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Exercise: 1A
Exercise: 1B
Exercise: 1C
Exercise: 1D
Multiple Choice Questions
Case Study based Questions
Assertion-reason Questions
Competency Focused Questions
Assertion- Reason Questions
Q1:Assertion (A): The number obtained on rationalising the denominator of \(\frac{1}{\sqrt5-2}\) is \(2+\sqrt5\).
Reason (R): If the product of two irrational numbers is rational, then each one is called the rationalising factor of the other.
i. Check Assertion (A)
Step 1: Rationalise the denominator of \(\frac{1}{\sqrt{5}-2}\) by multiplying by the conjugate \(\sqrt{5}+2\).
\(\frac{1}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2}\)
Step 2: Use the identity \((a-b)(a+b) = a^2 – b^2\) for the denominator.
Denominator: \((\sqrt{5})^2 – (2)^2 = 5 – 4 = 1\).
Step 3: The expression becomes \(\sqrt{5}+2\), which is the same as \(2+\sqrt{5}\).
Step 4: Therefore, Assertion (A) is true.
ii. Check Reason (R)
Step 1: By definition, if the product of two irrational numbers is a rational number, then each number is indeed called the rationalising factor (R.F.) of the other.
Step 2: Therefore, Reason (R) is true.
iii. Verify the Relationship
Step 1: Since both statements are true and Reason (R) correctly explains why we use the conjugate to rationalise the denominator in Assertion (A).
Answer: c. Both A and R are true
Q2: Assertion (A): Each of the numbers \(\sqrt[3]{2},\ \sqrt[3]{3},\ \sqrt[3]{4},\ \sqrt[3]{5},\ \sqrt[3]{6},\ \sqrt[3]{7}\) is irrational.
Reason (R): The cube roots of all natural numbers is irrational.
i. Check Assertion (A)
Step 1: Examine the numbers \(\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7}\).
Step 2: A cube root of a natural number \(n\) is rational only if \(n\) is a perfect cube (e.g., 1, 8, 27).
Step 3: Since none of the numbers 2, 3, 4, 5, 6, or 7 are perfect cubes, their cube roots are non-terminating and non-repeating.
Step 4: Therefore, Assertion (A) is true.
ii. Check Reason (R)
Step 1: Reason (R) states that the cube roots of *all* natural numbers are irrational.
Step 2: Test with a counter-example: Consider the natural number 8.
Step 3: \(\sqrt[3]{8} = 2\). Since 2 is a rational number, the statement in Reason (R) is incorrect.
Step 4: Therefore, Reason (R) is false.
iii. Conclusion
Step 1: Since Assertion (A) is true and Reason (R) is false, option (a) is the correct choice.
Answer: a. A is true, R is false



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