()h4(

4

1

5

1

2

1

""

""

= 0himG

!

)h2(16

h712

32

h1512

4

h12

41

5

1

2

1

"12

345

6"

12

345

6"12

345

6"

=0h

imG!

)h2(......64

h712

....32

h15

2

5

4

5

1

32

h312....

16

h

2

2

1

2

1

8

h12

22

"12

345

6""

1111

2

3

4444

5

6

"+,

-./

0+,

-./

0+,

-./

0

""

1111

2

3

4444

5

6

"++,

-../

0+,

-./

0+,

-./

0

""

=0h

imG!

.....132

7h

...256

9

64

1h

16

3

4

1h 2

"+,

-./

0

"+,

-./

0"+

,

-./

0

=1

32

716

3

4

1

= 252


33/150


(ii)0x

imG!

3

2x

x

2

xtanxsin1e !!!

3

2353432

0h x

...3

xx

2

1...

!5

x

!3

xx...

!4

x

!3

x

!2

xx

im++

,

-../

0""+

+,

-../

0""+

+,

-../

0""""

G

!

= 3

432

0h x

....3

1

!4

1x

6

1

6

1x

2

1

2

1x

im

"++,

-../

0"+

,

-./

0"+

,

-./

0

G!

=6

1+

6

1=

3

1

SECTION (E) :

E-2. (i)4

x

im8G

!(tan x)

tan2x

=

x2tan)1x(tan

4x

im

e

8G

!

=

xtan1

xtan2

4x

im

e

"8G

!

= e 1 =e

1

(ii)'Gx

im!

x

x31

x21+,

-./

0""

='Gx

im!

x

3

x

1

2x

1

++++

,

-

.

.

.

.

/

0

"

"

=

'

+,

-./

03

2= 0

(iii)1x

imG! L M 2

xsec

nx1

8

"! = 2x

sec).nx(1x

im

e

8

G!!

= 2x

cos

nx

1xim

e

8G

!!

(0

0form) = 2

xsin

2

x

1

1xlim

e

88G=

+,

-./

082

e

(iv) "G0xim! L M

2xx ((0)0form)

Let y = "G0xim! L M

2xx

& y = "G0xim

e! x2(!nx) & y =

2x

1

nx

0x

im

e

!!

"G

& y =3x

2

x

1

0x

im

e

"G!

= e0= 1

(v)

2x

im8

G

! (tan x)cosx ('0 form) & y =

)xtann.(xcos

2x

im

e

!!

8G

& y =xsec

xtann

2x

im

e

!!

8G

& y =

xtanxsec

x2sec

xtan

1

2x

im

e

\8

G

!

& y =x2sin

xcos

2x

im

e

8G

!

& y = e0= 1


34/150


(vi) !G1xim! ([x])1x

0himG! [1 h]1 (1 h)

0himG!

(0)h= 0

E-3.'Gn

im! 3n

]x)1n(.n[.....]x3.2[]x2.1[ """"

(1.2)x 1 < [1.2x] )(1. 2)x(2.3)x 1 < [2.3x] )(2.3)x % % %

n(n + 1) x 1 < [n (n + 1)x] )n(n + 1)xso (1.2)x + (2.3)x + ... n (n + 1)x n

< [1.2 x] + [2. 3 x] + .... [n(n +1)x]

)(1. 2) x + (2. 3) x + .... n(n +1)xx . (]n2 + ]n) n ) [1. 2x] +[2. 3x] +......[n(n+1)x] ) x (]n2 + ]n)

& 'Gnim!

3

n

n2

)1n(n

6

)1n2()1n(n.x 12

345

6 ""

""

< 3n n

]x)1n(n[.....]x3.2[]x2.1[im

""""'G

!

) 'Gnim!

3n

2

)1n(n

6

)1n2()1n(nx 12

345

6 ""

""

'Gnim! x 2

2

n

1

2

n

1

n

1

6

n

12

n

11.1

1111

2

3

4444

5

6+,

-./

0"

"+,

-./

0"+

,

-./

0"

3n n

]x)1n(n[.....]x3.2[]x2.1[im

""""=

'G!

) 'Gnim! x

1111

2

3

4444

5

6+,

-./

0"

"+,

-./

0"+

,

-./

0"

2n

1

n

1

6n

12

n

11.1

2

3

x

n

]x)1n(n[.....]x3.2[]x2.1[im

3

x3n

)""""

='G

!

so'Gn

im! 3n

]x)1n(n[.....]x3.2[]x2.1[ """"=

3

x

E-5. f(x) = 'Gnim!

1x

1xn2

n2

"

!

case (i) when x = 1 f(x) = 'Gnim!

11

11

"= 0

case (ii) when x > 1 f(x) = 'Gnim!

n2

n2

x

11

x

11

+,

-./

0"

+,

-./

0

=01

01

"= 1

case(iii) when x < 1 f(x) = 'Gnim!

1x

1xn2

n2

"

!=

10

10

" = 1

range of f(x) is {1, 0, 1}


35/150


PART - IISection (A) :

A-3*. f(x) =]x[x

20x9x2

!"!

!G5x

im!

]x[x

20x9x2

!

"!=

1

204525 "= 0

"G5xim!

]x[x

20x9x2

!"!

=]h5[h5

20)h5(9)h5(im

2

0h """""

G!

=h

20h945hh1025im

2

0h

"""G!

=h

hhim

2

0h

"G! =

h

)1h(him

0h

"G! = 1

" 5xim

G! f(x) $ "G5x

im! f(x) so5x

imG! f(x) does not exist

SECTION (B) :

B-2.0x

imG!

++,

-../

0"++

,

-../

0

!

3

x1n

p

xsin

)14(2

3x

!

=0x

imG!

++,

-../

0"

1111

2

3

4444

5

6++,

-../

0

++,

-../

0

++,

-../

0

3

x1n.

p

x

p

xsin

p

x

x

14x

2

3x

3

!

= 3p .0x

imG!

1111

2

3

4444

5

6++

,

-../

0

++

,

-../

0

p

x

p

xsin

x

14 3x

.

++

,

-../

0"

++

,

-../

0

3

x1n

3

x

2

2

!

= 3 p (!n 4)3

B-6.2

x

im8

G

! tan2 x +,-.

/0 ""!"" 2xsin6xsin4xsin3xsin2 22

=2

x

im8

G! tan2 x +++

,

-

.../

0

"""""""!""

2xsin6xsin4xsin3xsin2)2xsin6x(sin4xsin3xsin2

22

22

=2

x

im8

G

! tan2 x .261432

)2xsin3x(sin2

"""""

"

=2

x

im8

G

!112

3

445

6 "

xcos

2xsin3xsin

6

12

2

+,

-./

0form

0

0(use L'Hospital rule)

=6

1

2x

im8

G

! )xsin(xcos2

xcos3xcosxsin2

=6

1

2x

im8

G

!xsin2

3xsin2= +

,

-./

0+

,

-./

02

1

6

1=

12

1


36/150


B-7*. Let f(x) =|x|x

x2cos2cos2 !

