CBSE Class 11 Mathematics Binomial Theorem JEE Previous Year Questions Set 01

Answer: (a) \( {}^{56}C_{4} \)

Explanation:
1. Expression = \( {}^{50}C_{4} + {}^{50}C_{3} + {}^{51}C_{3} + {}^{52}C_{3} + {}^{53}C_{3} + {}^{54}C_{3} + {}^{55}C_{3} \)
2. \( {}^{50}C_{4} + {}^{50}C_{3} = {}^{51}C_{4} \); then \( {}^{51}C_{4} + {}^{51}C_{3} = {}^{52}C_{4} \), and so on
3. Finally \( {}^{55}C_{4} + {}^{55}C_{3} = {}^{56}C_{4} \)

Teacher's Note:
1. Write the sum in increasing order so that Pascal's rule \( {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r} \) can be applied repeatedly.
2. Each step raises the upper index by 1 while the lower index stays 4.

Answer: (a) an irrational number

Explanation:
1. In the difference, the terms with even powers of \( \sqrt{3} \) cancel and those with odd powers double.
2. Value = \( 2[{}^{2n}C_{1}(\sqrt{3})^{2n-1} + {}^{2n}C_{3}(\sqrt{3})^{2n-3} + \ldots + {}^{2n}C_{2n-1}\sqrt{3}] \)
3. Each term is a positive integer multiple of \( \sqrt{3} \), so the value is \( \sqrt{3} \times \) (positive integer), an irrational number.

Teacher's Note:
1. Subtracting conjugate expansions keeps only odd powers of the surd, so a factor \( \sqrt{3} \) remains.
2. Check with n = 1: \( (\sqrt{3} + 1)^2 - (\sqrt{3} - 1)^2 = 4\sqrt{3} \), which is irrational.

Answer: (c) 202

Explanation:
1. \( (1 + x)^{101}(1 - x + x^2)^{100} = (1 + x)[(1 + x)(1 - x + x^2)]^{100} = (1 + x)(1 + x^3)^{100} \)
2. \( (1 + x^3)^{100} \) has 101 terms: powers 0, 3, 6, …, 300.
3. Multiplying by (1 + x) gives powers 3k and 3k + 1, all distinct: 2 × 101 = 202 terms.

Teacher's Note:
1. Use the identity \( (1 + x)(1 - x + x^2) = 1 + x^3 \) before counting terms.
2. Terms can be counted directly only when no two products give the same power of x.

Answer: (c) \( \frac{1}{2}(3^{50} + 1) \)

Explanation:
1. Integral powers of x come from even powers of \( \sqrt{x} \). Let \( y = \sqrt{x} \): \( (1 - 2y)^{50} = a_0 + a_1y + \ldots + a_{50}y^{50} \)
2. y = 1: \( a_0 + a_1 + \ldots + a_{50} = (-1)^{50} = 1 \); y = −1: \( a_0 - a_1 + \ldots + a_{50} = 3^{50} \)
3. Even sum = \( \frac{1 + 3^{50}}{2} \)

Teacher's Note:
1. Replacing \( \sqrt{x} \) by y turns 'integral powers of x' into 'even powers of y'.
2. The even-power sum is always half of the sum of the values at y = 1 and y = −1.

Answer: (d) \( 2^{20} - 2^{10} \)

Explanation:
1. \( {}^{21}C_{0} + \ldots + {}^{21}C_{10} = \frac{2^{21}}{2} = 2^{20} \) (by symmetry), so \( {}^{21}C_{1} + \ldots + {}^{21}C_{10} = 2^{20} - 1 \)
2. \( {}^{10}C_{1} + \ldots + {}^{10}C_{10} = 2^{10} - 1 \)
3. Value = \( (2^{20} - 1) - (2^{10} - 1) = 2^{20} - 2^{10} \)

Teacher's Note:
1. For the odd index 21, the first eleven coefficients make exactly half of \( 2^{21} \).
2. Split the bracketed sum into two separate sums before using the \( 2^n \) results.

Answer: (a) 2

Explanation:
1. Sum = \( 2[x^5 + {}^{5}C_{2}x^3(x^3 - 1) + {}^{5}C_{4}x(x^3 - 1)^2] = 2[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x] \)
2. = 10x⁷ + 20x⁶ + 2x⁵ − 20x⁴ − 20x³ + 10x
3. Odd degree coefficients: 10 + 2 − 20 + 10 = 2

Teacher's Note:
1. In \( (a + b)^5 + (a - b)^5 \) only even powers of b survive, which removes every square root.
2. Write out the full polynomial before picking the odd-degree terms; do not just put x = 1.

Answer: (c) \( (12, (20)^4) \)

Explanation:
1. Rationalising (the denominators multiply to (5x³ + 1) − (5x³ − 1) = 2): the expression = \( (a + b)^8 + (a - b)^8 \) with \( a = \sqrt{5x^3 + 1},\ b = \sqrt{5x^3 - 1} \)
2. \( = 2[a^8 + {}^{8}C_{2}a^6b^2 + {}^{8}C_{4}a^4b^4 + {}^{8}C_{6}a^2b^6 + b^8] \), a polynomial in \( x^3 \) of degree 4 in \( x^3 \), so n = 12
3. Leading coefficient: \( 2(5^4)[1 + 28 + 70 + 28 + 1] = 2 \times 625 \times 128 = 160000 = (20)^4 \)
4. (n, m) = (12, (20)⁴)

Teacher's Note:
1. Since the product of the two denominators is 2, each bracket simplifies to the other surd sum.
2. Each of \( a^2 \) and \( b^2 \) has leading term \( 5x^3 \), so every surviving term contributes \( 5^4 x^{12} \) times its coefficient.

