JEE Mathematics Binomial Theorem for Positive Integral Index MCQs Set 06

Answer: (c) \( f(x) \) is a polynomial of degree 6 with coefficient of \( x^{6} \) equal to 64

Explanation:
1. Since \( (\sqrt{x^{2}+1})^{2}-(\sqrt{x^{2}-1})^{2}=2 \), we get \( \dfrac{2}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}=\sqrt{x^{2}+1}-\sqrt{x^{2}-1} \).
2. So \( f(x)=(A+B)^{6}+(A-B)^{6}=2\left(A^{6}+15A^{4}B^{2}+15A^{2}B^{4}+B^{6}\right) \) with \( A^{2}=x^{2}+1,\ B^{2}=x^{2}-1 \), which has only even powers of A and B, hence is a polynomial.
3. Substituting and simplifying gives \( f(x)=64x^{6}-48x^{2} \).
4. This is a polynomial of degree 6 with leading coefficient 64, and it has two terms.

Teacher's Note:
1. Rationalise the second term so that both parts look like \( (A\pm B)^{6} \).
2. In the sum of the two expansions the odd powers of B cancel.

Answer: (c) \( {}^{18}C_{6}\cdot{}^{12}C_{6} \)

Explanation:
1. The coefficient is \( \dfrac{18!}{8!\,6!\,4!} \).
2. \( {}^{18}C_{14}\cdot{}^{14}C_{8}=\dfrac{18!}{14!\,4!}\cdot\dfrac{14!}{8!\,6!} \), equal. \( {}^{18}C_{10}\cdot{}^{10}C_{6}=\dfrac{18!}{10!\,8!}\cdot\dfrac{10!}{6!\,4!} \), equal. \( {}^{18}C_{6}\cdot{}^{12}C_{8}=\dfrac{18!}{6!\,12!}\cdot\dfrac{12!}{8!\,4!} \), equal.
3. \( {}^{18}C_{6}\cdot{}^{12}C_{6}=\dfrac{18!}{6!\,6!\,6!} \), which is not equal to the required coefficient (the factorials \( 6!,6!,6! \) do not match \( 8!,6!,4! \)).

Teacher's Note:
1. Choose places for one letter at a time: the product of two binomials gives the multinomial.
2. Expand each product into factorials and compare.

Answer: (b) (i) and (iii) only

Explanation:
1. \( (1+x)^{n}\left(1-\tfrac1x\right)^{n}=\dfrac{(1+x)^{n}(x-1)^{n}}{x^{n}}=\dfrac{(x^{2}-1)^{n}}{x^{n}} \).
2. The term independent of x is the coefficient of \( x^{n} \) in \( (x^{2}-1)^{n}=\sum_k {}^{n}C_k x^{2k}(-1)^{n-k} \).
3. If n is odd, \( x^{n} \) would need \( 2k=n \), impossible, so the term is 0 and (i) is true, (ii) is false.
4. If n is even, \( k=n/2 \) and the coefficient is \( (-1)^{n/2}\,{}^{n}C_{n/2} \), so (iii) is true.

Teacher's Note:
1. Combine the two factors into \( (x^{2}-1)^{n}/x^{n} \).
2. Only even powers of x occur in \( (x^{2}-1)^{n} \).

Answer: (b) \( 5 \) or \( 15 \)

Explanation:
1. The \( (r+1) \)th term of \( \left(x+\tfrac1x\right)^{20} \) has coefficient \( {}^{20}C_r \).
2. Since \( x^{2}+2+\tfrac{1}{x^{2}}=\left(x+\tfrac1x\right)^{2} \), the second expression equals \( \left(x+\tfrac1x\right)^{20} \), whose 6th term has coefficient \( {}^{20}C_5 \).
3. \( {}^{20}C_r={}^{20}C_5 \) gives \( r=5 \) or \( r=20-5=15 \).

Teacher's Note:
1. Recognise the second expression as a perfect square.
2. \( {}^{n}C_r={}^{n}C_s \) means \( r=s \) or \( r+s=n \).

