RBSE Solutions Class 11 Maths Chapter 7 Binomial Theorem Exercise 7.1

Expand Each Expression in the Following Questions (Q. 1 to 5):

Question 1. \( (2-x)^3 \)
Answer: To expand \( (2-x)^3 \), we use the binomial theorem. Here, \( a=2 \), \( b=-x \), and \( n=3 \).
The expansion is given by:
\( \binom{3}{0} (2)^3 (-x)^0 + \binom{3}{1} (2)^2 (-x)^1 + \binom{3}{2} (2)^1 (-x)^2 + \binom{3}{3} (2)^0 (-x)^3 \)
Now, we calculate each term:
\( = 1 \times 8 \times 1 + 3 \times 4 \times (-x) + 3 \times 2 \times x^2 + 1 \times 1 \times (-x^3) \)
\( = 8 - 12x + 6x^2 - x^3 \)
So, the expanded form is \( 8 - 12x + 6x^2 - x^3 \).
In simple words: We break down the expression using a special math rule called the binomial theorem. We multiply out each part and then combine them to get the final expanded form.

🎯 Exam Tip: Remember the signs carefully when using the binomial theorem, especially when one of the terms is negative like \( (-x) \).

Question 2. \( \left(\frac {2}{x}-\frac {x}{ 2 } \right)^{5} \)
Answer: To expand \( \left(\frac {2}{x}-\frac {x}{ 2 } \right)^{5} \), we use the binomial theorem. Here, \( a=\frac{2}{x} \), \( b=-\frac{x}{2} \), and \( n=5 \).
The expansion is:
\( \binom{5}{0} \left(\frac{2}{x}\right)^5 \left(-\frac{x}{2}\right)^0 + \binom{5}{1} \left(\frac{2}{x}\right)^4 \left(-\frac{x}{2}\right)^1 + \binom{5}{2} \left(\frac{2}{x}\right)^3 \left(-\frac{x}{2}\right)^2 + \binom{5}{3} \left(\frac{2}{x}\right)^2 \left(-\frac{x}{2}\right)^3 + \binom{5}{4} \left(\frac{2}{x}\right)^1 \left(-\frac{x}{2}\right)^4 + \binom{5}{5} \left(\frac{2}{x}\right)^0 \left(-\frac{x}{2}\right)^5 \)
Now, we calculate and simplify each term:
\( = 1 \times \frac{32}{x^5} \times 1 + 5 \times \frac{16}{x^4} \times \left(-\frac{x}{2}\right) + 10 \times \frac{8}{x^3} \times \left(\frac{x^2}{4}\right) + 10 \times \frac{4}{x^2} \times \left(-\frac{x^3}{8}\right) + 5 \times \frac{2}{x} \times \left(\frac{x^4}{16}\right) + 1 \times 1 \times \left(-\frac{x^5}{32}\right) \)
\( = \frac{32}{x^5} - \frac{80x}{2x^4} + \frac{80x^2}{4x^3} - \frac{40x^3}{8x^2} + \frac{10x^4}{16x} - \frac{x^5}{32} \)
\( = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - \frac{5x}{1} + \frac{5x^3}{8} - \frac{x^5}{32} \)
In simple words: We used the binomial theorem to expand this expression. Each part of the expansion involves powers of the terms and binomial coefficients, which we then multiply and simplify to get the final answer.

🎯 Exam Tip: Pay close attention to the powers of x in the numerator and denominator, as they cancel out differently in each term. Keep track of negative signs as well.

