JEE Mathematics Binomial Theorem MCQs Set 15

Answer: (d) 24

Explanation:
1. The sum equals \( 2\left[{}^6C_0x^6+{}^6C_2x^4(x^3-1)+{}^6C_4x^2(x^3-1)^2+{}^6C_6(x^3-1)^3\right] \).
2. Expanding: \( 2\left[x^6+15x^7-15x^4+15x^8-30x^5+15x^2+x^9-3x^6+3x^3-1\right] \). The terms with even powers of \( x \) are \( x^6,\ -15x^4,\ 15x^8,\ 15x^2,\ -3x^6,\ -1 \).
3. Sum of their coefficients \( =2(1-15+15+15-3-1)=24 \). Answer: (d)

Teacher's Note:
1. Use \( (a+b)^n+(a-b)^n=2[{}^nC_0a^n+{}^nC_2a^{n-2}b^2+\cdots] \).
2. Pick out only the terms with even power of \( x \) after expansion.

Answer: (d) \( 8^2 \)

Explanation:
1. \( T_4={}^6C_3\left(\frac2x\right)^3\left(x^{\log_8x}\right)^3=20\cdot\frac8{x^3}\cdot x^{3\log_8x}=20\times8^7 \).
2. So \( x^{3\log_8x-3}=8^6 \), i.e. \( x^{\log_8x-1}=8^2 \). Taking \( \log_8 \): with \( u=\log_8x \), \( u(u-1)=2 \), so \( u^2-u-2=0 \) and \( u=2 \) or \( -1 \).
3. \( x=8^2 \) or \( 8^{-1} \). Only \( 8^2 \) is among the options. Answer: (d)

Teacher's Note:
1. Use the general term \( T_{r+1}={}^nC_ra^{n-r}b^r \).
2. Take \( \log_8 \) of both sides to get a quadratic.

Answer: (a) \( (28,315) \)

Explanation:
1. Coefficient of \( x^2 \): \( {}^{15}C_2\cdot9-3a\cdot15+b=0\Rightarrow945-45a+b=0 \), i.e. \( 45a-b=945 \).
2. Coefficient of \( x^3 \): \( -27\,{}^{15}C_3+9a\,{}^{15}C_2-3b\cdot15=0\Rightarrow-12285+945a-45b=0 \), i.e. \( 21a-b=273 \).
3. Subtracting: \( 24a=672\Rightarrow a=28 \), and \( b=45\cdot28-945=315 \). Answer: (a)

Teacher's Note:
1. Use \( (1-3x)^{15}=\sum{}^{15}C_r(-3x)^r \).
2. Set the two coefficients to zero and solve the pair of linear equations.

Answer: (a) 84

Explanation:
1. \( (1-x)(1+x+x^2)=1-x^3 \), so the product is \( (1+x)(1-x)(1-x^3)^9=(1-x^2)(1-x^3)^9 \).
2. Coefficient of \( x^{18} \) = coefficient of \( x^{18} \) in \( (1-x^3)^9 \) minus coefficient of \( x^{16} \) in \( (1-x^3)^9 \). \( (1-x^3)^9 \) has only powers that are multiples of 3, so the second is 0.
3. First: \( {}^9C_6(-1)^6=84 \). Answer: (a)

Teacher's Note:
1. Combine \( (1-x) \) with \( (1+x+x^2) \) to get \( 1-x^3 \).
2. Only multiples of 3 appear in \( (1-x^3)^9 \).

Answer: (d) 8

Explanation:
1. \( 2^{403}=8\cdot2^{400}=8\cdot16^{100} \), and \( 16=15+1 \).
2. \( 16^{100}=(1+15)^{100}=1+15\lambda \), so \( 2^{403}=8+120\lambda \).
3. \( \frac{2^{403}}{15}=8\lambda+\frac8{15} \), so the fractional part is \( \frac8{15} \) and \( k=8 \). Answer: (d)

Teacher's Note:
1. Write \( 2^{403} \) using powers of 16.
2. Use the binomial theorem on \( (1+15)^{100} \).

Answer: (b) \( \frac14 \)

Explanation:
1. \( T_3={}^5C_2\left(x^{\log_2x}\right)^2=10\,x^{2\log_2x}=2560 \), so \( x^{2\log_2x}=256 \).
2. Taking \( \log_2 \): \( 2(\log_2x)^2=8 \), so \( \log_2x=\pm2 \).
3. \( x=4 \) or \( x=\frac14 \). Only \( \frac14 \) is among the options. Answer: (b)

Teacher's Note:
1. Use \( T_3={}^5C_2(\ldots)^2 \).
2. Take \( \log_2 \) on both sides.