!

(A) )x(fim1x !G

! for x = 1 |x| = x f(x) =xx

x2cos2cos2 "

!

Nowxx

x2cos2cosim

21x "

!!G

! (0

0form) =

1x2

x2sin2im

1x "!G! = 2sin2

(B)xx

x2cos2cosim

21x !

!G

! (0

0form) =

1x2

x2sin2im

1x !G! = 2sin2

SECTION (C) :

C-1. sinh < h < tanh , h % +,

-./

0 82

,0

1sinh

h? & 1

hsin

h=

LHL =1111

2

3

4444

5

6

+,

-./

0!

8

8!!

8

Gh

2cos

2

h

2im0h

!= 1

234

56!

G sinhhim

0h! = 2

RHL =1111

2

3

4444

5

6

+,

-./

0 "8

8!"

8

Gh

2cos

2h

2im0h

!= 1

2

345

6!G sinh

him

0h! = 2 # LHL = RHL = 2

C-3.'Gn

im! n

n

)1(n4

)1(n3

!!

!"! =

'Gn

im!

n

1.)1(4

n

1)1(3

n

n

!!

!"!

=

04

03 "=

4

3

C-6*. (A) 'Gxim!

6x3

2x2

!

" = 'Gx

im!6x3

2x2

!

"= 'Gx

im!

x

63

x

21

2"

= 3

1

(B) 'Gxim!

6x3

2x2 "= 'Gx

im!

x

63

x

21

2"

=3

1

SECTION (D) :

D-4.0x

imG!

xsinx

xcose3

2

2x

=0x

imG!

112

3

445

6"

112

3

445

6"

112

3

445

6"

^^

.......!3

xxx

......!4

x

!2

x1......

!2.4

x

2

x1

33

44

=0x

imG!

112

3

445

6"

"12

345

612

345

6

.......!3

x1x

.....!6

1

!3.8

1x

!4

1

8

1x

24

64

= 12

345

624

1

8

1=

12

1


37/150


D-5*. For0x

imG!

2x

xcosa1"

for ++,

-../

0

00

form

1 + a = 0 & a = 1

for 0ximG

!3x

xsinb

= 0ximG

!

2x

b

so b = 0

Now != 0ximG!

2x

xcosa1"

0ximG! 3x

xsinb

=0x

imG!

2x

xcos1=

0ximG!

2x2x2sin2

=0x

imG!

42

2

2x

2xsin

++++

,

-

.

.

.

.

/

0

=2

1. (1)2=

2

1

# (a, b) = (1, 0) and !=2

1

SECTION (E) :

E-3.)x(tann

1

4x

])x[1(im !! "8

G=

)x(tann

1

4x

)1exact(im !!8

G= 1

E-7.'Gn

im! 1

4n

]xn[...]x2[]x1[ 333 """

13x 1 < [13x] ) 13x23x 1 < [23x] ) 23x

. . .

. . .

. . .

n3x 1 < [n3x] ) n3xAdding all these inequilities

(13+ 23+ 33...........+ n3) x n < [13x] + [23x] + ...........[n3x] ) (13+ 23+ ..........n3) x

4

22

n

nx4

)1n(n!

"

& L

1= 'Gx

im! x

xx

e

e)x(f)x(fe`

`` `>

& L = L1= 'Gx

im!Y Z

+,

-./

0`

`

`

1.e

dx

d

e)x(fdx

d

x

x

& L = 'Gxim!

++,

-../

0

`

`

`

x

x

e

e)x(f

& L ='Gx

im! `f(x) & L = `'Gx

im! f(x)

& L = ` L2

& L2=

`

L

12. 1 < 2 < n

1 < 3 < n1 < 4 < n1 < 5 < n. . .

. . .

. . .1 < n 1< nmultiply all these inequilities1 < 2 . 3. 4 . 5 ... (n1) < nn2

1 < (n 1)! < nn2

n < n! < nn 1

nn

n

< nn

!n

< n

1

& 'Gnim!

1nn

1

< 'Gnim!

nn

!n

< 'Gnim!

n

1

0 < 'Gnim! nn

!n< 0 # 'Gn

im!nn

!n= 0

14.CN

AN= tanX

CN = AN cotX

Area of aABC =2

1(AB) (CN) = AN.CN = AN. AN

XX

sin

cos

= r cosX.1123

4456

XX

sincosr 2

=X

Xsin

cosr 32


40/150


Area of aDEC =2

1(DE) (CM) = (DM) (CM) = (CM)2tanX= (OC r)2tanX

=

2

rsin

r+,

-./

0X

tanX CXX

X!cossin

)sin1(r 22

now 0ABimG

!

DEC

ABC

a

a

= 2

im8

GX

!

22

32

)sin1(r.sin

sin.cos.cosr

XX

XXX

=2

im8

GX

!2

4

)sin1(

cos

X

X=

2

im8

GX

! 2

22

)sin1(

)sin1(

X

X=

2

im8

GX

! (1 + sinX)2 = (2)2= 4

PART - II

1. 'Gxim!

x

n

e

x= 0 (n %integer)

case(i) when n = 0

then 'Gx im! xe

nx= 'Gx im! xe

1= 0

case(ii) when n is +ve integer

'Gxim!

xe

nx +,

-./

0

'' form

= 'Gxim!

xe

!n= 0

case(iii) when n is ve n = m where m %z+

'Gxim! xe

nx= 'Gx

im! xe

mx= 'Gx

im! xe.mx1

= 0

so 'Gxim!

x

n

e

x

= 0

3. 0ximG! f(g (h (x)))

L.H.L. x G0

0xim

G! h (x) = 0+

"G0xim! f(g(x))

then "G0xim! g(x) = 1+

"G1x im! f(x) = 1 1 = 0

R.H.L. x G0+

"G0xim! h (x) = 0+ so "G0x

im! f(g(x)) = 0 L.H.L. = R.H.L. = 0

5. 'Gxim! +

+,

-../

0"

+++

,

-

.

.

.

/

0

++,

-../

0

2x

1sinx1sinx

'Gxim!

++++

,

-

..

.

.

/

0

++,

-../

0"

++

,

-

.

.

/

0

2x

1sin

x1

x1sin

= 1 + 0 = 1


41/150


6. +++

,

-

.

.

.

/

0

12

345

6

G

3

ax

a

3|x|ax

im! (a > 0) # x = a h

=++++

,

-

..

.

.

/

0

112

3

445

6

G

3

aha

a

3|ha|

0him! =

++++

,

-

..

.

.

/

0

112

3

445

6

G

3

ah

1a

3|a|

0him! = a20 = a2

12. 'Gx im! sec

1 +,

-

./

0

" 1xx

replace x Gy

1& 'Gx

im! G 0yimG!

= 0yimG! sec1

++++

,

-

.

.

.

.

