JEE Mathematics Binomial Theorem MCQs Set 12

Answer: (c) depends on the ratio \( x/y \) and \( m \)

Explanation:
1. \( T_{m+1} = {}^{2m+1}C_m \left(\frac{x}{y}\right)^{2m+1-m}\left(\frac{y}{x}\right)^{m} \)
2. \( = {}^{2m+1}C_m \cdot \frac{x}{y} \)
3. The binomial coefficient depends on m and the factor x/y depends on the ratio of x and y.

Teacher's Note:
1. In (a+b)^n the (r+1)th term is nCr a^(n-r) b^r; always substitute r = m first and simplify the powers.
2. Check whether the powers of x and y survive after simplification before choosing.

Answer: (c) 51

Explanation:
1. The terms with odd powers of \( a \) cancel; only even powers of \( a \) remain.
2. Powers of \( a \) are \( 0,2,4,\dots,100 \), which gives 51 terms.

Teacher's Note:
1. Adding (x+a)^n and (x-a)^n doubles the even-indexed terms and cancels the odd-indexed ones.
2. For an even n, the number of surviving terms is n/2 + 1.

Answer: (a) 1

Explanation:
1. Since \( 18+7=25 \), the numerator is \( a^3+b^3+3ab(a+b)=(a+b)^3=25^3 \).
2. The denominator is the expansion of \( (3+2)^6=5^6=15625 \).
3. Also \( 25^3=15625 \), so the ratio is 1.

Teacher's Note:
1. Spot the pattern a^3+b^3+3ab(a+b) and the full binomial row 1, 6, 15, 20, 15, 6, 1 before computing anything.
2. Recognising a hidden binomial expansion saves a lot of arithmetic.

Answer: (a) 4

Explanation:
1. \( \dfrac{a}{\sqrt{a^{-1}}}=a\cdot a^{1/2}=a^{3/2} \).
2. \( T_2={}^nC_1\,(a^{1/13})^{n-1}a^{3/2}=14a^{5/2} \) gives \( n=14 \) and \( (n-1)/13=1 \), which is consistent.
3. \( \dfrac{{}^nC_3}{{}^nC_2}=\dfrac{n-2}{3}=\dfrac{12}{3}=4 \).

Teacher's Note:
1. Simplify the second factor to a power of a first, then equate the coefficient and the exponent separately.
2. The ratio nC(r+1)/nCr equals (n-r)/(r+1).

Answer: (a) 730

Explanation:
1. General term: \( {}^{6561}C_r\,7^{(6561-r)/3}\,11^{r/9} \).
2. \( r \) must be a multiple of 9, and then \( (6561-r)/3 \) is an integer because 6561 is divisible by 3.
3. \( r=0,9,18,\dots,6561 \) gives \( 6561/9+1=730 \) terms.

Teacher's Note:
1. A term is free from radicals only when every exponent is an integer.
2. Count multiples carefully: include both r = 0 and r = n.

Answer: (b) 1

Explanation:
1. The coefficients are \( {}^{10}C_{2m} \) and \( {}^{10}C_{4m+4} \).
2. \( 2m=4m+4 \) is impossible for positive m, so \( 2m+4m+4=10 \), which gives \( m=1 \).

Teacher's Note:
1. Remember nCx = nCy implies x = y or x + y = n.
2. The (r+1)th term has coefficient nCr, so subtract 1 from the term number.

Answer: (c) 154

Explanation:
1. \( \left(1+\frac1x\right)^8=\sum {}^8C_k x^{-k} \).
2. \( x^1 \) from \( -2x^3 \) needs \( x^{-2} \): coefficient \( -2\cdot{}^8C_2=-56 \).
3. \( x^1 \) from \( 3x^5 \) needs \( x^{-4} \): coefficient \( 3\cdot{}^8C_4=210 \); the term 1 contributes nothing.
4. Total: \( 210-56=154 \).

Teacher's Note:
1. For a product, take each term of the first factor and find the matching power in the second.
2. Do not forget the negative sign on the middle term.

Answer: (c) 1001

Explanation:
1. \( T_{r+1}={}^{15}C_r(-1)^r x^{(15-r)/3-r/2} \).
2. Exponent zero: \( 2(15-r)-3r=0 \), so \( r=6 \).
3. \( T_7={}^{15}C_6=5005=5m \), hence \( m=1001 \).

