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 c²is ruled out.
Hence g (x) =
Also g (1) = = 1.