JEE Mathematics Binomial Theorem MCQs Set 19

Answer: (a) 24

Explanation:
1. The sum equals \( 2\left[x^6+\binom62x^4(x^3-1)+\binom64x^2(x^3-1)^2+\binom66(x^3-1)^3\right] \), a polynomial in x.
2. Expand: \( 2\left[x^6+15x^7-15x^4+15x^8-30x^5+15x^2+x^9-3x^6+3x^3-1\right] \) = \( 2\left[x^9+15x^8+15x^7-2x^6-30x^5-15x^4+3x^3+15x^2-1\right] \).
3. Even-degree coefficients (powers 8, 6, 4, 2, 0): \( 2[15-2-15+15-1]=24 \).

Teacher's Note:
1. Expand the symmetric sum so odd powers of the root cancel, then add the coefficients of the even powers.

Answer: (a) 2

Explanation:
1. The sum is \( 2\left[x^5+10x^3(x^3-1)+5x(x^3-1)^2\right] \).
2. \( =2\left[x^5+10x^6-10x^3+5x^7-10x^4+5x\right] \) = \( 2\left[5x^7+10x^6+x^5-10x^4-10x^3+5x\right] \).
3. Odd powers: \( 2[5+1-10+5]=2 \).

Teacher's Note:
1. Only even powers of the square root survive, giving a polynomial.
2. Add the coefficients of the odd powers of x.

Answer: (a) 232

Explanation:
1. Let the coefficients be \( {}^nC_r,{}^nC_{r+1},{}^nC_{r+2} \). The ratios give \( \dfrac{{}^nC_{r+1}}{{}^nC_r}=\dfrac{15}2 \) and \( \dfrac{{}^nC_{r+2}}{{}^nC_{r+1}}=\dfrac{70}{15}=\dfrac{14}3 \).
2. \( \dfrac{n-r}{r+1}=\dfrac{15}2 \) and \( \dfrac{n-r-1}{r+2}=\dfrac{14}3 \) give \( r=1,\ n=16 \).
3. Coefficients: \( 16,120,560 \); average \( =\dfrac{696}{3}=232 \).

Teacher's Note:
1. Use the ratio of consecutive binomial coefficients, (n minus r) over (r plus 1), twice to solve for n and r.

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

Explanation:
1. \( T_4={}^6C_3\left(\dfrac2x\right)^3\left(x^{\log_8x}\right)^3=160\,x^{3\log_8x-3} \).
2. Set \( 160\,x^{3\log_8x-3}=20\cdot8^7 \), so \( x^{3(\log_8x-1)}=8^6 \). Let \( t=\log_8x \): \( 3t(t-1)=6 \), \( t^2-t-2=0 \), \( t=2 \) or \( -1 \).
3. So \( x=8^2 \) (or \( 8^{-1} \)); of the options, \( 8^2 \) is correct.

Teacher's Note:
1. Take logs base 8 and reduce the equation to a quadratic in log base 8 of x.

Answer: (a) (28, 315)

Explanation:
1. Coefficient of \( x^2 \): \( {}^{15}C_2\cdot9-3a\cdot15+b=0 \), i.e. \( 945-45a+b=0 \).
2. Coefficient of \( x^3 \): \( -{}^{15}C_3\cdot27+a\,{}^{15}C_2\cdot9-3b\cdot15=0 \), i.e. \( -12285+945a-45b=0 \).
3. Solving: \( a=28,\ b=315 \).

Teacher's Note:
1. Write the coefficients of x squared and x cubed in the product and solve the two linear equations.

Answer: (a) 38

Explanation:
1. General term: \( {}^nC_r\,x^{2(n-r)-3r}=x^{2n-5r} \). For the power of x to be 1: \( 2n-5r=1 \).
2. The coefficient \( {}^nC_r={}^nC_{23} \) means \( r=23 \) or \( r=n-23 \).
3. \( r=23 \): \( n=58 \). \( r=n-23 \): \( 2n-5n+115=1 \), so \( n=38 \). The smallest is 38.

Teacher's Note:
1. Two cases arise from the symmetry of binomial coefficients, and both give valid n.

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. The coefficient of \( x^{18} \): from \( (1-x^3)^9 \) at \( x^{18} \): \( {}^9C_6=84 \); the \( -x^2 \) term needs \( x^{16} \), which is not a multiple of 3.
3. So the coefficient is 84.

