Example.1
If for non-zero x, af(x) + b f
= 
where
a > b, then find 
.
Solution
af(x) + b f
= 
….
(1)
Integrating, we have
a
…. (2)
Replacing x by 
in (1), we get, af+ bf(x) = x –5
Integrating, we
have, a
…. (3)
Eliminating 
from (2) and (3) by multiplying (2) by
a and (3) by b and subtracting, we get (a2 – b2) 
&⇒ 
.
Example.2
Prove that 
=
2p.
Solution
Let I = real 
= real part of 
= real part 
= real part 
= real part 
= real part 
=
= 
= (2p + 0 + 0 + .....) – (0 + 0 + 0
+.....) = 2p.
Example.3
Evaluate 
Solution
Let tan-1x
= q. As 
, so –
Now 
= 
= cos-1(
sin2q) + tan-1(
tan2q) = cos-1
+ tan-1( tan2q)
= 
– 2q
+ 2q [ as 
and 
]
= p/2
Hence, I = 
= 
=
=
.
Example.4
If Un
= 
where n is a positive integer or zero,
then show that Un+2 + Un = 2 Un+1. Hence
deduce that 
Solution
Un+2 – Un+1 =
= 
= 
Un+2 –
Un+1 = 
Un+1 – Un = 
From (1) and (2), we get
(Un+2 – Un+1) – (Un+1
– Un) = 
Un+2 + Un – 2Un+1
= 
= 
= 
= 0
Un+2 + Un = 2 Un+1
Un, Un+1, Un+2
are in A.P.
Now, U0 = 
= 
=0,
U1 =
U1 – U0 = 
(Common difference)
Un = U0 + n.
Un = n
Now In = 
(
writing 2q = x )
= 
= 
Alternative:
We have Un+2
+ Un = 
= 2
= 
= 2Un+1 +
0. .
Example.5
Find a
function g: R → R, continuous in [0, ¥)
and positive in (0, ¥) satisfying g (1) =1 and 
.
Solution
Let F (x) =
Differentiating both
sides with respect to x, we get
&⇒
t2 – 4 t + 2 = 0 where 
&⇒
or 
&⇒
ln F (x) = (2 
) lnx + constant
&⇒
F (x) = C 
&⇒
where 
are
constants. But g is continuous on [0, ¥).
Then c²
is ruled out.
Hence g (x) = 
Also g (1) = 
= 1
.
