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Byju's Answer
Standard XII
Mathematics
First Derivative Test for Local Maximum
Using differe...
Question
Using differentials, find the approximate values of the following:
(i)
25
.
02
(ii)
0
.
009
1
3
(iii)
0
.
007
1
3
(iv)
401
(v)
15
1
4
(vi)
255
1
4
(vii)
1
(
2
.
002
)
2
(viii) log
e
4.04, it being given that log
10
4 = 0.6021 and log
10
e
= 0.4343
(ix) log
e
10.02, it being given that log
e
10 = 2.3026
(x) log
10
10.1, it being given that log
10
e
= 0.4343
(xi) cos 61°, it being given that sin60° = 0.86603 and 1° = 0.01745 radian
(xii)
1
25
.
1
(xiii)
sin
22
14
(xiv)
cos
11
π
36
(xv)
80
1
4
(xvi)
29
1
3
(xvii)
66
1
3
(xviii)
26
[CBSE 2000]
(xix)
37
[CBSE 2000]
(xx)
0
.
48
[CBSE 2002C]
(xxi)
82
1
4
[CBSE 2005]
(xxii)
17
81
1
4
(xxiii)
33
1
5
(xxiv)
36
.
6
(xxv)
25
1
3
(xxvi)
49
.
5
[CBSE 2012]
(xxvii)
3
.
968
3
2
[CBSE 2014]
(xxviii)
1
.
999
5
[NCERT EXEMPLAR]
(xxix)
0
.
082
[NCERT EXEMPLAR]
Open in App
Solution
(i)
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
25
x
+
∆
x
=
25
.
02
Then
,
∆
x
=
0
.
02
For
x
=
25
,
y
=
25
=
5
Let
:
d
x
=
∆
x
=
0
.
02
Now
,
y
=
x
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
25
=
1
10
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
10
×
0
.
02
=
0
.
002
⇒
∆
y
=
0
.
002
∴
25
.
02
=
y
+
∆
y
=
5
.
002
(ii)
Consider
the
function
y
=
f
x
=
x
3
.
Let
:
x
=
0
.
008
x
+
∆
x
=
0
.
009
Then
,
∆
x
=
0
.
001
For
x
=
0
.
008
,
y
=
0
.
008
=
0
.
2
Let
:
d
x
=
∆
x
=
0
.
001
Now
,
y
=
x
3
⇒
d
y
d
x
=
1
3
x
2
3
⇒
d
y
d
x
x
=
0
.
008
=
1
3
×
0
.
04
=
1
0
.
12
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
0
.
12
×
0
.
001
=
1
120
⇒
∆
y
=
1
120
=
0
.
008333
∴
0
.
009
1
3
=
y
+
∆
y
=
0
.
208333
(iii)
Consider
the
function
y
=
f
x
=
x
.
3
Let
:
x
=
0
.
008
x
+
∆
x
=
0
.
007
Then
,
∆
x
=
-
0
.
001
For
x
=
0
.
008
,
y
=
0
.
008
=
0
.
2
Let
:
d
x
=
∆
x
=
-
0
.
001
Now
,
y
=
x
3
⇒
d
y
d
x
=
1
3
x
2
3
⇒
d
y
d
x
x
=
0
.
008
=
1
3
×
0
.
04
=
1
0
.
12
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
0
.
12
×
0
.
001
=
1
120
⇒
∆
y
=
1
120
=
0
.
008333
∴
0
.
007
1
3
=
y
+
∆
y
=
0
.
191667
(iv).
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
400
x
+
∆
x
=
401
Then
,
∆
x
=
1
For
x
=
400
,
y
=
400
=
20
Let
:
d
x
=
∆
x
=
1
Now
,
y
=
x
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
400
=
1
40
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
40
×
1
=
1
40
⇒
∆
y
=
1
40
=
0
.
025
∴
401
=
y
+
∆
y
=
20
.
025
(v)
Consider
the
function
y
=
f
x
=
x
1
4
.
Let
:
x
=
16
x
+
∆
x
=
15
Then
,
∆
x
=
-
1
For
x
=
16
,
y
=
16
1
4
=
2
Let
:
d
x
=
∆
x
=
-
1
Now
,
y
=
x
1
4
⇒
d
y
d
x
=
1
4
x
3
4
⇒
d
y
d
x
x
=
16
=
1
32
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
32
×
-
1
=
-
1
32
⇒
∆
y
=
-
1
32
=
-
0
.
