MCQ

Rationalisation is the process of converting an irrational denominator into a rational number. This is typically done by multiplying both the numerator and the denominator by a suitable factor.

Key Techniques for Rationalisation

• Type 1: Denominator is a single surd (e.g., \(1/\sqrt{a}\))
• Multiply numerator and denominator by \(\sqrt{a}\).
• Example: \(1/\sqrt{2} = (1 \times \sqrt{2}) / (\sqrt{2} \times \sqrt{2}) = \sqrt{2}/2\).

• Type 2: Denominator is a binomial with one surd (e.g., \(1/(a \pm \sqrt{b})\))
• Multiply numerator and denominator by the conjugate of the denominator.
• The conjugate of \((a + \sqrt{b})\) is \((a - \sqrt{b})\) and vice-versa.
• Uses the identity \((x+y)(x-y) = x^2 - y^2\).
• Example: \(1/(2 + \sqrt{3}) = (1 \times (2 - \sqrt{3})) / ((2 + \sqrt{3})(2 - \sqrt{3})) = (2 - \sqrt{3}) / (2^2 - (\sqrt{3})^2) = (2 - \sqrt{3}) / (4 - 3) = 2 - \sqrt{3}\).

• Type 3: Denominator is a binomial with two surds (e.g., \(1/(\sqrt{a} \pm \sqrt{b})\))
• Multiply numerator and denominator by the conjugate of the denominator.
• The conjugate of \((\sqrt{a} + \sqrt{b})\) is \((\sqrt{a} - \sqrt{b})\) and vice-versa.
• Example: \(1/(\sqrt{5} - \sqrt{2}) = (1 \times (\sqrt{5} + \sqrt{2})) / ((\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})) = (\sqrt{5} + \sqrt{2}) / ((\sqrt{5})^2 - (\sqrt{2})^2) = (\sqrt{5} + \sqrt{2}) / (5 - 2) = (\sqrt{5} + \sqrt{2}) / 3\).

• Type 4: Denominator with three terms (e.g., \(1/(\sqrt{a} + \sqrt{b} + \sqrt{c})\))
• Group two terms and treat them as one. Apply the conjugate method twice.
• Example: To rationalise \(1/(\sqrt{2} + \sqrt{3} + \sqrt{5})\), first multiply by \((\sqrt{2} + \sqrt{3} - \sqrt{5})) / ((\sqrt{2} + \sqrt{3} - \sqrt{5}))\).
• This will result in a denominator of the form \((a + b)^2 - c^2\), which will still have a surd. Then rationalise again.

Important Identities for Rationalisation

• \((\sqrt{a})^2 = a\)
• \((a + b)(a - b) = a^2 - b^2\)
• \((\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b\)
• \((a + \sqrt{b})(a - \sqrt{b}) = a^2 - b\)

These laws are fundamental for simplifying expressions involving powers and roots. Remember that \(a^{m/n} = (a^m)^{1/n} = (a^{1/n})^m = \sqrt[n]{a^m} = (\sqrt[n]{a})^m\).

Summary of Laws of Exponents

Let \(a, b\) be positive real numbers and \(m, n\) be rational numbers.

1. Product Rule: \(a^m \times a^n = a^{m+n}\)

• Example: \(2^3 \times 2^4 = 2^{3+4} = 2^7 = 128\)

1. Quotient Rule: \(a^m / a^n = a^{m-n}\)

• Example: \(5^6 / 5^2 = 5^{6-2} = 5^4 = 625\)

1. Power Rule: \((a^m)^n = a^{mn}\)

• Example: \(( (3^2)^3 ) = 3^{2 \times 3} = 3^6 = 729\)

1. Product to a Power: \((ab)^m = a^m b^m\)

• Example: \((2 \times 3)^4 = 2^4 \times 3^4 = 16 \times 81 = 1296\)

1. Quotient to a Power: \((a/b)^m = a^m / b^m\)

• Example: \((4/2)^3 = 4^3 / 2^3 = 64 / 8 = 8\)

1. Zero Exponent: \(a^0 = 1\) (for \(a \neq 0\))

• Example: \(7^0 = 1\)

1. Negative Exponent: \(a^{-m} = 1/a^m\) (for \(a \neq 0\))

• Example: \(3^{-2} = 1/3^2 = 1/9\)

1. Fractional Exponent: \(a^{m/n} = \sqrt[n]{a^m}\) or \(( \sqrt[n]{a} )^m\)

• Example: \(8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\)

Connecting Exponents and Roots

• \(\sqrt{a} = a^{1/2}\)
• \(\sqrt[3]{a} = a^{1/3}\)
• \(\sqrt[n]{a} = a^{1/n}\)
• \(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\)
• \(\sqrt[n]{a/b} = \sqrt[n]{a} / \sqrt[n]{b}\)
• \(\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}\)

To compare surds (like \(\sqrt[n]{a}\) and \(\sqrt[m]{b}\)), convert them to a common order (or index) by finding the LCM of the orders, or by converting them to fractional exponents with a common denominator.

