JEE Mathematics Binomial Theorem MCQs Set 14

Answer: (c) 10

Explanation:
1. \( T_6={}^8C_5\left(x^{-8/3}\right)^3\left(x^2\log_{10}x\right)^5=56\,x^{-8}x^{10}(\log_{10}x)^5 \).
2. So \( 56x^2(\log_{10}x)^5=5600 \), i.e. \( x^2(\log_{10}x)^5=100 \).
3. \( x=10 \) gives \( 100\cdot1=100 \), which satisfies the equation.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (a) 2

Explanation:
1. The first part equals \( (4^x+44)^{1/5} \) and the second equals \( (2^{x-1}+7)^{-1/3} \).
2. \( T_4={}^8C_3\left[(4^x+44)^{1/5}\right]^5\left[(2^{x-1}+7)^{-1/3}\right]^3=56\dfrac{4^x+44}{2^{x-1}+7}=336 \).
3. So \( 4^x+44=6\cdot2^{x-1}+42=3\cdot2^x+42 \). With \( t=2^x \): \( t^2-3t+2=0 \), so \( t=1 \) or \( 2 \).
4. Hence \( x=0 \) or \( x=1 \): two values.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (d) 7.59

Explanation:
1. \( \dfrac{T_3}{T_2}=\dfrac{n-1}2\cdot\dfrac ax=3 \) and \( \dfrac{T_4}{T_3}=\dfrac{n-2}3\cdot\dfrac ax=\dfrac32 \).
2. Dividing: \( \dfrac{3(n-1)}{2(n-2)}=2 \), so \( 3n-3=4n-8 \) and \( n=5 \), then \( a/x=\frac32 \).
3. \( T_2=5x^4a=240 \) gives \( x^5\cdot\frac32\cdot5=240 \), so \( x^5=32 \), \( x=2 \), \( a=3 \).
4. The ratio is \( \dfrac{a^5}{x^5}=\dfrac{243}{32}\approx7.59 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (b) \( \dfrac12 \)

Explanation:
1. \( 2^n=256 \) gives \( n=8 \); the greatest binomial coefficient term is \( T_5=70\cdot3^{-x}\cdot3^{5x}=70\cdot3^{4x} \).
2. \( T_3=28\cdot3^{-3x/2}3^{5x/2}=28\cdot3^x \).
3. \( 4\cdot70\cdot3^{4x}-784\cdot3^{2x}=168 \). With \( u=3^{2x} \): \( 280u^2-784u-168=0 \), i.e. \( 5u^2-14u-3=0 \), so \( u=3 \).
4. \( 3^{2x}=3 \) gives \( x=\frac12 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (b) \( -\dfrac{19}2 \)

Explanation:
1. Multiply by \( 19! \): \( \sum_{k=1}^{19}{}^{19}C_k(-2)^k=(1-2)^{19}-1=-2 \).
2. So the sum equals \( -\dfrac2{19!}=-\dfrac2{19\cdot18!} \).
3. Equating with \( \dfrac1{k_0\cdot18!} \) gives \( k_0=-\dfrac{19}2 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (d) 50

Explanation:
1. Put \( y=\dfrac x{1+x} \). The expression is \( (1+x)^{1000}\sum_{j=0}^{1000}(j+1)y^j \), and terms with \( j>50 \) cannot affect the coefficient of \( x^{50} \), so we may sum the full series \( (1+x)^{1000}(1-y)^{-2}=(1+x)^{1002} \).
2. The coefficient of \( x^{50} \) is therefore \( {}^{1002}C_{50} \), so the smallest \( p \) is 50 (the other solution is \( p=952 \)).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (a) \( \dfrac{82}{3} \)

Explanation:
1. \( 3^4=81=82-1 \), so \( 3^{1001}=3\cdot3^{1000}=3\cdot81^{250}=3(82-1)^{250} \).
2. \( (82-1)^{250}=82m+1 \), hence \( 3^{1001}=246m+3 \) leaves remainder 3 on division by 82.
3. The fractional part is \( \dfrac3{82} \), so \( \lambda=\dfrac{82}3 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (c) \( \dfrac54 \)

