JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Analyzing +4 | -1

n

is the degree of the polynomial,

\left[\frac{2}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}\right]^8+\left[\frac{2}{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}\right]^8

and

\mathrm{m}

is the coefficient of

x^n

in it, then the ordered pair

(n, m)

is equal to

View answer and explanation

Correct option: B

\left(12,(20)^4\right)

Option B (

(12, (20)^4)

) is correct because after rationalizing and simplifying, the polynomial's degree is 12 and the coefficient of

x^{12}

(20)^4

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