Method 1: Common Order (LCM of indices)

1. Find the LCM of the orders (indices) of the surds.
2. Convert each surd to an equivalent surd with the LCM as its new order.

• If you have \(\sqrt[n]{a}\), and the LCM is \(L\), then \(\sqrt[n]{a} = \sqrt[L]{a^{L/n}}
• Example: Compare \(\sqrt{2}\) and \(\sqrt[3]{3}\).
• Orders are 2 and 3. LCM(2, 3) = 6.
• \(\sqrt{2} = \sqrt[6]{2^{6/2}} = \sqrt[6]{2^3} = \sqrt[6]{8}\)
• \(\sqrt[3]{3} = \sqrt[6]{3^{6/3}} = \sqrt[6]{3^2} = \sqrt[6]{9}\)
• Since \(9 > 8\), \(\sqrt[6]{9} > \sqrt[6]{8}\), so \(\sqrt[3]{3} > \sqrt{2}\).

Method 2: Fractional Exponents

1. Convert each surd into its fractional exponent form.
2. Find the LCM of the denominators of the fractional exponents.
3. Rewrite each fractional exponent with the LCM as the new denominator.
4. Compare the resulting numbers.

• Example: Compare \(\sqrt{2}\) and \(\sqrt[3]{3}\).
• \(\sqrt{2} = 2^{1/2}\)
• \(\sqrt[3]{3} = 3^{1/3}\)
• Denominators are 2 and 3. LCM(2, 3) = 6.
• \(2^{1/2} = 2^{3/6} = (2^3)^{1/6} = 8^{1/6}\)
• \(3^{1/3} = 3^{2/6} = (3^2)^{1/6} = 9^{1/6}\)
• Since \(9 > 8\), \(9^{1/6} > 8^{1/6}\), so \(\sqrt[3]{3} > \sqrt{2}\).

Comparing numbers with different roots

This method is crucial for problems like Q5 and Q9 in the NCERT exercise.

• Steps:

1. Express all numbers in exponential form (e.g., \(\sqrt[n]{a} = a^{1/n}\)).
2. Find the LCM of all the denominators of the exponents.
3. Convert all exponents to have this common denominator.
4. Compare the bases raised to the new numerators. The one with the largest value is the greatest.

Many algebraic identities can be applied to expressions involving surds. These are essential for simplifying complex expressions.

Common Identities Applied to Surds

Let \(x, y\) be real numbers or surds.

1. \((x + y)^2 = x^2 + 2xy + y^2\)

• Example: \((\sqrt{a} + \sqrt{b})^2 = (\sqrt{a})^2 + 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 = a + 2\sqrt{ab} + b\)

1. \((x - y)^2 = x^2 - 2xy + y^2\)

• Example: \((\sqrt{a} - \sqrt{b})^2 = (\sqrt{a})^2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 = a - 2\sqrt{ab} + b\)

1. \((x + y)(x - y) = x^2 - y^2\)

• Example: \((\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b\)
• Example: \((a + \sqrt{b})(a - \sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b\)

1. \((x + y)^3 = x^3 + y^3 + 3xy(x + y)\)

• Example: \((1 + \sqrt{2})^3 = 1^3 + (\sqrt{2})^3 + 3(1)(\sqrt{2})(1 + \sqrt{2}) = 1 + 2\sqrt{2} + 3\sqrt{2} + 3(2) = 1 + 5\sqrt{2} + 6 = 7 + 5\sqrt{2}\)

1. \((x - y)^3 = x^3 - y^3 - 3xy(x - y)\)

• Example: \((2 - \sqrt{3})^3 = 2^3 - (\sqrt{3})^3 - 3(2)(\sqrt{3})(2 - \sqrt{3}) = 8 - 3\sqrt{3} - 12\sqrt{3} + 18 = 26 - 15\sqrt{3}\)

1. \(x^3 + y^3 = (x + y)(x^2 - xy + y^2)\)
2. \(x^3 - y^3 = (x - y)(x^2 + xy + y^2)\)

Working with Expressions like \(x + 1/x\) and \(x^2 + 1/x^2\)

These types of problems frequently appear in exams, especially when \(x\) is a surd.

• If \(x = a + \sqrt{b}\), then \(1/x\) can be found by rationalising the denominator.
• Then, \(x + 1/x\) or \(x - 1/x\) can be calculated.
• To find \(x^2 + 1/x^2\), use the identity \((x + 1/x)^2 = x^2 + 2(x)(1/x) + (1/x)^2 = x^2 + 2 + 1/x^2\).
• So, \(x^2 + 1/x^2 = (x + 1/x)^2 - 2\).
• Similarly, \(x^2 + 1/x^2 = (x - 1/x)^2 + 2\).
• For \(x^3 + 1/x^3\), use \((x + 1/x)^3 = x^3 + 1/x^3 + 3(x)(1/x)(x + 1/x) = x^3 + 1/x^3 + 3(x + 1/x)\).
• So, \(x^3 + 1/x^3 = (x + 1/x)^3 - 3(x + 1/x)\).