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.
Access Some Additional Topics Questions and Solutions
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Multiple Choice Questions
(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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45. .
(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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79. .
(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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64. .
(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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(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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55. .
(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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65. .
(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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51. .
(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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63. .
(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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62. .
(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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70. .
(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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57. .
(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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(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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49. .
(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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51. .
(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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66. .
(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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Free study material for Binomial Theorem
Practice MCQs for JEE Mathematics Some Additional Topics
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FAQs
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