CBSE Class 11 Mathematics Binomial Theorem Competitive Exam MCQs Set 01

Answer: (d) \( {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n}C_{r+1} \)

Explanation:
1. (a) is the symmetry property and (b) is Pascal's rule, both true.
2. (c): \( r \cdot \frac{n!}{r!(n-r)!} = n \cdot \frac{(n-1)!}{(r-1)!(n-r)!} = n\,{}^{n-1}C_{r-1} \), true.
3. (d): by Pascal's rule the sum equals \( {}^{n+1}C_{r} \), not \( {}^{n}C_{r+1} \); e.g. n = 3, r = 1 gives 3 + 1 = 4 but \( {}^{3}C_{2} = 3 \).
4. Hence (d) is not true.

Teacher's Note:
1. Pascal's rule adds two neighbouring entries of row n to give an entry of row n + 1, never of the same row.
2. Testing a doubtful identity with small numbers such as n = 3, r = 1 exposes a false option within seconds.

Answer: (c) 1

Explanation:
1. \( {}^{n}C_{0} = \frac{n!}{0!\,n!} \)
2. Since 0! = 1, this equals 1.

Teacher's Note:
1. There is exactly one way to choose nothing from n objects, which is why \( {}^{n}C_{0} = 1 \).
2. Remember 0! = 1; it makes \( {}^{n}C_{0} \) and \( {}^{n}C_{n} \) both equal to 1.

Answer: (a) 17

Explanation:
1. \( {}^{2n}C_{3} = \frac{2n(2n-1)(2n-2)}{6} = \frac{2n(2n-1)(n-1)}{3} \) and \( {}^{n}C_{2} = \frac{n(n-1)}{2} \)
2. Ratio \( = \frac{2n(2n-1)(n-1)}{3} \times \frac{2}{n(n-1)} = \frac{4(2n-1)}{3} \)
3. \( \frac{4(2n-1)}{3} = 44 \) ⇒ 2n − 1 = 33 ⇒ n = 17

Teacher's Note:
1. Write both combinations as products and cancel the common factors n(n − 1) before equating.
2. Check: \( {}^{34}C_{3} = 5984 \) and \( {}^{17}C_{2} = 136 \), and 5984 ÷ 136 = 44.

Answer: (b) 5

Explanation:
1. \( \frac{(n+2)!}{(n-1)!} = (n+2)(n+1)n = 210 \)
2. 210 = 7 × 6 × 5, so n + 2 = 7
3. n = 5

Teacher's Note:
1. Dividing (n + 2)! by (n − 1)! leaves the three consecutive factors (n + 2)(n + 1)n.
2. Splitting the number into a product of consecutive integers (210 = 5 × 6 × 7) is faster than solving a cubic.

Answer: (a) 100

Explanation:
1. Multiply both sides by 10!: \( \frac{10!}{8!} + \frac{10!}{9!} = x \)
2. \( \frac{10!}{8!} = 10 \times 9 = 90 \) and \( \frac{10!}{9!} = 10 \)
3. x = 90 + 10 = 100

Teacher's Note:
1. Multiplying through by the largest factorial removes every fraction at once.
2. \( \frac{10!}{8!} \) is simply the product of the factors from 9 to 10.

Answer: (c) 5

Explanation:
1. \( {}^{n}C_{x} = {}^{n}C_{y} \) with x ≠ y gives x + y = n.
2. n = 3 + 2 = 5

Teacher's Note:
1. Equal combinations with different lower indices mean the lower indices are complementary.
2. Check: \( {}^{5}C_{3} = {}^{5}C_{2} = 10 \).

Answer: (c) 91

Explanation:
1. \( {}^{n}C_{8} = {}^{n}C_{6} \) ⇒ n = 8 + 6 = 14
2. \( {}^{14}C_{2} = \frac{14 \times 13}{2} = 91 \)

Teacher's Note:
1. First find n from the symmetry property, then evaluate the required combination.
2. \( {}^{n}C_{2} = \frac{n(n-1)}{2} \) is worth remembering as a direct formula.

Answer: (c) 12

Explanation:
1. Either 2n − 1 = n − 3, giving n = −2 (not possible),
2. or (2n − 1) + (n − 3) = 32 ⇒ 3n = 36
3. n = 12

Teacher's Note:
1. Always test both cases r₁ = r₂ and r₁ + r₂ = n, and reject the case that gives a negative or non-integer value.
2. With n = 12 the two lower indices are 23 and 9, which indeed add to 32.

