Test Index

Binomial Theorem Part 3

Section: Mathematics

© examsnet.com

Question : 29 of 53

Marks: +1, -0

The term independent of

x

in the expansion of

\left( \frac{x+1}{x^{2/3}+1-x^{1/3}} - \frac{x-1}{x-x^{1/2}} \right)^{10}, x>1

[2 Apr 2025 Shift 1]

120
240
210
150

Solution:

\;\text{ Consider }\left[ \frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}} \right]^{10}

= \left[ \frac{ (x^{1/3})^3+1^3 }{ x^{2/3}-x^{1/3}+1 } - \frac{ (\sqrt{x})^2-1 }{ \sqrt{x}(\sqrt{x}-1) } \right]^{10}

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

= \left[ (x^{1/3}+1) - \frac{ \sqrt{x}+1 }{ \sqrt{x} } \right]^{10} = (x^{1/3}-x^{-1/2})^{10}

∴ The general term is

T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r

= {}^{10}C_r(-1)^r x^{ \frac{10-r}{3} - \frac{r}{2} }

For independent of

x

, put

\frac{10-r}{3} - \frac{r}{2} = 0

\Rightarrow 20-2r-3r=0

\Rightarrow 20=5r \Rightarrow r=4

\therefore T_5 = {}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

© examsnet.com

Go to Question:

Prev Question

Next Question