Teacher's Note:
1. Set the power of x to zero to locate the independent term.
2. The sign is (-1)^r, which is +1 when r = 6.

Answer: (b) 0

Explanation:
1. Write the product as \( \left(x-\frac1x\right)\left(x^2-\frac1{x^2}\right)^3 \).
2. Every term of \( \left(x^2-x^{-2}\right)^3 \) has an even power of x, and multiplying by \( x-1/x \) makes every power odd.
3. An odd power can never be zero, so the independent term is 0.

Teacher's Note:
1. Combine factors with the same base using (a-b)(a+b) = a^2-b^2 to reduce the work.
2. Parity of exponents is a quick way to decide whether a constant term exists.

Answer: (b) 2

Explanation:
1. The middle term is \( T_5={}^8C_4(k/2)^4\,2^4=70k^4 \).
2. \( 70k^4=1120 \) gives \( k^4=16 \), so \( k=2 \).

Teacher's Note:
1. For even n the single middle term is T(n/2 + 1).
2. k^4 = 16 gives only k = 2 when k is positive.

Answer: (c) 8

Explanation:
1. \( 2^8=256=15\cdot17+1\equiv1\pmod{17} \).
2. \( 2003=8\cdot250+3 \), so \( 2^{2003}\equiv2^3=8 \).

Teacher's Note:
1. Find a power of the base that is one more or one less than a multiple of the divisor.
2. Then reduce the exponent modulo that cycle length.

Answer: (d) 01

Explanation:
1. \( 3^{400}=9^{200}=(10-1)^{200} \).
2. Mod 100, only the last two terms matter: \( {}^{200}C_{199}\cdot10\cdot(-1)^{199}+1=-2000+1\equiv1 \).
3. So the last two digits are 01.

Teacher's Note:
1. Write the base as (multiple of 10) plus or minus 1 and expand.
2. For the last two digits keep terms up to the first power of 10.

Answer: (a) 800

Explanation:
1. \( 10!=3628800 \).
2. The last three digits are 800.

Teacher's Note:
1. 10! contains the factors 2^8 and 5^2, so it ends in at least two zeros.
2. Direct computation is quickest here.

Answer: (a) \( 5(2n-9) \)

Explanation:
1. \( \dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{n-r+1}{r} \), so the term equals \( n-r+1 \).
2. \( \sum_{r=1}^{10}(n+1-r)=10(n+1)-55=10n-45=5(2n-9) \).

Teacher's Note:
1. Use nCr/nC(r-1) = (n-r+1)/r to remove the binomial coefficients at once.
2. Then the sum is just an arithmetic series.

Answer: (a) \( \dfrac n2 \)

Explanation:
1. \( {}^nC_r+{}^nC_{r+1}={}^{n+1}C_{r+1} \).
2. \( \dfrac{{}^nC_r}{{}^{n+1}C_{r+1}}=\dfrac{r+1}{n+1} \).
3. \( \sum_{r=0}^{n-1}\dfrac{r+1}{n+1}=\dfrac1{n+1}\cdot\dfrac{n(n+1)}2=\dfrac n2 \).

Teacher's Note:
1. Pascal's rule nCr + nC(r+1) = (n+1)C(r+1) is the key step.
2. Then use (n+1)C(r+1) = ((n+1)/(r+1)) nCr.

Answer: (a) \( {}^9C_6\cdot2^9\cdot(3/2)^{12} \)

Explanation:
1. \( (2+3x)^9=2^9(1+3x/2)^9 \) and \( 3x/2=9/4=(3/2)^2 \).
2. \( T_{r+1}/T_r=\dfrac{10-r}{r}\cdot\dfrac94\ge1 \) gives \( r\le 6.9 \), so the terms grow up to \( T_7 \) and then fall.
3. \( T_7={}^9C_6\,2^9(9/4)^6={}^9C_6\,2^9(3/2)^{12} \).

Teacher's Note:
1. Use the ratio of consecutive terms to find where the sequence stops increasing.
2. Substitute the given value of x before forming the ratio, and compare it with 1 to find the turning point.

Answer: (b) 197

Explanation:
1. \( (\sqrt2+1)^6+(\sqrt2-1)^6=2\left(8+60+30+1\right)=198 \).
2. \( 0<(\sqrt2-1)^6<1 \), so \( (\sqrt2+1)^6=198-(\text{a fraction}) \).
3. Hence the greatest integer not exceeding it is 197.

