The greatest value of the term independent of $x$ in the expansion of ${\left( {x\sin {\rm{\alpha }} + {x^{ - 1}}\cos {\rm{\alpha }}} \right)^{10}},\;{\rm{\alpha }} \in R$ is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

The greatest value of the term independent of

x

in the expansion of

{\left( {x\sin \mathrm{\alpha } + {x^{ - 1}}\cos \mathrm{\alpha }} \right)^{10}},\;\mathrm{\alpha } \in R

View answer and explanation

Correct option: C

\frac{1}{{{2^5}}}\;{(^{10}}{C_5})

Option C (

\frac{1}{{{{2^5}}}}\;{(^{10}}{C_5})

) is correct because the greatest value of the term independent of x is obtained when sin(2α) = 1.

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