Answer: (a) 52

Explanation:
1. Need a + 2b + 3c = 10 with 0 ≤ a ≤ 2, 0 ≤ b ≤ 3, 0 ≤ c ≤ 4; term contributes \( {}^{2}C_{a}\,{}^{3}C_{b}\,{}^{4}C_{c} \)
2. c = 2: (a, b) = (0, 2), (2, 1) → 1·3·6 + 1·3·6 = 36; c = 3: (1, 0) → 2·1·4 = 8
3. c = 1: a + 2b = 7 → (1, 3) → 2·1·4 = 8; c = 0 or 4: no solutions
4. Total = 36 + 8 + 8 = 52

Teacher's Note:
1. List the power each factor can supply and organise the cases by the factor with the largest step (here \( x^3 \)).
2. Multiply the binomial coefficients within a case and add across cases.

Answer: (c) 7

Explanation:
1. Expression = \( 2[x^5 + {}^{5}C_{2}x^3(x^3 - 1) + {}^{5}C_{4}x(x^3 - 1)^2] \)
2. The last term gives \( 5x(x^6 - 2x^3 + 1) = 5x^7 - 10x^4 + 5x \)
3. Highest power is \( x^7 \), so the degree is 7

Teacher's Note:
1. Only even powers of the square root survive in \( (a + b)^5 + (a - b)^5 \), so the result is a polynomial.
2. The largest power comes from \( x \cdot (x^3)^2 \), not from \( x^5 \).

Answer: (c) 12

Explanation:
1. \( (1 + x)^m(1 - x)^n = \left(1 + mx + \frac{m(m-1)}{2}x^2 + \ldots\right)\left(1 - nx + \frac{n(n-1)}{2}x^2 - \ldots\right) \)
2. Coefficient of x: m − n = 3
3. Coefficient of x²: \( \frac{m(m-1)}{2} + \frac{n(n-1)}{2} - mn = \frac{(m-n)^2 - (m+n)}{2} = -6 \) ⇒ 9 − (m + n) = −12 ⇒ m + n = 21
4. m = 12, n = 9

Teacher's Note:
1. Only the first three terms of each expansion are needed to get the coefficients of x and x².
2. Writing the x² coefficient in terms of m − n and m + n avoids solving a quadratic.

Answer: (d) \( {}^{n+2}C_{r} \)

Explanation:
1. Group as \( ({}^{n}C_{r} + {}^{n}C_{r-1}) + ({}^{n}C_{r-1} + {}^{n}C_{r-2}) \)
2. \( = {}^{n+1}C_{r} + {}^{n+1}C_{r-1} \) by Pascal's rule
3. \( = {}^{n+2}C_{r} \)

Teacher's Note:
1. Split the middle term 2 nC(r−1) into two parts so Pascal's rule can be used twice.
2. Check with n = 2, r = 2: 1 + 4 + 1 = 6 = \( {}^{4}C_{2} \).

Answer: (d) \( {}^{12}C_{6} + 2 \)

Explanation:
1. \( (1 + t^{12})(1 + t^{24}) = 1 + t^{12} + t^{24} + t^{36} \)
2. t²⁴ arises as \( t^{24} \cdot 1 \), \( t^{12} \cdot t^{12} \), \( t^0 \cdot t^{24} \) from \( (1 + t^2)^{12} \) times these terms
3. Coefficient = \( {}^{12}C_{12} + {}^{12}C_{6} + {}^{12}C_{0} = {}^{12}C_{6} + 2 \)

Teacher's Note:
1. In \( (1 + t^2)^{12} \) the coefficient of \( t^{2k} \) is \( {}^{12}C_{k} \).
2. Multiply out the two short factors first, then match powers with the binomial factor.

Answer: (d) \( (\sqrt{3}, 2] \)

Explanation:
1. \( {}^{n}C_{r+1} = \frac{n}{r+1}\,{}^{n-1}C_{r} \), so \( k^2 - 3 = \frac{r+1}{n} \)
2. Since 0 ≤ r ≤ n − 1, we have 0 < \( \frac{r+1}{n} \) ≤ 1
3. 3 < k² ≤ 4 ⇒ k ∈ [−2, −√3) ∪ (√3, 2]; among the options this is \( (\sqrt{3}, 2] \)

Teacher's Note:
1. The identity \( {}^{n}C_{r+1} = \frac{n}{r+1}\,{}^{n-1}C_{r} \) links combinations with consecutive upper indices.
2. Fix the range of \( \frac{r+1}{n} \) first, then solve the inequality for k.

Answer: (c) 1113

Explanation:
1. Need 2a + 3b + 4c = 11 (a ≤ 4, b ≤ 7, c ≤ 12); each case contributes \( {}^{4}C_{a}\,{}^{7}C_{b}\,{}^{12}C_{c} \)
2. b must be odd. b = 1: 2a + 4c = 8 → (a, c) = (0, 2), (2, 1), (4, 0) → 7(66 + 6·12 + 1) = 7 × 139 = 973
3. b = 3: 2a + 4c = 2 → (1, 0) → 4 × 35 = 140
4. Total = 973 + 140 = 1113

Teacher's Note:
1. Parity helps: 2a + 4c is even, so 3b must be odd, which forces b to be odd.
2. Organise cases by one variable to avoid missing or repeating combinations.

Answer: -1

Explanation:
1. Sum of coefficients = value at x = 1
2. \( (1 + 1 - 3)^{2163} = (-1)^{2163} \)
3. = −1, since 2163 is odd

Teacher's Note:
1. Putting x = 1 in a polynomial always gives the sum of its coefficients.
2. Look at the parity of the exponent to decide the sign of a power of −1.