JEE Mathematics Some Additional Topics MCQs Set 03

Practice MCQs for JEE Mathematics Some Additional Topics

Access targeted multiple-choice questions for Some Additional Topics designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: \( \frac1{0!}+\frac2{1!}+\frac3{2!}+\frac4{3!}+\cdots \) to ∞ is equal to
(a) 4e
(b) 3e
(c) 2e
(d) none of these
Show Answer & Explanation

Answer: (c) 2e

Explanation:
1. The nth term is \( \frac{n+1}{n!}=\frac1{(n-1)!}+\frac1{n!} \).
2. Summing: e + e = 2e (each part sums to e).

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Question 2: \( \frac1{3!}+\frac2{5!}+\frac3{7!}+\cdots \) to ∞ is equal to
(a) \( \frac12e \)
(b) \( \frac1{2e} \)
(c) \( \frac3{2e} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac1{2e} \)

Explanation:
1. The rth term is \( \frac r{(2r+1)!}=\frac12\left[\frac1{(2r)!}-\frac1{(2r+1)!}\right] \).
2. Summing over r ≥ 1: \( \frac12\left[(\cosh1-1)-(\sinh1-1)\right]=\frac12(\cosh1-\sinh1)=\frac1{2e} \).

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Question 3: \( (a^2-b^2)+\frac{a^4-b^4}{2!}+\frac{a^6-b^6}{3!}+\cdots \) to ∞ is equal to
(a) \( e^{a^2}-e^{b^2} \)
(b) \( e^{b^2}-e^{a^2} \)
(c) \( \frac{e^{a^2}}{e^{b^2}} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( e^{a^2}-e^{b^2} \)

Explanation:
1. Split into two series: \( \left(a^2+\frac{a^4}{2!}+\cdots\right)-\left(b^2+\frac{b^4}{2!}+\cdots\right) \).
2. Each is \( e^{t}-1 \) with t = a² and t = b²; the difference is \( e^{a^2}-e^{b^2} \).

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Question 4: \( \frac{a-b}a+\frac12\left(\frac{a-b}a\right)^2+\frac13\left(\frac{a-b}a\right)^3+\cdots \) to ∞ is equal to
(a) \( \log a+\log b \)
(b) \( \log\frac ba \)
(c) \( \log a-\log b \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \log a-\log b \)

Explanation:
1. The series is \( -\ln(1-z) \) with \( z=\frac{a-b}a \).
2. So the sum is \( -\ln\frac ba=\ln a-\ln b \).

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Question 5: \( \sum_{n=1}^\infty\frac1{2n-1}x^{2n} \) is equal to
(a) \( \frac x2\log\frac{1+x}{1-x} \)
(b) \( \frac12\log\frac{1+x^2}{1-x^2} \)
(c) \( x\log\frac{1+x}{1-x} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac x2\log\frac{1+x}{1-x} \)

Explanation:
1. \( \sum\frac{x^{2n-1}}{2n-1}=\frac12\ln\frac{1+x}{1-x} \) (the series for tanh⁻¹x).
2. Multiplying by x gives \( \frac x2\ln\frac{1+x}{1-x} \).

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Question 6: \( \sum_{n=1}^\infty\frac1{2n(2n+1)} \) is equal to
(a) \( \log_e\frac2e \)
(b) \( 1-\log_e2 \)
(c) \( 1-\log_e\frac1{2e} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( 1-\log_e2 \)

Explanation:
1. \( \frac1{2n(2n+1)}=\frac1{2n}-\frac1{2n+1} \).
2. The sum is \( \left(\frac12-\frac13\right)+\left(\frac14-\frac15\right)+\cdots=1-\ln2 \) (since \( \ln2=1-\frac12+\frac13-\cdots \)).

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Question 7: If \( x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots=y \) to ∞, then \( y+\frac{y^2}{2!}+\frac{y^3}{3!}+\cdots \) to ∞ is equal to
(a) −x
(b) x
(c) x + 1
(d) none of these
Show Answer & Explanation

Answer: (b) x

Explanation:
1. The first series is ln(1 + x) = y, so x = eʸ − 1.
2. The second series is eʸ − 1 = x.

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Question 8: The sum of the series \( \frac1{1\cdot2}-\frac1{2\cdot3}+\frac1{3\cdot4}-\cdots \) up to ∞ is equal to
(a) \( \log_e2-1 \)
(b) \( \log_e2 \)
(c) \( \log_e\frac4e \)
(d) \( 2\log_e2 \)
Show Answer & Explanation

Answer: (c) \( \log_e\frac4e \)

Explanation:
1. Write \( \frac1{n(n+1)}=\frac1n-\frac1{n+1} \).
2. Alternating sum \( =\sum\frac{(-1)^{n+1}}n-\sum\frac{(-1)^{n+1}}{n+1}=\ln2-(1-\ln2)=2\ln2-1=\ln\frac4e \).

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Question 9: If \( \frac{3x+5}{2x^2+x-6}=\frac A{2x-3}+\frac B{x+2} \) identically, then
(a) \( A=\frac17,\ B=\frac{19}7 \)
(b) \( A=-\frac{19}7,\ B=\frac17 \)
(c) \( A=\frac{19}7,\ B=\frac17 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( A=\frac{19}7,\ B=\frac17 \)

Explanation:
1. Since 2x² + x − 6 = (2x − 3)(x + 2), we need 3x + 5 = A(x + 2) + B(2x − 3).
2. x = −2: −1 = −7B so B = 1/7. x = 3/2: 19/2 = (7/2)A so A = 19/7.

