JEE Mathematics Some Additional Topics MCQs Set 04

Answer: (b) 7th term

Explanation:
1. \( \tfrac{27}{6}=\tfrac92 \). The coefficient of the \( (r+1) \)th term is \( \dfrac{\frac92\left(\frac92-1\right)\cdots\left(\frac92-r+1\right)}{r!} \).
2. The factors \( \tfrac92,\tfrac72,\tfrac52,\tfrac32,\tfrac12 \) are positive for \( r\le5 \), and the sixth factor \( \tfrac92-5=-\tfrac12 \) is negative.
3. So the coefficient first becomes negative at \( r=6 \), the 7th term.

Teacher's Note:
1. A term turns negative when a factor of the numerator first becomes negative.

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

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

Teacher's Note:
1. Separate the constant factor e cubed from the exponential in x.

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

Explanation:
1. \( 10^x=e^{x\ln10}=\sum\dfrac{(x\ln10)^n}{n!} \).
2. The coefficient of \( x^{10} \) is \( \dfrac{(\ln10)^{10}}{10!} \).

Teacher's Note:
1. Write the base as e raised to the natural log of the base.

Answer: (d) none of these

Explanation:
1. \( e^{-x}=1-x+\tfrac{x^2}2-\tfrac{x^3}6+\dots \) and \( \dfrac1{1+x}=1-x+x^2-x^3+\dots \).
2. The coefficient of \( x^3 \) in the product is \( (1)(-1)+(-1)(1)+\left(\tfrac12\right)(-1)+\left(-\tfrac16\right)(1)=-1-1-\tfrac12-\tfrac16=-\tfrac83 \).
3. The value \( -\tfrac83 \) is not among the options.

Teacher's Note:
1. Multiply the two series and collect all pairs of terms whose powers add up to 3.

Answer: (b) \( -\dfrac1n\log_{10}e \)

Explanation:
1. \( \log_e(1-x)=-\sum\dfrac{x^n}{n} \).
2. \( \log_{10}(1-x)=\log_{10}e\cdot\log_e(1-x) \).
3. So the coefficient of \( x^n \) is \( -\dfrac1n\log_{10}e \).

Teacher's Note:
1. Use the change of base formula from natural log to base 10.

Answer: (a) \( \tfrac32 \)

Explanation:
1. \( \log_e(1+x)=x-\dfrac{x^2}2+\dots \) and \( (1-x)^{-2}=1+2x+3x^2+\dots \).
2. The \( x^2 \) coefficient is \( 1\cdot2+\left(-\tfrac12\right)\cdot1=\tfrac32 \).
3. So the coefficient is \( \tfrac32 \).

Teacher's Note:
1. Pair the terms of the two series whose powers add up to 2.

Answer: (a) \( \tfrac12\left(e-e^{-1}\right) \)

Explanation:
1. \( \sinh x=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dots \).
2. At \( x=1 \): \( 1+\dfrac1{3!}+\dfrac1{5!}+\dots=\sinh1 \).
3. \( \sinh1=\tfrac12(e-e^{-1}) \).

Teacher's Note:
1. The odd-factorial series is the hyperbolic sine.

Answer: (b) \( \dfrac1{2e} \)

Explanation:
1. The nth term is \( \dfrac{n}{(2n+1)!}=\dfrac12\cdot\dfrac{(2n+1)-1}{(2n+1)!}=\dfrac12\left[\dfrac1{(2n)!}-\dfrac1{(2n+1)!}\right] \).
2. Summing for \( n\ge1 \): \( \tfrac12\left[(\cosh1-1)-(\sinh1-1)\right]=\tfrac12(\cosh1-\sinh1) \).
3. \( \cosh1-\sinh1=e^{-1} \), so the sum is \( \dfrac1{2e} \).

Teacher's Note:
1. Write n as half of (2n + 1) minus 1 to split the term.