Answer: (d) 11

Explanation:
1. \( \frac{(n+1)n(n-1)(n-2)}{24} = \frac{9\,n(n-1)}{2} \)
2. Cancel n(n − 1): (n + 1)(n − 2) = 108
3. n² − n − 110 = 0 ⇒ (n − 11)(n + 10) = 0 ⇒ n = 11

Teacher's Note:
1. Cancelling the common factor n(n − 1) turns a quartic into a simple quadratic.
2. Reject the negative root because n must be a natural number.

Answer: (c) 3

Explanation:
1. \( \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r} = \frac{84}{36} = \frac{7}{3} \) ⇒ 3n − 10r + 3 = 0
2. \( \frac{{}^{n}C_{r+1}}{{}^{n}C_{r}} = \frac{n-r}{r+1} = \frac{126}{84} = \frac{3}{2} \) ⇒ 2n − 5r − 3 = 0
3. Solving: n = 9, r = 3

Teacher's Note:
1. The ratio \( \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r} \) is the key tool whenever consecutive combinations are given.
2. Check: \( {}^{9}C_{2} = 36,\ {}^{9}C_{3} = 84,\ {}^{9}C_{4} = 126 \).

Answer: (c) 14

Explanation:
1. A.P. condition: \( 2\,{}^{n}C_{5} = {}^{n}C_{4} + {}^{n}C_{6} \)
2. Divide by \( {}^{n}C_{5} \): \( 2 = \frac{5}{n-4} + \frac{n-5}{6} \)
3. 12(n − 4) = 30 + (n − 5)(n − 4) ⇒ n² − 21n + 98 = 0
4. (n − 7)(n − 14) = 0 ⇒ n = 7 or 14; from the options n = 14

Teacher's Note:
1. Dividing by the middle combination converts the A.P. condition into simple ratios.
2. Both 7 and 14 work, so pick the one that appears among the options.

Answer: (b) a natural number

Explanation:
1. The finite expansion with \( {}^{n}C_{r} \) coefficients is the binomial theorem for positive integral index.
2. \( {}^{n}C_{r} \) is defined only when n is a natural number (0 ≤ r ≤ n).
3. Hence n must be a natural number.

Teacher's Note:
1. In Class 11 the binomial theorem is studied for positive integral (natural number) index only.
2. The expansion has exactly n + 1 terms, which only makes sense for a natural number n.

Answer: (b) 51

Explanation:
1. In the sum, terms with odd powers of a cancel and terms with even powers of a double.
2. Surviving terms: powers of a = 0, 2, 4, …, 100
3. Number of terms = 100/2 + 1 = 51

Teacher's Note:
1. Adding \( (x + a)^n \) and \( (x - a)^n \) keeps only the terms in which a has an even power.
2. For even n the count is \( \frac{n}{2} + 1 \); for odd n it is \( \frac{n+1}{2} \).

Answer: (a) 1.012

Explanation:
1. \( (1 + 0.003)^4 = 1 + 4(0.003) + 6(0.003)^2 + \ldots \)
2. = 1 + 0.012 + 0.000054 + …
3. ≈ 1.012

Teacher's Note:
1. When the second term is small, the first two terms of the expansion give a good approximation.
2. Here \( 6(0.003)^2 = 0.000054 \), which does not change the answer to three decimal places.

Answer: (c) 5

Explanation:
1. Adding the two expansions cancels the odd powers of x and doubles the even powers.
2. Remaining powers: x⁰, x², x⁴, x⁶, x⁸
3. Number of non-zero terms = 5

Teacher's Note:
1. In \( (1 + y)^n + (1 - y)^n \) only even powers of y survive.
2. For n = 9 the even powers run from 0 to 8, giving (9 + 1)/2 = 5 terms.

Answer: (b) 2, 4

Explanation:
1. \( (1 + ax)^m = 1 + m(ax) + \frac{m(m-1)}{2}(ax)^2 + \ldots \)
2. ma = 8 and \( \frac{m(m-1)}{2}a^2 = 24 \) ⇒ m(m − 1)a² = 48
3. m(m − 1)a² = (ma)(m − 1)a = 8(m − 1)a = 48 ⇒ (m − 1)a = 6
4. ma − a = 6 ⇒ 8 − a = 6 ⇒ a = 2 and m = 4

Teacher's Note:
1. Compare coefficients of x and x² term by term with the expansion \( 1 + {}^{m}C_{1}(ax) + {}^{m}C_{2}(ax)^2 + \ldots \)
2. Check: \( (1 + 2x)^4 = 1 + 8x + 24x^2 + 32x^3 + 16x^4 \).

