JEE Mathematics Some Additional Topics MCQs Set 05

Answer: (c) \( -1\lt x\lt 1 \)

Explanation:
1. The binomial series \( (1+u)^{m} \) for a non-integer m converges only for \( |u|\lt 1 \).
2. Here \( u=-x \), so we need \( |x|\lt 1 \), i.e. \( -1\lt x\lt 1 \).

Teacher's Note:
1. A non-integral index gives an infinite series, valid only when the variable part has modulus less than 1.
2. Identify the variable part, here \( -x \).

Answer: (b) \( -3\lt x\lt 3 \)

Explanation:
1. Write \( (3+x)^{p/q}=3^{p/q}\left(1+\dfrac{x}{3}\right)^{p/q} \).
2. The series in \( \dfrac{x}{3} \) converges when \( \left|\dfrac{x}{3}\right|\lt 1 \).
3. So \( |x|\lt 3 \), i.e. \( -3\lt x\lt 3 \).

Teacher's Note:
1. Take the larger number out of the bracket so that the bracket looks like \( 1+u \).
2. Then apply the condition \( |u|\lt 1 \).

Answer: (b) \( 36 \)

Explanation:
1. \( (1-x)^{-n}=\sum {}^{n+r-1}C_r\,x^{r} \).
2. For \( n=3 \), \( r=7 \): the coefficient is \( {}^{9}C_7=\dfrac{9\cdot 8}{2}=36 \).

Teacher's Note:
1. The coefficient of \( x^{r} \) in \( (1-x)^{-n} \) is \( {}^{n+r-1}C_r \).
2. Use \( {}^{9}C_7={}^{9}C_2 \).

Answer: (c) \( \dfrac{1155\sqrt2}{16384} \)

Explanation:
1. \( (2+x)^{-5/2}=2^{-5/2}\left(1+\dfrac{x}{2}\right)^{-5/2} \).
2. The coefficient of \( u^{4} \) in \( (1+u)^{-5/2} \) is \( \dfrac{(-\frac52)(-\frac72)(-\frac92)(-\frac{11}{2})}{4!}=\dfrac{5\cdot7\cdot9\cdot11}{16\cdot 24}=\dfrac{1155}{128} \).
3. With \( u=\dfrac{x}{2} \) we get a factor \( \left(\dfrac12\right)^{4}=\dfrac1{16} \), so the coefficient is \( 2^{-5/2}\cdot\dfrac{1155}{128}\cdot\dfrac1{16}=\dfrac{1155}{2048\cdot 2^{5/2}}=\dfrac{1155\sqrt2}{16384} \) (checked with a computer algebra system, about 0.0997).

Teacher's Note:
1. Factor out \( 2^{-5/2} \) before applying the binomial series.
2. Do not forget the factor \( (1/2)^{4} \) that comes from replacing x by \( x/2 \).

Answer: (d) \( a_p=a_q \) for all \( p,q\ge 2 \)

Explanation:
1. \( \dfrac{x^{2}+x}{1-x}=(x+x^{2})(1+x+x^{2}+\dots)=x+2x^{2}+2x^{3}+2x^{4}+\dots \).
2. So \( a_1=1 \) and \( a_k=2 \) for every \( k\ge 2 \).
3. Then \( a_1+a_2=3 \), \( a_1-a_2=-1 \) and \( a_1\neq a_2 \), so (a), (b), (c) are false. But \( a_p=a_q=2 \) for all \( p,q\ge2 \), so (d) is true.

Teacher's Note:
1. Multiply by the geometric series \( 1/(1-x) \).
2. Collect the coefficient of each power of x.

Answer: (d) \( \dfrac{e^{3}\,2^{n}}{n!} \)

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

Teacher's Note:
1. Separate the constant factor \( e^{3} \).
2. Use the exponential series for \( e^{2x} \).

Answer: (d) \( \dfrac{x}{2}\ln\dfrac{1+x}{1-x} \)

Explanation:
1. Write \( \dfrac{x^{2n}}{2n-1}=x\cdot\dfrac{x^{2n-1}}{2n-1} \).
2. So the sum is \( x\left(x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+\dots\right) \).
3. The series \( x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+\dots=\dfrac12\ln\dfrac{1+x}{1-x} \).
4. Hence the sum is \( \dfrac{x}{2}\ln\dfrac{1+x}{1-x} \).

Teacher's Note:
1. Take out one factor x to get the odd-power series.
2. Recall the series for \( \tfrac12\ln\frac{1+x}{1-x} \).

