limits and derivatives - · pdf filecos + − 9. x x 1 tan sin + 10. x 5 cosec x 11. ......
TRANSCRIPT
12
I PUC – MATHEMATICS
CHAPTER - 13
Limits and Derivatives
One mark question
I. Evaluate
16. 0→x
Lim 63
952
++
+−
xx
xx 7.
0→xLim
+
−
1
232x
x
17.
2
π→x
Lim π2tan3
2coscossin2
+
−+ xxx 8.
2
π→x
Lim x
xcos
18.
2
1→x
Lim x
xx
x −
−
−
+
22
22 9
2
π→x
Lim xsinx
19. 0→x
Lim 3
814
−
−
x
x 10.
0→θLim
θ
θθ 3sin4sin3 −
20. 0→x
Lim 1
15
+
+
x
x 11.
0→xLim
x
xo
tan
21. 0→x
Lim ax
ax
−
−
II. Differentiate the following w.r.t. ‘x’
1. x
xx 754−+
2. x
xx 132++
3. (2x +1) (3x – 5)
4. x2 cosx 5. 5 cotx – 3 cosecx 6. x
ecxsin2
2
cos3+
7. 242
tanx
x+
Two marks questions:
I. Evaluate :
1. 2→x
Lim 65
22
2
+−
−−
xx
xx 2.
5→xLim
25
15722
2
−
−−
x
xx
3. 0→θ
Lim θθ
θ
tan5sin
3sin2 2
4. 1→x
Lim 1
1
−
−q
p
x
x
5. 0→θ
Lim 23
2cos1
θ
θ− 6.
1→xLim
1
2 2
−
++
x
xx
7. 5→t
Lim 56
1252
3
+−
−
tt
t 8.
2→xLim
8
233
2
−
+−
x
xx
13
9. 0→y
Lim 2
33
53
)()(
yay
yaya
+
−−+ 10.
1→xLim
)1(
sin
+xx
xπ
11. 2−→x
Lim 32
85
3
+
+
x
x 12.
0→xLim
x
x
3cos1
5cos1
−
−
13. 0→x
Lim xx
xx
4
332
+
−−+ 14.
0→xLim
x
x 24 −+
15. 0→x
Lim x
xx 242−++
16. ax
Lim→
aaxx
ax
−
−
17. 0→x
Lim xx
xx
sin25
sin32
+
+ 18.
0→xLim
)2sin(tan
)4tan(sin
x
x
19. 0→x
Lim xx
xx
sin3
2tan
−
− 20.
0→xLim
xx
x
2tan
3sin 2
II.
1. If f(x) = x tanx then find f1
4
π
2. If f(x) = x5 + 4x
3 +
x
1then, find f
1(1)
3. f(x) = 3 sinx + 4tanx – secx, find f1(π)
4. f(x) = x2 - 2
1
x find f1(-1)
III. Differentiate the following w.r.t ‘x’
1. x
x
+
−
1
1 2.
2
sin
+x
x 3. x
xx cos
12
+ 4.
3
)4( −x
5.
+
xx
1
−
xx
1 6.
x
x
2cos1
2cos1
−
+ 7.
x
xtan
8. 4 cot x - xx
x
sin
4
cos
3
2
cos−+ 9.
x
x
tan1
sin
+ 10. x
5 cosec x
11. (3x2 – x) secx 12. secx cosec x
Three marks questions
I. Evaluate :
1. ax
Lim→
ax
ax
−
− sinsin 2.
2
πθ →
Lim 3
2
cos33cos
−
+
θπ
θθ
14
3.
4
π→x
Lim
x
x
−
−
4
tan1(
π 4
0→xLim
x
xaxa )sin()sin( −−+
5. 0→t
Lim t
t 125)5( 3−+
6. 2→x
Lim 22
83
−
−
xx
x
7. π→x
Lim x
x2
3
tan
cos1+ 8.
1→xLim
−−
−31
3
1
1
xx
9. 1→x
Lim
−+
− xx 1
1
1
22
10. 3→x
Lim 243
92
+−−
−
xx
x
11.
2
π→x
Lim (secθ - tanθ) 12.
2
π→x
Lim 18
1163
4
−
−
x
x
13. 0→x
Lim x
xaxa )cos()cos( −−+ 14.