Answer: (d) 60

Explanation:
1. \( (1 + x^2)^5 = 1 + 5x^2 + 10x^4 + \ldots \) and \( (1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 \)
2. x⁵ arises from x⁴ · x and x² · x³ (x⁰ · x⁵ is not possible since (1 + x)⁴ stops at x⁴)
3. Coefficient = 10 × 4 + 5 × 4 = 60

Teacher's Note:
1. For a product, pair up powers from each factor that add to the required power and add the products of their coefficients.
2. (1 + x²)⁵ contains only even powers, so the other factor must supply an odd power.

Answer: (c) 1

Explanation:
1. Sum of coefficients = value of the expression at x = 1
2. \( (1 + 1 - 3)^{3148} = (-1)^{3148} \)
3. = 1, since 3148 is even

Teacher's Note:
1. Putting x = 1 in any polynomial gives the sum of all its coefficients.
2. An even power of −1 is 1 and an odd power is −1.

Answer: (b) \( 2^{49} \)

Explanation:
1. Sum of coefficients of odd powers = \( C_1 + C_3 + C_5 + \ldots \)
2. This equals half of the total \( 2^{50} \)
3. \( = 2^{49} \)

Teacher's Note:
1. In \( (1 + x)^n \) the even and odd coefficient sums are equal, each \( 2^{n-1} \).
2. This follows by adding and subtracting the results at x = 1 and x = −1.

Answer: (d) 3

Explanation:
1. Put x = 1: \( (a^2 - 6a + 11)^{10} = 1024 = 2^{10} \)
2. a² − 6a + 11 = 2 (the value −2 is impossible since a² − 6a + 11 = (a − 3)² + 2 ≥ 2)
3. (a − 3)² = 0 ⇒ a = 3

Teacher's Note:
1. Sum of coefficients always means putting the variable equal to 1.
2. Completing the square shows the base is at least 2, so the base must equal exactly 2.

Answer: (b) \( 2^{58} \)

Explanation:
1. \( (1 + x)^{59} \) has 60 coefficients \( C_0, C_1, \ldots, C_{59} \), with total \( 2^{59} \).
2. By symmetry \( C_r = C_{59-r} \), so the first 30 and last 30 coefficients have equal sums.
3. Required sum = \( \frac{2^{59}}{2} = 2^{58} \)

Teacher's Note:
1. Coefficients equidistant from the two ends are equal, so the two halves of an even-length row have equal sums.
2. This trick works only when the number of terms (n + 1) is even.

Answer: (a) \( \frac{3^n + 1}{2} \)

Explanation:
1. x = 1: \( a_0 + a_1 + a_2 + \ldots + a_{2n} = (1 - 1 + 1)^n = 1 \)
2. x = −1: \( a_0 - a_1 + a_2 - \ldots + a_{2n} = (1 + 1 + 1)^n = 3^n \)
3. Adding: \( 2(a_0 + a_2 + \ldots + a_{2n}) = 3^n + 1 \) ⇒ sum \( = \frac{3^n + 1}{2} \)

Teacher's Note:
1. Adding the values at x = 1 and x = −1 isolates the even-indexed coefficients.
2. Subtracting them instead would give the odd-indexed sum \( \frac{1 - 3^n}{2} \).

Answer: (d) none of these

Explanation:
1. \( {}^{10}C_{0} + {}^{10}C_{1} + \ldots + {}^{10}C_{10} = 2^{10} = 1024 \)
2. Required sum = 1024 − \( {}^{10}C_{0} \) = 1024 − 1 = 1023
3. 1023 is not among (a), (b), (c), so the answer is none of these.

Teacher's Note:
1. When the sum starts from r = 1, subtract \( {}^{n}C_{0} = 1 \) from \( 2^n \).
2. Options like 512 and 511 are traps based on \( 2^{n-1} \).

Answer: (d) 0

Explanation:
1. \( (1 + x)^n = {}^{n}C_{0} + {}^{n}C_{1}x + \ldots + {}^{n}C_{n}x^n \)
2. Put x = −1: \( 0 = {}^{n}C_{0} - {}^{n}C_{1} + {}^{n}C_{2} - \ldots + (-1)^n\,{}^{n}C_{n} \)
3. Value = 0

Teacher's Note:
1. Alternating sums of binomial coefficients come from putting x = −1.
2. This shows that the even coefficients and the odd coefficients have the same total.