Explanation:
1. The coefficient of the \( r \)th term is proportional to \( {}^nC_{r-1}2^{r-1} \). Then \( \dfrac{T_{r+1}}{T_r}=\dfrac{2(n-r+1)}r \).
2. For \( T_9 \) to be the unique greatest we need \( \dfrac{T_9}{T_8}>1 \) and \( \dfrac{T_{10}}{T_9}<1 \): \( \dfrac{2(n-7)}8>1 \) gives \( n>11 \) and \( \dfrac{2(n-8)}9<1 \) gives \( n<12.5 \).
3. So \( n=12 \) and \( \dfrac{T_9}{T_8}=\dfrac{2\cdot5}{8}=\dfrac54 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (c) 34

Explanation:
1. Rearrange: \( {}^{39}C_{3r-1}+{}^{39}C_{3r}={}^{39}C_{r^2-1}+{}^{39}C_{r^2} \), i.e. \( {}^{40}C_{3r}={}^{40}C_{r^2} \).
2. So \( 3r=r^2 \) or \( 3r+r^2=40 \): \( r=3 \) (r = 0 is rejected) or \( r=5 \) (\( r=-8 \) is rejected).
3. Sum of squares \( =9+25=34 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (a) 1

Explanation:
1. The sum is \( \sum_{j=0}^{6}(-1)^j\,{}^6C_j\,{}^{12-j}C_6 \), the coefficient of \( x^6 \) in \( \sum(-1)^j\,{}^6C_j(1+x)^{12-j}=(1+x)^6\left[(1+x)-1\right]^6=x^6(1+x)^6 \).
2. The coefficient of \( x^6 \) in \( x^6(1+x)^6 \) is \( 1 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (d) \( \dfrac1{12} \)

Explanation:
1. \( \dfrac{C_k}{C_{k-1}}=\dfrac{n-k+1}k \), so the term is \( \dfrac{k(n-k+1)^2}{n(n+1)^2(n+2)} \).
2. \( \sum_{k=1}^nk(n+1-k)^2=\dfrac{n(n+1)^2(n+2)}{12} \) (check \( n=1 \): \( 1 \) and \( 1\cdot4\cdot3/12=1 \); \( n=2 \): \( 4+2=6 \) and \( 2\cdot9\cdot4/12=6 \)).
3. Hence the value is \( \dfrac1{12}\approx0.08 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (b) 1

Explanation:
1. The first sum is \( 2^{10} \).
2. The second sum is \( \left(1-\frac12\right)^{10}=2^{-10} \).
3. The product is 1.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (b) 8.25

Explanation:
1. \( {}^{100}C_m\,{}^mC_{97}={}^{100}C_{97}\,{}^{3}C_{m-97} \), so the sum is \( {}^{100}C_{97}\cdot2^3=161700\cdot8 \).
2. \( {}^{99}C_{96}=156849 \). Hence \( \lambda=\dfrac{1293600}{156849}=\dfrac{800}{97}\approx8.25 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (d) 3.50

Explanation:
1. Differentiate and put \( x=1 \): \( \sum ka_k=7\cdot\left(1+x+\cdots+x^7\right)^6\left(1+2x+\cdots+7x^6\right) \) at \( x=1 \), which is \( 7\cdot8^6\cdot28 \).
2. Dividing by \( 7\cdot8^7 \) gives \( \dfrac{28}{8}=3.5 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (a) \( \dfrac43 \)

Explanation:
1. \( \dfrac{2r+3}{r+1}=2+\dfrac1{r+1} \).
2. \( \sum2\,{}^nC_r=2^{n+1} \) and \( \sum\dfrac{{}^nC_r}{r+1}=\dfrac{2^{n+1}-1}{n+1} \).
3. Total \( =\dfrac{(n+1)2^{n+1}+2^{n+1}-1}{n+1}=\dfrac{(2n+4)2^n-1}{n+1} \), so \( 3k=4 \) and \( k=\dfrac43 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (c) 3.5

Explanation:
1. Using \( \dfrac{{}^nC_r}{(r+1)(r+2)(r+3)}=\dfrac{{}^{n+3}C_{r+3}}{(n+1)(n+2)(n+3)} \), the sum is \( \dfrac{1}{(n+1)(n+2)(n+3)}\sum(-1)^r{}^{n+3}C_{r+3} \).
2. Since \( \sum_{j=0}^{n+3}(-1)^j{}^{n+3}C_j=0 \), \( \sum_{r=0}^{n}(-1)^r{}^{n+3}C_{r+3}=1-(n+3)+{}^{n+3}C_2=\dfrac{(n+1)(n+2)}2 \).
3. Hence the sum is \( \dfrac1{2(n+3)}=\dfrac{1/2}{n+3} \), so \( a=\frac12 \), \( b=3 \) and \( a+b=3.5 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (c) 0

