If the ratio of the coefficient of $x^{10}$ in $\left(1-x^2\right)^{10}$ and term independent of $x$ in $\left(x-\frac{2}{x}\right)^{10}$ is $\frac{p}{q}$ (where ${p}$ and ${q}$ are relative prime numbers), then $\frac{{q}}{8}-{p}$ is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

If the ratio of the coefficient of

x^{10}

\left(1-x^2\right)^{10}

and term independent of

x

\left(x-\frac{2}{x}\right)^{10}

\frac{p}{q}

(where

{p}

and

{q}

are relative prime numbers), then

\frac{{q}}{8}-{p}

View answer and explanation

Correct option: A

Option A (3) is correct because the ratio of the coefficients simplifies to 1/32, leading to p=1, q=32, and (q/8)-p = 3.

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