Teacher's Note:
1. Combine the factors using (1 minus x)(1 plus x plus x squared) equals 1 minus x cubed.

Answer: (c) \(-36\)

Explanation:
1. General term of the second factor: \( {}^6C_r(2x^2)^{6-r}\left(-\dfrac3{x^2}\right)^r=(-3)^r2^{6-r}{}^6C_r\,x^{12-4r} \).
2. Independent term from \( \tfrac1{60} \): \( 12-4r=0 \), \( r=3 \): \( \tfrac1{60}\cdot20\cdot(-27)\cdot8=-72 \). From \( -\tfrac{x^8}{81} \): need \( x^{-8} \): \( 12-4r=-8 \), \( r=5 \): \( -\tfrac1{81}\cdot6\cdot(-243)\cdot2=36 \).
3. Total \( =-72+36=-36 \).

Teacher's Note:
1. Find the two values of r that give a net power of zero when multiplied by each part of the first factor.

Answer: (a) 8

Explanation:
1. \( 2^{403}=8\cdot2^{400}=8\cdot16^{100} \).
2. \( 16\equiv1\pmod{15} \), so \( 16^{100}\equiv1 \) and \( 2^{403}\equiv8\pmod{15} \).
3. The fractional part is \( 8/15 \), so \( k=8 \).

Teacher's Note:
1. Reduce the power of 2 modulo 15 using 16 congruent to 1.

Answer: (a) 15

Explanation:
1. \( \left(\dfrac{1-t^6}{1-t}\right)^3=(1-t^6)^3(1-t)^{-3} \).
2. For \( t^4 \) only the term 1 of \( (1-t^6)^3 \) contributes.
3. Coefficient of \( t^4 \) in \( (1-t)^{-3} \) is \( {}^{3+4-1}C_4={}^6C_4=15 \).

Teacher's Note:
1. Low powers of t come only from the first term of (1 minus t to the sixth) cubed.

Answer: (a) \( \dfrac14 \)

Explanation:
1. \( T_3={}^5C_2\left(x^{\log_2x}\right)^2=10\,x^{2\log_2x} \).
2. \( x^{2\log_2x}=256 \). Let \( t=\log_2x \): \( 2^{2t^2}=2^8 \), so \( t^2=4 \) and \( t=\pm2 \).
3. \( x=4 \) or \( \dfrac14 \); \( \dfrac14 \) is among the options.

Teacher's Note:
1. Take logarithms to base 2 to turn the equation into a quadratic in log base 2 of x.

Answer: (a) 4

Explanation:
1. General term: \( x^2\,{}^{10}C_r\,x^{(10-r)/2}\lambda^rx^{-2r}=\,{}^{10}C_r\lambda^rx^{2+(10-r)/2-2r} \).
2. For \( x^2 \): \( \dfrac{10-r}2-2r=0 \), so \( r=2 \).
3. \( {}^{10}C_2\lambda^2=45\lambda^2=720 \), so \( \lambda^2=16 \) and \( \lambda=4 \).

Teacher's Note:
1. Match the exponent of x to find r first, then solve for lambda.

Answer: (a) \( 2^{25} \)

Explanation:
1. \( {}^{50}C_r\,{}^{50-r}C_{25-r}=\dfrac{50!}{r!\,(25-r)!\,25!}={}^{50}C_{25}\,{}^{25}C_r \).
2. \( \sum_{r=0}^{25}{}^{25}C_r=2^{25} \).
3. \( K=2^{25} \).

Teacher's Note:
1. Factor the product of binomial coefficients as 50C25 times 25Cr, then use the sum of binomial coefficients.

Answer: (a) 0

Explanation:
1. The middle term is \( T_5={}^8C_4\left(\dfrac{x^3}{3}\right)^4\left(\dfrac3x\right)^4=70\,x^8 \).
2. \( 70x^8=5670 \) gives \( x^8=81 \), so \( x=\pm\sqrt3 \).
3. The sum of the real values is 0.

Teacher's Note:
1. The middle term of an even power is the (n/2 + 1)th term.
2. Solving x to the 8 gives plus and minus roots, which cancel in the sum.