JEE Mathematics Binomial Theorem MCQs Set 13

Answer: (c) \( \dfrac58 \)

Explanation:
1. \( T_{r+1}={}^5C_r\left(\frac2x\right)^{5-r}\left(-\frac x2\right)^r \) has the power \( x^{2r-5} \).
2. For \( x^3 \) we need \( r=4 \): \( {}^5C_4\cdot\frac2x\cdot\frac{x^4}{16}=\frac{10}{16}x^3 \).
3. The sign is positive because \( (-1)^4=1 \), so the coefficient is \( 5/8 \).

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) \( y^8+8y^5+24y^2+\dfrac{32}{y}+\dfrac{16}{y^4} \)

Explanation:
1. Coefficients are \( {}^4C_r2^r=1,8,24,32,16 \).
2. Powers of \( y \) are \( 2(4-r)-r=8-3r \), that is \( 8,5,2,-1,-4 \).
3. Hence the expansion is \( y^8+8y^5+24y^2+32/y+16/y^4 \).

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) 9

Explanation:
1. The 7th term from the end is the 7th term from the beginning of the expansion taken in reverse order.
2. \( \dfrac{T_7}{T_7'}=\dfrac{{}^nC_6(2^{1/3})^{n-6}(3^{-1/3})^6}{{}^nC_6(3^{-1/3})^{n-6}(2^{1/3})^6}=\left(\sqrt[3]{2\cdot3}\right)^{n-12}=6^{(n-12)/3} \).
3. \( 6^{(n-12)/3}=6^{-1} \) gives \( n-12=-3 \), so \( n=9 \).

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) 7

Explanation:
1. Odd powers of the radical cancel, so the sum is \( 2\left[x^5+{}^5C_2x^3(x^3-1)+{}^5C_4x(x^3-1)^2\right] \).
2. The middle term gives \( 10x^3\cdot x^3=10x^6 \) and the last term gives \( 5x\cdot x^6=5x^7 \), which is the highest power.
3. Hence the degree is 7.

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) \( -2^7\cdot{}^{12}C_7 \)

Explanation:
1. \( T_{r+1}={}^{12}C_ra^{12-r}(-2b)^r \).
2. For \( b^7 \) take \( r=7 \): coefficient \( {}^{12}C_7(-2)^7=-2^7\cdot{}^{12}C_7=-101376 \).

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) \( ab=1 \)

Explanation:
1. In the first expansion \( T_{r+1}={}^{11}C_ra^{11-r}b^{-r}x^{22-3r} \); \( 22-3r=7 \) gives \( r=5 \) and the coefficient \( {}^{11}C_5\dfrac{a^6}{b^5} \).
2. In the second \( T_{r+1}={}^{11}C_r(-1)^ra^{11-r}b^{-r}x^{11-3r} \); \( 11-3r=-7 \) gives \( r=6 \) and the coefficient \( {}^{11}C_6\dfrac{a^5}{b^6} \) (sign \( + \)).
3. Since \( {}^{11}C_5={}^{11}C_6 \), equality gives \( \dfrac{a^6}{b^5}=\dfrac{a^5}{b^6} \), so \( ab=1 \).

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: (a) \( \dfrac{17}{54} \)

Explanation:
1. General term of the bracket: \( {}^9C_r\left(\frac32\right)^{9-r}\left(-\frac13\right)^rx^{18-3r} \).
2. We need \( x^0 \) from \( 1\cdot T \), \( x^{-1} \) from \( x\cdot T \) and \( x^{-3} \) from \( 2x^3\cdot T \); the powers \( 18-3r \) are multiples of 3, so only the first (r = 6) and third (r = 7) can contribute.
3. \( r=6 \): \( {}^9C_6\left(\frac32\right)^3\left(\frac13\right)^6=\frac{84\cdot27}{8\cdot729}=\frac{7}{18} \). \( r=7 \): \( 2\cdot{}^9C_7\left(\frac32\right)^2\left(-\frac13\right)^7=-\frac{2\cdot36\cdot9}{4\cdot2187}=-\frac{2}{27} \).
4. Sum \( =\frac7{18}-\frac2{27}=\frac{21-4}{54}=\frac{17}{54} \).

