JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, curve y=f(x) passes through (1,2) and satisfies
‌x‌

dy

dx

+y‌‌=bx4
⇒‌x‌

dy

dx

+y‌‌=bx4
⇒‌‌

dy

dx

+‌

y

x

‌‌=bx3
‌‌‌‌ IF ‌‌‌=ej‌

1

x

⋅dx=x
‌‌‌yx‌‌=∫bx4dx=‌

bx5

5

+C
‌∴‌‌y‌‌‌=‌

bx4

5

+‌

C

x

‌‌=f(x). . . (i)
∵ This curve passes through (1,2).
∴‌‌2×1=‌

b×(1)5

5

+C
⇒‌‌2=‌

b

5

+C . . . (ii)
Also,

2

∫

1

f(x)dx=‌

62

5

⇒‌‌

2

∫

1

(‌

bx4

5

+‌

C

x

)dx=‌

62

5

⇒‌‌[b×‌

x5

25

+C‌log‌x]12=‌

62

5

⇒‌‌[(‌

b×32

25

+C‌log‌2)−(‌

b

25

+c‌log‌1)]=‌

62

5

⇒‌

b×32

25

+C‌log‌2−‌

b

25

=‌

62

5

+0‌log‌2
[from Eq. (i)] On comparing, we get
‌

b

25

×31‌‌=‌

62

5

‌ and ‌c=0
b‌‌=‌

62×25

31×5

b‌‌=10
Hence, the required value of b=10.