JEE Main Binomial Theorem Questions with Solutions - Download PDF

JEE Main Binomial Theorem questions that are highly expected in JEE Mains 2024 are based on the Binomial Coefficients, Binomial Theorem Statement, Pascal's Triangle, Factorial Notation and Combinatorial Interpretation. The candidates must thoroughly know the concept and trend of questions to score good marks in the JEE Main examination.

JEE Main 2024 Session 2 will be taking place from Apr 4 to Apr 15, 2024. The JEE Main 2024 admit cards for the exams on Apr 4, Apr 5, and Apr 6 have been released on the official website of the National Testing Agency, on Mar 31, 2024. The admit cards for the other exams are expected to be released shortly.

JEE Main Binomial Questions with Answers - Download PDF

The PDF containing all the top 10 JEE Main Binomial Theorem questions with answers is listed below for students to download. Students can download the PDF by clicking on it.

Also Practise: JEE Mains Differentiation Questions with Answer

Top JEE Main Binomial Theorem Questions with Solutions

JEE Main Binomial Theorem questions with answers are listed below for students to practice to prepare the Binomial Theorem topics.

Question 1: If (1 + a^x)ⁿ = 1 + 8x + 24x² + —, then the value of a and n are ?

Answer: As given (1 + ax)ⁿ = 1 + 8x + 24x² + —---?

On comparing the same coefficients, we have

= ⁿa (n − 1) a = 48

8 (8 − a) = 48

8 − a = 6

= a = 2

= n = 4

Also Read: JEE Main Fluid Mechanics Questions with Solutions

Question 2: If the coefficient of x⁷ in (ax² + [1/bx])11 is equal to the coefficient of x−7 in (ax − 1/bx²)11, then ab = ___.

Answer: In the expansion of (ax² + [1/bx])¹¹, the general term is

T^r + 1 = ¹¹C_r (ax²)11 ^{− r} (1/bx)^r

= ¹¹C_r * a^{11 − r} * [ 1/br ] * x^{22 − 3r}

For x⁷, we must have 22 – 3r = 7

r = 5, and the coefficient of x⁷ = ¹¹C₅ * a¹¹⁻⁵ * [1/b⁵] = ¹¹C₅ * a⁶ * b⁵

Similarly, in the expansion of (ax−1bx2)11, the general term is

T^{r + 1} = ₁₁C^r * (−1)^r * [a^{11 − r}/b^r ] * x^{11 − 3r}

For x^-7 we must have, 11 – 3r = -7 , r = 6, and the coefficient of x−7 is ₁₁C⁶ * [a⁵/b⁶] = ₁₁C⁵ * a⁵ * b⁶.

As given, ₁₁C⁵ [a⁶/b⁵] = ₁₁C⁵ * [a⁵/b⁶]

= ab = 1

Also Check: 9 Most Difficult Questions in JEE Main 2024

Question 3: In the polynomial (x − 1) (x − 2) (x − 3) . . . . . (x − 100), what is the coefficient of x99?

Answer: (x − 1) (x − 2) (x − 3) . . . . . (x − 100)

Number of terms = 100

Coefficient of x99 = (x − 1) (x − 2) (x − 3) . . . . . (x − 100)

= (−1 −2 −3 − —  −100)

= − (1 + 2 + — +100)

= − [100 * 101 / 2

= – 5050

Question 4: In the expansion of (x + a)n, the sum of odd terms is P and sum of even terms is Q, then the value of (P2 − Q2) will be _________.

Answer: (x + a)ⁿ = xⁿ + ⁿC₁xⁿ⁻¹ a + . . . . . = (xⁿ + ⁿC₂xⁿ⁻²a² + . . . . . . . + (_nC¹xⁿ⁻¹a + _nC³xⁿ⁻³a³ + —) = P+Q

(x − a)ⁿ = P − Q

As the terms are altered,

P² − Q² = (P + Q) (P − Q) = (x + a)ⁿ (x − a)ⁿ

P² − Q² = (x² − a²)ⁿ

Also Read: JEE Mains Complex Number Questions with Solutions 

Question 5: If the sum of the coefficients in the expansion of (1 − 3x + 10x²)ⁿ is a and if the sum of the coefficients in the expansion of (1 +x²)ⁿ is b, then what is the relation between a and b?

Answer: We have a = sum of the coefficient in the expansion of (1 − 3x + 10x2)n = (1 − 3 + 10)n = (8)n

=(1 − 3x + 10x2)ⁿ = (2)3ⁿ, [Putting x = 1] .