/

0

" 1y

1

y

1

= 0yimG! sec 1 ++

,

-../

0

"1y1

when y G 0+;1y

1

" < 1

so 0yimG! sec 1 ++

,

-../

0" 1y1

does not exist.

14. (i) != L Mxsin1xsinimx

!"'G

!

& != ++,

-../

0 !"++,

-../

0 ""'G 2

x1xsin.

2

x1xcos2im

x!

& ! = L M++,

-../

0

""

!"++,

-../

0 ""'G x1x2

x1xsin.

2

x1xcos2im

x!

& != L M++,

-../

0

""++,

-../

0 ""'G 1xx2

1sin

2

1xxcos2im

x!

& != (oscillating value 1 to 1) 0 = 0

(ii) m = xsin1xsinimx

!"!'G

!

when x G !'

then x undefined

m is undefined

16. To find 0x imG! f(x)

L.H.L. = 0xim

G! f(x)

= 0xim

G! }x{cot}x{ =

0himG! )h1(cot)h1( = 1cot

R.H.L. = "G0xim! f(x) = "G0x

im!22

2

]x[x

]x[tan=

0himG!

22

2

]h0[)h0(

]h0[tan

""

"=

0himG! 2

2

h

0tan= 0

" L.H.L. $R.H.L. so0x

imG! f(x) does not exist. f(x) is not continuous at x = 0.

Now cot1

2

0x

)x(fim +,

-./

0

G

!

= cot1 2)1cot( = cot1(cot 1) =1


42/150


18. sinX< X< tanX , X% +,

-./

0 82

,0

XX

==X

X tan1

sin

X

X==

X

X tannn

sinn

0imGX! ++

,

-../

012

345

6X

X"1

2

345

6X

X tannsinn; n % N

L.H.L. = !GX 0im! ++

,

-../

012

345

6X

X"1

2

345

6X

X tannsinn= n 1 + n = 2n 1

R.H.L. = "GX 0im!

++,

-../

012

345

6X

X"1

2

345

6X

X tannsinn= n 1 + n = 2n 1 # L.H.L. = R.H.L = 2n 1

20. 'Gnim!

++

,

-..

/

0"

"""

""

"n2n

1.............2n

1

1n

1

n

12222

using sandwitch theorem

2n

1)

n

1

1n

1

2 ")

n

1

% %

n2n1

2 " )

n1

adding all these inequilities

n2n

1.............

2n

1

1n

1

n

1

2222 """

""

"" )

n

n2

Taking both side 'Gnim!

'Gnim! +

+

,

-

.

.

/

0

"""

""

""

n2n

1.............

2n

1

1n

1

n

1

2222= 2

21. f(x)= 0tim

G! +

,

-./

08

!2

1

t

xcot

x2

Case-I : when x = 0f(x) = 0

Case-II :when x > 0

f(x) = +,

-./

08

!

G 21

0t t

xcot

x2im! = )(cot

x2 1 '\8

!=

8x2

. 0 = 0

Case-III when x < 0

f(x) = +,

-

./

0

8

!

G 2

1

0t t

x

cot

x2

im! = 8x2

cot

1(

') = 8

x2

. 8= 2x

& f(x) = 2x


43/150


25. 0yimG!

+++++

,

-

.

.

.

.

.

/

0++,

-../

0+,

-./

0 "!+,-

./0 +

,-.

/0 "

'G y

1nxexp1nxexp

imxyb

xya

x

!!

! = 0yimG!

+++++

,

-

.

.

.

.

.

/

0+,-

./0 "!+

,-.

/0 "

'G y

11

im

xx

x

xyb

xya

!

by expansion = 0yimG!

++++++

,

-

..

.

.

..

/

0

++,

-../

0"""!++

,

-../

0"""

'G y

.....!2

)1

x(xby1......!2

)1

x(xay1

im2

22

2

22

x

xyb

xya

!

= 0yimG!

1111

2

3

4444

5

6""

y

.....)ba(2

y)ba(y 22

2

= a b

29.Ax

)1ax(im n

n

x "

"'G!

(A) If n %N

'Gxim!

n

n

x

A1

x

1a

"

+,

-./

0"

=01

)0a( n

""

= an

(B) If n %Z& a = A = 0

then 'Gxim!

Ax

)1ax(n

n

"

"= 'Gx

im! nx

1= ' n %Z

(C) If n = 0

then 'Gxim!

Ax

)1ax(n

n

"

"= 'Gx

im! A1

1

"=

A1

1

"

(D) If n %Z, A = 0 & a $0

then 'Gxim!

Ax

)1ax(n

n

"

"= 'Gx

im!n

n

x

)1ax( "= 'Gx

im!

n

x

1a +

,

-./

0" = (a + 0)n= an

EXERCISE # 3

1. (A) 0ximG!

xsin

x]etan[x]etan[2

2222 !!

= 0ximG!

2

22

22

2222

22

2222

x

xsinx

x]e[

x]etan[x]e[

x]e[

x]etan[x]e[

!

!!

= [e2] [e2] = 15

(B) 0ximG! L M 1

2

345

6""

x

xsin)6t4tmin( 2 = 0x

imG!

12

345

6x

xsin2

" sin x < x &x

xsin2< 2 & 1

23

456

x

xsin2= 1 So 0x

imG!

123

456

x

xsin2= 1


44/150


(C) 0ximG!

2

4

13/12

xx

)x21()x1(

"

"= 0x

imG!

2

22

xx

....x4!2

14

1

4

1

2

x1....

3

x1

"

++++

,

-

.

.

.

.

/

0

"+

,

-./

0

"++

,

-../

0""

=2

1

(D) 0ximG! =

xsin

xcos122

"!= 0x

imG!

xsin

4xsin22

2

2

= 0ximG!

2

22

2

x

xsin.

16

x.16

4

xsin22

=8

2

Comprehension # 1(For Q.No. 3 to 5)

f(x) = 'Gnim!

n

n

xcos ++

,

-../

0=

n.1n

xcosim

ne

++,

-../

0!

'G!

Substituting, n = 2t

1& f(x) =

+,

-./

0 !G 20t t

1txcosim

e!

=+,

-./

0 !!

G 222

0t xt

txcos1xim

e!

=2x

2

1

e!

g(x) = b4x & b = 'Gx

im! +,-.

/0 "!"" 1x1xx 22 = 'Gx

im!++

,

-

.

.

/

0

""""

!!""

1x1xx

1x1xx

22

22

=2

1

# g(x) = 21

.4

x = x2

By observation, graphs of f(x) and g(x) intersect each other at two points# Number of solutions is 2.

9. Statement -1 is true as

12

345

6G x

xsinim

0x! = 0 and 1

2

345

6G x

xsinim

0x! = 1

Statement - 2 is true as

))x(g(himaxG

! = h ))x(g(im(axG

! if h(x) is continuous at x = g(a).

10. Statement -1

0ximG!

x

2

x2cos1

= 0ximG!

x

|xsin|& L.H.L. = 1 & R.H.L. = 1

Statement -2 is true

14. True indeterminate form of type ''

18. 1ximG!