Question 3. \( \left(\frac {x}{ 3 } -\frac { 1 }{ x } \right)^{ 6 } \)
Answer: To expand \( \left(\frac {x}{ 3 } -\frac { 1 }{ x } \right)^{ 6 } \), we use the binomial theorem. Here, \( a=\frac{x}{3} \), \( b=-\frac{1}{x} \), and \( n=6 \).
The expansion is:
\( \binom{6}{0} \left(\frac{x}{3}\right)^6 \left(-\frac{1}{x}\right)^0 + \binom{6}{1} \left(\frac{x}{3}\right)^5 \left(-\frac{1}{x}\right)^1 + \binom{6}{2} \left(\frac{x}{3}\right)^4 \left(-\frac{1}{x}\right)^2 + \binom{6}{3} \left(\frac{x}{3}\right)^3 \left(-\frac{1}{x}\right)^3 + \binom{6}{4} \left(\frac{x}{3}\right)^2 \left(-\frac{1}{x}\right)^4 + \binom{6}{5} \left(\frac{x}{3}\right)^1 \left(-\frac{1}{x}\right)^5 + \binom{6}{6} \left(\frac{x}{3}\right)^0 \left(-\frac{1}{x}\right)^6 \)
Now, we calculate and simplify each term:
\( = 1 \times \frac{x^6}{729} \times 1 + 6 \times \frac{x^5}{243} \times \left(-\frac{1}{x}\right) + 15 \times \frac{x^4}{81} \times \left(\frac{1}{x^2}\right) + 20 \times \frac{x^3}{27} \times \left(-\frac{1}{x^3}\right) + 15 \times \frac{x^2}{9} \times \left(\frac{1}{x^4}\right) + 6 \times \frac{x}{3} \times \left(-\frac{1}{x^5}\right) + 1 \times 1 \times \left(\frac{1}{x^6}\right) \)
\( = \frac{x^6}{729} - \frac{6x^5}{243x} + \frac{15x^4}{81x^2} - \frac{20x^3}{27x^3} + \frac{15x^2}{9x^4} - \frac{6x}{3x^5} + \frac{1}{x^6} \)
\( = \frac{x^6}{729} - \frac{2x^4}{81} + \frac{5x^2}{27} - \frac{20}{27} + \frac{5}{3x^2} - \frac{2}{x^4} + \frac{1}{x^6} \)
In simple words: We used the binomial theorem for this expansion. We calculated each term by applying the binomial coefficients and powers, and then simplified the algebraic fractions to get the final expanded expression.

🎯 Exam Tip: When simplifying terms with x in both the numerator and denominator, remember the rules of exponents, e.g., \( \frac{x^m}{x^n} = x^{m-n} \).

Question 5. \( \left(\sqrt { \frac {x}{ a } } -\sqrt {\frac { a }{ x } } \right)^{6} \)
Answer: To expand \( \left(\sqrt { \frac {x}{ a } } -\sqrt {\frac { a }{ x } } \right)^{6} \), we use the binomial theorem. Here, let \( A = \sqrt{\frac{x}{a}} = \left(\frac{x}{a}\right)^{1/2} \) and \( B = -\sqrt{\frac{a}{x}} = -\left(\frac{a}{x}\right)^{1/2} \), and \( n=6 \).
The expansion is:
\( \binom{6}{0} A^6 B^0 + \binom{6}{1} A^5 B^1 + \binom{6}{2} A^4 B^2 + \binom{6}{3} A^3 B^3 + \binom{6}{4} A^2 B^4 + \binom{6}{5} A^1 B^5 + \binom{6}{6} A^0 B^6 \)
Substitute A and B back:
\( = \binom{6}{0} \left(\frac{x}{a}\right)^{6/2} - \binom{6}{1} \left(\frac{x}{a}\right)^{5/2} \left(\frac{a}{x}\right)^{1/2} + \binom{6}{2} \left(\frac{x}{a}\right)^{4/2} \left(\frac{a}{x}\right)^{2/2} - \binom{6}{3} \left(\frac{x}{a}\right)^{3/2} \left(\frac{a}{x}\right)^{3/2} + \binom{6}{4} \left(\frac{x}{a}\right)^{2/2} \left(\frac{a}{x}\right)^{4/2} - \binom{6}{5} \left(\frac{x}{a}\right)^{1/2} \left(\frac{a}{x}\right)^{5/2} + \binom{6}{6} \left(\frac{a}{x}\right)^{6/2} \)
Now, we calculate and simplify each term:
\( = 1 \times \left(\frac{x}{a}\right)^3 - 6 \times \left(\frac{x^5}{a^5}\right)^{1/2} \left(\frac{a}{x}\right)^{1/2} + 15 \times \left(\frac{x}{a}\right)^2 \left(\frac{a}{x}\right)^1 - 20 \times \left(\frac{x^3}{a^3}\right)^{1/2} \left(\frac{a^3}{x^3}\right)^{1/2} + 15 \times \left(\frac{x}{a}\right)^1 \left(\frac{a}{x}\right)^2 - 6 \times \left(\frac{x}{a}\right)^{1/2} \left(\frac{a^5}{x^5}\right)^{1/2} + 1 \times \left(\frac{a}{x}\right)^3 \)
\( = \frac{x^3}{a^3} - 6 \frac{x^2}{a^2} + 15 \frac{x}{a} - 20 + 15 \frac{a}{x} - 6 \frac{a^2}{x^2} + \frac{a^3}{x^3} \)
In simple words: We used the binomial theorem to expand this expression with square roots. We converted the square roots to powers of 1/2, then expanded using the formula, carefully simplifying the exponents of x and a in each term.

🎯 Exam Tip: When dealing with fractional exponents or square roots in binomial expansions, convert them to fractional powers and apply the exponent rules carefully. Simplify common bases by adding or subtracting exponents.

