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Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia Binomial Theorem PYQ

Jamia Millia Islamia PYQ
The total number of terms in the expansion of $(x+a)^{100} + (x-a)^{100}$ after simplification is:





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

$(x+a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}a^r$ $(x-a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}(-a)^r$ Adding, odd powers of $a$ cancel and even powers remain. Even $r$: total even $r$ from 0 to 100 → 51 terms.
Jamia Millia Islamia PYQ
The middle term in the expansion of $\left(1 + \dfrac{1}{x^2}\right)\!\left(1 + x^2\right)^n$ is —





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

Expand $\left(1 + x^2\right)^n = \sum_{k=0}^{n} {}^{n}C_{k}x^{2k}.$ Multiply by $\left(1 + \dfrac{1}{x^2}\right)$: $= \sum_{k=0}^{n} {}^{n}C_{k}x^{2k} + \sum_{k=0}^{n} {}^{n}C_{k}x^{2k-2}.$ To find middle term → powers of $x$ that are equal when $2k = 2n - (2k-2)$. Simplifying gives $k = n.$ So middle term = ${}^{2n}C_{n}.$
Jamia Millia Islamia PYQ
The sum of $(n+1)$ terms of the series $\dfrac{C_0}{2} - \dfrac{C_1}{3} + \dfrac{C_2}{4} - \dfrac{C_3}{5} + \dots$ is





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Using binomial relation and telescoping pattern, the series reduces to $\dfrac{1}{(n+1)(n+2)}$
Jamia Millia Islamia PYQ
What will be the value of $(102)^5$?





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Using binomial expansion: $(100 + 2)^5 = 100^5 + 5(100^4)(2) + 10(100^3)(2^2) + 10(100^2)(2^3) + 5(100)(2^4) + 2^5$ $= 10^{10} + 10^{8}(10) + 10^{6}(40) + 10^{4}(80) + 10^{2}(80) + 32$ $= 11040603032.$
Jamia Millia Islamia PYQ
What will be an approximation of $(0.99)^5$ using the first three terms of expansion?





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Using $(1 - x)^n \approx 1 - nx + \dfrac{n(n-1)}{2}x^2$ For $x = 0.01, n = 5$: $(1 - 0.01)^5 \approx 1 - 5(0.01) + 10(0.01)^2 = 1 - 0.05 + 0.001 = 0.951.$
Jamia Millia Islamia PYQ
The coefficient of the middle term in the binomial expansion of $(1 + ax)^4$ and of $(1 - ax)^6$ is the same, if $a$ is equal to...





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Middle term of $(1 + ax)^4$ ⇒ $\text{Coefficient} = \binom{4}{2} a^2 = 6a^2.$ Middle term of $(1 - ax)^6$ ⇒ $\text{Coefficient} = \binom{6}{3} (-a)^3 = -20a^3.$ Given equal ⇒ $6a^2 = 20a^3 \Rightarrow a = \dfrac{3}{10}.$ Since signs are opposite ⇒ $a = -\dfrac{3}{10}.$
Jamia Millia Islamia PYQ
The coefficient of $x^8 y^{10}$ in $(x + y)^{18}$ is …





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

General term = ${}^{18}C_r x^{18-r} y^r$. We need $x^8 y^{10} \Rightarrow 18 - r = 8 \Rightarrow r = 10$. Coefficient = ${}^{18}C_{10}$.
Jamia Millia Islamia PYQ
The coefficient of $y$ in the expansion of $\left(y^2+\dfrac{c}{y}\right)^5$ is:





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

General term $T_k=\binom{5}{k}(y^2)^{5-k}\left(\dfrac{c}{y}\right)^k =\binom{5}{k}c^k\,y^{10-3k}$. For power of $y^1$: $10-3k=1\Rightarrow k=3$. Coefficient $=\binom{5}{3}c^3=10c^3$.
Jamia Millia Islamia PYQ
The coefficient of $x^4$ in expansion of $(1 + x + x^2 + x^3)^{11}$ is …





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

We can write generating function $f(x) = (1 + x + x^2 + x^3)^{11}$. Coefficient of $x^4 =$ number of ways to get power 4 as sum of 11 terms each $0,1,2,3$. By multinomial expansion, coefficient of $x^4 = {}^{11}C_4 + 10{}^{11}C_3 + 6{}^{11}C_2 + {}^{11}C_1 = 990$.
Jamia Millia Islamia PYQ
In the expansion of $(1+x)^{50}$, the sum of coefficients of odd powers of $x$ is:





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

Sum of odd coefficients $=\dfrac{(1+1)^{50}-(1-1)^{50}}{2} =\dfrac{2^{50}-0}{2}=2^{49}$.
Jamia Millia Islamia PYQ
The coefficient of the middle term in the expansion of $(2 + 3x)^4$ is





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

Total terms = $4 + 1 = 5$ Middle term = $\dfrac{5 + 1}{2} = 3^\text{rd}$ term $\text{T}_3 = \binom{4}{2} (2)^{2} (3x)^{2} = 6 \times 4 \times 9x^2 = 216x^2$ Coefficient = 216
Jamia Millia Islamia PYQ
If A and B are coefficients of $x^n$ in $(1+x)^{2n}$ and $(1+x)^{2n-1}$ respectively, then $\dfrac{A}{B}$ equals





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

$A = \binom{2n}{n},\ B = \binom{2n-1}{n} \Rightarrow \dfrac{A}{B} = 2.$

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