JEE Mathematics Some Additional Topics MCQs Set 02

Multiple Choice Questions (MCQs) for JEE Mathematics: Some Additional Topics

Review structured MCQ sets for JEE Mathematics Some Additional Topics. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Some Additional Topics MCQs for JEE Mathematics

Access the complete set of multiple-choice questions for Some Additional Topics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: \( (1-x)^{3/2} \) can be expanded in ascending powers of x if
(a) −1 < x < 1
(b) x < −1
(c) x > 1
(d) none of these
Show Answer & Explanation

Answer: (a) −1 < x < 1

Explanation:
1. A binomial series \( (1+z)^n \) with n not a non-negative integer converges only for |z| < 1.
2. Here z = −x, so |x| < 1.

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Question 2: \( (3+x)^{p/q} \) can be expanded in ascending powers of x if
(a) −1 < x < 1
(b) x > 3
(c) −3 < x < 3
(d) x < −3
Show Answer & Explanation

Answer: (c) −3 < x < 3

Explanation:
1. Write \( 3^{p/q}\left(1+\frac x3\right)^{p/q} \).
2. The series converges for |x/3| < 1, i.e. −3 < x < 3.

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Question 3: If x is positive, the first negative term in the expansion of \( (1+x)^{27/6} \) is the
(a) 5th term
(b) 7th term
(c) 6th term
(d) 8th term
Show Answer & Explanation

Answer: (b) 7th term

Explanation:
1. Here n = 27/6 = 9/2. The coefficient of x^r is \( \frac{n(n-1)\cdots(n-r+1)}{r!} \).
2. The factors are 9/2, 7/2, 5/2, 3/2, 1/2, −1/2, ...; the first negative factor appears for r = 6 (the factor n − 5 = −1/2).
3. So the first negative term is the term with r = 6, the 7th term.

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Question 4: In the expansion of \( (1-x)^{-3} \), |x| < 1, the coefficient of \( x^7 \) is
(a) 36
(b) \( {}^8C_7 \)
(c) 45
(d) none of these
Show Answer & Explanation

Answer: (a) 36

Explanation:
1. \( (1-x)^{-3}=\sum\binom{r+2}{2}x^r \).
2. For r = 7: \( \binom92=36 \).

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Question 5: In the expansion of \( (2+x)^{-5/2} \), the coefficient of \( x^4 \), if it exists, is
(a) \( \frac{5\cdot7\cdot9\cdot11}{2^{5/2}\cdot4!}\left(\frac12\right)^4 \)
(b) \( \frac{5\cdot7\cdot9\cdot11}{4!}\left(\frac12\right)^4 \)
(c) \( \frac{5\cdot7\cdot9\cdot11}{4!\,(\sqrt2)^{21}} \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac{5\cdot7\cdot9\cdot11}{4!\,(\sqrt2)^{21}} \)

Explanation:
1. \( (2+x)^{-5/2}=2^{-5/2}\left(1+\frac x2\right)^{-5/2} \).
2. Coefficient of x⁴: \( 2^{-5/2}\binom{-5/2}{4}\frac1{2^4}=2^{-5/2}\cdot\frac{5\cdot7\cdot9\cdot11}{4!\,2^4}\cdot\frac1{2^4} \).
3. This equals \( \frac{5\cdot7\cdot9\cdot11}{4!\,2^{21/2}}=\frac{5\cdot7\cdot9\cdot11}{4!\,(\sqrt2)^{21}} \).

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Question 6: The coefficient of \( x^5 \) in the expansion of \( \frac{1+x^2}{1+x} \), |x| < 1, is
(a) −1
(b) 2
(c) 0
(d) −2
Show Answer & Explanation

Answer: (d) −2

Explanation:
1. \( \frac1{1+x}=\sum(-1)^kx^k \).
2. Coefficient of x⁵ in (1 + x²) times this: \( (-1)^5+(-1)^3=-2 \).

