Let alpha 0, beta 0 be such that alpha^{3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The general term $T_{r+1}$ in the expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (\alph

Vedclass · Accepted Answer

$336$

Vedclass · Answer

(B) The general term $T_{r+1}$ in the expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (\alpha x^{\frac{1}{9}})^{10-r} (\beta x^{-\frac{1}{6}})^{r} = {}^{10}C_{r} \alpha^{10-r} \beta^{r} x^{\frac{10-r}{9} - \frac{r}{6}}$.For the term to be independent of $x$,the exponent of $x$ must be zero:$\frac{10-r}{9} - \frac{r}{6} = 0$ $\Rightarrow 2(10-r) - 3r = 0$ $\Rightarrow 20 - 5r = 0$ $\Rightarrow r = 4$.The independent term is $T_{5} = {}^{10}C_{4} \alpha^{6} \beta^{4} = 210 \alpha^{6} \beta^{4}$.We are given $\alpha^{3} + \beta^{2} = 4$. By $AM \geq GM$ inequality for positive numbers:$\frac{\frac{\alpha^{3}}{2} + \frac{\alpha^{3}}{2} + \frac{\beta^{2}}{2} + \frac{\beta^{2}}{2}}{4} \geq \sqrt[4]{\frac{\alpha^{3}}{2} \cdot \frac{\alpha^{3}}{2} \cdot \frac{\beta^{2}}{2} \cdot \frac{\beta^{2}}{2}}$$\frac{4}{4} \geq \sqrt[4]{\frac{\alpha^{6} \beta^{4}}{16}}$ $\Rightarrow 1 \geq \frac{\alpha^{6} \beta^{4}}{16}$ $\Rightarrow \alpha^{6} \beta^{4} \leq 16$.The maximum value of $T_{5}$ is $210 	imes 16 = 3360$.Given $10k = 3360$,we get $k = 336$.

Let $\alpha > 0, \beta > 0$ be such that $\alpha^{3} + \beta^{2} = 4$. If the maximum value of the term independent of $x$ in the binomial expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is $10k$,then $k$ is equal to

Similar Questions

The sum of all those terms which are rational numbers in the expansion of $(2^{1/3} + 3^{1/4})^{12}$ is:

If for positive integers $r > 1$ and $n > 2$,the coefficients of the $(3r)^{th}$ and $(r + 2)^{th}$ powers of $x$ in the expansion of $(1 + x)^{2n}$ are equal,then:

If the $9^{th}$ and $10^{th}$ terms are the numerically greatest terms in the expansion of $(5x - 6y)^n$ when $x = 2/5$ and $y = 1/2$,then the absolute value of the middle term of that expansion is:

The term independent of $x(x>0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$ is:

If the $k^{\text{th}}$ term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is independent of $x$,then the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^k$ when $x = \frac{2}{3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module