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Question 10: If f(x) is a function of x such that \( \frac1{(1+x)(1+x^2)}=\frac A{1+x}+\frac{f(x)}{1+x^2} \) for all x ∈ R, then f(x) is
(a) \( \frac{1-x}2 \)
(b) \( \frac{x+1}2 \)
(c) 1 − x
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac{1-x}2 \)

Explanation:
1. Multiply by (1 + x): at x = −1, A = 1/2.
2. Then \( f(x)(1+x)=1-\frac12(1+x^2)=\frac{(1-x)(1+x)}2 \), so f(x) = (1 − x)/2.

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Question 11: If |x| < 1, the coefficient of \( x^n \) in the expansion of \( \frac1{x^2-3x+2} \) is
(a) \( 1-\frac1{2^n} \)
(b) \( 1-\frac1{2^{n+1}} \)
(c) \( 1-2^n \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( 1-\frac1{2^{n+1}} \)

Explanation:
1. \( \frac1{(x-1)(x-2)}=\frac1{1-x}-\frac12\cdot\frac1{1-x/2} \).
2. Coefficient of xⁿ: \( 1-\frac12\cdot\frac1{2^n}=1-\frac1{2^{n+1}} \).

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Question 12: If |x| < 1/2, the coefficient of \( x^4 \) in the expansion of \( \frac1{(1+2x)(1-x^2)} \) is
(a) 1
(b) 2
(c) 21
(d) none of these
Show Answer & Explanation

Answer: (c) 21

Explanation:
1. \( \frac1{1+2x}=1-2x+4x^2-8x^3+16x^4 \) and \( \frac1{1-x^2}=1+x^2+x^4 \).
2. Coefficient of x⁴: 16 + 4 + 1 = 21.

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Question 13: The sum to infinite terms of the series \( \frac1{(1+a)(2+a)}+\frac1{(2+a)(3+a)}+\frac1{(3+a)(4+a)}+\cdots \), where a is a constant, is
(a) \( \frac1{1+a} \)
(b) \( \frac2{1+a} \)
(c) ∞
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac1{1+a} \)

Explanation:
1. Each term is \( \frac1{n+a}-\frac1{n+1+a} \).
2. The sum telescopes to \( \frac1{1+a}-0=\frac1{1+a} \).

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Question 14: If x = 1 + √2 − √3, then the reciprocal of x is
(a) \( \frac12-\frac14(\sqrt6+\sqrt2) \)
(b) \( \sqrt3-\sqrt2-1 \)
(c) \( \frac12+\frac14(\sqrt6+\sqrt2) \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac12+\frac14(\sqrt6+\sqrt2) \)

Explanation:
1. Multiply by the conjugate: \( \frac1{(1+\sqrt2)-\sqrt3}=\frac{(1+\sqrt2)+\sqrt3}{(1+\sqrt2)^2-3}=\frac{1+\sqrt2+\sqrt3}{2\sqrt2} \).
2. Multiply by √2/√2: \( \frac{\sqrt2+2+\sqrt6}{4}=\frac12+\frac14(\sqrt6+\sqrt2) \).

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Question 15: The value of \( \frac1{\sqrt{40}+\sqrt{20}+\sqrt{10}-\sqrt{80}} \) is equal to
(a) \( \frac1{70}(3\sqrt{10}+2\sqrt5) \)
(b) \( \frac{3\sqrt{10}-2\sqrt5}{70} \)
(c) \( \frac{3\sqrt{10}+2\sqrt5}{50} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac1{70}(3\sqrt{10}+2\sqrt5) \)

Explanation:
1. Simplify: √40 = 2√10, √20 = 2√5, √80 = 4√5, so the denominator is \( 3\sqrt{10}-2\sqrt5 \) (since 2√5 − 4√5 = −2√5).
2. Rationalise: \( \frac{3\sqrt{10}+2\sqrt5}{90-20}=\frac{3\sqrt{10}+2\sqrt5}{70} \).

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Question 16: The square root of \( 2x+2\sqrt{x^2-1} \) is
(a) \( \sqrt{x+1}-\sqrt{x-1} \)
(b) \( \sqrt{x+1}+\sqrt{x-1} \)
(c) \( \sqrt{x-1}-\sqrt{x+1} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \sqrt{x+1}+\sqrt{x-1} \)

Explanation:
1. \( (\sqrt{x+1}+\sqrt{x-1})^2=(x+1)+(x-1)+2\sqrt{x^2-1}=2x+2\sqrt{x^2-1} \).

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Practice MCQs for JEE Mathematics Some Additional Topics

Chapter MCQs with Answers for JEE Mathematics

Test your conceptual understanding of Some Additional Topics with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Mathematics, these problem sets build accuracy and prepare students for objective exams.

Expert Practice Material for JEE Mathematics

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Complete Your Chapter Revision

Follow up your worksheet practice by attempting the interactive online Mathematics MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Mathematics Some Additional Topics MCQs Set 03?

You can get most exhaustive JEE Mathematics Some Additional Topics MCQs Set 03 for free on StudiesToday.com. These MCQs for JEE Mathematics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics JEE material?

Yes, our JEE Mathematics Some Additional Topics MCQs Set 03 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in JEE exams?

By solving our JEE Mathematics Some Additional Topics MCQs Set 03, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for JEE Mathematics Some Additional Topics MCQs Set 03?

Yes, Mathematics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Mathematics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Mathematics Some Additional Topics MCQs Set 03 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.