03125
∴
15
1
4
=
y
+
∆
y
=
1
.
96875
(vi)
Consider
the
function
y
=
f
x
=
x
1
4
.
Let
:
x
=
256
x
+
∆
x
=
255
Then
,
∆
x
=
-
1
For
x
=
256
,
y
=
256
1
4
=
4
Let
:
d
x
=
∆
x
=
-
1
Now
,
y
=
x
1
4
⇒
d
y
d
x
=
1
4
x
3
4
⇒
d
y
d
x
x
=
256
=
1
256
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
256
×
-
1
=
-
1
256
⇒
∆
y
=
-
1
256
=
-
0
.
003906
∴
255
1
4
=
y
+
∆
y
=
3
.
99609
≈
3
.
9961
(vii)
Consider
the
function
y
=
f
x
=
1
x
2
.
Let
:
x
=
2
x
+
∆
x
=
2
.
002
Then
,
∆
x
=
-
0
.
002
For
x
=
2
,
y
=
1
2
2
=
1
4
Let
:
d
x
=
∆
x
=
0
.
002
Now
,
y
=
1
x
2
⇒
d
y
d
x
=
2
x
3
⇒
d
y
d
x
x
=
2
=
1
4
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
4
×
-
0
.
002
=
-
0
.
0005
⇒
∆
y
=
-
0
.
0005
∴
1
2
.
002
2
=
y
+
∆
y
=
0
.
2495
(viii)
Consider
the
function
y
=
f
x
=
log
e
x
.
Let
:
x
=
4
x
+
∆
x
=
4
.
04
Then
,
∆
x
=
0
.
04
For
x
=
4
,
y
=
log
e
4
=
log
10
4
log
10
e
=
0
.
6021
0
.
4343
=
1
.
386368
Let
:
d
x
=
∆
x
=
0
.
04
Now
,
y
=
log
e
x
⇒
d
y
d
x
=
1
x
⇒
d
y
d
x
x
=
4
=
1
4
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
4
×
0
.
04
=
0
.
01
⇒
∆
y
=
0
.
01
∴
log
e
4
.
04
=
y
+
∆
y
=
1
.
396368
(ix)
Consider
the
function
y
=
f
x
=
log
e
x
.
Let
:
x
=
10
x
+
∆
x
=
10
.
02
Then
,
∆
x
=
0
.
02
For
x
=
,
y
=
log
e
10
=
2
.
3026
Let
:
d
x
=
∆
x
=
0
.
02
Now
,
y
=
log
e
x
⇒
d
y
d
x
=
1
x
⇒
d
y
d
x
x
=
10
=
1
10
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
10
×
0
.
02
=
0
.
002
⇒
∆
y
=
0
.
002
∴
log
e
10
.
02
=
y
+
∆
y
=
2
.
3046
(x)
Consider
the
function
y
=
f
x
=
log
10
x
.
Let
:
x
=
10
x
+
∆
x
=
10
.
1
Then
,
∆
x
=
0
.
1
For
x
=
,
y
=
log
10
10
=
1
Let
:
d
x
=
∆
x
=
0
.
1
Now
,
y
=
log
10
x
=
log
e
x
log
e
10
⇒
d
y
d
x
=
1
2
.
3025
x
⇒
d
y
d
x
x
=
10
=
0
.
04343
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
0
.
04343
×
0
.
1
=
0
.
004343
⇒
∆
y
=
0
.
004343
∴
log
10
10
.
1
=
y
+
∆
y
=
1
.
004343
(xi)
Consider
the
function
y
=
f
x
=
cos
x
°
.
Let
:
x
=
60
°
x
+
∆
x
=
61
°
Then
,
∆
x
=
1
°
=
0
.
01745
For
x
=
60
°
,
y
=
cos
60
°
=
0
.
5
Let
:
d
x
=
∆
x
=
0
.
01745
Now
,
y
=
cos
x
⇒
d
y
d
x
=
-
sin
x
⇒
d
y
d
x
x
=
60
=
-
0
.
86603
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
-
0
.
