JEE Binomial Theorem Previous Year Question Paper With Solutions

# JEE Binomial Theorem Previous Year Question Paper With Solutions  Infinity Learn NEET Festival
Infinity Learn IIT JEE Festival 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.

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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

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

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

Solution:

The coefficients of xp and xq in the expression (1 + x) p + q

We know, yp + 1 = p + qCp xp and yq + 1 = p + qCq xq

Coefficient of xp = p + qCp = [p + q]! / P! Q!

and

Coefficient of xq = p + qCq = [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

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

Solution:

Binomial Expansion: (x + y) n = C0 xn + C1 xn-1 y + C2 xn-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

nCn / 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

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 + x55C1 x4y + 5C2 x3y25C3 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) 66C1 (2x) 5 (3) + 6C2 (2x) 4 (3) 2 6C3 (2x) 3 (3) 3 + 6C4 (2x) 2 (3) 46C5 (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 – 2916x + 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. !

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