1)x2(

2)x5(

!!

!!

1ximG!

2)x5(

1)x2(1)x2(4)x5(

""\ = 1x

imG!

4

2

x1

x1 \ =2

1


45/150


21. 'Gxim! L Mx)bx()ax( !""

= 'Gxim!

x)bx()ax(

x)bx()ax( 2

"""

""

= 'Gxim!

112

3

445

6""""

""

1xab

x)ba(1x

abx)ba(

2

=11

ba

""

=2

ba"


1. 0ximG!

2)x2cos1(

xtanx2x2tanx

!

!= 0x

imG!

22 )xsin2(

x

STU

:; 00x

imG! L M

xsin

x

1

x

1xsin +

,

-./

0"

xsin

0x

x

1

0x x

1im)x(sinim +

,

-./

0"

GG!! = 0 +

xsin

0x x

1im +

,

-./

0G! =

+,

-./

0G x

1n).x(sinim

0xe !!

= xeccos)x(n

im0x

e

!!

G +,

-./

0'

'form

=)xcotecx(cos

x

1

0xim

e G!

= xcosx

xsinim2

0x

e G!

= e = 1

7*. L =4

222

0x x

4

xxaa

im!!!

G! =

0ximG!

4

2

4

4

2

2

x

4

x....

a

x.

2

12

1

2

1

a

x.

2

11.aa !

1111

2

3

4444

5

6

!+,

-./

0 !"!!

("a > 0)

=0x

imG!

4

2

3

42

x

4

x......

a

x.

8

1

a2

x!""

Since L is finite & 2a = 4 & a = 2 # L =0x

im

G

! 3a.8

1=

6

8. x)2b1(nx

0xelim

"

G

!

= 1 + b2= 2b sin2X & sin2X=2

1 +

,

-./

0"

b

1b

We know b +b

1*2 & sin2X * 1 but sin2X ) 1

& sin2X =1 & X = 2

8

9. ++

,

-..

/

0

"""

'Gbax

1x

1xxim

2

x! = 4

'Gxim! +

+,

-../

0

"""

1x

)b1()ba1(x)a1(x2

= 4

Limit is finite it exists when 1 a = 0 &a = 1

then++++

,

-

.

.

.

.

/

0

"

"

'G

x

11

x

b1ba1

imx!

= 4 # 1 a b = 4 & b = 4

10. ((1 + a)1/31)x2+ ((a+1)1/2 1)x + ((a+1)1/61) = 0let a + 1 = t6

# (t21)x2+ (t31)x + (t 1) = 0(t + 1)x2+ (t2+ t + 1)x +1 = 0

As a G0 , t G1 2x2+ 3x + 1 = 0 &x = 1 and x = 2

1

PART - II

1.0x

imG!

x2

x2cos1

0ximG!

x2

xsin2 2

= 0ximG!

x

|xsin|

LHL = 0himG!

1

h

hsin

C , RHL = 0himG!

1h

hsin

CLHL $RHLSo limit does not exist


47/150


2. 'Gxim!

x

2x

3x+,

-./

0"

(1'form)

=x1

2x

3xim

xe

+,

-./

0"'G

!

= 2xx5

imxe "'G!

= x2

1

5im

x

e"'G

!

= e5

3. f(2) = 4, f>(2) = 4

2ximG!

2x

)x(f2)2(fx +,

-./

0form

0

0= 2x

imG!

1

)x(f2)2(f >= f(2) 2f>(2) = 4 8 = 4

4. f(1) = 1, f>(1) = 2

1ximG!

1x

1)x(f

!

!+

,

-./

0form

0

0=

1ximG!

)x(f2

x2)x(f>= 2

1

2

)1(f

)1(fCC

>

5. 'Gxim!

]x[

]x[xn n!

= 'Gxim!

]x[

nxn! 'Gx

im! ]x[

]x[= 'Gx

im! ]x[

nxn! 'Gx

im! 1 = 0 1 = 1

6. 'Gxim!

x

2

2

3xx

3x5x++

,

-../

0

""

""

= 'Gxim!

x

2 3xx

1x41 +

,

-./

0

""

"" (1' form) = 3xx

)1x4(xim

2xe """

'G!

= e4

7.2

x im8G!3)x2(

2

xtan1

)xsin1(

2

xtan1

8+,

-./

0"

+

,

-.

/

0

=2

x im8G! 3)x2(

)xsin1(

2

x

4

tan

8

+

,

-.

/

08

put x = y2

"8

y G 0+

= 0yimG! 3)y2(

)ycos1(2

ytan +

,

-./

0

= 0yimG!

3y8

)ycos1(2

ytan +

,

-./

0

= 0yimG!

++++

,

-

.

.

..

/

0

2

y2ytan

64

2

4

y

2

ysin

2

2

++++

,

-

.

.

..

/

0

= (1) (1) 32

1

32

1C

8.0x

imG! k

x

)x3(n)x3(nC

" !!

&0x

imG! k

x

3

x1n

3

x1n

C+,

-./

0+,

-./

0 " !!

&0x

imG!

1111

2

3

4444

5

6

+,

-./

0

+,

-./

0

+,

-./

0+,

-./

0 "

3

1

3

x

3

x1n

3

1

3

x

3

x1n !!

= k

& k3

1

3

1C" & k =

3

2


48/150


Alt.

0ximG! k

x

)x3(n)x3(nC

" !!

Apply L' Hospital rule

&0x

imG! k

x3

1

x3

1C+

,

-./

0"

" & k31

3

1C"

& k =32

9.ax

imG! 4

)x(f)x(g

)a(g)x(f)a(g)a(f)x(g)a(fC

"

Apply L' Hospital rule

&ax

imG! ++

,

-../

0>>

>>

)x(f)x(g

)x(f)a(g)x(g)a(f= 4

&ax

imG! 4

)x(f

)x(g

)x(f)x(gk C++

,

-..

/

0

>>

>>

& k = 4

10. 'Gxim!

2x2

2e

x

b

x

a1 C+

,

-./

0""

& 2x2xb

x

aim

ee2x C

+,

-./

0"

'G! & 'Gx

im! 2x

2)bax(C

"

& 'Gxim! 2

x

b2a2 C" # b %R, a = 1

11. ax2+ bx + c = 0

a(x _) (x b) = 0

2

2

x )x(

)cbxaxcos(1im

_!

""!_G

!

=2

22

x )x(

2

cbxaxsin2

im_!

++,

-../

0 ""

_G! = _Gx

im! 2

2

)x(

)x()x(2

asin2

_

+,

-./

0b_

= _Gx im!

4

)x(a2

2

)x()x(a

2

)x()x(asin 22

2

b

+++++

,

-

..

.

.

.

/

0

b_

+,

-./

0 b_

= (1) 2

)(a

4

)(a2 2222 b_C

b_

12.0x

limG )x(f

)x3(f= 1

f(x) < f(2x) < f(3x) Divide by f(x)

)x(f

)x3(f

)x(f

)x2(f1 ==

using sandwitch theorem

& 'Gxlim )x(f)x2(f

= 1


49/150


13.2lim

2xG )2x(

)2x(sin

!