86603
×
0
.
01745
=
-
0
.
01511
⇒
∆
y
=
-
0
.
01511
∴
cos
61
°
=
y
+
∆
y
=
0
.
48488
≈
0
.
48489
(xii)
Consider
the
function
y
=
f
x
=
1
x
.
Let
:
x
=
25
x
+
∆
x
=
25
.
1
Then
,
∆
x
=
0
.
1
For
x
=
,
y
=
1
25
=
0
.
2
Let
:
d
x
=
∆
x
=
0
.
1
Now
,
y
=
1
x
⇒
d
y
d
x
=
-
1
2
x
3
2
⇒
d
y
d
x
x
=
25
=
-
0
.
004
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
-
0
.
004
×
0
.
1
=
-
0
.
0004
⇒
∆
y
=
-
0
.
0004
∴
1
25
.
1
=
y
+
∆
y
=
0
.
1996
(xiii)
Consider
the
function
y
=
f
x
=
sin
x
.
Let
:
x
=
22
7
x
+
∆
x
=
22
14
Then
,
∆
x
=
-
22
14
For
x
=
π
,
y
=
sin
22
7
=
0
Let
:
d
x
=
∆
x
=
sin
-
22
14
=
-
sin
π
2
=
-
1
Now
,
y
=
sin
x
⇒
d
y
d
x
=
cos
x
⇒
d
y
d
x
x
=
22
7
=
-
1
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
-
1
×
-
1
=
1
⇒
∆
y
=
1
∴
sin
22
14
=
y
+
∆
y
=
1
(xiv)
Consider
the
function
y
=
f
x
=
cos
x
.
Let
:
x
=
π
3
x
+
∆
x
=
11
π
36
Then
,
∆
x
=
-
π
36
=
-
5
°
For
x
=
π
3
,
y
=
cos
π
3
=
0
.
5
Let
:
d
x
=
∆
x
=
-
sin
5
°
=
-
0
.
08716
Now
,
y
=
cos
x
⇒
d
y
d
x
=
-
sin
x
⇒
d
y
d
x
x
=
π
3
=
-
0
.
86603
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
-
0
.
86603
×
-
0
.
08716
=
0
.
075575
⇒
∆
y
=
0
.
075575
∴
cos
11
π
36
=
y
+
∆
y
=
0
.
5
+
0
.
075575
=
0
.
575575
(xv)
Consider
the
function
y
=
f
x
=
x
1
4
.
Let
:
x
=
81
x
+
∆
x
=
80
Then
,
∆
x
=
-
1
For
x
=
81
,
y
=
81
1
4
=
3
Let
:
d
x
=
∆
x
=
-
1
Now
,
y
=
x
1
4
⇒
d
y
d
x
=
1
4
x
3
4
⇒
d
y
d
x
x
=
81
=
1
108
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
108
×
-
1
=
-
0
.
009259
⇒
∆
y
=
-
0
.
009259
∴
80
1
4
=
y
+
∆
y
=
2
.
99074
(xvi)
Consider
the
function
y
=
f
x
=
x
1
3
.
Let
:
x
=
27
x
+
∆
x
=
29
Then
,
∆
x
=
2
For
x
=
27
,
y
=
27
1
3
=
3
Let
:
d
x
=
∆
x
=
2
Now
,
y
=
x
1
3
⇒
d
y
d
x
=
1
3
x
2
3
⇒
d
y
d
x
x
=
27
=
1
27
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
27
×
2
=
0
.
074
⇒
∆
y
=
0
.
074
∴
29
1
3
=
y
+
∆
y
=
3
.
074
(xvii)
Consider
the
function
y
=
f
x
=
x
1
3
.
Let
:
x
=
64
x
+
∆
x
=
66
Then
,
∆
x
=
2
For
x
=
64
,
y
=
64
1
3
=
4
Let
:
d
x
=
∆
x
=
2
Now
,
y
=
x
1
3
⇒
d
y
d
x
=
1
3
x
2
3
⇒
d
y
d
x
x
=
64
=
1
48
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
48
×
2
=
0
.
042
⇒
∆
y
=
0
.
042
∴
66
1
3
=
y
+
∆
y
=
4
.