Answer: (b) \( 2\ln 2-1 \)

Explanation:
1. The general term is \( \dfrac{(-1)^{n+1}}{n(n+1)}=(-1)^{n+1}\left(\dfrac1n-\dfrac1{n+1}\right) \).
2. \( \sum\dfrac{(-1)^{n+1}}{n}=\ln 2 \) and \( \sum\dfrac{(-1)^{n+1}}{n+1}=\dfrac12-\dfrac13+\dfrac14-\dots=1-\ln 2 \).
3. The sum is \( \ln2-(1-\ln2)=2\ln2-1 \approx 0.386 \).

Teacher's Note:
1. Split each term using partial fractions.
2. Use the series for \( \ln(1+x) \) at \( x=1 \).

Answer: (b) \( \dfrac92 \)

Explanation:
1. \( x=ky \), so \( 2=3k \) and \( k=\dfrac23 \).
2. For \( x=3 \): \( 3=\dfrac23 y \), so \( y=\dfrac92 \).

Teacher's Note:
1. Write the variation as an equation with a constant k.
2. Find k from the first pair of values.

Answer: (d) (i) and (iv) only

Explanation:
1. For \( |x|\lt 1 \): \( (1+x)^{-3}=1-3x+6x^{2}-\dots \). The third term is \( 6x^{2}={}^{4}C_2x^{2} \), so (i) is true.
2. At \( x=\tfrac12 \) the third term is \( 6\cdot\tfrac14=\tfrac32 \) (positive), so (iii) is false.
3. For \( |x|\gt 1 \): \( (1+x)^{-3}=x^{-3}\left(1+\tfrac1x\right)^{-3} \), whose third term is \( x^{-3}\cdot{}^{4}C_2\,x^{-2}=\dfrac{6}{x^{5}} \), not \( {}^{4}C_2x^{-2} \). So (ii) is false.
4. At \( x=-3 \): \( \dfrac{6}{(-3)^{5}}=-\dfrac{6}{243}=-\dfrac{2}{81} \), so (iv) is true.

Teacher's Note:
1. For \( |x|\gt 1 \) take x out of the bracket first.
2. Substitute the values carefully, watching the signs.

Answer: (b) \( a=\dfrac87,\ b=-\dfrac{319}{343} \) or \( a=-\dfrac87,\ b=\dfrac{319}{343} \)

Explanation:
1. \( \sqrt{4+ax}=2\left(1+\dfrac{ax}{4}\right)^{1/2}=2+\dfrac{ax}{4}-\dfrac{a^{2}x^{2}}{64}+\dfrac{a^{3}x^{3}}{512}-\dots \).
2. \( \dfrac{2}{\sqrt{1-ax}}=2\left(1+\dfrac{ax}{2}+\dfrac{3a^{2}x^{2}}{8}+\dfrac{5a^{3}x^{3}}{16}+\dots\right) \).
3. The left side becomes \( \dfrac{3ax}{4}+2+\dfrac{ax}{4}-\dfrac{a^{2}x^{2}}{64}+\dfrac{a^{3}x^{3}}{512}-2-ax-\dfrac{3a^{2}x^{2}}{4}-\dfrac{5a^{3}x^{3}}{8} \), so the x term vanishes, and the \( x^{2} \) coefficient is \( -\dfrac{49a^{2}}{64} \) and the \( x^{3} \) coefficient is \( -\dfrac{319a^{3}}{512} \).
4. \( -\dfrac{49a^{2}}{64}=-1 \) gives \( a=\pm\dfrac87 \).
5. \( b=-\dfrac{319a^{3}}{512} \): for \( a=\dfrac87 \), \( b=-\dfrac{319}{343} \); for \( a=-\dfrac87 \), \( b=\dfrac{319}{343} \).

Teacher's Note:
1. Expand both square-root terms by the binomial series up to \( x^{3} \).
2. Match the coefficients of \( x^{2} \) and \( x^{3} \).

Answer: (d) (ii) and (iii) only

Explanation:
1. \( e^{ax}=1+ax+\dfrac{a^{2}x^{2}}{2}+\dfrac{a^{3}x^{3}}{6}+\dots \) and \( e^{-bx}=1-bx+\dfrac{b^{2}x^{2}}{2}-\dfrac{b^{3}x^{3}}{6}+\dots \).
2. Adding: \( p_0=2,\ p_1=a-b,\ p_2=\dfrac{a^{2}+b^{2}}{2},\ p_3=\dfrac{a^{3}-b^{3}}{6} \).
3. So (ii) and (iii) are true, while (i) and (iv) are false.

Teacher's Note:
1. Expand both exponentials and add the coefficients of equal powers.
2. Watch the sign of odd powers of b.

