JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

Assuming to be small that

{x^2}

and higher powers of

x

can be neglected, then

\frac{{{{\left( {1 + \frac{3}{4}x} \right)}^{ - 4}}{{\left( {16 - 3x} \right)}^{1/2}}}}{{{{\left( {8 + x} \right)}^{2/3}}}}

is approximately equal to

View answer and explanation

Correct option: B

1 - \frac{{305}}{{96}}x

Option B (

1 - \frac{{305}}{{96}}x

) is correct because using the binomial approximation and neglecting higher powers of x leads to this result.

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