JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

x

is so small that

{x^3}

and higher powers of

x

may be neglected,then

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

may be approximated as

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by using binomial expansion and neglecting terms of

x^3

and higher, the expression simplifies to this value.

More Questions from BINOMIAL THEOREM

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n