JEE Main 26 Feb 2021 Shift 1 Solved Paper

Using Binomial expansion, its (r+1) th term be,
Tr+1‌‌=‌10Ct(tx1∕5)10−r{‌

(1−x)1∕10

t

}r
‌‌=‌10Cr‌

(t)10−r

(t)t

(x1∕5)10−t(1−x)r∕10
‌‌=‌10Cr(t)10−2r(x)‌

10−r

5

(1−x)r∕10
If this term is independent of ' t′, then we havel 0−2r=0 gives, r=5
∴‌‌T6=‌10C5(x)1(1−x)1∕2
∴‌‌T6=‌10C5(x)1(1−x)1∕2
Let f(x)=x(1−x)1∕2, to obtain its maximum value, we have to differentiate it and equate it to 0 . i.e. f′(x)=0⇒‌

x

2√1−x

(−1)+√1−x=0
⇒−x+2(1−x)=0⇒−3x+2=0
⇒‌‌x=‌

2

3

and f′′(‌

2

3

)<0 (Maximum value)
Thus, greatest term will be
T6=‌10C5(‌

2

3

)(1−‌

2

3

)1∕2=‌10C5‌

2

3√3

=‌

10!⋅2

(5!)2(3√3)