Using Binomial Theorem Find the Values of Following (Q. 6 to 9):

Question 6. \( (96)^3 \)
Answer: We can write 96 as \( (100-4) \). This makes it easier to use the binomial theorem.
So, \( (96)^3 = (100 - 4)^3 \)
This can also be written as \( [100 + (-4)]^3 \).
By using the binomial theorem, with \( a=100 \), \( b=-4 \), and \( n=3 \):
\( (100 - 4)^3 = \binom{3}{0} (100)^3 (-4)^0 + \binom{3}{1} (100)^2 (-4)^1 + \binom{3}{2} (100)^1 (-4)^2 + \binom{3}{3} (100)^0 (-4)^3 \)
Now, we calculate each term:
\( = 1 \times 1,000,000 \times 1 + 3 \times 10,000 \times (-4) + 3 \times 100 \times 16 + 1 \times 1 \times (-64) \)
\( = 1,000,000 - 120,000 + 4,800 - 64 \)
\( = 880,000 + 4,800 - 64 \)
\( = 884,800 - 64 \)
\( = 884,736 \)
Therefore, \( (96)^3 = 884,736 \).
In simple words: We changed 96 to (100 - 4) to make it easy to use the binomial rule. Then, we worked out each part of the sum and added or subtracted them to find the final number.

🎯 Exam Tip: When calculating powers of numbers close to 10, 100, or 1000, use the binomial theorem by expressing them as \( (10^k \pm x)^n \). This simplifies calculations by avoiding large multiplications.

Question 7. \( (101)^4 \)
Answer: We can write 101 as \( (100+1) \). This helps us use the binomial theorem easily.
So, \( (101)^4 = (100+1)^4 \).
By using the binomial theorem, with \( a=100 \), \( b=1 \), and \( n=4 \):
\( (100+1)^4 = \binom{4}{0} (100)^4 (1)^0 + \binom{4}{1} (100)^3 (1)^1 + \binom{4}{2} (100)^2 (1)^2 + \binom{4}{3} (100)^1 (1)^3 + \binom{4}{4} (100)^0 (1)^4 \)
Now, we calculate each term:
\( = 1 \times 100,000,000 \times 1 + 4 \times 1,000,000 \times 1 + 6 \times 10,000 \times 1 + 4 \times 100 \times 1 + 1 \times 1 \times 1 \)
\( = 100,000,000 + 4,000,000 + 60,000 + 400 + 1 \)
\( = 104,060,401 \)
Therefore, \( (101)^4 = 104,060,401 \).
In simple words: We broke down 101 into (100 + 1) to use the binomial rule. We calculated each part of the sum and then added them all up to find the final answer.

🎯 Exam Tip: For expressions like \( (100+1)^4 \), the powers of 1 remain 1, simplifying the calculation of each term considerably. Be careful with the number of zeros for powers of 100.

Question 8. \( (99)^5 \)
Answer: We can write 99 as \( (100-1) \). This simplifies using the binomial theorem.
So, \( (99)^5 = (100-1)^5 \).
This can also be written as \( [100 + (-1)]^5 \).
By using the binomial theorem, with \( a=100 \), \( b=-1 \), and \( n=5 \):
\( (100-1)^5 = \binom{5}{0} (100)^5 (-1)^0 + \binom{5}{1} (100)^4 (-1)^1 + \binom{5}{2} (100)^3 (-1)^2 + \binom{5}{3} (100)^2 (-1)^3 + \binom{5}{4} (100)^1 (-1)^4 + \binom{5}{5} (100)^0 (-1)^5 \)
Now, we calculate each term:
\( = 1 \times 10,000,000,000 \times 1 + 5 \times 100,000,000 \times (-1) + 10 \times 1,000,000 \times 1 + 10 \times 10,000 \times (-1) + 5 \times 100 \times 1 + 1 \times 1 \times (-1) \)
\( = 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1 \)
\( = 9,500,000,000 + 10,000,000 - 100,000 + 500 - 1 \)
\( = 9,510,000,000 - 100,000 + 500 - 1 \)
\( = 9,509,900,000 + 500 - 1 \)
\( = 9,509,900,499 \)
Therefore, \( (99)^5 = 9,509,900,499 \).
In simple words: We expressed 99 as (100-1) and used the binomial rule. We carefully added and subtracted the results from each part of the expansion to get the final answer. Remember to handle negative signs correctly.

🎯 Exam Tip: When \( b \) is negative in \( (a+b)^n \), the signs of the terms alternate. Make sure to correctly apply the powers to \( (-1) \) to get the right sign for each term.