2
π→x
Lim 2)2(
2cos1
θπ
θ
−
+
15. 0→x
Lim x
xx 2121 −−+ 16.
0→xLim
6
24 4sin)2tan(
x
xx
17.
3
π→x
Lim
3
cos3sin
π−
−
x
xx 18. f(x)
>−
<+
138
114
xifx
xifx , find
1→xLim f(x)
II. Differentiate the following w.r.t ‘x’ from first principles.
1. x
1 2. x x 3. ax2 + bx + c 4. tan3x
5. x2sinx 6. 1
1
+
−
x
x
III. Differentiate the following w.r.t. ‘x’
1. x
x
sin23
cos32
−
+ 2.
64
235
5
++
+−
xx
xx 3.
x
x
−
+
2
2 4. x
2(3x + 2) cosec x
Five marks questions
1. Evaluate 0→x
Lim x
xecx cotcos −
2. If f(x) =
>+
<<+
<+
6711
6512
51102
xwhenx
xwhenx
xwhenx
Evaluate 5→x
Lim f(x) and 6→x
Lim f(x)
15
3. f(x) =
=
≠
−
−
4
4
4
cossin
π
π
π
xwhenK
xwhen
x
xx
and if
4
π→x
Lim f(x) = f
4
π, then find the value of K.
4. f(x) =
>+
=
<+−
15
13
1432
xwhenbx
xwhen
xwhenxax
find the values of ‘a’ and ‘b’ if 1→x
Lim f(x) = f(1)
5. Evaluate 0→x
Lim x
x || if its exists.
6. f(x) =
>−
−
<−
0)1(
)1(4
02
4cos1
2
2
xforx
x
xforx
x
Evaluate 0→x
Lim f(x)
7. Show that f(x) = ||1 x
x
+ is differentiable at x = 0
8. Differentiate )cos1(
sin)23( 2
xx
xx
+
+ w.r.t. ‘x’
9. If y = 1sectan
1sectan
+−
−+
xx
xx prove that
dx
dy = secx (secx + tanx)
10. Differentiate xcos w.r.t. ‘x’ from first principles.
11. Differentiate )3(
cot2
4
+x
xx w.r.t. ‘x’
16
Limits and Derivatives Solutions to one mark questions
1. 0→x
Lim 63
952
++
+−
xx
xx
= 2
3
60
90=
+
+
2.
2
π→x
Lim π2tan3
2coscossin2
+
−+ xxx
= 03
cos02
+
−+ π
= 3
)1(2 −−
= 1
3.
2
1→x
Lim x
xx
x −
−
−
+
22
22
=
2
12
2
12
−
+
=
2
12
2
12
−
+
= 1
3
= 3
4. 0→x
Lim 3
814
−
−
x
x
= 3
lim→x
3
344
−
−
x
x
= 4(3)4-1
= 4 x 33
= 108
5. 0→x
Lim 1
15
+
+
x
x
= 1
lim−→x
)1(
)1( 55
−−
−−
x
x
= 5 (-1)5-1
= 5 x 1
= 5
6. 0→x
Lim ax
ax
−
−
= ax→
lim ax
ax
−
− 21
21
17
= 21
1
2
1
2
1
2
1 −
−
= aa
= a2
1
7. 0→x
Lim
+
−
1
232x
x
=
3
1
2
−
= -8
8.
2
π→x
Lim x
xcos
=
2
2cos
π
π
=
2
0
π
= 0
9.
2
π→x
Lim xsinx
= 2
π sin
2
π
= 2
π x 1
= 2
π
10. 0→θ
Lim θ
θθ 3sin4sin3 −
= 0
lim→θ
θ
θ3sin
= 0
lim→θ
33
3sin×
θ
θ
= 1 × 3
= 3
11. 0→x
Lim x
x otan
= 0
lim→x
x
x
180tan
π
= 0
lim→x
180
180
180tan
π
π
π
×
x
x
18
= 1 × 180
π
= 180
π
II.