042
(xviii)
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
25
x
+
∆
x
=
26
Then
,
∆
x
=
1
For
x
=
25
,
y
=
25
=
5
Let
:
d
x
=
∆
x
=
1
Now
,
y
=
x
1
/
2
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
25
=
1
10
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
10
×
1
=
0
.
1
⇒
∆
y
=
0
.
1
∴
26
=
y
+
∆
y
=
5
.
1
(xix)
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
36
x
+
∆
x
=
37
Then
,
∆
x
=
1
For
x
=
36
,
y
=
36
=
6
Let
:
d
x
=
∆
x
=
1
Now
,
y
=
x
1
2
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
36
=
1
12
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
12
×
1
=
0
.
0833
⇒
∆
y
=
0
.
0833
∴
37
=
y
+
∆
y
=
6
.
0833
(xx)
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
0
.
49
x
+
∆
x
=
0
.
48
Then
,
∆
x
=
-
0
.
01
For
x
=
0
.
49
,
y
=
0
.
49
=
0
.
7
Let
:
d
x
=
∆
x
=
0
.
01
Now
,
y
=
x
1
2
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
0
.
49
=
1
1
.
4
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
1
.
4
×
-
0
.
01
=
-
0
.
007143
⇒
∆
y
=
-
0
.
007143
∴
0
.
48
=
y
+
∆
y
=
0
.
693
(xxi)
Consider
the
function
y
=
f
x
=
x
1
4
.
Let
:
x
=
81
x
+
∆
x
=
82
Then
,
∆
x
=
1
For
x
=
81
,
y
=
81
1
4
=
3
L
e
t
:
d
x
=
∆
x
=
1
Now
,
y
=
x
1
4
⇒
d
y
d
x
=
1
4
x
3
4
⇒
d
y
d
x
x
=
81
=
1
108
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
108
×
1
=
0
.
009259
⇒
∆
y
=
0
.
009259
∴
82
1
4
=
y
+
∆
y
=
3
.
009259
(xxii)
Consider
the
function
y
=
f
x
=
x
1
4
.
Let
:
x
=
16
81
x
+
∆
x
=
17
81
Then
,
∆
x
=
1
81
For
x
=
16
81
,
y
=
16
81
1
4
=
2
3
Let
:
d
x
=
∆
x
=
1
81
Now
,
y
=
x
1
4
⇒
d
y
d
x
=
1
4
x
3
4
⇒
d
y
d
x
x
=
16
81
=
27
32
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
27
32
×
1
81
=
1
96
=
0
.
01042
⇒
∆
y
=
0
.
01042
∴
17
81
1
4
=
y
+
∆
y
=
0
.
6771
(xxiii)
Consider
the
function
y
=
f
x
=
x
1
5
.
Let
:
x
=
32
x
+
∆
x
=
33
Then
,
∆
x
=
1
For
x
=
33
,
y
=
32
1
5
=
2
Let
:
d
x
=
∆
x
=
1
Now
,
y
=
x
1
5
⇒
d
y
d
x
=
1
5
x
4
5
⇒
d
y
d
x
x
=
32
=
1
80
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
80
×
1
=
0
.
0125
⇒
∆
y
=
0
.
0125
∴
33
1
5
=
y
+
∆
y
=
2
.
0125
(xxiv)
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
36
x
+
∆
x
=
36
.
6
Then
,
∆
x
=
0
.
6
For
x
=
36
,
y
=
36
=
6
Let
:
d
x
=
∆
x
=
0
.
6
Now
,
y
=
x
1
2
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
36
=
1
12
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
12
×
0
.
6
=
0
.
05
⇒
∆
y
=
0
.
05
∴
36
.
6
=
y
+
∆
y
=
6
.
05
(xv)
Consider
the
function
y
=
f
x
=
x
1
3
.
Let
:
x
=
27
x
+
∆
x
=
25
Then
,
△
x
=
-
2
For
x
=
27
,
y
=
27
1
3
=
3
Let
:
d
x
=
∆
x
=
-
2
Now
,
y
=
x
1
3
⇒
d
y
d
x
=
1
3
x
2
3
⇒
d
y
d
x
x
=
27
=
1
27
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
27
×
-
2
=
-
0
.
07407
⇒
∆
y
=
-
0
.
07407
∴
25
1
3
=
y
+
∆
y
=
2
.
