Important Questions for Class 11 Maths Chapter 8 - Binomial Theorem

Important questions for class 11 Maths Chapter 8 Binomial Theorem are provided here. While calculating the higher powers, the multiplication process becomes complex. To avoid the difficulty of repeated multiplication, the binomial theorem was introduced. In class 11 Maths Chapter 8, students can learn about the binomial theorem for positive integral indices. Also, it covers the concept of middle terms. Here, all the important questions from chapter 8 are provided with solutions. These questions are referred from the previous year questions papers and NCERT textbooks. Access, all the chapters important questions for class 11 Maths at BYJU’S.

Class 11 Maths Chapter 8 – Binomial Theorem describe the following important concepts:

• Introduction to binomials
• Binomial theorem for positive integral indices
  • For Positive integers “n” and special cases
• General and middle terms

Also, Check:

• Important 1 Mark Questions for CBSE Class 11 Maths
• Important 4 Marks Questions for CBSE Class 11 Maths
• Important 6 Marks Questions for CBSE Class 11 Maths

Important Questions for Class 11 Maths Chapter 8 Binomial Theorem with Solutions

Practice the given below important questions for class 11 Maths Chapter 8  Binomial Theorem. Solving the binomial theorem problem is a little bit time-consuming task. But, the repeated practise of these problems helps you to manage the time in the examination and allows you to solve the problems without any mistake.

Question 1:

Expand the expression (2x-3)⁶ using the binomial theorem.

Solution:

Given Expression: (2x-3)⁶

By using the binomial theorem, the expression (2x-3)⁶ can be expanded 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 – 2916x + 729

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

Question 2:

Evaluate (101)⁴ using the binomial theorem

Solution:

Given: (101)^4.

Here, 101 can be written as the sum or the difference of two numbers, such that the binomial theorem can be applied.

Therefore, 101 = 100+1

Hence, (101)⁴ = (100+1)⁴

Now, by applying the binomial theorem, we get:

(101)⁴ = (100+1)⁴ = ⁴C₀(100)⁴ +⁴C₁ (100)³(1) + ⁴C₂(100)²(1)² +⁴C₃(100)(1)³ +⁴C₄(1)⁴

(101)⁴ = (100)⁴+4(100)³+6(100)²+4(100) + (1)⁴

(101)⁴ = 100000000+ 4000000+ 60000+ 400+1

(101)⁴ = 104060401

Hence, the value of (101)⁴ is 104060401.

Question 3:

Using the binomial theorem, show that 6ⁿ–5n always leaves remainder 1 when divided by 25

Solution:

Assume that, for any two numbers, say x and y, we can find numbers q and r such that x = yq + r, then we say that b divides x with q as quotient and r as remainder. Thus, in order to show that 6ⁿ – 5n leaves remainder 1 when divided by 25, we should prove that 6ⁿ – 5n = 25k + 1, where k is some natural number.

We know that,

(1 + a)ⁿ = ⁿC₀ + ⁿC₁ a + ⁿC₂ a² + … + ⁿC_n aⁿ

Now for a=5, we get:

(1 + 5)ⁿ = ⁿC₀ + ⁿC₁ 5 + ⁿC₂ (5)² + … + ⁿC_n 5ⁿ

Now the above form can be weitten as:

6ⁿ = 1 + 5n + 5^{2 n}C₂ + 5^{3 n}C₃+ ….+ 5ⁿ

Now, bring 5n to the L.H.S, we get

6ⁿ – 5n = 1 + 5^{2 n}C₂ + 5^{3 n}C₃+ ….+ 5ⁿ

6ⁿ – 5n = 1 + 5² (ⁿC₂ + 5 ⁿC₃+ ….+ 5^n-2)

6ⁿ – 5n = 1 + 25 (ⁿC₂ + 5 ⁿC₃+ ….+ 5^n-2)

6ⁿ – 5n = 1 + 25 k (where k =ⁿC₂ + 5 ⁿC₃+ ….+ 5^n-2)

The above form proves that, when 6ⁿ–5n is divided by 25, it leaves the remainder 1.

Hence, the given statement is proved.

Question 4:

Find the value of r, If the coefficients of (r – 5)^th and (2r – 1)^th terms in the expansion of (1 + x)³⁴ are equal.

Solution:

For the given condition, the coefficients of (r – 5)^th and (2r – 1)^th terms of the expansion (1 + x)³⁴ are ³⁴C_r-6 and ³⁴C_2r-2 respectively.

Since the given terms in the expansion are equal,

³⁴C_r-6 = ³⁴C_2r-2

From this, we can write it as either

r-6=2r-2

(or)

r-6=34 -(2r-2) [We know that, if ⁿC_r = ⁿC_p , then either r = p or r = n – p]

So, we get either r = – 4 or r = 14.

We know that r being a natural number, the value of r = – 4 is not possible.

Hence, the value of r is14.

Important Questions for Class 11 Maths Chapter 8 Binomial Theorem for Practice

Practising the important questions for class 11 Maths Chapter 8 Binomial Theorem help you score good marks in the final examination.

1. Expand (2a – 3b)⁴ by binomial theorem.
2. Using Binomial theorem, expand (a + 1/b)¹¹.
3. Write the general term in the expansion of (a² – b )⁶.
4. Find x, if T₁₁ and T₁₂ in the expansion of (2+ x)⁵⁰ are equal.
5. The coefficients of three consecutive terms in the expansion of (1 + a)ⁿ are in the ratio 1:7:42. Find n.
6. Find the coefficient of x in the expansion of (1 – 3x + 1x²)( 1 -x)¹⁶.
7. Find the value of r, if the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)¹⁸ are equal.
8. If the integers r > 1, n > 2 and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then;
   (i) n=2r
   (ii) n=3r
   (iii) n=2r+1
   (iv) none of these
9. The ratios of the coefficient of x^p and x^q in the expansion of (1+x)^p+q is                              .

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