JEE Mains Differentiation Questions with Answers - Download PDF

The top 10 JEE Main Binomial Theorem questions that are highly expected in JEE Mains 2024 cover every concept of the Binomial Coefficients, Binomial Theorem Statement, Pascal's Triangle, Factorial Notation and Combinatorial Interpretation, Finding Particular Term in the Expansion, Middle Term and Symmetry, Multinomial Theorem.

JEE Main 2024 Session 2 will be conducted from Apr 4 to Apr 15, 2024. The JEE Main 2024 admit card for the exams on Apr 4, Apr 5, and Apr 6 have been made available on the official website by the National Testing Agency. The admit cards for the other exams till Apr 15, 2024, shall be available 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 directly download the PDF by clicking on it.

Top 10 JEE Mains Binomial Theorem Questions with Answers

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Also Read: JEE Main Complex Number Questions with Solutions PDF

Top 10 JEE Main Binomial Theorem Questions with Solutions

The top 10 JEE Main Binomial Theorem questions with answers are listed below for students to practise 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 + a^x)ⁿ = 1 + 8x + 24x² + —---?

On comparing same coefficients, we have

= ⁿa (n − 1) a = 48

8 (8 − a) = 48

8 − a = 6

= a = 2

= n = 4

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} = ¹¹Cr (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} = ¹¹Cr * (−1)^r * [a¹¹ − 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: JEE Main Most Difficult Question Papers

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

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

Also Check: JEE Main Model Question Paper

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 − nx − 1 is divisible by x².

Trick: (1 + x)ⁿ − nx − 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.

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 = 3r and 4n − 5 = 9r

4n − 5 / n + 1 = 3

= n = 8

Also Check: JEE Main Important Formulas

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} = 2²ⁿ⁺¹C1 (5√5)²ⁿ * 11 + 2^{n + 1}C³(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)2^{n + 1}

= 42ⁿ⁺¹

Also Check: JEE Main Math Important Formulas 2024

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, _nC¹ 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: Tips to Prepare Mathematics for JEE Main 2024