Answer: (d) 4

Explanation:
1. \( T_{r+1}=x^2\,{}^{10}C_r\,x^{\frac{10-r}2}\lambda^rx^{-2r} \); its power of \( x \) is \( 2+\frac{10-r}2-2r \).
2. For \( x^2 \): \( \frac{10-r}2-2r=0\Rightarrow r=2 \). Coefficient \( ={}^{10}C_2\lambda^2=45\lambda^2=720 \).
3. \( \lambda^2=16 \), so the positive value is \( \lambda=4 \). Answer: (d)

Teacher's Note:
1. Include the outer factor \( x^2 \) when counting powers.
2. Solve for \( r \) first.

Answer: (b) 0

Explanation:
1. With \( n=8 \) the middle term is \( T_5={}^8C_4\left(\frac{x^3}3\right)^4\left(\frac3x\right)^4=70x^8 \).
2. \( 70x^8=5670\Rightarrow x^8=81\Rightarrow x=\pm\sqrt3 \).
3. The sum of the real values is \( \sqrt3+(-\sqrt3)=0 \). Answer: (b)

Teacher's Note:
1. For \( n=8 \) the middle term is the 5th term.
2. Both signs of \( x \) satisfy an even power.

Answer: (a) 12.25

Explanation:
1. Only even powers survive: \( a_0=2\cdot10^{50} \) and \( a_2=2\,{}^{50}C_2\cdot10^{48} \).
2. \( \frac{a_2}{a_0}=\frac{1225\cdot10^{48}}{10^{50}}=\frac{1225}{100}=12.25 \). Answer: (a)

Teacher's Note:
1. Add the two expansions so that odd terms cancel.
2. Compute \( {}^{50}C_2=1225 \).

Answer: (c) 14

Explanation:
1. AP condition: \( 2\,{}^nC_5={}^nC_4+{}^nC_6 \). Dividing by \( {}^nC_5 \): \( 2=\frac5{n-4}+\frac{n-5}6 \).
2. \( 12(n-4)=30+(n-4)(n-5) \) gives \( n^2-21n+98=0 \), i.e. \( (n-7)(n-14)=0 \).
3. \( n=7 \) or \( 14 \); only 14 is among the options. Answer: (c)

Teacher's Note:
1. Use \( \frac{{}^nC_{r+1}}{{}^nC_r}=\frac{n-r}{r+1} \).
2. Solve the quadratic in \( n \).

Answer: (d) 2

Explanation:
1. The expression equals \( 2\left[x^5+10x^3(x^3-1)+5x(x^3-1)^2\right]=2\left(x^5+10x^6-10x^3+5x^7-10x^4+5x\right) \).
2. Odd-degree terms: \( x^5,\ -10x^3,\ 5x^7,\ 5x \), with coefficients \( 1,-10,5,5 \), sum \( =1 \); with the factor 2 the sum is 2. Answer: (d)

Teacher's Note:
1. Use \( (a+b)^n+(a-b)^n \) to keep only the even-index terms.
2. Add the coefficients of the odd powers of \( x \) only.

Answer: (d) 729

Explanation:
1. The number of terms in the expansion of a trinomial to the power \( n \) is \( {}^{n+2}C_2=\frac{(n+2)(n+1)}2 \).
2. \( \frac{(n+2)(n+1)}2=28\Rightarrow(n+2)(n+1)=56=8\times7\Rightarrow n=6 \).
3. The sum of the coefficients is the value at \( x=1 \): \( (1-2+4)^6=3^6=729 \). Answer: (d)

Teacher's Note:
1. Count the terms of a trinomial expansion.
2. The sum of coefficients is found by putting \( x=1 \).

Answer: (d) \( \left(16,\frac{272}3\right) \)

Explanation:
1. Coefficient of \( x^3 \): \( -8\,{}^{18}C_3+4a\,{}^{18}C_2-2b\,{}^{18}C_1=0 \), which simplifies to \( 51a-3b=544 \).
2. Coefficient of \( x^4 \): \( 16\,{}^{18}C_4-8a\,{}^{18}C_3+4b\,{}^{18}C_2=0 \), which simplifies to \( 544a-51b=4080 \) (and the first condition is \( 51a-3b=544 \)).
3. Solving the two equations: \( a=16 \) and \( b=\frac{272}3 \). Answer: (d)

Teacher's Note:
1. Use \( (1-2x)^{18}=\sum{}^{18}C_r(-2x)^r \).
2. Form two linear equations in \( a \) and \( b \).

