**Miscellaneous Exercise**

Binomial Theorem

**Question and Answers**

**Class 11 – Maths**

Binomial Theorem

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Binomial Theorem |

Chapter No. | Chapter 7 |

Exercise | Miscellaneous Exercise |

Category | Class 11 Maths NCERT Solutions |

**Question 1** If a and b are distinct integers, prove that a – b is a factor of a ^{n} – b^{ n} , whenever n is a positive integer.

[Hint write a^{n} = (a – b + b)^{ n} and expand]

**Answers**

a = a − b + b

So, a^{n} = [a − b +b]^{n
}= [(a − b) + b]^{n
}= ^{n}C_{0} (a − b)^{n} + ^{n}C_{1} (a − b)^{n−1}b^{1} + ^{n}C_{2} (a − b)^{n−2}b^{2} + …….+ ^{n}C_{n−1} (a − b)b^{n−1} + ^{n}C_{n} (b^{n})

⇒ a^{n} − b^{n} = (a − b)_{n} + ^{n}C_{1} (a − b)^{n−1}b + ^{n}C_{2} (a − b)^{n−2}b^{2} + …….+ ^{n}C_{n−1} (a − b)b^{n−1
}= (a − b) [(a − b)^{n−1} + ^{n}C_{1} (a − b)^{n−2}b + ^{n}C_{2} (a − b)^{n−3}b^{2} + ……+ ^{n}C_{n−1} b^{n−1}]
= (a – b)[a^{n} integer]
⇒ a^{n} – b^{n} is divisible by (a – b)

**Question 2 Evaluate (√3 + √2) ^{6} – ( √3 – √2 )^{6}**

**Answers**

$( √3 + √2 )6−(√3 −√2 )6$= ((√3+ √2)^{2 }− (√3−√2)^{2})^{3 }+ 3 (√3+√2)^{2}( √3−√2)^{2 }((√3+√2)^{2}−(√3+√2)^{2}))

= (4√6)^{3 }+ 4√6(5+2√6) (5−2√6)

= 384√6+4√6

= 388√6

**Question 3 Find the value of**

** Answers**

**Question 4 Find an approximation of (0.99) ^{5} using the first three terms of its expansion**

**Answer **0.99 can be written as

0.99 = 1 – 0.01

Now, by applying the binomial theorem, we get

(o. 99)^{5} = (1 – 0.01)^{5
}= ^{5}C_{0 }(1)^{5} – ^{5}C_{1 }(1)^{4} (0.01) + ^{5}C_{2 }(1)^{3} (0.01)^{2
}= 1 – 5 (0.01) + 10 (0.01)^{2
}= 1 – 0.05 + 0.001

= 0.951

**Question 5 Expand using Binomial Theorem**

**Answer** Using the binomial theorem, the given expression can be expanded as

**Question 6 Find the expansion of (3x ^{2} – 2ax + 3a^{ 2} ) ^{3} using binomial theorem.**

**$ Answer$**(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}Putting a = 3x^{2
} b = -a (2x-3a), we get

$[3x2 + (-a (2x-3a))]3$= (3x^{2})^{3}+3(3x^{2})^{2}(-a (2x-3a)) + 3(3x^{2}) (-a (2x-3a))^{2} + (-a (2x-3a))^{3
}= 27x^{6} – 27ax^{4 }(2x-3a) + 9a^{2}x^{2 }(2x-3a)^{2} – a^{3}(2x-3a)^{3
}= 27x^{6} – 54ax^{5} + 81a^{2}x^{4} + 9a^{2}x^{2 }(4x^{2}-12ax+9a^{2}) – a^{3 }[(2x)^{3} – (3a)^{3} – 3(2x)^{2}(3a) + 3(2x)(3a)^{2}]
= 27x^{6} – 54ax^{5} + 81a^{2}x^{4} + 36a^{2}x^{4} – 108a^{3}x^{3} + 81a^{4}x^{2} – 8a^{3}x^{3} + 27a^{6} + 36a^{4}x^{2} – 54a^{5}x

= 27x^{6} – 54ax^{5}+ 117a^{2}x^{4} – 116a^{3}x^{3} + 117a^{4}x^{2} – 54a^{5}x + 27a^{6
}Thus, (3x^{2} – 2ax + 3a^{2})^{3
}= 27x^{6} – 54ax^{5}+ 117a^{2}x^{4} – 116a^{3}x^{3} + 117a^{4}x^{2} – 54a^{5}x + 27a^{6}

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