Ratio, Proportion, Indices, LogarithmMCQPYQ Dec. 21Question 882 of 305
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Let a=5+35−3\displaystyle a = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} and b=5−35+3\displaystyle b = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}. What is the value of a2+b2\displaystyle a^2+b^2?

Options

A64
B62
C60
D254
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Correct Answer

✅ Option b — 62

All Options:

  • A64
  • B62
  • C60
  • D254

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Detailed Solution & Explanation

We are given:
a=5+35−3andb=5−35+3a = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \quad \text{and} \quad b = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}
Let us first rationalize the denominator of a\displaystyle a by multiplying the numerator and denominator by (5+3)\displaystyle (\sqrt{5}+\sqrt{3}):
a=(5+3)(5+3)(5−3)(5+3)=(5+3)25−3a = \frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{(\sqrt{5}+\sqrt{3})^2}{5 - 3}
Using the algebraic identity (x+y)2=x2+y2+2xy\displaystyle (x+y)^2 = x^2 + y^2 + 2xy:
a=5+3+2152=8+2152=4+15a = \frac{5 + 3 + 2\sqrt{15}}{2} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}
Similarly, since b\displaystyle b is the reciprocal of a\displaystyle a (b=1/a\displaystyle b = 1/a), we have:
b=4−15b = 4 - \sqrt{15}
Now, let us calculate the sum (a+b)\displaystyle (a+b) and the product (ab)\displaystyle (ab):
- Sum: a+b=(4+15)+(4−15)=8\displaystyle a + b = (4 + \sqrt{15}) + (4 - \sqrt{15}) = 8
- Product: ab=(4+15)(4−15)=42−(15)2=16−15=1\displaystyle ab = (4 + \sqrt{15})(4 - \sqrt{15}) = 4^2 - (\sqrt{15})^2 = 16 - 15 = 1
We want to find the value of a2+b2\displaystyle a^2 + b^2. Using the algebraic identity a2+b2=(a+b)2−2ab\displaystyle a^2 + b^2 = (a+b)^2 - 2ab:
a2+b2=82−2(1)=64−2=62a^2 + b^2 = 8^2 - 2(1) = 64 - 2 = 62
This matches **Option B**.
Note: The textbook answer key incorrectly specifies **Option A** (64\displaystyle 64) as correct (likely due to omitting the subtraction of 2ab=2\displaystyle 2ab = 2). However, our mathematical proof shows the exact correct answer is 62\displaystyle 62 (Option B).
Hence, **Option B** is the correct answer.

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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