!

# does not exist

14. 0|5x|

9))x(f(im

2

5xC

G!

0]9))x(f[(im 25x

CG!

5ximG! f(x) = 3

ADVANCE LEVEL PROBLEM

1.0x

imG!

x

1

x2

x1

x2

x1

bb

aa

++

,

-

.

.

/

0

"

"=

x

11

bb

aaim

x2

x1

x2

x1

0x

e++

,

-

.

.

/

0!

"

"G!

=++

,

-

.

.

/

0

"++

,

-

.

.

/

0 !!

!!

!"

!G x

2x

1

x2

x1

x2

x1

0x bb

1

x

)1b(

x

)1b(

x

1a

x

1aim

e!

=)blogblogaloga(log

2

12e1e2e1e

e!!"

=++,

-../

0

21

21e

bb

aalog

2

1

e =21

21

bb

aa

2.1x

imG!

qpqp

qp

xxx1

pxqxqp""!!

!"! +,

-./

0form

0

0 =

1ximG! 1qp1q1p

1q1p

x)qp(qxpx

pqxpqx!"!!

!!

""!!

! +

,

-./

0form

0

0

=1x

imG! 2qp2q2p

2q2p

x)1qp)(qp(x)1q(qx)1p(p

x)1q(pqx)1p(pq!"!!

!!

!"""!!!!

!!!=

2

qp !

3.0t

im

G

! t

11t1t

ab

ab

+

+

,

-

.

.

/

0

!

! ""

=0t

imG!

t

11

ab

ab 1t1t

e++

,

-../

0!

!! ""

=+,

-./

0!+

+,

-../

0 !!

!G ab

1

t

)1a(a

t

)1b(bim

tt

0t

e!

= abnaanbb

e !! !!

=

ab

1

a

b

a

b !

++,

-../

0

4. LHL = "G0him! f(0 h) = "G0h

im! f(h) = "G0him! L MB A

2h1 =1

RHL = "G0him! f(0 + h) = "G0h

im! f(h) = "G0him! +

,

-./

0

"'G nn h1

1im! = 1

5. 0ximG!

)xtanx(sin

xk2xn

cos2ee 2xnxn

!

!!"!

=

++,

-../

0"""!+

+,

-../

0"!

!112

3

445

6!"!!

112

3

445

6"""

G.......x

15

2

3

xx........

!3

xx

xk........!4.16

xn

!2.4

xn12.....

!4

xn

!2

xn12

im5

33

244224422

0x!

=..........

3

1

!3

1x

!4.16

n2

!4

n2xk

4

nnx

im3

444

222

0x"+

,

-./

0!!

++,

-../

0!"+

+,

-../

0!"

G! limit exists, if coff. of x2is zero.


50/150


& n2+4

n2k = 0 & 4k = 5n2

so the possible value match that is n = 2, k = 5

6. L =

2

1x

)x4x3(cosim31

2

1x !

!!

G

!=

2

1x

)x3x4(cosim31

2

1x !

!!8 !

G

!

Let X= cos1x " As x G2

1# XG

3

8

& L =

2

1cos

)3(coscosim1

3 !X

X!8 !8

GX

!

Now LHL =

2

1cos

)3(coscosim1

3 !X

X!8 !8

GX!

!=

2

1cos

3im

3 !X

X!8!8

GX

! +,

-./

0form

0

0

=

X!!

!8GX sin

3im

3

! = 32

Also, RHL =

2

1cos

)3(coscosim1

3 !X

X!8 !8

GX"

! =

2

1cos

))32(cos(cosim1

3 !X

X!8!8 !

8GX

"! =

2

1cos

)32(im

3!X

X!8!8"8

GX

!

=

2

1cos

3im

3 !X

8!X"8

GX

! +,

-./

0form

0

0 =X!

"8GX sin

3im

3

! = 32

& LHL $RHL # Limit does not exist

7.

I

I

C

C

'G

"!

n

1r

3

n

1r

2

n

r

)1rn(r

im! = I

II

C

CC

'G

!"

n

1r

3

n

1r

3n

1r

2

n

r

rr)1n(

im!

=++++

,

-

.

.

.

.

/

0

!"

"""

'G1

4

)1n(n

6

)1n2()1n()n()1n(

im22n

! =4/1

3/11 =

3

11

3

4C!

8. 'Gnim!

++,

-../

0!"!! ...

n6

C

n2

CCn

n

23

x2

x

1x1x

1999

=2000

1

the limit obviously exists if 2000 x = 0 & x = 2000

9. Let X= sin1 x " as2

1xG #

4

8GX

2

1sin

)cossin2(cosim1

4 !X

XX!8

GX

!=

2

1sin

)2(sincosim1

4 !X

X!8

GX

!=

2

1sin

22

coscosim

1

4 !X

++,

-../

0+,

-./

0X!

8!

8GX

!

Left hand limit =

2

1sin

22coscosim

1

4 !X

++,

-../

0+,

-./

0X!

8!

8GX

!! =

2

1sin

22im

4 !X

X!8

!8GX

! +,-.

/0 form

0

0


51/150


= X

!

8GX

cos

2im

4

! = 22!

Right hand limit ="8

GX4

im!

2

1

sin

22

coscos 1

!X

++,

-../

0+,

-./

0X!

8!

="8

GX4

im!

2

1

sin

22coscos

!X

++,

-../

0+,

-./

0 8!X

="8

GX4

im!

2

1sin

22

!X

8!X

= "8GX

4

im! Xcos

2= 22 & LHL $RHL # Limit does not exist

10. 0ximG!

x

k

xf...

3

xf

2

xf)x(f +

,

-./

0""+

,

-./

0"+

,

-./

0"

+,

-./

0form

0

0

= 0ximG!

12

3

45

6

+,

-./

0>""+,

-./

0>"+,

-./

0>"> k

x

fk

1

...3

x

f3

1

2

x

f2

1

)x(f

= 1 +2

1+

3

1+ .... +

n

1= c !""""

1

0

1n2 dx)x...xx1( = c !!

1

0

n

dxx1

x1= c

!!1

0

n

dxx

)x1(1

= nC1

2

C2n

+3

C3n

........ + (1)n1.n

Cnn

PART - II

1. Let Ln+1

=0x

imG!

21n321

x

)xacos().......xa(cos).xa(cos.)xacos(1 "

& Ln+1

=0x

imG!

21n211n1n

x

)xa(cos.....)xa(cos.)xa(cos)xa(cos)xa(cos1 """ "

& Ln+1

=0x

imG!