1) Y = x
xx 754−+
Y = xx
x
x
x 754
−+
Y = x3 + 5 - x
7
Diff. w.r.t ‘x’
dx
dy = 3x
2 + 0 – 7
−
2
1
x
= 3x2 +
2
7
x
2) Y = x
xx 132++
Y = xx
x
x
x 132
++
Y = 21
23
3−
++ xxx
Diff . w.r.t ‘x’
dx
dy=
12
1
12
3
2
1
2
3
2
3
+
−
−+
xx
x
= 2
3
2
1
2
3
2
3
xx
x−+
3) Y = (2x + 1) (3x – 5)
Diff w.r.t ‘x’
dx
dy=(2x +1)
dx
d (3x-5) + (3x – 5)
dx
d (2x+1)
= (2x+1) (3) + (3x – 5) (2)
= 6x + 3 + 6x – 10
= 12x – 7
4) Y = x2 cosx
dx
dy = x
2 dx
d(cosx) + cosx
dx
d(x
2)
= -x2sinx + 2x cosx
5) Y = 5 cot x – 3 cosecx
dx
dy= - 5 cosec
2 x + 3 cosec x cot x
19
6) Y = 2
3cosec x + 2 sin x
dx
dy = -
2
3cosec x cot x + 2 cos x
7) Y = 2
tan x + 4x
2
dx
dy =
2
sec2 x + 8 x
Solutions to two marks questions
1. = 2
lim→x
)3)(2(
)1)(2(
−−
+−
xx
xx
= 1
3
)32(
)12(
−=
−
+
= - 3
2. = 5
lim→x
)5)(5(
)5)(32(
−+
−+
xx
xx
= 55
3)5(2
+
+
= 10
13
3. 0
lim→θ
θθ
θθ
θ
θ
θθ
θ
×
××
×
tan5
5
5sin
93
3sin2 2
2
= 151
912 2
××
×× =
5
18
4. = 1
lim→x
1
1
1
1
−
−
−
−
x
x
x
x
q
p
= 1
1
)1.(
)1.(−
−
q
p
q
p =
q
p
5. = 0
lim→θ
2
2
3
sin2
θ
θ
= 0
lim→θ
2sin
3
2
θ
θ
= 213
2×
20
= 3
2
6. 1
lim→x
22
22
1
22
2
22
+++
+++×
−
−++
xx
xx
x
xx
= 1
lim→x
( )
]22)[1(
22
2
22
2
+++−
−++
xxx
xx
= 1
lim→x
]22)[1(
42
2
2
+++−
−++
xxx
xx
= 1
lim→x
]22)[1(
2
2
2
+++−
−+
xxx
xx
= 1
lim→x
]22)[1(
)2)(1(
2+++−
+−
xxx
xx
= 22
3
2112
21
+=
+++
+
= 4
3
7. = 5
lim→t
)1)(5(
)55)(5( 22
−−
++−
tt
ttt
= 15
252525
−
++
= 4
75
8. = 2
lim→x
)42)(2(
)1)(2(2
++−
−−
xxx
xx
= 444
12
++
−
= 12
1
9. =0
lim→y ]53[
])())(())][([ 222
yay
yayayayayaya
+
−+−+++−−+
= 0
lim→y
]53[
])())(()[(2 22
yay
yayayayay
+
−+−+++
= 03
]))(([2 22
+
++
a
aaaa
= a
a
3
)3(2 2
= 2a
10. = 1
lim→x
1
sin
+×
xx
x π
π
π
21
= 1 × 211
ππ=
+
11. = 2
lim−→x
)2(
)2(
)2(
)2(
55
33
−−
−−
−
−−
x
x
x
x
= 15
13
)2(5
)2(3−
−
−
− =
165
43
×
×
= 20
3
12. = 0
lim→x
2
32sin2
2
5sin2 2
x
x
= [∴ 0
lim→x
2
2
2
2
sin
sin
n
m
nx
mx= ]
= 9
25
2
3
2
5
2
2
=
13. = 0
lim→x
xx
xx
xx
xx
−++
−++×
+
−−+
33
33
4
332
= 0
lim→x
]33)[4(
)3()3(2
xxxx
xx
−+++
−−+
= 0
lim→x
]33)[4(
2
xxxx
x
−+++
= )32(4
2
]33[4
2=
+
= 34
1
14. = 0
lim→x
)24(
)24)(24(
++
++−+
xx
xx
= 0
lim→x
)24(
2)4( 22
++
−+
xx
x
= 0
lim→x
)24( ++ xx
x
= 204
1
++
22
= 4
1
15. = 0
lim→x
]24[
2)4(
2
222
+++
−++
xxx
xx
= 0
lim→x
]24[ 2
2
+++
+
xxx
xx
= 0
lim→x
24[
)1(
2+++
+
xxx
xx
= 240
1
++ =
4
1
16. = ax→
lim 2
32
3
ax
ax
−
−
= 2
112
33
2
][2
3
1
