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8/13/2019 2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)
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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
Copyright reserved 2012-13.All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to only the enrolledstudent of RESONANCE. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only.
MATHEMATICSCLASS : XII
CONTENT
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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
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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
EXERCISE # 2PART - I
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)
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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
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SOLUTIONS (XII) # 11
(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.
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SOLUTIONS (XII) # 12
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
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SOLUTIONS (XII) # 13
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
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SOLUTIONS (XII) # 14
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}.
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SOLUTIONS (XII) # 20
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
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SOLUTIONS (XII) # 22
& 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
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SOLUTIONS (XII) # 24
(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
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SOLUTIONS (XII) # 28
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.
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SOLUTIONS (XII) # 29
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
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SOLUTIONS (XII) # 30
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
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SOLUTIONS (XII) # 31
(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
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SOLUTIONS (XII) # 32
(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
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SOLUTIONS (XII) # 33
(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}
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SOLUTIONS (XII) # 34
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
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SOLUTIONS (XII) # 35
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
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SOLUTIONS (XII) # 36
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
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SOLUTIONS (XII) # 39
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
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SOLUTIONS (XII) # 40
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
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SOLUTIONS (XII) # 41
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
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SOLUTIONS (XII) # 42
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
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SOLUTIONS (XII) # 43
(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
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SOLUTIONS (XII) # 44
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"
EXERCISE # 4PART - I
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
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SOLUTIONS (XII) # 46
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
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SOLUTIONS (XII) # 47
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
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SOLUTIONS (XII) # 48
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.
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SOLUTIONS (XII) # 49
& 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
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SOLUTIONS (XII) # 50
= 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!
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SOLUTIONS (XII) # 51
& 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
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SOLUTIONS (XII) # 56
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!
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SOLUTIONS (XII) # 60
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
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SOLUTIONS (XII) # 61
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) =
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SOLUTIONS (XII) # 75
(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) =
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SOLUTIONS (XII) # 76
" 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
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SOLUTIONS (XII) # 77
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
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SOLUTIONS (XII) # 78
= = = 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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SOLUTIONS (XII) # 79
METHOD OF DIFFERENTIATION
EXERCISE # 1PART - I
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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SOLUTIONS (XII) # 80
(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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SOLUTIONS (XII) # 84
again = exsin x + excos x
= exsin x + excos x + excos x exsin x
= 2v
similarly other options can be checked.
EXERCISE # 2PART - I
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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SOLUTIONS (XII) # 85
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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SOLUTIONS (XII) # 86
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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SOLUTIONS (XII) # 87
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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SOLUTIONS (XII) # 88
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
EXERCISE # 4PART - I
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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SOLUTIONS (XII) # 89
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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SOLUTIONS (XII) # 90
= # 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