2n211n1n

x

)}xa(cos).....xacos().xacos(1{)xa(cos)xacos(1 "" "

& Ln+1

=0x

imG! 2

1n

x

)xacos(1 "+

0ximG! cos (a

n+1x) . L

n

& Ln+1

=L M

2

a2

1n" + Ln

L1= 2

a21

, L2= 2

a22

+ L1= 2

a22

+ 2

a21

, L3= 2

a23

+ L2= 2

a23

+ 2

a22

+ 2

a21

, Ln= 2

1

ICn

1i

2ia

2. '!Gxim! x x a x b2 1! " " !0

/. -

,+ = 0

+,-.

/0 !!""

'Gbax1xxim 2

x! = 0

Let x =h

10

h

bhahh1im

2

0hC

!!"""G

! " limit exists

So 1 a = 0 & a = 1

0h

bh1hh1im2

0hC!!""

"G!


52/150


& 20h hh12

)1h2(im

""

""G

! b = 0 &2

1b = 0 & b =

2

1

So a = 1, b =2

1

3. " 'Gxim! xnf(x) = p

& 'Gxim!

x

)x(fx 1n"= p

using L- Hospital rule, we get

'Gxim!

1

)x(fx)x(fx)1n( 1nn >"" "= p &(n + 1) p + 'Gx

im! xn+1.f>(x) = p & 'Gxim! xn+1f>(x) = np.

4. Let f(x) =3

2x

x

x1xe "

Since limit exists & "G0him! f(0 + h) = "G0h

im! f(0 h) = L (say)

L = "G0him!

3

2h

h

h1he "

Also, L = "G0him!

3

2h

h

h1he ""&2L = "G0h

im! 3

hh

h

h2ee

Put h = 3t

&2L = "G0tim! 3

t3t3

t27

t6ee &54 L = "G0t

im! 3

ttt2t23t3t

t

t6)ee(3)ee(3)1e()1e( "

& 54 L = "G0t im! 112

3

445

6

"++,

-

../

0

"++,

-

../

03

tt

3

t2t23

t3

t

t

)t2ee(3t

)t4ee(3t

1e

t

1e

& 54 L = "G0tim! 1

1

2

3

44

5

6

++

,

-../

0++

,

-../

0"+

+,

-../

0"+

+,

-../

03

tt

3

t2t23

t3

t

t

t2ee3

)t2(

t4ee24

t

1e

t

1e

& 54 L = 1 + 1 + 48 L 6L &12 L = 2 &L =6

1

5. 'Gnim! n

n! 2 ( ) .......n n n nn

n

" "0/.

-,+

"0/.

-,+ "

0/.

-,+

1 & when m >1 continuous and derivableif 0 < m )1 continuous but not derivable

Section (D) :

D-2. y = f(x)

Section (E) :

E-2. f : R GR and f(x + y) = f(x) f(y) J x, y %RPut x = y = 0 f(0) = f2(0) since f(0) $0

& f(0) = 1

f '(x) =h

)x(f)hx(flim

0h

!"G

=B A

h

1)h(f)x(flim

0h

!G

= f(x) f '(0)

Let f(x) = y &dx

dy= y.f '(0)

On solving !ny = x f '(0) + cy = f(x) = ec. ex f '(0) "f(0) = 1 & c = 0

Thus f(x) = ex.f '(0) J x %R

E-3._ f(x) is continuous and3

7% [f(2), f(2)], by intermediate value theorem (IVT), there exists a point

c % (2, 2) such that f(c) =3

7

PART - II

Section (A) :

A-2.* (A) f(x) is continuous no where

(B) g(x) is continuous at x = 1/2(C) h(x) is continuous at x = 0(D) k(x) is continuous at x = 0


57/150


Section (B) :

B-4. y =2tt

12 !"

, where t =1x

1

!, y = f(x) is discontinuous at x = 1, where t is discontinuous and y =

)1t)(2t(

1

!" at t = 2 and t = 1

& 1x 1! & 2x + 2 = 1, x = 21

1 1x

1

! & x = 2 f(g(x)) is discontinuous at x =

2

1, 2,1

Section (C) :

C-9. !G2xLim

f(x) =5

3= f(2) $ "G2x

Limf(x) = 1

f(x) is not continous at x = 2

!G3xLim

f(x) = "G3xLim

f(x) = f(3) =2

9

Now LHD (x =3) is 0hLim

G h

2

9))h3()h3((

4

1 23

!

!!!!

C 0hLim

G 421

4

21h8h2C

"!

and RHD (x = 3) is 0hLim

G 0h

2

9|)h1||1h(|

4

9

C!!!"!

# f(x) is not differentiable at x =2 and x = 3

Section (D) :

D-3.* 0x

0x

0

x/1cos)x(sin)x(fy

21

C

$

9:

9;(1) =0h

LimG

h

)ba(b)h1(a8)h1(3)h1(

])h1sin[( 32

2

!

"!"!""!!!

8!

=0h

LimG h

a)h1(a 3

!!!

+,-

./0

form0

0 =

0hLim

G 1

)h1(a3 2!

f>(1) = 3a

f>(1+) =0h

LimG h

ba)h1(tan)h1cos(2 1 !!""8" !=

0hLim

G h

)ba)h1(tanhcos2( 1 !!""8! !

Function is differentiable

# 2 +4

8= a + b .....(1)

=0h

LimG h

2/2)h1(tanhcos2 1 8"!""8! !=

0hLim

G 28sin 8h + 2)h1(1

1

""=

2

1

Now f>(1) = f>(1+) 3a =2

1

a =6

1....(2) by (1) and (2) b =

48

6

13

PART - II

3. f +,

-./

082

=]h41[n

(cosh)n

h4

cosh1lim

220h "

!G !

!

=)h41(n

h4.

2/h

2/hsin

1616

2lim

2

2

2

2

0h "++,

-../

0

\G !.

2/hsin2

)2/hsin21(n2

2!!.

2/h

2/hsin22

2

=641 . 1. 1.(1) .1 =

641!


61/150


5. f(x) = [sin[x]] =

99

99

:

9999

;

(0) = 4

0xlimG 2x

)x4(f)x2(f3)x(f2 "!12

345

60

0form

using L>Hospital rule

0xlimG x2

)x4(f4)x2(f6)x(f2 >">!>12

345

60

0form

using L>Hospital rule

0xlimG 2

)x4(f16)x2(f12)x(f2 >>">>!>>=

2

4.164.124.2 "!= 12

17. f(x) = [n + psinx] , x %(0, 8)graph of y = n + p sinx


62/150


obviouslyf(x) = [n +psinx] is discontinous at points mark in above curve

& number of such points & (p 1) + 1 + p 1 = 2p 1

21. f(x) = [x] + }x{

Curve of y = f(x) = 9

:

9;

(1) = Rf>(1).Therefore, function is differentiableat x = 1.

Again Lf>(2) =

= = (4 1) (2 1) sin 2 = 3 sin 2

and R f>(2) = = = (221) sin 2 = 3 sin 2

So L f>(2) $R f> (2), f is not differentiable at x = 2Therefore, (d) is the answer.