aa
=−
= a3
2
17. = 0
lim→x
]sin
25[
]sin
32[
x
xx
x
xx
+
+
= 125
132
×+
×+
= 7
5
18. 0
lim→x
)2sin(tan
)4tan(sin
x
x
= 0
lim→x
xx
x
x
x
xx
x
x
x
22
2tan
2tan
)2sin(tan
44
4sin
4sin
)4tan(sin
××
××
= =××
××
211
4112
19. = 0
lim→x
−
−
x
xx
x
xx
sin3
12tan
23
= 0
lim→x
−
−×
x
xx
x
sin3
122
2tan
= 13
121
−
−× =
2
1
20. = 0
lim→x
xx
x
2tan
3sin 2
= 0
lim→x
xx
xx
xx
x
22
2tan
93
3sin 2
2
×
×
= 0
lim→x
21
91
×
×
= 2
9
II.
1. f(x) = x tan x
f1(x) = x )(tan x
dx
d+ tanx )(x
dx
d
= x sec2 x + tanx
f1
4
π =
2
)2(4
π + 1 =
2
π+ 1
2. f(x) = x5 + 4x
3 +
x
1
f1(x) = 5x
4 + 12x
2 -
2
1
x
f1(1) = 5 + 12 – 1
= 16
3. f(x) = 3sin x + 4 tan x – sec x
f1(x) = 3 cosx + 4 sec2x – secx tanx
f1(π) = 3cos π + 4sec
2 π - secπ tanπ
= 3 (-1) + 4(-1)2 – 0
= - 3 + 4
= -1
4. f(x) = x2 -
2
1
x
f1(x) = 2x -
−3
2
x
= 2x + 3
2
x
f1(-1) = 2(-1) +
3)1(
2
−
24
= -2 – 2
= - 4
III.
1. y = x
x
+
−
1
1
Diff. w.r.t. ‘x’
2)1(
)1()1()1()1(
x
xdx
dxx
dx
dx
dx
dy
+
+−−−+
=
= 2)1(
)1)(1()1)(1(
x
xx
+
−−−+
= 2)1(
11
x
xx
+
+−−−
∴ dx
dy =
2)1(
2
x+
−
2. y = 2
sin
+x
x
diff w.r.t. ‘x’
dx
dy =
2)2(
)2(sin)(sin)2(
+
+−+
x
xdx
dxx
dx
dx
dx
dy =
2)2(
sincos)2(
+
−+
x
xxx
3. y = xx
x cos1
2
+
y = xx
x cos21
2
2
++
diff w.r.t. ‘x’
dx
dy =
++ 2
12
2
xx (- sinx) + cos x
−
3
22
xx
= - sinx
++ 2
12
2
xx + 2 cos x
−
3
1
xx
4. y = 3
cot)4( xx −
diff w.r.t ‘s’
dx
dy =
3
1
−+− )4(cot)(cot)4( x
dx
dxx
dx
dx
= 3
1 [ ])1(cotcos)4( 2 xxecx +−−
25
= 3
cotcos)4( 2xxecx +−
5. y =
+
xx
1
−
xx
1
y = x x - x . x
1+
x
1.x -
xx
1
y = 2
32
3 111
x
x −+−
diff w.r.t. ‘x’
dx
dy =
12
3
12
3
2
3
1
2
3
+
−
−
x
x
= 2
5
3
2
2
3
x
x−
6. y = x
x
2cos1
2cos1
−
+
y = x
x2
2
sin2
cos2
y = cotx
∴dx
dy = - cosec
2x
7. y = x
xtan
dx
dy =
2
2 )1(tan).(sec
x
xxx −
= 2
2 tansec
x
xxx −
8. y = 4 cotx - xx
x
sin
4
cos
3
2
cos−+
y = 4 cot x - ecxxx
cos4sec32
cos−+
∴dx
dy = - 4cosec
2x +
2
sin x+ secx tanx + 4 cosecx cotx
9. y = (3x2 – x) secx
dx
dy = (3x
2 – x) secx tanx + secx(6x – 1)
= secx [x(3x-1) tanx + 6x – 1]
10. y = x
x
tan1
sin
+
dx
dy =
2
2
)tan1(
)(secsincos)tan1(
x
xxxx
+
−+
26
= 2
2
)tan1(
cos
1sincostancos
x
xxxxx
+
−+
dx
dy =
2)tan1(
tansecsincos
x
xxxx
+
−+
11. y = secx cosecx
dx
dy = secx (-cosecx cotx) + cosecx(secx tanx)
= - x
x
xxx
x
xx cos
sin
cos
1
sin
1
sin
cos
sin
1
cos
1××+××
= - cosec2
x + sec2
x
Solutions to three marks questions I.
