If a and b are positive real numbers. Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Identities | Relation | ||
| (A) | \((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\) | (I) | \(=a-b\) |
| (B) | \((a+\sqrt{b})(a-\sqrt{b})\) | (II) | \(=a^{2}-b\) |
| (C) | \((\sqrt{a}+\sqrt{b})^{2}\) | (III) | \(=a-b^{2}\) |
| (D) | \((\sqrt{a})^{2}-(\sqrt{b})^{4}\) | (IV) | \(=a+2\sqrt{ab}+b\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
Each expression is simplified with \((x+y)(x-y)=x^{2}-y^{2}\), \((x+y)^{2}=x^{2}+2xy+y^{2}\) and \((\sqrt{a})^{2}=a\) (true because \(a\) and \(b\) are positive).
(A) \((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\) — this matches (I) \(=a-b\). It equals \((\sqrt{a})^{2}-(\sqrt{b})^{2}=a-b\).
(B) \((a+\sqrt{b})(a-\sqrt{b})\) — this matches (II) \(=a^{2}-b\). It equals \(a^{2}-(\sqrt{b})^{2}=a^{2}-b\); here the first term is \(a\) itself, not \(\sqrt{a}\).
(C) \((\sqrt{a}+\sqrt{b})^{2}\) — this matches (IV) \(=a+2\sqrt{ab}+b\). It equals \((\sqrt{a})^{2}+2\sqrt{a}\sqrt{b}+(\sqrt{b})^{2}=a+2\sqrt{ab}+b\).
(D) \((\sqrt{a})^{2}-(\sqrt{b})^{4}\) — this matches (III) \(=a-b^{2}\). Since \((\sqrt{b})^{4}=\big((\sqrt{b})^{2}\big)^{2}=b^{2}\), it equals \(a-b^{2}\).