Explanation:
1. Let \( \omega=i\sqrt3 \), so \( \omega^2=-3 \) and \( (-3)^k=\omega^{2k}=\omega\cdot\omega^{2k-1} \). The required sum is \( \omega\times \) (the odd-index part of \( (1+\omega)^{6n} \)).
2. \( 1+\omega=2e^{i\pi/3} \), so \( (1+\omega)^{6n}=2^{6n}e^{2n\pi i}=2^{6n} \) is real. The odd-index terms are purely imaginary, so their sum must be zero.
3. The required sum is 0.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (a) \( \dfrac12 \)

Explanation:
1. The product is \( \left(1-\dfrac{y^2}{x^2}\right)^{-1/2} \).
2. Neglecting powers higher than \( y^2/x^2 \), \( \left(1-t\right)^{-1/2}\approx1+\dfrac t2 \) with \( t=\dfrac{y^2}{x^2} \), so \( k=\dfrac12 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (d) All four (1, 2, 3 and 4)

Explanation:
1. \( T_{r+1}={}^{20}C_r\,4^{(20-r)/3}\,6^{-r/4} \). Writing the term as a product of powers of 2 and 3, it is rational only if \( r \) is a multiple of 4 and \( 20-r \) is a multiple of 3.
2. \( r\in\{8,20\} \), so there are 2 rational terms and 19 irrational ones (statements 1 and 3).
3. The middle term is \( T_{11} \) (r = 10), irrational; the 9th term is \( T_9 \) (r = 8), rational. All four are correct.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (b) 3 and 4 only

Explanation:
1. \( \left(\dfrac{1+x}{1-x}\right)^2=(1+x)^2(1-x)^{-2} \) with \( (1-x)^{-2}=\sum(r+1)x^r \).
2. Coefficient of \( x^4 \): \( 5+2\cdot4+3=16 \).
3. \( 10+{}^4C_2=16 \) too, so options 3 and 4 are correct.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (b) 1 and 3 only

Explanation:
1. \( 7^9+9^7=40353607+4782969=45136576 \).
2. \( 45136576=64\times705259 \) and is divisible by 16 and 64, but not by 3 (digit sum 37), so not by 24 or 72.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (d) 1 and 3 only

Explanation:
1. \( \sum_{r=1}^n(-1)^{r-1}{}^nC_r=1 \) and \( \sum_{r=1}^n(-1)^{r-1}r\,{}^nC_r=0 \) for \( n\ge2 \).
2. So the sum equals \( a \): \( 5 \) when \( a=5 \) and \( -5 \) when \( a=-5 \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (a) 3 and 4 only

Explanation:
1. \( \dfrac{a_{n+1}}{a_n}=\dfrac{1000}{n+1} \), which is greater than 1 for \( n<999 \), equal to 1 at \( n=999 \) and less than 1 afterwards.
2. So \( a_{999}=a_{1000} \) is the greatest value.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (c) 1 and 3 only

Explanation:
1. \( \sum(r+1)\,{}^nC_r(-3)^r=(1-3)^n+(-3n)(1-3)^{n-1}=(-2)^{n-1}(-2-3n)=(-2)^n\left(1+\dfrac{3n}2\right) \).
2. For even \( n \) this is \( 2^n\left(\frac{3n}2+1\right) \) and for odd \( n \) it is \( -2^n\left(\frac{3n}2+1\right) \).

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.

Answer: (c) 1, 3 and 4

Explanation:
1. By Pascal's rule the left side is \( {}^{20}C_r \) and \( {}^{20}C_{13}={}^{20}C_7 \).
2. \( {}^{20}C_r\ge{}^{20}C_7 \) holds for \( 7\le r\le13 \), so 9, 7 and 10 qualify but 5 does not.

Teacher's Note:
1. Work out the value yourself and then match it with the options; do not guess from the pattern.
2. Check small cases (for example n = 1, 2, 3) when a result involves a general n.