1. ax→
lim ax
ax
−
− sinsin
= ax→
lim )(
2sin
2cos2
ax
axax
−
−
+
= ax→
lim 2 cos 2
1
2
)(
2sin
2×
−
−
×
+
ax
ax
ax
= cos 12
×
+ aa
= cos a
2.
2
limπ
θ →
3
2
cos33cos
−
+
θπ
θθ
=
2
limπ
θ →
3
2
cos3cos33cos4
−
+−
θπ
θθθ
=
2
limπ
θ →
3
3
2
cos4
−θ
π
θ put θ
π−
2 = t As
2
πθ → = t → 0
=
2
limπ
θ →
3
3
)2
cos(4
t
t
−
π
= 0
lim→t
3
3sin4
t
t
= 4 x 1
27
= 4
3.
4
limπ
→x
−
−
x
x
4
tan1
π
=
4
limπ
→x
−
+×
+
−
x
x
x
x
4
tan1
tan1
tan1
π
=
4
limπ
→x
−
+
−
x
xx
4
)tan1(4
tan
π
π
= 1 ×
+
4tan1
π ∴ As
4
π→x = 0
4→− x
π
= 1 + 1
= 2
4. 0
lim→x
x
xaxa )sin()sin( −−+
= 0
lim→x
x
xa )sin()cos(2
= 2 cos a × 1
= 2cos a
5. 0
lim→x
x
xaxa )cos()cos( −−+
= 0
lim→x
- x
xaxaxaxa
+−+
−++
2sin
2sin2
= 0
lim→x
x
xa sinsin2 ⋅−
= - 2sina . 0
lim→x
x
xsin
= -2 sinax1 = -2 sina
6.
2
limπ
→x
2)2(
2cos1
θπ
θ
−
+
=
2
limπ
→x
2
2
)2(
cos2
θπ
θ
−
=
2
limπ
→x
2
2
2
22
2sin2
−
−⋅
θπ
θπ
28
= 2
1
2
limπ
→x
2
2
22
2sin
−
−
θπ
θπ
= 2
1 × 1 =
2
1
7. 0
lim→x
x
xx 2121 −−+
= 0
lim→x
]2121[
)21()21(
xxx
xx
−++
−−+
= 0
lim→x
xx 2121
4
−++
= 22
4=
8. 0
lim→x
6
24 )4(sin)2tan(
x
xx
= 0
lim→x
2. 4
4
2
2tan
x
x ×
2
44
4sin
×
x
x
= 2 × 16 = 32
9.
3
limπ
→x
3
cos3sin
π−
−
x
xx
=
2
limπ
→x
3
cos2
3sin
2
1.2
π−
−
x
xx
=
2
limπ
→x
2.