Answer: (a) an irrational number

Explanation:
1. Expand both by the binomial theorem; terms with even powers of \( \sqrt3 \) cancel in the difference and the odd powers add.
2. So the difference is \( 2\left[{}^{2n}C_1(\sqrt3)^{2n-1}+{}^{2n}C_3(\sqrt3)^{2n-3}+\cdots\right] \), every term of which is a rational multiple of \( \sqrt3 \) with a positive coefficient.
3. Hence the value is \( \sqrt3 \) times a positive rational number, an irrational number. Answer: (a) (Check: \( n=1 \) gives \( 4\sqrt3 \).)

Teacher's Note:
1. Look at which terms survive in the difference.
2. Odd powers of \( \sqrt3 \) are irrational.

Answer: (c) A is true, but R is false.

Explanation:
1. \( S_1=\sum j(j-1)\frac{10!}{j!(10-j)!}=90\sum_{j=2}^{10}\frac{8!}{(j-2)!(10-j)!}=90\times2^8 \).
2. \( S_2=\sum j\,{}^{10}C_j=10\times2^9 \), not \( 10\times2^8 \). So the Reason is false.
3. \( S_3=S_1+S_2=90\times2^8+10\times2^9=110\times2^8=55\times2^9 \), so the Assertion is true. Answer: (c)

Teacher's Note:
1. Use \( j\,{}^nC_j=n\,{}^{n-1}C_{j-1} \).
2. Write \( j^2=j(j-1)+j \).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Multiply \( \sum{}^nC_rx^r=(1+x)^n \) by \( x \) and differentiate: \( \sum(r+1){}^nC_rx^r=(1+x)^n+nx(1+x)^{n-1} \). So the Reason is true.
2. Put \( x=1 \): \( \sum(r+1){}^nC_r=2^n+n2^{n-1}=(n+2)2^{n-1} \), which is the Assertion. So the Reason explains the Assertion. Answer: (a)

Teacher's Note:
1. Multiply by \( x \) and differentiate the binomial expansion.
2. Substitute \( x=1 \).

Answer: (b) \( \frac12\,{}^{20}C_{10} \)

Explanation:
1. Put \( x=-1 \) in \( (1+x)^{20} \): the alternating sum of all terms is 0. By symmetry the terms \( {}^{20}C_k \) and \( {}^{20}C_{20-k} \) have equal sign, so \( 2\left({}^{20}C_0-{}^{20}C_1+\cdots-{}^{20}C_9\right)+{}^{20}C_{10}=0 \).
2. Hence \( {}^{20}C_0-{}^{20}C_1+\cdots-{}^{20}C_9=-\frac12{}^{20}C_{10} \), and adding \( {}^{20}C_{10} \) gives \( \frac12\,{}^{20}C_{10} \). Answer: (b)

Teacher's Note:
1. Use the alternating sum of all coefficients, which is 0.
2. Use the symmetry of the coefficients.

Answer: (c) \( (35,45) \)

Explanation:
1. \( a_1=n-m=10 \) and \( a_2={}^mC_2+{}^nC_2-mn=10 \).
2. \( \frac{m(m-1)+n(n-1)}2-mn=10\Rightarrow(m-n)^2-(m+n)=20 \), so \( 100-(m+n)=20 \) and \( m+n=80 \).
3. With \( n-m=10 \): \( m=35 \), \( n=45 \). Answer: (c)

Teacher's Note:
1. Find the coefficients of \( y \) and \( y^2 \) in the product.
2. Use \( n-m=10 \) to simplify the second equation.

Answer: (b) \( m^2-m(4r+1)+4r^2-2=0 \)

Explanation:
1. The coefficients are \( {}^mC_{r-1},\ {}^mC_r,\ {}^mC_{r+1} \), so \( {}^mC_{r-1}+{}^mC_{r+1}=2\,{}^mC_r \).
2. Dividing by \( {}^mC_r \): \( \frac r{m-r+1}+\frac{m-r}{r+1}=2 \), i.e. \( r(r+1)+(m-r)(m-r+1)=2(m-r+1)(r+1) \).
3. Simplifying: \( m^2-m(4r+1)+4r^2-2=0 \). Answer: (b) (Check: \( m=7,\ r=2 \) gives \( 49-63+16-2=0 \), and the coefficients 7, 21, 35 are in AP.)

