# Important Questions Class 11 Maths Chapter 8

## Important Questions Class 11 Mathematics Chapter 8

### Important Questions for CBSE Class 11 Mathematics Chapter 8 – Binomial Theorem

The algebraic expression is known as a binomial because it contains only two terms. It is also known as a two-term polynomial. It is the sum or difference of two or more monomials. It is the most basic polynomial form.

Extramarks: Class 11 Mathematics Chapter 8 Important Questions are compiled by subject matter experts in a concise manner by referring to NCERT Books. It is critical to practise and answer all questions because they cover a wide range of subjects and concepts, providing students with a good understanding of the types of questions that may be set in those areas. These questions also teach them how different questions are set on the same topic and which questions are most likely to be asked in board exams.

### Study Important Questions for Class 11 Mathematics Chapter-8 – Binomial Theorem

Given below are Important Questions for Class 11 Mathematics Binomial Theorem each for 2, 4 and 6 marks.

2 Marks Questions

Q1. Expand the expression (1-2x)5 using binomial theorem.

Ans. We know that

Considering the following elements, x = 1, y = -2x and n = 5

(1-2x)5 = 5C015 + 5C114(-2x)1 + 5C213(-2x)2 + 5C312(-2x)3 + 5C411(-2x)4 + 5C010(-2x)5

⇒ 1 – 5(2x) + 10(4×2) – 10(8×3) + 5(16×4) – (32×5)

⇒ 1 – 10x + 40x – 80×3 + 80×4 – 32×5

Therefore, the expanded form of (1-2x)5 is 1 – 10x + 40x – 80×3 + 80×4 – 32×5

Q2. Demonstrate how the expression (x2 + 1)5 can be expanded by the help of binomial theorem.

Ans. The binomial theorem formula states that:

(x+y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + nC3xn-3y3 +……..+ nCrxn-ryr +….. + nC0x0yn

Here, x = x2 , y = 1 and n = 5

Substituting the following values of x, y and n in the above equation we get,

⇒ 5C0(x2)5 + 5C1(x2)4 + 5C2(x2)3 + 5C3(x2)2 + 5C4(x2)1 + 5C5

⇒ 1.x10 + 5.x8 + 10.x6 + 10.x4 + 5.x2 + 1

⇒ x10 + 5.x8 + 10.x6 + 10.x4 + 5.x2 + 1

⇒ x10 + 5.×8 + 10.×6 + 10.×4 + 5.×2 + 1×10 + 5.×8 + 10.×6 + 10.×4 + 5.×2 + 1

Therefore, the binomial expansion of (x2 + 1)5 is x10 + 5.×8 + 10.×6 + 10.×4 + 5.×2 + 1×10 + 5.×8 + 10.×6 + 10.×4 + 5.×2 + 1.

4 Marks Questions

Q1. Find the coefficient of x5 in binomial expansion of (1 + 2x)5 (1 – x)7.

Ans. Using binomial theorem we will expand both the terms.

We know that,

(x+y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + nC3xn-3y3 +……..+ nCrxn-ryr +….. + nC0x0yn

Applying the formula we get,

(1 + 2x)5 (1 – x)7 = (1 + 6C1(2x) + 6C2(2x)2 + 6C3(2x)3 + 6C4(2x)4 + 6C5(2x)5 + 6C6(2x)6)

(1 – 7C1x + 7C2(x)2 + 7C3(x)3 + 7C4(x)4 + 7C5(x)5 + 7C6(x)6 + 7C7(x)7)

= (1 + 12x + 60×2 + 160×3 + 240×4 + 192×5 + 64×6) (1 – 7x + 21×2 – 35×3 + 35×4 – 21×5 + 7×6 – x7)

Clearly, it can determined that the coefficient of x5 is

⇒ 1 * (-21) + 12 * 35 + 60(-35) + 160 * 21 + 240 * (-7) + 192 * 1

⇒ 171

Therefore, the coefficient of x5 in (1 + 2x)5 (1 – x)7 is 171.

Q2. Determine the expression’s middle term: (2x – x2/4)9?

Ans. It is obvious that n = 9 in the equation (2x – x2/4)9.

Given that n is odd, there should be two middle terms: (n+1)/2nd term and (n+3)/2nd term.

Here, n = 9, As a result, the two middle terms are (9+1)/2nd term and (9+3)/2nd term, or the 5th and 6th terms. So, let us find the fifth and sixth terms of (2x – x2/4)9.

T5 = T4+1 = 9C4(2x)9-4.(-x2/4)4

= (63/4)x13.

T6 = T5+1 = 9C5(2x)9-5.(-x2/4)5

= -9C4(2)4.x4[(x)10/(4)5]

= (-63/32)x14

The middle terms of (2x – x2/4)9 are thus (63/4)x13 and (-63/32)x14.

6 Marks Questions

Q1. The product of the coefficients of the first three terms in the expansion of (x- 3/x2)m, where m is the natural number 559. Determine the coefficient of expansion containing x3.

Ans. The coefficients of the first three terms of (x- 3/x2)m are mC0, (-3) mC1, and 9 mC2.