)3
(
3sin
π
π
−
−
x
x
= 2 × 1 = 2
10. f(x) = 4x + 1 if x< 1
= 8x – 3 if x > 1
LHL = −→1
limx
f(x) = −→1
limx
(4x + 1) = 5
RHL = +→1
limx
f(x) = +→1
limx
(8x – 3) = 5
As LHL = RHL . Limit of the function at x = 1 exists and 1
lim→x
f(x) = 5
29
11. 0→t
Lim t
t 125)5( 3−+
= 5→x
Lim 5
533
−
−
x
x put 5 + t = x
As t → 0 = x → 5
= 3(5)3-1
= 3 × 25
= 75
12. 2→x
Lim 22
83
−
−
xx
x
= 2→x
Lim
2
2
2
2
23
23
33
−
−
−
−
x
x
x
x
= 1
23
13
)2(2
3
)2(3
−
−
= 2
1
2
42 ⋅ = 4 2
13. π→x
Lim x
x2
3
tan
cos1+
= π→x
Lim
x
x
xxx
2
2
2
cos
sin
]coscos1)[cos1( +++
= π→x
Lim xxx
xxx 22
cos)cos1)(cos1(
)coscos1)(cos1(×
−+
+++
= π
ππ
cos1
)coscos1( 2
−
++
= )1(1
)1)(111( 2
−−
−+−
= 2
1
14. 1→x
Lim 31
3
1
1
xx −−
−
= 1→x
Lim x−1
1
++−
21
31
xx
= 1→x
Lim x−1
1
++
−+21
)1)(2(
xx
xx
= 1→x
Lim 21
)2(
xx
x
++
+− =
3
3− = -1
15. 1→x
Lim
−+
− xx 1
1
1
22
30
= 1→x
Lim
−+
+− xxx 1
1
)1)(1(
2
= 1→x
Lim
−1(
1
x
−
+1
1
2
x
= 1→x
Lim 1
1
)1(
1
+
−×
− x
x
x
= 1→x
Lim 2
11
1−=
+
−
x
16. 3→x
Lim 243
92
+−−
−
xx
x
= 3→x
Lim )2()43(
]243)[3)(3(
+−−
++−−+
xx
xxxx
= 3→x
Lim )3(2
)243)(3)(3(
−
++−−+
x
xxxx
= 2
526 ×
= 6 5
17.
2
πθ →
Lim (secθ - tanθ)
=
2
πθ →
Lim
−
θ
θ
θ cos
sin
cos
1
=
2
πθ →
Lim θ
θ
θ
θ
sin1
sin1
cos
sin1
+
+×
−
=
2
πθ →
Lim )sin1(cos
sin1 2
θθ
θ
+
−
=
2
πθ →
Lim θ
θ
sin1
cos
+
= 1
0 = 0
18. 2
1→x
Lim 18
1163
4
−
−
x
x
= 2
1→x
Lim )124)(12(
)14)(14(2
22
++−
+−
xxx
xx
= 2
1→x
Lim )124)(12(
)14)(12)(12(2
2
++−
++−
xxx
xxx
= 3
4111
22=
++
×
II.
31
1. f(x) = x
1
f1(x) =
0→∆xLim
x
xfxxf
∆
−∆+ )()(
= 0→∆x
Lim x
xxx
∆
−∆+
11
= 0→∆x
Lim xxxx
xxx
∆∆+
∆+−
)(
)(
= 0→∆x
Lim xxxx
x
∆∆+
∆−
)(
= - )0(
1
+xx
= - 2
1
x
2. f(x) = x x
f(x) = 23
x
f1(x) =
0→∆xLim
x
xfxxf
∆
−∆+ )()(
f1(x) = 0→∆x
Lim xxx
xxx
−∆+
−∆+
)(
)( 23
23
= 1
23
)(2
3 −x
f1(x) = 2
1
2
3x
f1(x) =
2
3 x
3. f(x) = ax2 + bx + c
f(x) = 0→∆x
Lim x
xfxxf
∆
−∆+ )()(
= 0→∆x
Lim x
cbxaxcxxbxxa
∆
++−+∆++∆+ }{)()( 22
= 0→∆x
Lim x
bxaxxxbxxa
∆
−−∆++∆+22 )()(
= 0→∆x
Lim x
bxaxxbbxxaxxaax
∆
−−∆++∆+∆+22 2)(2
= 0→∆x
Lim x
baxxax
∆
++∆∆ }2){(
= a (0) + 2ax + b
= 2ax+b
4. f(x) = tan3x
32
f1(x) =