Teacher's Note:
1. The sum of conjugate powers is an integer, a standard trick for floor-value questions.
2. Check the sign of the small conjugate term to decide between 197 and 198.

Answer: (b) \( \dfrac{2^{11}-1}{6} \)

Explanation:
1. \( \dfrac{{}^{11}C_k}{k+1}=\dfrac{{}^{12}C_{k+1}}{12} \).
2. Sum \( =\dfrac1{12}\left({}^{12}C_1+\cdots+{}^{12}C_{11}\right)=\dfrac{2^{12}-2}{12}=\dfrac{2^{11}-1}{6} \).

Teacher's Note:
1. The identity nCk/(k+1) = (n+1)C(k+1)/(n+1) converts such sums into row sums of a larger binomial.
2. Watch which end terms (r = 0 and r = 12) are missing.

Answer: (c) \( \dfrac1{3(n+1)} \)

Explanation:
1. Factor out \( 1/3 \): the sum is \( \dfrac13\sum(-1)^r\dfrac{{}^nC_r}{r+1} \).
2. \( \sum(-1)^r\dfrac{{}^nC_r}{r+1}=\int_0^1(1-x)^ndx=\dfrac1{n+1} \).
3. So the value is \( \dfrac1{3(n+1)} \).

Teacher's Note:
1. Alternating sums with 1/(r+1) come from integrating the binomial expansion between 0 and 1.
2. Take out constant factors first.

Answer: (c) \( {}^{52}C_4 \)

Explanation:
1. The sum is \( {}^{51}C_3+{}^{50}C_3+{}^{49}C_3+{}^{48}C_3+{}^{47}C_3 \).
2. Add \( {}^{47}C_4+{}^{47}C_3={}^{48}C_4 \), then \( {}^{48}C_4+{}^{48}C_3={}^{49}C_4 \), and so on.
3. The final result is \( {}^{52}C_4 \).

Teacher's Note:
1. Apply nCr + nC(r-1) = (n+1)Cr step by step to merge the terms.
2. Always arrange the terms so the lower index of the first matches r.

Answer: (b) \( \dbinom{100}{51} \)

Explanation:
1. The sum is \( \sum_{k=0}^{49}\binom{50}{k}\binom{50}{k+1}=\sum\binom{50}{k}\binom{50}{49-k} \).
2. By Vandermonde's identity this equals \( \dbinom{100}{49}=\dbinom{100}{51} \).

Teacher's Note:
1. Use nCr = nC(n-r) to turn a product of neighbouring coefficients into a Vandermonde convolution.
2. The indices of the two coefficients must add up to a constant.

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

Explanation:
1. \( 1+x+x^2+\cdots=(1-x)^{-1} \), so the expression is \( (1-x)^{-2} \).
2. \( (1-x)^{-2}=\sum(r+1)x^r \), so the coefficient of \( x^n \) is \( n+1 \).

Teacher's Note:
1. Sum the geometric series first, then expand with the negative-index theorem.
2. Alternatively count the pairs (i, j) with i + j = n.

Answer: (d) \( {}^{12}C_3+3\,{}^{13}C_3+{}^{14}C_4 \)

Explanation:
1. Using the multinomial theorem, \( x^4 \) arises from \( (a,b,c) \) copies of \( 1,\,-x,\,2x^2 \) with \( b+2c=4 \).
2. \( (b,c)=(4,0),(2,1),(0,2) \) gives \( \dfrac{12!}{8!4!}+\dfrac{12!}{9!2!1!}\cdot(-1)^2\cdot2+\dfrac{12!}{10!0!2!}\cdot4=495+1320+264 \).
3. Total \( =2079 \), which equals \( {}^{12}C_3+3\,{}^{13}C_3+{}^{14}C_4=220+858+1001 \).

Teacher's Note:
1. List all combinations (b, c) with b + 2c = 4 and apply the multinomial coefficient to each.
2. Verify option sums numerically when options are written as combinations.

Answer: (a) \( 2^{10} \)

Explanation:
1. Put \( x=i \): \( (1+i)^{10}=(a_0-a_2+\cdots)+i(a_1-a_3+\cdots) \).
2. The required sum of squares is \( |(1+i)^{10}|^2=\left(\sqrt2\right)^{20}=2^{10} \).

Teacher's Note:
1. Substituting x = i separates the even and odd coefficients with alternating signs.
2. The sum of squares of real and imaginary parts is the squared modulus.