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) 171

Explanation:
1. The coefficient of \( x^5 \) is \( \sum_{i+j=5}{}^6C_i2^i\,{}^7C_j(-1)^j \).
2. With \( (1+2x)^6 \) coefficients \( 1,12,60,160,240,192 \) and \( (1-x)^7 \) coefficients \( 1,-7,21,-35,35,-21 \), the sum over \( i+j=5 \) is \( -21+420-2100+3360-1680+192=171 \).

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) -438

Explanation:
1. Write \( (1+2x)^4=1+8x+24x^2+32x^3+16x^4 \) and \( (2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5 \).
2. Coefficient of \( x^4 \): \( 1\cdot10+8\cdot(-40)+24\cdot80+32\cdot(-80)+16\cdot32=-438 \).

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) 15

Explanation:
1. \( T_{r+1}={}^nC_r(-1)^rx^{3n-5r} \).
2. \( 3n-5r=5 \) and \( 3n-5r=10 \) give \( r=\frac{3n-5}5 \) and \( r'=\frac{3n-10}5 \), so \( r'=r-1 \).
3. The coefficients are \( (-1)^r{}^nC_r \) and \( (-1)^{r-1}{}^nC_{r-1} \); their sum is zero when \( {}^nC_r={}^nC_{r-1} \), i.e. \( r+r-1=n \).
4. Then \( 2\cdot\frac{3n-5}5-1=n \) gives \( 6n-10-5=5n \), so \( n=15 \) (with \( r=8 \)).

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) \( -\dfrac{35x}{y} \) and \( \dfrac{35y}{x} \)

Explanation:
1. For \( n=7 \) there are two middle terms, \( T_4 \) and \( T_5 \).
2. \( T_4={}^7C_3\left(\frac xy\right)^4\left(-\frac yx\right)^3=-35\frac xy \).
3. \( T_5={}^7C_4\left(\frac xy\right)^3\left(\frac yx\right)^4=35\frac yx \).

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) \( (-1)^n\dfrac{(2n)!}{n!\,n!}x^n \)

Explanation:
1. \( (1-2x+x^2)^n=(1-x)^{2n} \), which has \( 2n+1 \) terms.
2. The middle term is \( T_{n+1}={}^{2n}C_n(-x)^n=(-1)^n\dfrac{(2n)!}{n!\,n!}x^n \).

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) 4

Explanation:
1. \( 7^{98}=49^{49}=(50-1)^{49} \).
2. All terms except the last are multiples of 50, and the last term is \( (-1)^{49}=-1 \).
3. So the remainder is \( -1+5=4 \).

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) 803

Explanation:
1. \( 27^{27}=3^{81} \), so square repeatedly modulo 1000.
2. \( 3^{10}\equiv 49 \), \( 3^{20}\equiv401 \), \( 3^{40}\equiv801 \), \( 3^{80}\equiv601 \) (mod 1000).
3. \( 3^{81}\equiv601\cdot3=1803\equiv803 \) (mod 1000).

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) \( 101^{50} \)

Explanation:
1. \( 101^{50}=(100+1)^{50}=100^{50}+50\cdot100^{49}+\cdots \).
2. \( 99^{50}=(100-1)^{50}=100^{50}-50\cdot100^{49}+\cdots \).
3. Subtracting, \( 101^{50}-99^{50}=2\left[50\cdot100^{49}+{}^{50}C_3100^{47}+\cdots\right]=100^{50}+\text{positive terms} \), because \( 2\cdot50\cdot100^{49}=100^{50} \).
4. Therefore \( 101^{50}>99^{50}+100^{50} \).

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) \( T_4 \) and \( T_5 \)

Explanation:
1. At \( x=\frac15 \) the expression is \( (3-1)^{15} \) term by term: \( T_{r+1}={}^{15}C_r3^{15-r}(-1)^r \).
2. \( \left|\frac{T_{r+1}}{T_r}\right|=\frac{16-r}{r}\cdot\frac13=1 \) when \( r=4 \).
3. So \( |T_4|=|T_5| \) is the greatest magnitude, \( 455\cdot3^{12} \), with \( T_4 \) negative and \( T_5 \) positive.

