If -2/3 [AP EAPCET 23 May 2025 S1]

Correct Answer is : 3576823{35}{768[3]{2}}7683235

Correct Answer is : 3576823{35}{768[3]{2}}7683235

Test Index

Binomial Theorem

Section: Mathematics

Question : 7 of 33

Marks: +1, -0

-\frac{2}{3} < x < \frac{2}{3}

5^{\text{th}}

rac{1}{\sqrt[3]{2-3x}}

x=\frac{1}{2}

Solution:

Given,

-\frac{2}{3} < x < \frac{2}{3}

\Rightarrow -2 < 3x < 2 \Rightarrow -1 < \frac{3x}{2} < 1

Now,

rac{1}{3\sqrt{2-3x}} = \frac{1}{(2-3x)^{1/3}}

= \frac{1}{2^{1/3}} \left(1-\frac{3x}{2}\right)^{-1/3}

= \frac{1}{3\sqrt{2}} \left[1 + \left(\frac{1}{3}\right)\left(\frac{3x}{2}\right) + \frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)}{2\times 1} \left(\frac{3x}{2}\right)^2 + \frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right)}{3\times 2\times 1} \left(\frac{3x}{2}\right)^3 + \frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right)\left(\frac{1}{3}+3\right)}{4\times 3\times 2\times 1} \left(\frac{3x}{2}\right)^4 + \cdots \right]

\therefore 5

th term when

x=\frac{1}{2}

is given by

\frac{1}{\sqrt[3]{2}} \left[\frac{1}{3} \times \frac{4}{3} \times \frac{7}{3} \times \frac{10}{3} \times \frac{1}{24} \times \frac{81}{16} \times \frac{1}{16}\right]

= \frac{1}{3\sqrt{2}} \left[\frac{7\times 5}{24\times 2\times 6}\right] = \frac{35}{768\sqrt[3]{2}}

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