12. f(x) = [x sin 8x]f(0) = 0

f(0 ) = [h sin8(h)] = 0 ; f(0+) = [h sin 8h] = 0

Here f(0) = f(0+) = f(0) = 0So continuous at x = 0Since graph of f(x) is as shown in the figureNow all options can be checked from graph.

PART - II

1. f(x) =

(i) for 0 < sin x < 1,

f(x) = = 1

(ii) for sin x = 0,

f(x) =


76/150


(iii) for 1 )sinx < 0,

f(x) = # f(x) =

# f(x) is discontinuous at integral multiples of 8

2.

= =

= =

for g(x) to be continuous (!na)2= (!n2a)2 & (!na + !n2a) = 0

& a = # g(0) = (!n 2)2

3. " f(x) =

by definition of g(x)

g(x) =

g>(x) =

clearly g(x) is discontinuous at x = 0 and not differentiable at x = 0, 2

4. f(x) =

graph of f(x) is as shown is figure

#f(x) is continuous for all x butnon-differentiable for integral points.

5. " f(x) =


77/150


" f(x) = = 0 and f(x) = = 0

and f (0) = 0 # f(x) is continuous at x = 0

" f(x) = = and f(x) = G '

#f(x) is discontinuous for x = 1

Similarly we can check that f(x) is discontinuous at x =

1

" L.H.D. at (x = 0) is = = = 1

" R.H.D. at (x = 0) is = = = 1

& L.H.D. = R.H.D.# at x = 0, f(x) is derivable.

6. is 1 or 0 according to x is a rational number or an irrational number. As m!8x will

become integral multiple of 8when x is rational, then cos (m!8x) = 1. And when m!8x is not an integral multipleof 8i.e. when x is irrational then 1 < cos (m!8x) < 1

#f(x) =

# f(x) is discontinuous and non-differentiable at every real number.

7. f(x) + f(y) =

f(x) = f

f (x) = f

put x = 0f (0) = f (y) (1+y2)

= f (y)

Integrating both sidesf(0) tan1y = f(y) + c ..... (1)

Now put x = 0 & y = 0 in f(x) + f(y) = f

we get 2f(0) = f(0)

#f(0) = 0 ............. (2) and " = 2

& = 2 & f> (x) = 2 # f>(0) = 2 ........... (3)

from (2) & (3) & (1)2tan-1y = f(y) + cnow put y = 1, we get 2tan-11 = f(1) + c

= + c & c = 0

# f(x) = 2tan-1x


78/150


8. " R.H.L. = f(0 + h) =

= .

(putting 1h2= cos2X)

= = =

" L.H.L = f(0 h) =

= = = =

since R.H.L. $ L.H.L, therefore no value of f(0) can make f continuous at x = 0

9. As f is continuous on R, so f(0) = f(x)

Thus f(0) = f=

= 0 + 1 = 1

10. " f(1) = 0

R.H.L. = f(1 + h) = cos1 = cos1 = 0

L.H.L. = f(1 h) = cos1 = cos1(0) =

" f(1) = 0# f(x) is discontinuous hence non-derivable at x = 1

" f>(1+) = = 0

and f>(1) = = 0

& f>(1+) = f>(1) = 0# f(x) is derivable at x = 1

11. " f>(x) = =

= f(x) = &f>(x) =

so dx = =

12. we have

f(x) = (sin(h) + cos(h))cosec(-h) = (cosh sinh)cosech


79/150


= = = e

Now we have

f(x) = = =

Iffis continuous at x = 0 , then

e = a = gives a = e and b = 1

13. Given that

f(xy) = exyxy(eyf(x) + ex f(y)) x,y %R+

putting x = y = 1 , we getf(1) = e1(ef(1) + ef(1))

&f(1) = 0

now " f>(x) = =

= = f(x) +

& f>(x) = f(x) + & = & =

Integrating both sides w.r.t. x, we get

!n |x| + c =

or f(x) = ex(!n|x| + c)

since f(1) = 0 & c = 0#f(x) = ex!n|x|

14. Given f(x + y3) = f(x) + [f(y)]3

and f(0) > 0putting x = y = 0, we getf(0) = f(0) + (f(0))3 & f(0) = 0

also f>(0) = =

Let L = f(0) = = = L3

or L = L3

or L = 0 , 1, 1 as f>(0) > 0 &f>(0) = 0, 1

Thus f(x) =

= & f(x) = & f(x) = 0 , 1

Integrating both sides, we getf(x) = 0 or f(x) = x + cAs f(0) = 0 , we have f(x) = 0 or f(x) = x

Now f(x) = 0 is imposible as f(x) is not identically zero# f(x) = x and f(10) = 10


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METHOD OF DIFFERENTIATION


Section (A) :

A-1. (i) f(x) = sinx2

f(x) = =

= = . cos

= (2x + h) . cos & (f(x)) = (1) . (2x) . cosx2

f>(x) = 2x cosx2

(ii) f(x) = e2x + 3

f(x + h) = e2(x + h) + 3

(f(x)) = = = e2x + 3 . 2

= 2e2x + 3 & f(x) = 2e2x + 3. 1 & f>(x) = 2e2x + 3

A2. (i) y = x2/3+ 7e + 7 tan x

= x1/3+ + 7 sec2x

(ii) y = x2.!n x.ex

= 2x.!n x .ex+ x.ex+ x2.!nx.ex

(iii) y = !n

= . sec2 = . = = sec x

(iv) y =

=

=


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(v) y = tan

y =

y = tan & = sec2

Section (B) :B3. (i) ax2+ 2hxy + by2+ 2gx + 2fy + c = 0

& = & =

(ii) xy + xey + y. exx2= 0

= & =

Section (C) :

C3. (i) y = tan1 + tan1

y = tan1 + tan1

y = tan15x tan1x + tan-1 + tan1x

=

(ii) y = sin1 , 0 < x < 1

y = cos1

Let tan1x = X, X%

y = cos1(cos 2X)

0 < 2X(ex).ex

= ex f>(ex) + e2x f>>(ex)

E5.* u = exsin x, v = excos x

v

u = v(ex

cos x + ex

sin x)

u(ex

cos x

ex

sin x)

= exsin x(v + u) + excos x( v u)= u(v + u) + v(v u) = v2+ u2


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again = exsin x + excos x

= exsin x + excos x + excos x exsin x

= 2v

similarly other options can be checked.


3. x = at3 and y = bt2

= 3at2, = 2bt

= .

= . =

. .