Answer: (b) (i), (ii) and (iii)

Explanation:
1. Here \( a_k={}^{2n}C_k \). Putting \( x=1 \) and \( x=-1 \) gives \( \sum a_{\text{even}}=\sum a_{\text{odd}}=2^{2n-1} \), which is half of the total \( 2^{2n} \). So (i) is true.
2. The binomial coefficients of \( (1+x)^{2n} \) increase up to the middle term \( a_n \) and then decrease, so \( a_{n+1}\lt a_n \): (ii) is true.
3. By symmetry \( {}^{2n}C_{k}={}^{2n}C_{2n-k} \), so \( a_{n-3}=a_{n+3} \): (iii) is true.

Teacher's Note:
1. Use x = 1 and x = -1 for the sum of even-indexed coefficients.
2. The middle coefficient of an even power is the largest.

Answer: (c) \( a_1,a_2,a_3 \) are in AP and \( n=7 \)

Explanation:
1. \( a_{n-3},a_{n-2},a_{n-1} \) are \( {}^{n}C_3,\,{}^{n}C_2,\,{}^{n}C_1 \). AP means \( 2\,{}^{n}C_2={}^{n}C_3+{}^{n}C_1 \).
2. \( n(n-1)=\dfrac{n(n-1)(n-2)}{6}+n \), so \( 6(n-1)=(n-1)(n-2)+6 \), giving \( n^{2}-9n+14=0 \), so \( n=7 \) or \( n=2 \).
3. \( n=2 \) is rejected because \( a_{n-3} \) must exist (\( n\ge 3 \)). So \( n=7 \).
4. Then \( a_1,a_2,a_3=7,21,35 \), which are in AP with common difference 14, not in HP.

Teacher's Note:
1. Convert \( a_{n-3},a_{n-2},a_{n-1} \) to \( {}^{n}C_3,{}^{n}C_2,{}^{n}C_1 \) using symmetry.
2. Solve the resulting quadratic and discard the value that makes the terms meaningless.

Answer: (d) (ii) and (iii) only

Explanation:
1. Let \( f=(8-3\sqrt7)^{20} \). Since \( (8+3\sqrt7)(8-3\sqrt7)=1 \), we have \( 0\lt 8-3\sqrt7\lt 1 \), so \( 0\lt f\lt 1 \) and \( f=\dfrac{1}{(8+3\sqrt7)^{20}} \).
2. \( R+f=2\left[8^{20}+\binom{20}{2}8^{18}(3\sqrt7)^{2}+\dots\right] \) is an even integer N.
3. So \( [R]=N-1 \) is odd, and \( R-[R]=R-N+1=1-f=1-\dfrac{1}{(8+3\sqrt7)^{20}} \).
4. Hence (ii) and (iii) are true and (i) is false.

Teacher's Note:
1. Add the conjugate power so that irrational parts cancel.
2. The conjugate power is a small positive fraction.

Answer: (c) \( \dfrac{2^{n-1}}{n!} \) for all \( n\in\mathbb N \)

Explanation:
1. Multiply and divide by \( n! \): the sum is \( \dfrac{1}{n!}\left({}^{n}C_1+{}^{n}C_3+{}^{n}C_5+\dots\right) \).
2. The sum of the binomial coefficients at odd places is \( 2^{n-1} \) for every natural n.
3. Hence the sum equals \( \dfrac{2^{n-1}}{n!} \) for all n.

Teacher's Note:
1. Write each term as a binomial coefficient divided by \( n! \).
2. Odd-place binomial coefficients always add up to half of \( 2^{n} \).

Answer: (c) (i) and (ii) only

Explanation:
1. A general term is \( \dfrac{25!}{(25-r)!\,(r-k)!\,k!}x^{25-r}y^{r-k}z^{k}={}^{25}C_r\cdot{}^{r}C_k\,x^{25-r}y^{r-k}z^{k} \), so (i) is true.
2. In \( x^{8}y^{9}z^{9} \) the powers add to 26, not 25, so no such term occurs and its coefficient is 0: (ii) is true.
3. The number of terms is the number of non-negative solutions of \( p+q+s=25 \), namely \( {}^{27}C_2=351 \), not 325, so (iii) is false.

Teacher's Note:
1. Use the multinomial coefficient for the general term.
2. The number of terms in \( (x+y+z)^{n} \) is \( {}^{n+2}C_2 \).