0→hLim
h
xfhxf )()( −+
= 0→h
Lim h
xhx 3tan)33tan( −+
= 0→h
Lim h
xhxxhx }3tan)33tan(1){333tan( ++−+
= 0→h
Lim 33
}3tan)33tan(1{3tan×
++
h
hxh
= 1 × {1 + tan3x tan3x} × 3
= 3 (1 + tan23x)
f1(x) = 3 sec
23x
5. f(x) = x2sinx
f1(x) =
0→hLim
h
xfhxf )()( −+
= 0→h
Lim h
xxhxhx sin)sin()( 22−++
= 0→h
Lim h
xxhxxhhxhhxx sin)sin(2)sin()sin( 222−+++++
= 0→h
Lim h
hxxhxhhxhxx )]sin(2)sin([]sin)[sin(2++++−+
= 0→h
Lim
++++
−+
++×
h
hxxhxhh
h
xhxxhxx
)sin(2)sin([2sin
2
)cos22
= 0→h
Lim xxh
hhxx
sin2
22
2sin
2
2cos2
2
+
×
+
= x2 cosx × 1 + 2x sinx
= x2 cosx + 2 x sinx
6. f(x) = 1
1
+
−
x
x = 1 -
1
2
+x
f1(x) = 0→h
Lim h
xfhxf )()( −+
= 0→h
Lim h
xhx
+−−
++−
1
21
1
21
= 0→h
Lim h
hxx 1
2
1
2
++−
+
= 0→h
Lim )1)(1(
)]1(1[2
+++
+−++
hxxh
xhx
33
= 0→h
Lim )1)(1(
2
+++ hxxh
h
= 2)1(
2
)1)(1(
2
+=
++ xxx
III.
1. y = x
x
sin23
cos32
−
+
Diff w.r.t. ‘x’
dx
dy =
2)sin23(
)sin23()cos32()cos32()sin23(
x
xdx
dxx
dx
dx
−
−+−+−
= 2)sin23(
)cos2)(cos32()sin3)(sin23(
x
xsxxx
−
−+−−−
= 2
22
)sin23(
cos6cos4sin6sin9
x
xxxx
−
+++−
= dx
dy =
2)sin23(
sin9cos46
x
xx
−
−+
2. y = 64
235
5
++
+−
xx
xx
Diff w.r.t. ‘x’
dx
dy =
25
4545
)64(
)45)(23()35)(64(
++
++−−−++
xx
xxxxxx
25
45594559
)64(
8101215451830122035
++
−−++−−−+−+−
xx
xxxxxxxxxx
= 25
45
)64(
262028
++
−+
xx
xx
= 25
45
)64(
]131014[2
++
−+
xx
xx
3. y = x
x
−
+
2
2
dx
dy =
2)2(
2
1)2(
2
1)2(
x
xx
xx
−
−+−
−
= 2)2(
]22[2
1
x
xxx
−
++−
= ( )2
222
22
−x
34
= 2)2(2
2
x−
4. y = x2(3x +2) cosecx
dx
dy = x2 (3x+2) (-cosecx) + x2 cosec x(3) + (3x+2) cosecx (2x)
= - x2 (3x+2) cosecx + 3x2 cosec x+ 2x(3x+2) cosecx
Solution to five marks questions
1. 0→x
Lim x
xecx cotcos −
= 0→x
Lim x
x
x
x sin
cos
sin
1−
= 0→x
Lim xx
x
sin
cos1−
= 0→x
Lim xx
x
sin
cos1− ×
x
x
cos1
cos1
+
+
= 0→x
Lim )cos1(sin
cos1 2
xxx
x
+
−
= 0→x
Lim )cos1(sin
sin 2
xxx
x
+
= 0→x
Lim )cos1(
sin
xx
x
+
= ocos1
1
+ (∴
0→xLim
x
xsin = 1)
= 2
1
2. LHL = −
→5x
Lim f(x) = −
→5x
Lim 10x+1 = 10 (5) + 1 = 51
RHL = +
→5x
Lim f(x) = +
→5x
Lim (2x2+1) = 2(5)2 + 1 = 51
LHL = RHL ∴ 5→X
Lim f(x) = 51
LHL = −
→6X
Lim f(x) = −
→6X