Question 10. Using Binomial theorem find which number is greater \( (1.1)^{100000} \) or 1000.
Answer: We need to compare \( (1.1)^{100000} \) with 1000.
We can write \( 1.1 \) as \( (1+0.1) \).
So, \( (1.1)^{100000} = (1+0.1)^{100000} \).
Using the binomial theorem, the expansion starts with:
\( (1+0.1)^{100000} = \binom{100000}{0} (1)^{100000} (0.1)^0 + \binom{100000}{1} (1)^{99999} (0.1)^1 + \dots \)
Let's calculate the first two terms:
\( = 1 \times 1 \times 1 + 100000 \times 1 \times 0.1 + \dots \)
\( = 1 + 10000 + \dots \)
\( = 1001 + \dots \)
Since all the other terms in the binomial expansion for \( (1+0.1)^{100000} \) are positive, \( (1.1)^{100000} \) will be greater than \( 1001 \).
We know that \( 1001 > 1000 \).
Therefore, \( (1.1)^{100000} > 1000 \).
So, \( (1.1)^{100000} \) is greater.
In simple words: We changed 1.1 to (1 + 0.1) and used the binomial rule. Even just the first two parts of the expansion added up to more than 1000, and all the other parts are positive, so the total will be even bigger.

🎯 Exam Tip: For comparing numbers of the form \( (1+x)^n \) with a constant, use the first few terms of the binomial expansion. If \( x \) is positive, all subsequent terms will also be positive, so you only need to show the sum of the first terms exceeds the constant.

Question 11. Expand \( (a + b)^4 – (a – b)^4 \). Using this find the value of \( (\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} - \sqrt{2})^4 \).
Answer: First, let's expand \( (a+b)^4 \) and \( (a-b)^4 \) using the binomial theorem.
\( (a+b)^4 = \binom{4}{0} a^4 b^0 + \binom{4}{1} a^3 b^1 + \binom{4}{2} a^2 b^2 + \binom{4}{3} a^1 b^3 + \binom{4}{4} a^0 b^4 \) ......(1)
\( (a-b)^4 = \binom{4}{0} a^4 b^0 - \binom{4}{1} a^3 b^1 + \binom{4}{2} a^2 b^2 - \binom{4}{3} a^1 b^3 + \binom{4}{4} a^0 b^4 \) ......(2)
Now, we subtract equation (2) from equation (1):
\( (a + b)^4 – (a – b)^4 = [\binom{4}{0} a^4 + \binom{4}{1} a^3 b + \binom{4}{2} a^2 b^2 + \binom{4}{3} a b^3 + \binom{4}{4} b^4] - [\binom{4}{0} a^4 - \binom{4}{1} a^3 b + \binom{4}{2} a^2 b^2 - \binom{4}{3} a b^3 + \binom{4}{4} b^4] \)
When we subtract, the terms with even powers of b will cancel out, and terms with odd powers of b will be doubled:
\( = \binom{4}{0} a^4 + \binom{4}{1} a^3 b + \binom{4}{2} a^2 b^2 + \binom{4}{3} a b^3 + \binom{4}{4} b^4 - \binom{4}{0} a^4 + \binom{4}{1} a^3 b - \binom{4}{2} a^2 b^2 + \binom{4}{3} a b^3 - \binom{4}{4} b^4 \)
\( = 2 \times \binom{4}{1} a^3 b + 2 \times \binom{4}{3} a b^3 \)
\( = 2ab [\binom{4}{1} a^2 + \binom{4}{3} b^2] \)
Since \( \binom{4}{1} = 4 \) and \( \binom{4}{3} = 4 \):
\( = 2ab [4a^2 + 4b^2] \)
\( = 2ab \times 4(a^2 + b^2) \)
\( = 8ab (a^2 + b^2) \)
So, \( (a + b)^4 – (a - b)^4 = 8ab (a^2 + b^2) \).
Now, we use this result to find the value of \( (\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} - \sqrt{2})^4 \).
We compare this with \( (a + b)^4 – (a - b)^4 \), so we can substitute \( a = \sqrt{3} \) and \( b = \sqrt{2} \).
\( (\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} - \sqrt{2})^4 = 8 (\sqrt{3}) (\sqrt{2}) [(\sqrt{3})^2 + (\sqrt{2})^2] \)
\( = 8 \sqrt{6} [3 + 2] \)
\( = 8 \sqrt{6} \times 5 \)
\( = 40 \sqrt{6} \)
In simple words: First, we expanded the two binomials and subtracted them, which simplified to \( 8ab (a^2 + b^2) \). Then, we put in \( \sqrt{3} \) for 'a' and \( \sqrt{2} \) for 'b' into this simplified formula to get the final answer.

🎯 Exam Tip: Recognize patterns in binomial expansions, especially when subtracting \( (a-b)^n \) from \( (a+b)^n \). Only terms with odd powers of b will remain and be doubled, simplifying the problem significantly.