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Question 7: If \( \frac{x^2+x}{1-x}=a_1x+a_2x^2+\cdots \) to ∞, |x| < 1, then
(a) a₁ + a₂ = 4
(b) a₁ − a₂ = 3
(c) a_p = a_q for all p, q ≥ 2
(d) none of these
Show Answer & Explanation

Answer: (c) a_p = a_q for all p, q ≥ 2

Explanation:
1. \( \frac{x(1+x)}{1-x}=(x+x^2)(1+x+x^2+\cdots)=x+2x^2+2x^3+\cdots \).
2. So a₁ = 1 and a_r = 2 for every r ≥ 2; hence a_p = a_q for p, q ≥ 2. a₁ + a₂ = 3 and a₁ − a₂ = −1.

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Question 8: The coefficient of \( x^n \) in the expansion of \( e^{2x+3} \) is
(a) \( \frac{2^n}{n!} \)
(b) \( \frac{e^3\cdot2^n}{n!} \)
(c) \( \frac{e^2\cdot3^n}{n!} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac{e^3\cdot2^n}{n!} \)

Explanation:
1. \( e^{2x+3}=e^3\cdot e^{2x}=e^3\sum\frac{(2x)^n}{n!} \).
2. The coefficient is \( \frac{e^3\,2^n}{n!} \).

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Question 9: The coefficient of \( x^{10} \) in the expansion of \( 10^x \) in ascending powers of x is
(a) \( \frac{(\log_e10)^{10}}{10!} \)
(b) \( \frac1{10!} \)
(c) \( \frac{(\log_{10}e)^{10}}{10!} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac{(\log_e10)^{10}}{10!} \)

Explanation:
1. \( 10^x=e^{x\ln10}=\sum\frac{(x\ln10)^r}{r!} \).
2. For r = 10 the coefficient is \( \frac{(\ln10)^{10}}{10!} \).

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Question 10: In the expansion of \( \frac{e^x-1-x}{x^2} \) in ascending powers of x, the fourth term is
(a) \( \frac1{5!}x^3 \)
(b) \( \frac1{4!}x^4 \)
(c) \( \frac1{3!}x^3 \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac1{5!}x^3 \)

Explanation:
1. The series is \( \frac12+\frac x{3!}+\frac{x^2}{4!}+\frac{x^3}{5!}+\cdots \).
2. The fourth term is \( \frac{x^3}{5!} \).

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Question 11: The constant term in the expansion of \( \frac{3^x-2^x}{x^2} \) is
(a) \( \log_e3 \)
(b) \( \log_e6\cdot\log_e\frac32 \)
(c) \( \frac12\log_e6\cdot\log_e\frac32 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac12\log_e6\cdot\log_e\frac32 \)

Explanation:
1. \( 3^x-2^x=x\ln\frac32+\frac{x^2}2(\ln^23-\ln^22)+\cdots \).
2. Dividing by x², the constant term is \( \frac12(\ln3-\ln2)(\ln3+\ln2)=\frac12\ln\frac32\cdot\ln6 \).

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Question 12: If |x| < 1, the coefficient of \( x^3 \) in the expansion of \( \frac1{e^x(1+x)} \) is
(a) \( \frac{17}6 \)
(b) \( -\frac{17}6 \)
(c) \( -\frac{11}6 \)
(d) \( -\frac83 \)
Show Answer & Explanation

Answer: (d) \( -\frac83 \)

Explanation:
1. \( e^{-x}=1-x+\frac{x^2}2-\frac{x^3}6 \) and \( \frac1{1+x}=1-x+x^2-x^3 \).
2. Coefficient of x³ = 1·(−1) + (−1)·1 + (1/2)·(−1) + (−1/6)·1 = −1 − 1 − 1/2 − 1/6 = −8/3.

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Question 13: The constant term in the expansion of \( \frac{x+\log_e(1-x)}{x^3} \) is
(a) \( -\frac13 \)
(b) 0
(c) \( -\frac12 \)
(d) \( \frac13 \)
Show Answer & Explanation

Answer: (a) \( -\frac13 \)

Explanation:
1. \( \log_e(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\cdots \), so the numerator is \( -\frac{x^2}2-\frac{x^3}3-\cdots \).
2. Dividing by x³: \( -\frac1{2x}-\frac13-\frac x4-\cdots \). The constant term is −1/3.

