JEE Binomial Theorem Previous Year Question Paper With Solutions

Question 1: If the total coefficients of all equal forces of x in a product (1 + x + x² + … + x²ⁿ) (1 – x + x² – x³ +… + x²ⁿ) are 61, get the value n.

Solution:

Allow (1 + x + x² +… + x²ⁿ) (1 – x + x² – x³ +… + x²ⁿ) = a⁰ + ax + a²x² + a³x³ + ……

Set x = 1 => 2n + 1 = a⁰ + a¹ + a² + a³ + ……… (i)

Put x = -1 => 2n + 1 = a⁰ – a¹ + a² – a³ + ……… .. (ii)

Add (i) and (ii), we get

2n + 1 = 61

or n = 30

Question 2: If α and β are coefficients of x⁴ and x² respectively in addition (x + √ (x2 – 1))) ⁶ + (x – √ (x2 – 1) ⁶), then -ke:

(a) α + β = -30

(b) α− β = −132

(c) α− β = 60

(d) α + β = 60

Solution: (x + √ (x² – 1)) ⁶ + (x – √ (x² – 1) ⁶

= 2 [6C⁰ x⁶ + 6C2x⁴ (x² – 1) + 6C4x2 (x2 – 1) 2 + 6C⁶ (x2 – 1) ³]

= 2 [32x⁶ – 48x⁴ + 18x² – 1]

=> Now, α=- 96 and Β=-36

=> α – β = -132

Question 3: Find the coefficient of x4 in addition to (1 + x + x²) ¹⁰.

Solution:

The general term for a given speech is,

{10!} \ {P! Q! R!} x ^{{q + 2r}}

 

Here, q + 2r = 4

At p = 6, q = 4, r = 0, coefficient = 10! / [6! x 4!] = 210

At p = 7, q = 2, r = 1, coefficient = 10! / [7! x2! x 1!] = 360

At p = 8, q = 0, r = 2, coefficient = 10! / [8! x2!] = 45

Therefore, total = 615

Question 4: If x is positive, the first negative word in addition to

(1 + x) 27/5 by

(a) term 5 (b) term 6 (c) term 7 (d) term 8

Answer: (d)

Solution:

We know, T_{{r + 1}} = {n (n-1)… (n-r + 1)} \ {r!} X ^r

Therefore, n – r + 1 <0

=> 27/5 + 1 <r

or r> 6

Here r = 7

Thus, Tr + 1 <0

So Tr + 1 = T7 + 1 = T8 = term 8 is negative.

Question 5: Deposit number greater than (1 + 0.0001) 10000

(a) 4 (b) 5 (c) 2 (d) 3

Answer: (d)

Solution:

(1 + 0.0001) 10000 similar form (1 + 1 / n) n where n = 10000.

Using binomial extensions, we have

(1 + 0.0001) 10000 = (1 + 1 / n) n

= 1 + n x 1 / n + n (n-1) / 2! x 1 / n2 + [n (n-1) (n-2)] / 3! X 1 / n3 + …… ..

= 1 + 1 + 1/2! (1 – 1 / n) + 1/3! (1 – 1 / n) + (1 – 2 / n) + ……

<1 + 1/1! + 1/2! + 1/3! + ….. + 1 / (9999)!

= 1 + 1/1! + 1/2! + 1/3! + …… .∞

= e <3

Question 6: The coefficients of xp and xq in the expression (1 + x) ^{p + q}

(a) Equal

(b) It is equal to the opposite sign

(c) Reconciliation with each other

(d) None of this

Answer: (a)

Solution:

The coefficients of x^p and x^q in the expression (1 + x) ^{p + q}

We know, y^p + 1 = p + qCp x^p and y^q + 1 = p + qC^q x^q

Coefficient of x^p = p + qC^p = [p + q]! / P! Q!

and

Coefficient of x^q = p + qC^q = [p + q]! / P! Q!