Answer: (c) (i), (iii) and (iv) only

Explanation:
1. \( \ln\dfrac{1+x}{1-x}=2\left(x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+\dots\right) \).
2. So \( a_0=0,\ a_1=2,\ a_2=0,\ a_3=\dfrac23,\ a_5=\dfrac25 \). (i) is true and (ii) is false; \( a_0+a_1+a_2=2 \) so (iii) is true.
3. The reciprocals of \( a_1,a_3,a_5 \) are \( \dfrac12,\dfrac32,\dfrac52 \), which are in AP, so \( a_1,a_3,a_5 \) are in HP: (iv) is true.

Teacher's Note:
1. Write the logarithm as the difference of two standard series; the even powers cancel.
2. A set of numbers is in HP when their reciprocals are in AP.

Answer: (c) (i) and (iv) only

Explanation:
1. \( \dfrac{1+x}{1+x^{2}}=(1+x)\left(1-x^{2}+x^{4}-\dots\right)=1+x-x^{2}-x^{3}+x^{4}+x^{5}-\dots \).
2. For even n the coefficient is \( (-1)^{n/2} \): (i) is true and (iii) is false.
3. For odd n it is \( (-1)^{(n-1)/2} \), which is never 0: (iv) is true and (ii) is false.

Teacher's Note:
1. Expand \( \dfrac{1}{1+x^{2}} \) as a geometric series first.
2. Multiply by \( 1+x \) and read off the pattern of coefficients.

Answer: (d) (i), (ii) and (iv) only

Explanation:
1. Rationalise: \( a=\dfrac{1}{\sqrt5+2},\ b=\dfrac{1}{\sqrt7+\sqrt6},\ c=\dfrac{1}{\sqrt{13}+\sqrt{12}} \).
2. The denominators satisfy \( \sqrt5+2\approx4.236\lt \sqrt7+\sqrt6\approx5.095\lt \sqrt{13}+\sqrt{12}\approx7.070 \).
3. A larger denominator gives a smaller fraction, so \( a\gt b\gt c \).
4. Therefore (i), (ii) and (iv) are true; (iii) is false.

Teacher's Note:
1. Write each difference of roots as its conjugate fraction with numerator 1.
2. Compare the denominators, remembering the larger denominator gives the smaller number.

Answer: (a) (i) and (iv) only

Explanation:
1. Write \( x+y=k_1\left(z+\dfrac1z\right) \) and \( x-y=k_2\left(z-\dfrac1z\right) \). At \( z=2,\ x=3,\ y=1 \): \( 4=k_1\cdot\dfrac52 \), so \( k_1=\dfrac85 \), and \( 2=k_2\cdot\dfrac32 \), so \( k_2=\dfrac43 \).
2. Adding: \( 2x=\dfrac85\left(z+\dfrac1z\right)+\dfrac43\left(z-\dfrac1z\right)=\dfrac{44}{15}z+\dfrac{4}{15z} \), so \( x=\dfrac{22}{15}z+\dfrac{2}{15z}=\dfrac{2}{15}\left(11z+\dfrac1z\right) \): (i) is true and (ii) is false.
3. Subtracting: \( 2y=\dfrac{4}{15}z+\dfrac{44}{15z} \), so \( y=\dfrac{2}{15}\left(z+\dfrac{11}{z}\right) \): (iv) is true and (iii) is false.

Teacher's Note:
1. Introduce separate constants of variation for the two relations.
2. Add and subtract the two equations to separate x and y.

Answer: (a) (i), (ii) and (iii)

Explanation:
1. Let \( 2x+3y=k\sqrt{xy} \). Put \( t=\sqrt{x/y} \): \( 2t^{2}+3=kt \), a quadratic with constant coefficients, so t takes fixed values and \( x/y \) is a constant, say \( m \).
2. Then \( x^{2}+y^{2}=(m^{2}+1)y^{2} \) and \( xy=my^{2} \), so \( x^{2}+y^{2}=\dfrac{m^{2}+1}{m}\,xy \): (i) is true.
3. Similarly \( x^{3}+y^{3}=(m^{3}+1)y^{3} \) and \( (xy)^{3/2}=m^{3/2}y^{3} \): (ii) is true; and \( x^{4}+y^{4}=(m^{4}+1)y^{4} \) with \( x^{2}y^{2}=m^{2}y^{4} \): (iii) is true.

Teacher's Note:
1. Divide by y to see that x/y must be a constant.
2. Once the ratio is constant every homogeneous expression is proportional to the matching power of y.