JEE Mains 25-July-2022 Shift 1 Solved Paper

General term of (t2x‌

1

5

+‌

(1−x) 1 10

t

)15 is
Tr+1=‌15Cr⋅(t2x‌

1

5

)15−r⋅(‌

(1−x) 1 10

t

)r
=‌15Cr⋅t30−2r⋅x‌

15−r

5

⋅(1−x)‌

r

10

⋅t−r
=‌15Cr⋅t30−3r⋅x‌

15−r

5

⋅(1−x)‌

r

10

Term will be independent of t when 30−3r=0⇒r=10
∴T10+1=T11 will be independent of t
∴T11=‌15C10⋅x‌

15−10

5

⋅(1−x)‌

10

10

=‌15C10⋅x1⋅(1−x)1
T11 will be maximum when x(1−x) is maximum.
Let f(x)=x(1−x)=x−x2
f(x) is maximum or minimum when f′(x)=0
∴f′(x)=1−2x
For maximum ∕ minimum f′(x)=0
∴1−2x=0
⇒x=‌

1

2

Now, f′′(x)=−2<0
∴ At x=‌

1

2

,f(x) maximum
∴ Maximum value of T11 is
=‌15C10⋅‌

1

2

(1−‌

1

2

)
=‌15C10⋅‌

1

4

Given K=‌15C10⋅‌

1

4

Now, 8K=2(‌15C10)
=6006