9259
(xxvi)
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
49
x
+
∆
x
=
49
.
5
Then
,
∆
x
=
0
.
5
For
x
=
49
,
y
=
49
=
7
Let
:
d
x
=
∆
x
=
0
.
5
Now
,
y
=
x
1
2
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
49
=
1
14
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
1
14
×
0
.
5
=
0
.
0357
⇒
∆
y
=
0
.
0357
∴
49
.
5
=
y
+
∆
y
=
7
.
0357
(xxvii)
Consider
the
function
y
=
f
x
=
x
3
2
.
Let
:
x
=
4
x
+
∆
x
=
3
.
968
Then
,
∆
x
=
-
0
.
032
For
x
=
4
,
y
=
4
3
2
=
8
Let
:
d
x
=
∆
x
=
-
0
.
032
Now
,
y
=
x
3
2
⇒
d
y
d
x
=
3
x
2
⇒
d
y
d
x
x
=
4
=
3
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
3
×
-
0
.
032
=
-
0
.
096
⇒
∆
y
=
-
0
.
096
∴
3
.
968
3
2
=
y
+
∆
y
=
7
.
904
(xxviii)
Consider
the
function
y
=
f
x
=
x
5
.
Let
:
x
=
2
x
+
∆
x
=
1
.
999
Then
,
∆
x
=
-
0
.
001
For
x
=
2
,
y
=
2
5
=
32
Let
:
d
x
=
∆
x
=
-
0
.
001
Now
,
y
=
x
5
⇒
d
y
d
x
=
5
x
4
⇒
d
y
d
x
x
=
2
=
80
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
80
×
-
0
.
001
=
-
0
.
08
⇒
∆
y
=
-
0
.
08
∴
1
.
999
5
=
y
+
∆
y
=
31
.
92
(xxix)
Consider
the
function
y
=
f
x
=
x
.
Let
:
x
=
0
.
0841
x
+
∆
x
=
0
.
082
Then
,
∆
x
=
-
0
.
0021
For
x
=
0
.
0841
,
y
=
0
.
0841
=
0
.
29
Let
:
d
x
=
∆
x
=
-
0
.
0021
Now
,
y
=
x
1
2
⇒
d
y
d
x
=
1
2
x
⇒
d
y
d
x
x
=
0
.
0841
=
1
0
.
58
=
50
29
∴
∆
y
=
d
y
=
d
y
d
x
d
x
=
50
29
×
-
0
.
0021
=
-
0
.
0036
⇒
∆
y
=
-
0
.
0036
∴
0
.
082
=
y
+
∆
y
=
0
.
2864
Suggest Corrections
0
Similar questions
Q.
Using differentials, find the approximate values of the following:
(i)
25
.
02
(ii) (0.009)
1/3
(iii) (0.007)
1/3
(iv)
401
(v) (15)
1/4
(vi) (255)
1/4
(vii)
1
(
2
.
002
)
2
(viii) log
e
4.04, it being given that log
10
4 = 0.6021 and log
10
e = 0.4343.
(ix) log
e
10.02, it being given that log
e
10 = 2.3026.
(x) log
10
10.1, it being given that log
10
e = 0.4343.
(xi) cos 61°, it being given that sin 60° = 0.86603 and 1° = 0.01745 radian.
(xii)
1
25
.
1
(xiii)
sin
22
14
(xiv)
cos
11
π
36
(xv) (80)
1/4
(xvi) (29)
1/3
(xvii) (66)
1/3
(xviii)
26
(xix)
37
(xx)
0
.
48
(xxi) (82)
1/4
(xxii)
17
81
1
/
4
(xxiii) (33)
1/5
(xxiv)
36
.
6
(xxv) 25
1/3
Q.
Find
log
e
(
4.04
)
, if
log
10
4
=
0.6021
,
log
10
e
=
0.4343
Q.
Find the approximate value of log
10
1005, given that log
10
e = 0.4343.
Q.
Find the value of
x
, given that
2
log
10
(
2
x
−
1
)
=
log
10
2
+
log
10
(
2
x
+
3
)
Q.
Using the formula,
cos
A
=
1
+
cos
2
A
2
, find the value of cos 30°, it being given that cos 60° =
1
2
.
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