If for non-zero x, af(x) + b f= where a > b, then find .

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 (a^{2} – b^{2})

&⇒ .

Prove that = 2p.

Let I = real

= real part of = real part

= real part

= real part

= real part

=

=

= (2p + 0 + 0 + .....) – (0 + 0 + 0 +.....) = 2p.

Evaluate

Let tan^{-1}x
= q. As , so –

Now

=

= cos^{-1}(
sin2q) + tan^{-1}(
tan2q) = cos^{-1}+ tan^{-1}( tan2q)

= – 2q + 2q [ as and ]

= p/2

Hence, I =

= ==.

If U_{n}
= where n is a positive integer or zero,
then show that U_{n+2} + U_{n} = 2 U_{n+1}. Hence
deduce that

U_{n+2} – U_{n+1} =

= =

U_{n+2} –
U_{n+1} =

U_{n+1} – U_{n} =

From (1) and (2), we get

(U_{n+2} – U_{n+1}) – (U_{n+1
}– U_{n}) =

U_{n+2} + U_{n }– 2U_{n+1}
=

= = = 0

U_{n+2} + U_{n} = 2 U_{n+1}

U_{n}, U_{n+1}, U_{n+2}
are in A.P.

Now, U_{0} = = =0,
U_{1 }=

U_{1} – U_{0} = (Common difference)

U_{n} = U_{0} + n.

U_{n} = n

Now I_{n} = (
writing 2q = x )

= =

**Alternative:**

We have U_{n+2}
+ U_{n} =

= 2=

= 2U_{n+1} +
0. .

Find a function g: R → R, continuous in [0, ¥) and positive in (0, ¥) satisfying g (1) =1 and .

Let F (x) =

Differentiating both sides with respect to x, we get

&⇒
t^{2} – 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.