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.
CBSE Class 11 Mathematics Chapter-8 Important Questions
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.