If the maximum value of the term independent of t in the expansion of ( t 2 x 1 5 + ( 1 − x ) 1 1 0 t ) 1 5 , x ≥ , is K, then 8 K is equal to…………..

General term = 1 5 C r ( t 2 x 1 5 ) 1 5 − r ( ( 1 − x ) 1 1 0 t ) r for term independent on t 2 (15 – r) – r = 0 ⇒ r = 1 0 ∴ T 1 1 = 1 5 C 1 0 * ( 1 − x ) Maximum value of x (1 – x) occur at x = 1 2

If the maximum value of the term independent of t in the expansion of ${(t^{2} x^{\frac{1}{5}} + \frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{15}$ , x $\geq$ , is K, then 8 K is equal to…………..

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Vishal Baghel | Contributor-Level 10

6 months ago

General term

$=^{15} C_{r} {(t^{2} x^{\frac{1}{5}})}^{15 - r} {(\frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{r}$

for term independent on t

2 (15 – r) – r = 0

$\Rightarrow r = 10$

$∴ T_{11} =^{15} C_{10} * (1 - x)$

Maximum value of x (1 – x) occur at

x = $\frac{1}{2}$

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