Now, b = sum of the coefficients in the expansion of (1 + x²)ⁿ = (1 + 1)ⁿ = 2ⁿ.

Clearly, a = b³

Question 6: (1 + x)ⁿ − n^{x − 1} is divisible by?

1. A) 2x
2. B) x²
3. C) 2x³
4. D) All of these

Answer: (1 + x)ⁿ = 1 + n^x + ([n (n − 1)] / [2!]) * x² + ([n (n − 1) (n − 2)] / [3!]) * x³ + —

=(1 + x)ⁿ − n^x − 1 = x² [([n (n − 1)] / [2!]) + ([n (n − 1) (n − 3)] / [3!]) * x + —]

=From above it is clear that (1 + x)ⁿ − n^x − 1 is divisible by x².

Trick: (1 + x)ⁿ − n^x − 1, put n = 2 and x = 3;

Then 42 − 2 * 3 − 1 = 9 is not divisible by 6, 54 but divisible by 9, which is given by option (b) i.e., x² = 9.

Also Read: JEE Main Fluid Mechanics Question Papers  

Question 7: If the three consecutive coefficients in the expansion of (1 + x)ⁿ are 28, 56 and 70, then the value of n is __?

Answer: Let the three consecutive coefficients be _nC^r−1 = 28, _nC^r = 56 and _nC^r+1 = 70, so that

_nC^r / _nC^r−1 = [n − r + 1] / [r] = 56 / 28 = 2 and _nC^r+1 / _nC^r = [n − r] / [r + 1] = 70 / 56 = 5 / 4

This gives n + 1 = 3^r and 4n − 5 = 9r

4n − 5 / n + 1 = 3

= n = 8

Question 8: Let R = (5√5 + 11)2n + 1 and f = R − [R], where [.] denotes the greatest integer function. The value of R * f is

Answer: Since (5√5 − 11) (5√5 + 11) = 4

5√5 − 11 = 4 / (5√5 + 11), Because 0 < 5√5 − 11 < 1 = 0 < (5√5 − 11)2ⁿ + 1 < 1, for positive integer n.

Again, (5√5 + 11)2ⁿ + 1 − (5√5 − 11)2ⁿ + 1 = 2 {2ⁿ⁺¹C₁ (5√5)2n * 11 + 2n + ₁C³(5√5)2ⁿ − 2 × 113 + — +

2n + ₁C²ⁿ⁺¹ 11²ⁿ⁺¹}

= 2 {2ⁿ+₁C¹(125)ⁿ * 11 + 2ⁿ⁺¹C₃(125)ⁿ⁻¹ 113 + — + 2ⁿ⁺¹C²ⁿ⁺¹ 11²ⁿ⁺¹}

= 2k, (for some positive integer k)

Let f′= (5√5 − 11)2n + 1 , then [R] + f − f′=2k

f − f′ = 2k − [R]

= f − f′ is an integer.

But, 0 ≤ f < 1; 0 < f′<1

= −1 < f − f′ < 1

f − f′ = 0 (integer)

f = f′

Therefore, Rf = Rf′ = (5√5 + 11)²ⁿ + 1 * (5√5 − 11)²ⁿ + 1

= ([5√5]2 + 112)²ⁿ + 1

= 42n+1

Also Read: JEE Mains Most Unique Question Papers 

Question 9: The digit in the unit place of the number (183!) + 3183 is __?

Ans: We know that n! terminates in 0 for n³ at 5 and 34n terminator in 1, (because 34 = 81)

Therefore, 3180 = (34)45 terminates in 1.

Also 33 = 27 terminates in 7

3183 = 3180 * 33 terminates in 7.

183! + 3183 terminates in 7 i.e. the digit in the unit place = 7.

Question 10: If the coefficients of pth, (p + 1)^th, and (p + 2)^th terms in the expansion of (1 + x )ⁿ are in A.P., then find the equation in terms of n.

Answer: Coefficient of p^th, (p + 1)^th  and (p + 2)^th terms in expansion of (1 + x )ⁿ  are _nC^p−1, _nC^p ,_nC^p+1. Then _2nC^p = _nC^p−1 + _nC^p+1

=n² − n (4p + 1) + 4p² − 2 = 0

Trick: Let p = 1, hence _nC⁰, _nC^1, and _nC² are in A.P.

= 2 * _nC¹ = _nC⁰ + _nC²

= 2n = 1 + [n (n − 1)] / [2]

= 4n = 2 + n² − n

= n² − 5n + 2 = 0

Also Check: JEE Main Math Important Formulas 2024