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

Correct Answer: 6006

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
∴ T₁₁ = ¹⁵C₁₀x(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