Therefore, the coefficients of xp and xq in the expression (1 + x) p + q are equal.

Question 7: If an=√7+(√7+(√7+……
having n radical signs then in the form of true mathematical inventions

(a) <7 for all n ≥ 1

(b) an> 7 for all n ≥ 1

(c) 3 for all n ≥ 1

(d) <4 for all n ≥ 1

Answer: (c)

Solution: an=√7+(√7+(√7+……

=> i = √ (7 + an)

=> an2 – an – 7 = 0

Solving above the quadratic equation, we find

i = [1 ± √29] / 2

But> 0, therefore, is = [1 + √29] / 2> 3

=> i> 3

Question 8: If the sum of the coefficients in the extension (a + b) ⁿ is 4096, then the largest coefficient in the extension is

1. 1594
2. 924
3. 792
4. 2924

Answer: (b)

Solution:

Binomial Expansion: (x + y) n = C⁰ xⁿ + C¹ x^n-1 y + C² x^n-2y² + …. + Cⁿyⁿ

setting x=y=1,then

2n = C⁰ + C¹ + C² +…. + Cⁿ = 4096 = 212

=> n = 12

As n is present even here, so the coefficient of the largest term states

ⁿC_n / 2 = ¹²C₁₂ / 2 = ¹²C₆

= 12/6 x 11/5 x 10/4 x 9/3 x8 / 2 x 7/1

= 924

Question 9: The co-efficient amount of all the unusual term terms in the extension of

(x + √ (x³-1) ⁵ + (x – √ (x³-1) ⁵, (x> 1)

(a) 0 (b) 2 (c) 1 (d) -1

Answer:

Solution:

Allow y = √ (x³-1)

Thus, the given expression is reduced to (x + y) ⁵ + (x – y) ⁵

Using binomial extensions, we have

(x + y) ⁵ + (x – y) ⁵ = x⁵ + ⁵C₁ x⁴y + ⁵C₂ x³y² + ⁵C₃ x²y³ + ⁵C₄ xy⁴ + ⁵C₅ y⁵ + x⁵ – ⁵C₁ x⁴y + ⁵C₂ x³y² – ⁵C₃ x²y³ + ⁵C₅

= 2 (x⁵ + 10 x³y² + 5xy⁴)

= 2 (x⁵ + 10 x³ (x³ – 1) + 5x (x³ – 1) ²)

= 2 {10x⁶ + 5x⁷ + x5 – 10x³ – 10x⁴ + 5x}

Total coefficient of unconventional x: 2 {5 + 1 – 10 + 5} = 2

Question 10: Extend quote (2x-3) ⁶ using the binomial theorem.

Solution:

Presentation: (2x-3) ⁶

Using the binomial theorem, expression (2x-3) 6 can be extended as follows:

(2x-3) ⁶ = ⁶C₀ (2x) ⁶ –⁶C₁ (2x) ⁵ (3) + ⁶C₂ (2x) ⁴ (3) ² – ⁶C₃ (2x) ³ (3) ³ + ⁶C₄ (2x) ² (3) ⁴ – ⁶C₅ (2x) (3) ⁵ + ⁶C₆ (3) ⁶

(2x-3) ⁶ = 64x⁶ – 6 (32x⁵) (3) +15 (16x⁴) (9) – 20 (8x³) (27) +15 (4x²) (81) – 6 (2x) (243) + 729

(2x-3) ⁶ = 64x⁶ -576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916^x + 729

Thus, the binomial extension of the given expression (2x-3) ⁶ is 64x⁶ -576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729.

FAQs

Is binomial theorem easy for JEE?

Some basic introduction to the Binomial Theorem introduction is included in the eighth-grade syllabus as well. The concept of the Binomial Theorem becomes easier to understand once students become familiar with its output.

Is the binomial theorem important to JEE?

Binomial Theorem is one of the most important chapters of Algebra in the JEE syllabus. !