According to the issue,

559 = mC0 – 3 mC1 + 9 mC2

⇒ 1 – 3m + (9m(m-1)/2) = 559

m = 12 after simplifying

After determining the third term of (x- 3/x2)m,

12Cr(x)12-r(-3/x2)r = Tr+1

⇒ 12Cr(x) (x)

12-r(-3)

r.(x-2r)

⇒ 12Cr(x)12-3r(-3)r

Because we want to find the term with x3, 12 – 3r = 3, i.e., r = 3.

Using r = 3 as a value

⇒ 12C3(x)9(-3)3 = -5940×3

As a result, the coefficient of x3 is -5940.

Q2. The 2nd, 3rd and 4th terms in the binomial expansion (x+a)n are 240, 720 and 1080 respectively. Find x, a and n.

Ans. Given T2 = 240

nC1xn-1.a = 240 — (i)

nC1xn-2.a2 = 720 — (ii)

nC1xn-3.a3 = 1080 – (iii)

(ii)/(i) and (iii)/(ii) we get,

a/x = 6/n-1 and a/x = 9/[2(n-2)]

Putting of n = 5 we get,

x = 2 and a = 3.

Therefore, x = 2, a = 3 and n = 5.

Q.1 The sum of the binomial coefficients in the expansion of (1 + x)n is

Marks:1

Ans

2.

2n-1.

2n.

2n+1.

We have,
(1 + x)n = C0 + C1x + C2x2 + C3x3 ++ Cn-1xn-1 + Cnxn.
Putting x = 1 in this expansion, we get
2n = C0 + C1 + C2 + C3 ++ Cn-1 + Cn.

Q.2 The largest coefficient in the expansion of (1 + x)24 is

Marks:1

Ans

We know that, if n is even then nCr is greatest for r = n/2.

Therefore, 24Cr is greatest for r = 12.

Hence, the greatest coefficient is 24C12.

Q.3 Find the term independent of x in the expansion of

x+1×6.

Marks:2

Ans

Suppose (r+1) the term is independent of x in the expansion of x+1×6.Tr+1=6Crx6r1xrTr+1=6Crx6rrTo be independent of x, we have62r=0r=3T3+1=6C3x631x3=20 Hence, 4thterm is independent of x1 i.e., 20.

Q.4 Find the middle term in the expansion of

3x13x20.

Marks:1

Ans

Here, n=20, which is an even number. So, 202+1th  term =11th  term is the middle term Hence, the middle term =T11=T10+1=20C103x201013x10=20C10

Q.5 Find a if the 17th and 18th terms of the expansion (2 + a)50 are equal.

Marks:3

Ans

We know Tr+1=n Crxnryr…1 Here x=2andy=a Putting r=16andr=17in1,

we get T16+1=T17=50C1625016a16 and T17+1=T18=50C1725017d17T17=T18

given 50C1625016a16 =50C1725017a1750C162501650C1725017 =a17a16a =5016501623450175017233=1750171650162=1733342a=17234=1

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### 1. Define the binomial theorem.

The process of algebraically expanding the power of sums of two or more binomials is defined as the binomial theorem. Binomial coefficients are the coefficients of binomial terms during the expansion process. The first sections of these chapters contain proper definitions of various aspects of the binomial theorem.

### 2. How did the binomial theorem originate, and what are its applications?

The binomial theorem has been known to mankind since the 4th century BC. This is when the Greek mathematician Euclid applied the binomial theorem to the exponent of two. However, value three was first used in India in the 6th century. The Binomial theorem is used to find the reminder, the digit of a number, the test of divisibility, and number comparisons.

### 3. State some properties of the Binomial Expression.

The following are the Binomial Expression properties:

• The total number of terms in the expansion of (x + a) n is (n+1).
• The sum of the indices a and x in each term is n.
• Binomial expression is the correct expansion for complex numbers.
• The coefficients for the terms that are equidistant from the ends are the same. These are referred to as different binomial coefficients.
• As you progress through the expansion problem, you will notice that the values first increase and then decrease.

### 4. What are the different applications of the Binomial Theorem?

The following are some applications of the Binomial Theorem:

Using the Binomial Theorem to Determine the Remainder

For instance, find the remainder when 7105 is divided by 25.

Finding a Number’s Digits

For instance, find the last two digits of the number (13)10.

The Relationship Between Two Numbers

Find the greater of 9950 + 10050 and 10150.

Ex: Demonstrate that 119 + 911 is divisible by 10.

### 5. What are the Properties of Binomial Coefficients?

The various Properties of Binomial Coefficients as follows:

C0 + C1 + C2 + … + Cn = 2n

C0 + C2 + C4 + … = C1 + C3 + C5 + … = 2n-1

C0 – C1 + C2 – C3 + … +(−1)n . nCn = 0

nC1 + 2.nC2 + 3.nC3 + … + n.nCn = n.2n-1

C1 − 2C2 + 3C3 − 4C4 + … +(−1)n-1 Cn = 0 for n > 1

C02 + C12 + C22 + …Cn2 = [(2n)!/ (n!)2]