Binomial Theorem PYQs - Last 10 Years
TS EAMCET / Mathematics / Algebra / 60 recent questions
MathematicsAlgebra2016-2025
Practice 60 TS EAMCET Mathematics questions from Binomial Theorem. Use the year-wise and type-wise breakdown to prioritize recent PYQs, then continue into the question list below.
60
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Mathematics / Algebra
2020-2025
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47
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60
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2016-2025
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47 in last 5 years60 in last 10 years
Last 10 Years Binomial Theorem Questions
Showing 10 of 60 filtered questions.
1Binomial Theorem
If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{\left(1+\frac{2 x}{3}\right)^{-4}(4+5 x)^{1 / 2}}{(9+x)^{3 / 2}}$, when $x=\frac{6}{371}$, is
MCQ+1 / -02020
2Binomial Theorem
For $z \in \mathbf{C}$, if $(1+z)^n=1+{ }^n C_1 z+{ }^n C_2 z^2+\ldots{ }^n C_n z^n$ and $\sum_{r=0}^{100} 100 c_r(\sin r x)=\left(2 \cos \frac{x}{2}\right)^{100} \sin k x$, then $k=$
MCQ+1 / -02020
3Binomial Theorem
Suppose $1, m, n$ respectively represent the coefficient of $x^{10}$, the constant term and the coefficient of $x^{-10}$ in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$. If $\frac{l}{m}+\frac{m}{n}=\frac{26}{11}$, then $a^2: b^2...
MCQ+1 / -02020
4Binomial Theorem
The approximate value of $\left(3 \sqrt{126}+\sin 61^{\circ}\right)$ correct to three decimal places, obtained by taking $1^{\circ}=0.0174$ radians, is
MCQ+1 / -02020
5Binomial Theorem
If sum of the coefficients of $x^r(r=0,1,2, \ldots, 2 n)$ in the expansion of $\left(1+3 x-2 x^2\right)^n$ is 128 , then $\sum_{r=1}^{2 n} r \frac{(2 n)_{C_r}}{(2 n)_{C_{r-1}}}=$
MCQ+1 / -02020
6Binomial Theorem
If ${ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_n$ respectively are the binomial coefficients in the expansion of $(1+x)^n$, then when $n=10, \sum_{r=1}^{10}{ }^n C_r \cdot r(r-4)=$
MCQ+1 / -02020
7Binomial Theorem
For $0 < x < 1$, the expansion of $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ is
MCQ+1 / -02020
8Binomial Theorem
Let $x \in \mathbf{R}$ be so small that the powers of $x$ beyond two are insignificant and negligibly small. For such $x$, if $(1-x)^3(2+x)^6$ is approximated by $a+b x+c x^2$, then $a+b+c=$
MCQ+1 / -02020
9Binomial Theorem
If the 9th and 10th terms are the numerically greatest terms in the expansion of $(5 x-6 y)^n$ when $x=2 / 5$ and $y=1 / 2$, then the absolute value of the middle terms of that expansion is
MCQ+1 / -02020
10Binomial Theorem
\(1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3+\ldots\)
MCQ+1 / -02020