JEE Advanced Model Paper 10 with solutions for online practice

∵ x

x

∫

0

f(t) dt = (x + 1)

x

∫

0

t . f(t) dt , x > 0
x f(x) +

x

∫

0

f(t) dt =

x

∫

0

t . f(t) dt + (x + 1) . x f(x)
again on differentiating w.r.t. x we get
f(x) = 3x f(x) + x2 f'(x)
∴

dy

dx

=

(1−3x)y

x2

∴ ∫

1−3x

x2

dx = ∫

dy

y

⇒ −

1

x

- 3 ln x = ln y + c
∵ f(1) =

1

e

⇒ c = 0
∴ ln y x2 = −

1

x

∴ f(x) =

1

x3e1∕x

now , ∫ x . f(x) dx = ∫

1

x2e1∕x

dx
let

1

x

= t ⇒

1

x2

dt = -dt
∫ x . f(x) dx = ∫

−dt

et

=

1

et

=

1

e1∕x

+ c
∴

(ln5)−1

∫

(ln10)−1

x f(x) =

1

eln5

-

1

eln10

=

1

5

−

1

10

=

1

10

and

lim

x→0+

x3 f(x) =

lim

h→0

1

e1∕h

= 0