Question

If the maximum value of the term independent of t in the expansion of (t2x$^{\frac{1}{5}}$+$\frac{(1−x)^{\frac{1}{10}}}{t}$),x≥0 is K , then 8 K is equal to__________.

Answer 1

General Term
=15Cr($t^2x^{\frac{1}{5}}$)15−r($\frac{(1−x)^{\frac{1}{10}}}{t}$)r
for term independent on t
2(15 – r) – r = 0
⇒ r = 10
∴ T 11 = 15 C 10 x(1 – x)
Maximum value of x (1 – x) occur at
x=$\frac{1}{2}$
i.e.,(15(1−x))max=$\frac{1}{4}$
⇒K=15C10×$\frac{1}{4}$
⇒8K=2(15C10)=6006