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) the 22nd term

Explanation:
1. \( (2x+5y)^{34}=(6+10)^{34} \), so the ratio of consecutive terms is \( \dfrac{T_{r+1}}{T_r}=\dfrac{35-r}{r}\cdot\dfrac{10}{6} \).
2. This is \( \ge1 \) when \( 5(35-r)\ge3r \), that is \( r\le21.875 \), so the terms increase up to \( r=21 \).
3. The greatest term is \( T_{22} \).

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) \( T_5 \) and \( T_6 \) (equal in magnitude)

Explanation:
1. Magnitudes of \( T_{r+1}={}^6C_r2^{6-r}5^r \) are \( 64,960,6000,20000,37500,37500,15625 \).
2. The largest value is \( 37500 \), shared by \( T_5 \) and \( T_6 \).

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) \( T_5 \) and \( T_6 \)

Explanation:
1. The signs alternate: the terms are \( 64x^6,-960x^5,6000x^4,-20000x^3,37500x^2,-37500x,15625 \) at \( x=1 \).
2. The greatest value is \( +37500 \) (\( T_5 \)) and the least is \( -37500 \) (\( T_6 \)).

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) 1020

Explanation:
1. Use \( {}^nC_r+{}^nC_{r+1}={}^{n+1}C_{r+1} \) repeatedly: \( \sum_{k=r}^{m}{}^kC_r={}^{m+1}C_{r+1} \).
2. Sum \( ={}^{20}C_3-{}^{10}C_3=1140-120=1020 \).

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) \( \dfrac{15015}{16} \)

Explanation:
1. \( (1-2x)^{-5/2}=\sum\dfrac{(5/2)(7/2)\cdots\left(\frac52+r-1\right)}{r!}(2x)^r \).
2. For \( r=6 \): \( \dfrac{\frac52\cdot\frac72\cdot\frac92\cdot\frac{11}2\cdot\frac{13}2\cdot\frac{15}2}{720}\cdot64=\dfrac{15015}{16} \).

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) 142

Explanation:
1. The series of \( (1+x)^{-3} \) has coefficient \( (-1)^r\frac{(r+1)(r+2)}2 \) for \( x^r \), giving \( 91,-78,66 \) for \( r=12,11,10 \).
2. The coefficient of \( x^{12} \) is \( 4\cdot91+2\cdot(-78)-1\cdot66=364-156-66=142 \).

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) -197

Explanation:
1. \( (1-x)^{-2}=\sum(r+1)x^r \).
2. Coefficient of \( x^{100} \): \( 3\cdot101-5\cdot100=303-500=-197 \).

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) 280

Explanation:
1. Choose \( p \) factors \( bc \), \( q \) factors \( ca \), \( r \) factors \( ab \) with \( p+q+r=8 \): the term is \( a^{q+r}b^{p+r}c^{p+q} \).
2. \( q+r=5 \), \( p+r=4 \), \( p+q=7 \) gives \( p=3,q=4,r=1 \).
3. Coefficient \( =\dfrac{8!}{3!\,4!\,1!}=280 \).

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) \( 2^5 \)

Explanation:
1. Let \( f(x)=(9x^2+x-8)^6 \). The sum of the odd-power coefficients is \( \dfrac{f(1)-f(-1)}2 \).
2. \( f(1)=(9+1-8)^6=2^6 \) and \( f(-1)=(9-1-8)^6=0 \).
3. So the sum is \( \dfrac{64-0}2=32=2^5 \).

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) 20

Explanation:
1. Use the multinomial theorem with \( 1^{a}(-2x)^b(x^3)^c \), \( a+b+c=5 \), \( b+3c=7 \).
2. \( (b,c)=(1,2) \): \( a=2 \), coefficient \( \frac{5!}{2!1!2!}(-2)=-60 \). \( (b,c)=(4,1) \): \( a=0 \), coefficient \( \frac{5!}{0!4!1!}(16)=80 \).
3. Sum \( =-60+80=20 \).

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.