= . . = =

4. y = 1+ + +

= + + = +

= =

& !ny = !nx !n(x c1) + !nx !n(x c2) + !nx !n(x c3)Differentiating both sides w.r.t. x , we get

& = =

& =

10. y = x!n & = !n

& = ...(1) & =

& = &(1 + x) = from equation (1)

&(1 + x) = +

&(1 + x) = x + y 1 &(1 + x) + x = y 1


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12. let g(x) = &g(x) = + 0 + 0

" g(_) = g>(_) = 0i.e. _is the repeated root of g(x) = 0 and h(x) is a polynomial of degree 3" f(x) = 0 has repeated root _

hence g(x), is divisible by f(x)

PART - II : OBJECTIVE QUESTIONS

3. x + y = 0

x2(1 + y) = y2(1 + x)x2y2+ x2y y2x = 0(x + y) (x y) + xy (x y) = 0

(x y) (x + y + xy) = 0 " x $y & y =

& = =

5. y = sin1 + sin1x

= + = + &P =

6. f>(x) = , g(2) = a & g>(2) = ?

g is inverse of ff(g(x)) = xDifferentiating w.r.t. xf> (g(x)) . g>(x) = 1

g>(x) = & g>(2) = = & g> (2) =

10. u = ax + bLet y = f(ax + b)

= a f >(ax + b)

=a2f>>(ax + b)

= a3f>>>(ax + b)

= anfn(ax + b) & f(ax + b) = anfn(u) = an f(u)


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12. y = f & f>(x) = sin x

& = f> . = sin .

14. y2

= P(x)

& 2y = P>(x) & =

& 2 = & 2y3 = y2P>>(x) P>(x) . y

& 2y3 = P(x) P>>(x) " y2= P(x) & y =

& 2 = P>(x) . P>>(x) + P(x) . P>>>(x) 2

# 2 = P(x) . P>>>(x)

17. f>>(x) = f(x) .... (i)f>(x) = g(x) .... (ii)h>(x) = (f(x))2+ (g(x))2 .... (iii)h(0) = 2, h(1) = 4Differentiating equation (ii) w.r.t. xf>>(x) = g>(x) = f(x)Differentiating equation (iii) w.r.t. xh>>(x) = 2f(x) . f>(x) + 2 g(x) . g>(x)

= 2f(x) . f>(x)

2f>(x) . f(x) = 0 {" g>(x) =

f(x)}& h>(x) is constant& h(x) is linear function" h(0) = 2 & h(x) not passing through (0, 0)Let y = h(x) = ax + bat x = 0

y = 2 = b & y = ax + 2at x = 1

a + 2 = 4a = 2 & curve is y = 2x + 2

20. y = cos1

=

& = & =

& = when x < 0 & =


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when x > 0 & =

Alternate :

put x = tanX

tan1

x = X % & y = cos1

= cos1

y = cos1(cosX/2) & y = & =

EXERCISE # 3

2. (A) y = f(x3)

# = f>(x3) . 3x2 # = f>(1) . 3 = 9

(B) f(xy) = f(x) + f(y)f(1) = f(1) + f(1)

# f(1) = 0 " f(1) = f(e) + f # f(e) + f = 0

(C) ff(x) = f(x), f>(x) = g (x)# g>(x) = f>>(x) = f(x)

h (x) = (f(x))2+ (g(x))2

# h>(x) = 2f(x) . f>(x) + 2g(x) . g>(x) = 2f(x) . g (x) + 2g(x) (f (x)) = 0

# h (x) = c, x %R # h (10) = h (5) = 9

(D) y = tan1(cot x) + cot1(tan x), < x < 8

= + . sec2 x

= 1 1 = 2Comprehension # 2 (6, 7, 8)

= = = at (1, 1)

& = = & =

For question 8

Slope of normal at (1, 1) = =

Equation of normal

y 1 = (x 1) & 5y 5 = 8x 8 & 8x 5y 3 = 0


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11_. y = x2

= 2x

= 2

again 2x = 1

2 + 2x = 0 &x = & = & $1

Statement-2 :

" = & = =

15. " x = 1/2 # 0 < x < 1

then tan1 x = , 0 < (x

0) = 21F(x

0), f>(x

0) = 4f(x

0), g(x

0) = 7g(x

0)

h>(x0) = kh(x

0)

F>(x) = f>(x).g(x).h(x) + f(x).g>(x).h(x) + f(x).h>(x).g(x)at x = x

0

21F(x0) = 4f(x

0)g(x

0)h(x

0) 7f(x

0).g(x

0).h(x

0) + kf(x

0)g(x

0).h(x

0)

21(f(x0).g(x

0).h(x

0)) = (4 7 + k) f(x

0).g(x

0).h(x

0)

&k = 24


1. We have x2+ y2= 1Differentiating w.r.t. x.2x + 2yy>= 0Again diff. w.r.t. x.

1 + y>y>+ yy>>= 0yy>>+ (y>)2+ 1 = 0


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2. p(x) = a0+ a

1x + a

2x2+ ........ + a

nxn

& p>(x) = a1+ 2a

2x + ......... + na

nxn 1

& p>(1) = a1+ 2a

2+ ........ + na

n.......(i)

Now | p(x) | )|ex 11| Jx *0 (given)& |p(1)| )|e01| = |1 1| = 0But |

p(1)

| *0 & p(1) = 0

Now for 1 < h < ', h $0, 1 + h > 0and |p(x)| )|ex 11| Jx *0

& |p(1 + h)| )|eh1| Jh > 1, h $0& |p(1 + h) p(1)| )|eh1| " p(1) = 0

& )

Taking limit as h G0, we get

) & |p>(1)| )1

& |a1+ 2a

2+ 3a

3+ ........ na

n| )1 {from (i)}

Hence proved

3. f : R GR, f(1) = 3, f>(1) = 6

y =

& !ny = !n & !ny = By L.H. Rule

& !ny = = = 2 & y = e2

4. At x = 0, !ny = 0 & y = 1

Also on differentiating w.r.t x on both sides, we get (1 + y>) = 2y + 2xy>

Putting x = 0, we get, = 2y..

& y>= 2y21 = 1.

5. Let P(x) = bx2+ ax + c b $0P(0) = 0 c = 0

P(1) = 1 a + b = 1

P(x) = (1 a) x2+ ax P>(x) = 2(1 a)x + a

Now P>(x) > 0 for x %[0, 1]P>(0) > 0 & P>(1) > 0a > 0 & 2(1 a) + a > 0

0 < a < 2

# S = {(1 a) x2+ ax; 0 < a < 2}

6. x sin y + y cos x = 8# x = 0, y = 8

x cos y + sin y + y (sin x) + cosx = 0


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= # y>(0) = 0

=

# = 8

7. " g(x) = f>(x) ..... (i)& g>(x) = f>>(x) = f(x) ..... (ii)

F(x) = +

& F>(x) = 2f(x/2) . f>(x/2) . + 2 g(x/2) . g>(x/2) . & F>(x) = + g g>

& F>(x) = f . f> f> f {Using (i) & (ii)}

& F>(x) = 0 & F(x) = constant & F(5) = F(10) = 5

8. =

= . =

9. g(x + 1) = log [f(x + 1)] = log [xf(x)] = log x + log [f(x)] = log x + g(x)

& g(x + 1) g(x) = log x

& g>>(x + 1) g>>(x) = & g>> g>> = 4

& g>> g>> = & g>> g>> =

Adding all, g>> g>> = 4

PART - I

2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)

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Transcript of 2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)