Teacher's Note:
1. The \( r \)th term has coefficient \( {}^mC_{r-1} \).
2. Use ratios of consecutive coefficients.

Answer: (b) \( -\frac38x^2 \)

Explanation:
1. \( (1+x)^{3/2}=1+\frac32x+\frac38x^2+\cdots \) and \( \left(1+\frac x2\right)^3=1+\frac32x+\frac34x^2+\cdots \).
2. The numerator is \( \frac38x^2-\frac34x^2=-\frac38x^2 \) (higher powers dropped).
3. Dividing by \( (1-x)^{1/2}\approx1 \) at this order gives \( -\frac38x^2 \). Answer: (b)

Teacher's Note:
1. Expand each bracket up to \( x^2 \).
2. The denominator is 1 to leading order.

Answer: (b) \( (-1)^n(1-n) \)

Explanation:
1. Coefficient of \( x^n \) in \( (1-x)^n \) is \( (-1)^n \), and coefficient of \( x^{n-1} \) in \( (1-x)^n \) is \( (-1)^{n-1}n \).
2. The product with \( (1+x) \): coefficient \( =(-1)^n+(-1)^{n-1}n=(-1)^n(1-n) \). Answer: (b)

Teacher's Note:
1. Multiply \( (1-x)^n \) by \( 1 \) and by \( x \) and collect \( x^n \).
2. Check with \( n=3 \): the coefficient is 2.

Answer: (a) \( \frac n2 \)

Explanation:
1. Since \( {}^nC_r={}^nC_{n-r} \), \( S_n=\sum\frac1{{}^nC_{n-r}} \), so \( nS_n=\sum\frac n{{}^nC_{n-r}}=\sum\frac{(n-r)+r}{{}^nC_{n-r}} \).
2. \( \sum\frac{n-r}{{}^nC_{n-r}}=\sum_{k}\frac k{{}^nC_k}=t_n \) (relabel \( k=n-r \)) and \( \sum\frac r{{}^nC_{n-r}}=\sum\frac r{{}^nC_r}=t_n \).
3. So \( nS_n=2t_n \) and \( \frac{t_n}{S_n}=\frac n2 \). Answer: (a)

Teacher's Note:
1. Use the symmetry \( {}^nC_r={}^nC_{n-r} \).
2. Write \( n=(n-r)+r \).

Answer: (b) 33

Explanation:
1. \( T_{r+1}={}^{256}C_r\,3^{\frac{256-r}2}\,5^{\frac r8} \). It is an integer when \( \frac{256-r}2 \) and \( \frac r8 \) are both non-negative integers.
2. So \( r \) is a multiple of 8 (then \( 256-r \) is even): \( r=0,8,16,\ldots,256 \), which is \( 33 \) values. Answer: (b)

Teacher's Note:
1. Both exponents must be integers.
2. Count the multiples of 8 from 0 to 256.

Answer: (c) 8th term

Explanation:
1. \( T_{r+1}=\frac{\frac{27}5\left(\frac{27}5-1\right)\cdots\left(\frac{27}5-r+1\right)}{r!}x^r \).
2. The term turns negative when the last factor \( \frac{27}5-r+1 \) is negative: \( r\gt\frac{32}5=6.4 \), so the least \( r \) is 7.
3. \( T_{r+1}=T_8 \). Answer: (c)

Teacher's Note:
1. Look at the sign of the last factor in the numerator.
2. The first factor that goes negative is the one for the least integer \( r\gt6.4 \).

Answer: (d) None of these

Explanation:
1. \( 1+2x+3x^2+\cdots=(1-x)^{-2} \), so \( (1+2x+3x^2+\cdots)^{-3/2}=(1-x)^3 \).
2. \( (1-x)^3 \) is a polynomial of degree 3, so the coefficient of \( x^5 \) is 0, which is none of 21, 25, 26. Answer: (d)

Teacher's Note:
1. Identify the series as a binomial series.
2. A cubic has no \( x^5 \) term.

Answer: (d) \( n+1 \)

Explanation:
1. \( 1+x+x^2+\cdots=(1-x)^{-1} \), so the square is \( (1-x)^{-2} \).
2. The coefficient of \( x^n \) in \( (1-x)^{-2} \) is \( {}^{n+1}C_1=n+1 \). Answer: (d)

Teacher's Note:
1. Sum the geometric series first.
2. Use the expansion of \( (1-x)^{-2} \).