The maximum value of the term independent of 't' in the expansion of ( tx (1/5)+((1- x )(1/10)/ t ))10 where x ∈(0,1) is

Question

The maximum value of the term independent of 't' in the expansion of ( tx (1/5)+((1- x )(1/10)/ t ))10 where x ∈(0,1) is (A) (10 !/√3(5 !)2) (B) (2.10 !/3 √3(5 !)2) (C) (2.10 !/3(5 !)2) (D) (10 !/3(5 !)2)

Tardigrade Admin · Accepted Answer

Term independent of t will be the middle term due to exect same magnitude but opposite sign powers of t in the binomial expression given so T 6= 10 C 5( tx 2 5)5(((1- x )(1/10)/ t ))5 T 6=f( x )= 10 C 5( x √1- x ) ; for maximum f prime(x)=0 ⇒ x=(2/3) f prime prime((2/3))< 0 so f(x) max = 10 C5((2/3)) ⋅ (1/√3)

Q. The maximum value of the term independent of 't' in the expansion of $(t x^{\frac{1}{5}} + \frac{( 1 - x ) ^{\frac{1}{10}}}{t})^{10}$ where $x \in (0, 1)$ is

$\frac{10 !}{3 ( 5 ! ) ^{2}}$

$\frac{2.10 !}{3 3 ( 5 ! ) ^{2}}$

$\frac{2.10 !}{3 ( 5 ! ) ^{2}}$

$\frac{10 !}{3 ( 5 ! ) ^{2}}$

Q. The maximum value of the term independent of 't' in the expansion of (tx51​+t(1−x)101​​)10 where x∈(0,1) is

3​(5!)210!​

33​(5!)22.10!​

3(5!)22.10!​

3(5!)210!​

Q. The maximum value of the term independent of 't' in the expansion of $(t x^{\frac{1}{5}} + \frac{( 1 - x ) ^{\frac{1}{10}}}{t})^{10}$ where $x \in (0, 1)$ is

$\frac{10 !}{3 ( 5 ! ) ^{2}}$

$\frac{2.10 !}{3 3 ( 5 ! ) ^{2}}$

$\frac{2.10 !}{3 ( 5 ! ) ^{2}}$

$\frac{10 !}{3 ( 5 ! ) ^{2}}$