1. Answer to selected Miscellaneous Exercises
0.6 n
8. 0.0001
1 0.6
0.6 n 4 10 5
log 0.6 n log 4 10 5
5
n log 0.6 log 4 10 (note: log 0.6 0)
5
log 4 10
n
log 0.6
n 19.8
Hence the least integral value of n 20
ln p ln q ln r
9. Given: ln t
a b c
ln p
Proof:  ln t
a
ln p a ln t
ln p ln t a
p ta
Similarly, q tb ; r tc

2. q2
LHS=
pr
t 2b
= a c
t t
= t 2b a c
=RHS (Thus proved)
d
14. ln( x x2 a2 )
dx
1 x
= (1 )
x x2 a2 x2 a2
1 x2a2 x
= ( )
x x2 a2 x2 a2
1
=
x2 a2
1
Hence, dx ln( x x2 a2 ) C
2 2
x a
3
3 dx 2
ln( x x 1)
2
x2 1 2
ln(3 8 ) ln( 2 3)
3 8
ln( )
2 3

3. 16. y ln(1 sin x)
dy cos x
dx 1 sin x
d2y sin x(1 sin x) cos x cos x sin x sin 2 x cos2 x 1
dx 2 (1 sin x) 2 (1 sin x) 2 1 sin x
d3y cos x
dx 3 (1 sin x) 2
cos x 1 cos x
LHS= ( )( ) 0 (thus proved)
(1 sin x) 2 1 sin x 1 sin x
20.
d 3x
(e sin 2 x) 3e 3 x sin 2 x 2e 3 x cos 2 x
dx
e 3 x (3 sin 2 x 2 cos 2 x)
2
13e 3 x sin(2 x tan 1 )
3
2
a 13 b tan 1
3
d 2 3x d 2
(e sin 2 x) ( 13e 3 x sin(2 x tan 1 ))
dx 2 dx 3
2 2
3 13e 3 x sin(2 x tan 1 ) 2 13e 3 x cos(2 x tan 1 )
3 3
2 2
13e 3 x [3 sin(2 x tan 1 ) 2 cos(2 x tan 1 )]
3 3
2
13e 3 x [ 13 (sin 2 x 2 tan 1 )]
3
2
( 13 ) 2 e 3 x sin(2 x 2 tan 1 )
3

4. d n 3x 2
(e sin 2 x) ( 13 ) n e 3 x sin(2 x n tan 1
)
dx n 3
22.
a2 b2 7ab
a2 2ab b 2 9ab
( a b) 2 9ab
a b 2
( ) ab
3
a b
2 log( ) log ab
3
a b 1
log( ) log ab
3 2
a b 1
log( ) (log a log b)
3 2
24.
nt
S ( A Bt)e
e nt S ( A Bt)
de nt S d ( A Bt)
dt dt
dS
ne nt S e nt B
dt
d dS dB
(ne nt S e nt )
dt dt dt
dS dS d 2S
n 2 e nt S ne nt ne nt ent
0
dt dt dt 2
2
d S dS
2
2n n2S 0
dt dt

5. 29.
1
1 log a x
y a
1
log a y  (1)
1 log a x
Similarly,
1
log a z  (2)
1 log a y
log a y (1 log a x) 1
from (1)
log a y log a y log a x 1
1 log a y
log a x (3)
log a y
substitute (2) into (3)
1
log a z 1
log a x
1 log a z 1 log a z
log a z
1
1 log a z
x a