Answer: (a) \( 2^{2n} - 1 \)

Explanation:
1. \( (1 + x)^{2n+1} \) has 2n + 2 coefficients with total \( 2^{2n+1} \); by symmetry \( C_0 + C_1 + \ldots + C_n = \frac{2^{2n+1}}{2} = 2^{2n} \)
2. Required sum (from r = 1) = \( 2^{2n} - {}^{2n+1}C_{0} \)
3. \( = 2^{2n} - 1 \)

Teacher's Note:
1. For odd index 2n + 1, the coefficients \( C_0 \) to \( C_n \) form exactly half of the row.
2. Watch the starting value r = 1: it removes \( C_0 = 1 \).

Answer: (b) \( 2^8 - 2 \)

Explanation:
1. Each of \( {}^{7}C_{1}, \ldots, {}^{7}C_{6} \) appears twice, while \( {}^{7}C_{0} \) and \( {}^{7}C_{7} \) appear once.
2. Sum = \( 2(2^7) - {}^{7}C_{0} - {}^{7}C_{7} \)
3. = 256 − 2 = \( 2^8 - 2 \)

Teacher's Note:
1. Alternatively, each bracket is \( {}^{8}C_{r} \) by Pascal's rule, giving \( {}^{8}C_{1} + \ldots + {}^{8}C_{7} = 2^8 - 2 \).
2. Count carefully how many times the end coefficients appear.

Answer: (a) 1

Explanation:
1. Put a = b = 1 to get the sum of coefficients.
2. \( (6 - 5)^n = 1^n = 1 \)

Teacher's Note:
1. With two variables, put every variable equal to 1 to find the sum of the coefficients.
2. Do not confuse this with the sum of binomial coefficients \( {}^{n}C_{r} \), which is \( 2^n \).

Answer: (d) none of these

Explanation:
1. Let S = \( C_0 + C_1 + \ldots + C_{10} \). By symmetry \( C_{11} + \ldots + C_{20} = C_0 + \ldots + C_9 = S - C_{10} \).
2. Total: \( S + (S - C_{10}) = 2^{20} \) ⇒ \( 2S = 2^{20} + {}^{20}C_{10} \)
3. \( S = 2^{19} + \frac{1}{2}\,{}^{20}C_{10} = 2^{19} + \frac{1}{2} \cdot \frac{20!}{(10!)^2} \), which matches none of (a), (b), (c).

Teacher's Note:
1. For an even index the middle coefficient has no partner, so it must be handled separately when halving the total.
2. Option (c) forgets the factor \( \frac{1}{2} \) on the middle coefficient and option (b) has the wrong sign.

Answer: (c) \( 2^{12} \)

Explanation:
1. Put x = y = 1.
2. \( \left(1 + \frac{1}{3} + \frac{2}{3}\right)^{12} = 2^{12} \)

Teacher's Note:
1. The sum of coefficients is the value of the expression when every variable is 1.
2. Here all coefficients are positive, so 'numerical coefficients' gives the same answer.

Answer: (b) 197

Explanation:
1. \( (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 2[(\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1] = 2[8 + 60 + 30 + 1] = 198 \)
2. Since 0 < \( \sqrt{2} - 1 \) < 1, we have 0 < \( (\sqrt{2} - 1)^6 \) < 1.
3. So \( (\sqrt{2} + 1)^6 = 198 - (\text{a number between 0 and 1}) \), whose integral part is 197.

Teacher's Note:
1. Adding the conjugate expansion cancels all irrational terms and gives an integer.
2. In fact \( (\sqrt{2} + 1)^6 = 99 + 70\sqrt{2} ≈ 197.99 \).

Answer: (b) an odd number

Explanation:
1. Let \( (7 + 4\sqrt{3})^n = I + f \) (0 < f < 1) and \( f' = (7 - 4\sqrt{3})^n \), where 0 < f' < 1.
2. \( I + f + f' = (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n = 2[7^n + {}^{n}C_{2}7^{n-2}(48) + \ldots] \), an even integer.
3. Since 0 < f + f' < 2 and f + f' is an integer, f + f' = 1.
4. So I = even − 1, which is odd.

Teacher's Note:
1. \( 7 - 4\sqrt{3} ≈ 0.07 \), so its powers lie between 0 and 1.
2. In the conjugate sum the terms with \( \sqrt{3} \) cancel and the rest double, giving an even integer.