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  • Solve it, - Binomial theorem and its simple applications - JEE Main-3

If x is so small that x^{3} and higher powers of x may be neglected, then  \frac{(1+x)^{3/2}-(1+\frac{1}{2}x)^{3}}{(1-x)^{1/2}}   may be approximated as

  • Option 1)

    3x+\frac{3}{8}x^{2}\;

  • Option 2)

    \; 1-\frac{3}{8}x^{2}\;

  • Option 3)

    \; \; \frac{x}{2}-\frac{3}{8}x^{2}\;

  • Option 4)

    \; -\frac{3}{8}x^{2}

 

Answers (1)

As we learnt in

Binomial Theorem for Rational index -

\left ( 1+x \right )^{n}= 1+nx+\frac{n\left ( n-1 \right )x^{2}}{2!}+\frac{n\left ( n-1 \right )\left ( n-2 \right )x^{3}}{3!}+-----
 

- wherein

use ^{n}c_{r}= \frac{n!}{r!(n-r)!}

n> 0

 

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

\Rightarrow\ \;\frac{1+\frac{3x}{2}+\frac{3}{2}.\frac{1}{2}.\frac{x^{2}}{2!}-1-\frac{3x}{2}-\frac{3\times2}{2!}\frac{x^{2}}{4}}{(1-x)^{1/2}}=\left(\frac{3x^{2}}{8}-\frac{3x^{2}}{4} \right )\left(1-x \right )^{-1/2}=\frac{-3x^{2}}{8}(1-x)^{-1/2}

On approximation (1-x)^{-1/2}\approx1

=\frac{-3x^{2}}{8}

Correct option is 4.

 


Option 1)

3x+\frac{3}{8}x^{2}\;

This is an incorrect option.

Option 2)

\; 1-\frac{3}{8}x^{2}\;

This is an incorrect option.

Option 3)

\; \; \frac{x}{2}-\frac{3}{8}x^{2}\;

This is an incorrect option.

Option 4)

\; -\frac{3}{8}x^{2}

This is the correct option.

Posted by

Vakul

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