JEE Advanced Model Paper 3 with solutions for online practice

x = c , y = ln x , y = −

x

2

+ 2 , y = 0

Point of intersection of two curves occur at x = c
ln x = −

x

2

+ 2
⇒ 2 ln x + x = 4
Thus , 2 ln c + c = 4

c

∫

1

ln xdx =

4

∫

c

(−

x

2

+2) dx
⇒ [xnx−x]1c = [−

x2

4

+2x]c4
(c lnc - c) - (-1) = (−

16

4

+8) - (

c2

4

+2c)
⇒ c lnc = 3 +

c2

4

- c ..... (ii)
From eq (i)
c

(4−c)

2

= 3 +

c2

4

- c ⇒ c2 - 4c + 4 = 0
⇒ (c−2)2 = 0
⇒ c = 2