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Solved Subjective Question on Definite Integral Set 4

Posted on - 04-01-2017

Math

IIT JEE

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.

 
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