Lim 2x2 + 1 = 2(6)
2+ 1 = 72 + 1 = 73
RHL = +
→6X
Lim f(x) = +
→6X
Lim (11x + 7) = [11(6) +7] = 73
LHL = RHL
∴ −
→6X
Lim f(x) = 73
3. Given
4
π→X
Lim f(x) = f
4
π
35
→
4
π→X
Lim
−
−
x
xx
4
cossin
π = K
→
4
π→X
Lim
−
−
x
xx
4
cos2
1sin
2
12
π = K
→
4
π→X
Lim
−
−
x
x
4
4sin2
π
π
= k
→ 2 × 1 = K
∴ K = 2
4. Given −
→1X
Lim f(x) = +
→1X
Lim f(x) = f(1)
Consider −
→1X
Lim f(x) = f(1)
a(1) – 3(1) + 4 = 3
� a = 2
also +
→1X
Lim f(x) = f(1)
b(1) + 5 = 3
b = -2
5. | x | =
<−
=
>
0
00
0
xifx
xif
xifx
RHL = +
→0X
Lt x
x || =
+→0X
Lt x
x = 1
LHL = −
→0X
Lt x
x || =
−→0X
Lt x
x= = 1
LHL ≠ RHL
∴ 0→X
Lt x
x || doesn’t exist
6. LHL = −
→0X
Lim f(x) = −
→0X
Lim 22
4cos1
x
x− =
0→X
Lim 2
2
2
sin2
x
x
= 0→X
Lim 42
2sin2
×
x
x
= 1 × 4 = 4
RHL = +
→0X
Lim f(x) = +
→0X
Lim )1(
)1(4 2
−
−
X
x
= +
→0X
Lim )1(
)1)(1(4
−
−+
x
xx = 4 (0 +1)
36
∴ 0→X
Lim f(x) = 4
7. f(x) =
<−
=
>+
=+
01
00
01
||1xif
x
x
xif
xifx
x
x
x
Rf1(0) =
+→0X
Lt 0
)0()(
−
−
x
fxf
= +
→0X
Lt x
x
x0
1−
+ = 1
Lf1(0) =
−→0X
Lt 0
)0()(
−
−
x
fxf
= −
→0X
Lt x
x
x
−1
= −
→0X
Lt x−1
1
= 1
Rf1(0) = Lf
1(0) ∴ f
1(0) = 1
8. f(x) = xx
xx
cos1
sin)23( 2
+
+
f1(x) =
( )2
22
)cos1(
sin)23()cos1()cos1(.sin)23(
xx
xxdx
dxxxx
dx
dxx
+
++−++
= ( )
2
22
)cos1(
sin6cos)23()cos1(]cossin[sin)23(
xx
xxxxxxxxxxx
+
+++−+−+
9. y = 1sectan
1sectan
+−
−+
xx
xx
y = )1sec(tan
)tan)(sectan(sec)sec(tan
+−
−+−+
xx
xxxxxx
= )1sec(tan
]tansec1)[tan(sec
+−
+−+
xx
xxxx
Y = secs + tanx
Diff w.r.t. ‘x’
dx
dy secx tanx + sec
2 x
= secx [tanx + secx]
10. y = xcos
dx
dy =
0→hLim
h
xfhxf )()( −+
= 0→h
Lim h
xhx cos)cos( −+
37
= 0→h
Lim ( )[ ]
[ ]xhxh
xhxxhx
cos)cos(
cos)cos(cos)cos(
++
++−+
= 0→h
Lim[ ]xhxh
xhx
cos)cos(
cos)cos(
++
−+
= 0→h
Lim [ ]xhxh
hhx
cos)cos(
2sin
2
2sin2
++
+−
= 0→h
Lim
[ ] 22
cos)cos(
2sin
2
2sin2
×
++
+−
hxhx
hhx
= [ ]xx
x
coscos
1sin
+
×
= x
x
cos2
sin
11. y = 3(
cot2
4
+x
xx
diff w.r.t ‘x’
dx
dy =
22
2442
)3(
)3(cot)cot()3(
+
+−+
x
xdx
dxxxx
dx
dx
= 22
43242
)3(
)2(cot]cot4cos)[3(
+
−+−+
x
xxxxxxecxx
dx
dy =
22
5223
)3(
cot2]cot4cos)[3(
+
−+−+
x
xxxxecxxx
***********