# Important Questions for Class 11 Chapter 8 - Binomial Theorem

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**Important questions based on NCERT syllabus for Class 11 Chapter 8 - Binomial Theorem:**

*Question 1*: Expand the follwoing expression: (x + 1/x)^5

*Solution*: We know that

(x+y)^n = summation k = 0 to n (n k) a^)n-k) b^k = (n 0)x^n y^0 + (n 1)x^(n-1)y^1 + (n 2)x^(n-2)y^2 + ..... + (n n-1)x^1y^(n-1) + (n n)x^0y^n

Hence by binomial theorem we can expand the expression

(x + 1/x)^5 = 5C_0 x^5(1/x)^0 + 5C_1 x^(4)(1/x)^1 + 5C_2 x^(3)(1/x)^2 + 5C_3x^(2)(1/x)^3 + 5C_4x^(1)(1/x)^4 + 5C_5x^(0)(1/x)^5

= x^5 + 5x^(4)(1/x) + 10x^3 (1/x^2) + 10x^2 1/x^3 + 5x(1/x^4) + 1/x^5

(x + 1/x)^5 = x^5 + 5x^3 + 10x + 10 (1/x) + 5(1/x^3) + (1/x^5)

*Question 2*: Find (x+1)^6 + (x-1)^6 and hence evaluate (√3 + 1)^6 + (√3 -1)^6)

*Solution*: By binomial Theorem we can Write the expression as

6C_0 x^6 + 6C_1 x^5 + 6C_2 x^4 +6C_3 x^3 + 6C_4 x^2 + 6C_5 x^1 +6C_6 x^0 + [ 6C_0 x^6 - 6C_1 x^5 + 6C_2 x^4 - 6C_3 x^3 + 6C_4 x^2 - 6C_5 x^1 + 6C_6 x^0 ]

2[6C_0 x^6 + 6C_2 x^4 + 6C_4 x^2 + 6C_6 x^0]

2[x^6 + 15x^4 + 15 x^2 + 1]

Thus (√3 + 1)^6 + (√3 -1)^6) = 2[(√3)^6 + 15(√3)^4 + 15(√3)^2 + 1]

= 2[3 + 15(3) + 15 + 1]

= 2[3 + 45 + 16]

= 2[63]

= 126

*Question 3*: Find the coefficient of x^5 in (x+3)^9

*Solution*: T_(r+1) = 9C_r(x^(9-r) (3)^r

Put 9 - r = 5

r = 4

T_5 = 9C_4(x^5)(3^4)

Hence the coefficient of x^5 is 9C_4 (3^4)