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

Explanation:
1. \( -\log_e(1-t)=t+\dfrac{t^2}2+\dfrac{t^3}3+\dots \) with \( t=\dfrac{a-b}a \).
2. \( 1-t=\dfrac ba \), so the sum is \( -\log\dfrac ba=\log\dfrac ab \).
3. This equals \( \log a-\log b \).

Teacher's Note:
1. Recognise the series for minus log of (1 - t).

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

Explanation:
1. \( \dfrac1{2n(2n+1)}=\dfrac1{2n}-\dfrac1{2n+1} \).
2. The sum is \( \dfrac12-\dfrac13+\dfrac14-\dfrac15+\dots \).
3. Since \( \log_e2=1-\dfrac12+\dfrac13-\dfrac14+\dots \), the sum equals \( 1-\log_e2 \).

Teacher's Note:
1. Split into partial fractions and compare with the series of log 2.

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

Explanation:
1. Multiply by \( (1+x)(1+x^2) \): \( 1=A(1+x^2)+f(x)(1+x) \).
2. Put \( x=-1 \): \( 1=2A \), so \( A=\tfrac12 \).
3. Then \( f(x)=\dfrac{1-\tfrac12(1+x^2)}{1+x}=\dfrac{1-x^2}{2(1+x)}=\dfrac{1-x}2 \).

Teacher's Note:
1. Find A by substituting x = -1, then solve for f(x).

Answer: (b) \( 1-\dfrac1{2^{n+1}} \)

Explanation:
1. \( \dfrac1{x^2-3x+2}=\dfrac1{(1-x)(2-x)}=\dfrac1{1-x}-\dfrac1{2-x} \).
2. \( \dfrac1{1-x}=\sum x^n \) and \( \dfrac1{2-x}=\dfrac12\sum\dfrac{x^n}{2^n} \).
3. The coefficient of \( x^n \) is \( 1-\dfrac1{2^{n+1}} \).

Teacher's Note:
1. Use partial fractions and expand each as a geometric series.

Answer: (c) 21

Explanation:
1. \( \dfrac1{1+2x}=1-2x+4x^2-8x^3+16x^4-\dots \) and \( \dfrac1{1-x^2}=1+x^2+x^4+\dots \).
2. The \( x^4 \) coefficient is \( 16\cdot1+4\cdot1+1\cdot1=21 \).
3. So the coefficient is 21.

Teacher's Note:
1. Multiply the two geometric series and collect the pairs whose powers add up to 4.

Answer: (a) \( \dfrac1{1+a} \)

Explanation:
1. \( \dfrac1{(k+a)(k+1+a)}=\dfrac1{k+a}-\dfrac1{k+1+a} \).
2. The sum to N terms telescopes to \( \dfrac1{1+a}-\dfrac1{N+1+a} \).
3. As \( N\to\infty \) the sum tends to \( \dfrac1{1+a} \).

Teacher's Note:
1. Break each term into a difference so that the series telescopes.

Answer: (c) \( \tfrac12+\tfrac14(\sqrt6+\sqrt2) \)

Explanation:
1. \( \dfrac1{1+\sqrt2-\sqrt3}=\dfrac{(1+\sqrt2)+\sqrt3}{(1+\sqrt2)^2-3}=\dfrac{1+\sqrt2+\sqrt3}{2\sqrt2} \).
2. Multiply by \( \dfrac{\sqrt2}{\sqrt2} \): \( \dfrac{\sqrt2+2+\sqrt6}{4}=\dfrac12+\dfrac{\sqrt2+\sqrt6}4 \).
3. Numerically both are about 1.4659, which matches \( 1/x \).

Teacher's Note:
1. Rationalise using (A - B)(A + B) with A = 1 + root 2 and B = root 3.

Answer: (a) \( \dfrac1{70}\left(3\sqrt{10}+2\sqrt5\right) \)

Explanation:
1. Simplify: \( \sqrt{40}=2\sqrt{10},\ \sqrt{20}=2\sqrt5,\ \sqrt{80}=4\sqrt5 \). The denominator is \( 3\sqrt{10}-2\sqrt5 \).
2. Multiply by the conjugate: \( \dfrac{3\sqrt{10}+2\sqrt5}{(3\sqrt{10})^2-(2\sqrt5)^2}=\dfrac{3\sqrt{10}+2\sqrt5}{90-20} \).
3. So the value is \( \dfrac{3\sqrt{10}+2\sqrt5}{70} \).