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Question 14: In the expansion of \( \log_{10}(1-x) \), |x| < 1, the coefficient of \( x^n \) is
(a) \( -\frac1n \)
(b) \( -\frac1n\log_{10}e \)
(c) \( \frac1n \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( -\frac1n\log_{10}e \)

Explanation:
1. \( \log_{10}(1-x)=\log_{10}e\cdot\ln(1-x)=-\log_{10}e\sum\frac{x^n}n \).
2. The coefficient is \( -\frac1n\log_{10}e \).

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Question 15: If |x| < 1, the coefficient of \( x^2 \) in the expansion of \( \frac{\log_e(1+x)}{(1-x)^2} \) is
(a) \( \frac32 \)
(b) \( {}^2C_1 \)
(c) \( -\frac12 \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac32 \)

Explanation:
1. \( \ln(1+x)=x-\frac{x^2}2+\cdots \) and \( (1-x)^{-2}=1+2x+3x^2+\cdots \).
2. Coefficient of x²: 1·2 + (−1/2)·1 = 3/2.

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Question 16: The sum of the series \( {}^4C_0+{}^5C_1x+{}^6C_2x^2+{}^7C_3x^3+\cdots \) to ∞ is
(a) \( (1-x)^{-4} \)
(b) \( \frac1{(1-x)^5} \)
(c) \( (1+x)^{-5} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac1{(1-x)^5} \)

Explanation:
1. The general term is \( \binom{4+r}rx^r \), the coefficient of x^r in \( (1-x)^{-5} \).
2. So the sum is \( (1-x)^{-5} \).

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Question 17: The sum of the series \( {}^2C_0-{}^3C_1x^2+{}^4C_2x^4-{}^5C_3x^6+\cdots \) to ∞ is
(a) \( \frac1{(1+x^2)^3} \)
(b) \( (1-x^2)^{-3} \)
(c) \( \frac1{(1+x)^3} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac1{(1+x^2)^3} \)

Explanation:
1. The general term is \( (-1)^r\binom{r+2}rx^{2r} \).
2. This is the expansion of \( (1+x^2)^{-3} \).

Teacher's Note:
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Question 18: The sum of the series \( 1+\frac1{1!}\cdot\frac14+\frac{1\cdot3}{2!}\left(\frac14\right)^2+\frac{1\cdot3\cdot5}{3!}\left(\frac14\right)^3+\cdots \) to ∞ is
(a) \( \sqrt2 \)
(b) 2
(c) \( \frac1{\sqrt2} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \sqrt2 \)

Explanation:
1. \( (1-z)^{-1/2}=1+\frac z2+\frac{1\cdot3}{2\cdot4}z^2+\cdots=\sum\frac{1\cdot3\cdots(2r-1)}{r!}\left(\frac z2\right)^r \).
2. With z/2 = 1/4, z = 1/2: the sum is \( \left(1-\frac12\right)^{-1/2}=\sqrt2 \).

Teacher's Note:
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Question 19: The sum of the series \( {}^3C_0-{}^4C_1\cdot\frac12+{}^5C_2\left(\frac12\right)^2-{}^6C_3\left(\frac12\right)^3+\cdots \) to ∞ is
(a) 16
(b) 8
(c) \( \frac{16}{81} \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac{16}{81} \)

Explanation:
1. The general term is \( (-1)^r\binom{r+3}r\left(\frac12\right)^r \), the expansion of \( (1+x)^{-4} \) at x = 1/2.
2. Sum \( =\left(\frac32\right)^{-4}=\frac{16}{81} \).

Teacher's Note:
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Question 20: \( 1+\frac1{3!}+\frac1{5!}+\frac1{7!}+\cdots \) to ∞ is equal to
(a) \( \frac12(e-e^{-1}) \)
(b) \( \frac12(e+e^{-1}) \)
(c) \( \frac12e \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac12(e-e^{-1}) \)

Explanation:
1. The series is \( \sinh1=\frac{e-e^{-1}}2 \) (the sum of odd-factorial reciprocals), since 1 = 1/1! is its first term.

Teacher's Note:
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