Teacher's Note:
1. Reduce every surd to its simplest form and then rationalise.

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

Explanation:
1. \( \left(\sqrt{x+1}+\sqrt{x-1}\right)^2=(x+1)+(x-1)+2\sqrt{(x+1)(x-1)} \).
2. This equals \( 2x+2\sqrt{x^2-1} \).
3. So the square root is \( \sqrt{x+1}+\sqrt{x-1} \) (positive for \( x\ge1 \)).

Teacher's Note:
1. Try the sum of two square roots and square it to compare.

Answer: (a) \( x=-1,\ y=12 \)

Explanation:
1. \( 7+4\sqrt3=(2+\sqrt3)^2 \), so \( \sqrt{7+4\sqrt3}=2+\sqrt3 \).
2. \( \dfrac{4+3\sqrt3}{2+\sqrt3}=(4+3\sqrt3)(2-\sqrt3)=8-4\sqrt3+6\sqrt3-9=-1+2\sqrt3 \).
3. \( 2\sqrt3=\sqrt{12} \), so \( x=-1,\ y=12 \).

Teacher's Note:
1. Write the radicand as a perfect square of a surd.

Answer: (a) 1

Explanation:
1. \( \dfrac{1+x}{1+\sqrt{1+x}}=\dfrac{(1+x)(\sqrt{1+x}-1)}{x} \) and \( \dfrac{1-x}{1-\sqrt{1-x}}=\dfrac{(1-x)(1+\sqrt{1-x})}{x} \).
2. The sum is \( \dfrac{(1+x)^{3/2}+(1-x)^{3/2}-2x}{x} \). For \( x=\tfrac{\sqrt3}2 \), \( \sqrt{1\pm x}=\dfrac{\sqrt3\pm1}{2} \), so \( (1+x)^{3/2}+(1-x)^{3/2}=\dfrac{(6\sqrt3+10)+(6\sqrt3-10)}{8}=\dfrac{3\sqrt3}2 \).
3. The sum is \( \dfrac{\frac{3\sqrt3}2-\sqrt3}{\frac{\sqrt3}2}=1 \).

Teacher's Note:
1. Rationalise each fraction and use the surd forms of the roots at this value of x.

Answer: (b) z is a constant

Explanation:
1. \( x=k_1yz \) and \( y=k_2xz \).
2. Substituting, \( x=k_1z(k_2xz)=k_1k_2xz^2 \), so \( z^2=\dfrac1{k_1k_2} \).
3. Hence z is a constant.

Teacher's Note:
1. Substitute one relation into the other and cancel x.

Answer: (a) 30

Explanation:
1. \( A=kBC \). From the first set, \( 2=k\cdot\tfrac35\cdot\tfrac{10}{27}=k\cdot\tfrac29 \), so \( k=9 \).
2. Then \( A=9\cdot2\cdot\tfrac53=30 \).

Teacher's Note:
1. Joint variation means A is proportional to the product of B and C.

Answer: (c) \( 5x+\dfrac{36}{x^2} \)

Explanation:
1. Write \( y=\alpha x+\dfrac\beta{x^2} \).
2. \( 19=2\alpha+\dfrac\beta4 \) and \( 19=3\alpha+\dfrac\beta9 \). Subtracting, \( \alpha=\beta\left(\tfrac14-\tfrac19\right)=\tfrac{5\beta}{36} \).
3. Then \( 19=\tfrac{10\beta}{36}+\tfrac{9\beta}{36}=\tfrac{19\beta}{36} \), so \( \beta=36 \) and \( \alpha=5 \): \( y=5x+\dfrac{36}{x^2} \).

Teacher's Note:
1. Combine the constants of variation into two unknowns and use both conditions.