All Exams Test series for 1 year @ ₹349 only
Question

If \({\rm{x}} = \frac{{\sqrt {{\rm{a}} + {\rm{b\;}}} -{\rm{\;}}\sqrt {{\rm{a}} - {\rm{b}}} }}{{\sqrt {{\rm{a}} + {\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - {\rm{b\;}}} }}\) , then what is bx 2– 2ax + b equal to (b ≠ 0)?

This question was previously asked in
CDS I 2017 General Knowledge Previous Year Paper (05-Feb-2017)
The correct answer is

0

Simplifying the Mathematical Expression for x

The problem asks us to find the value of the expression \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\), given a specific value for \({\rm{x}}\): \[{\rm{x}} = \frac{{\sqrt {{\rm{a}} + {\rm{b\;}}} -{\rm{\;}}\sqrt {{\rm{a}} - {\rm{b}}} }}{{\sqrt {{\rm{a}} + {\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - {\rm{b\;}}} }}\] We need to perform algebraic manipulation to simplify this. A helpful technique for expressions involving ratios of square roots is the Componendo and Dividendo rule.

Step-by-Step Simplification of the expression for x

First, let's consider the reciprocal of \({\rm{x}}\): \[ \frac{1}{{\rm{x}}} = \frac{{\sqrt {{\rm{a}} + {\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - {\rm{b\;}}} }}{{\sqrt {{\rm{a}} + {\rm{b\;}}} -{\rm{\;}}\sqrt {{\rm{a}} - {\rm{b}}} }} \] Now, we apply the Componendo and Dividendo rule, which states that if \( \frac{p}{q} = \frac{r}{s} \), then \( \frac{p+q}{p-q} = \frac{r+s}{r-s} \). We apply this to \( \frac{1/x}{1} \): \[ \frac{{1/{\rm{x}}} + 1}{{1/{\rm{x}}} - 1} = \frac{(\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) + (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}})}{(\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) - (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}})} \] Simplifying the numerator and the denominator: Numerator: \( (\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) + (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}}) = 2\sqrt {{\rm{a}} + {\rm{b}}} \) Denominator: \( (\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) - (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}}) = 2\sqrt {{\rm{a}} - {\rm{b}}} \) So, the equation becomes: \[ \frac{\frac{1}{{\rm{x}}} + 1}{\frac{1}{{\rm{x}}} - 1} = \frac{2\sqrt {{\rm{a}} + {\rm{b}}}}{2\sqrt {{\rm{a}} - {\rm{b}}}} = \sqrt{\frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}}} \] We can rewrite the left side as \( \frac{1+x}{1-x} \). Therefore: \[ \frac{1+x}{1-x} = \sqrt{\frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}}} \] Squaring both sides to eliminate the square root: \[ \left(\frac{1+x}{1-x}\right)^2 = \frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}} \] \[ \frac{(1+x)^2}{(1-x)^2} = \frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}} \] \[ \frac{1 + 2x + x^2}{1 - 2x + x^2} = \frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}} \] Now, we apply Componendo and Dividendo again on this equation: \[ \frac{(1 + 2x + x^2) + (1 - 2x + x^2)}{(1 + 2x + x^2) - (1 - 2x + x^2)} = \frac{({\rm{a}} + {\rm{b}}) + ({\rm{a}} - {\rm{b}})}{({\rm{a}} + {\rm{b}}) - ({\rm{a}} - {\rm{b}})} \] Simplify the numerator and denominator on both sides: Left side numerator: \( (1 + 2x + x^2) + (1 - 2x + x^2) = 2 + 2x^2 \) Left side denominator: \( (1 + 2x + x^2) - (1 - 2x + x^2) = 4x \) Right side numerator: \( ({\rm{a}} + {\rm{b}}) + ({\rm{a}} - {\rm{b}}) = 2{\rm{a}} \) Right side denominator: \( ({\rm{a}} + {\rm{b}}) - ({\rm{a}} - {\rm{b}}) = 2{\rm{b}} \) The equation becomes: \[ \frac{2 + 2x^2}{4x} = \frac{2{\rm{a}}}{2{\rm{b}}} \] \[ \frac{2(1 + x^2)}{4x} = \frac{\rm{a}}{\rm{b}} \] \[ \frac{1 + x^2}{2x} = \frac{\rm{a}}{\rm{b}} \] Cross-multiplying gives: \[ {\rm{b}}(1 + x^2) = 2{\rm{a}}x \] \[ {\rm{b}} + {\rm{b}}{{\rm{x}}}^2 = 2{\rm{a}}x \]

Evaluating the Target Expression

We need to evaluate \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\). From the previous step, we derived the relationship:

\[ {\rm{b}} + {\rm{b}}{{\rm{x}}}^2 = 2{\rm{a}}x \]

To find the value of \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\), we can rearrange the derived equation. Subtract \( 2{\rm{a}}x \) from both sides:

\[ {\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}x + {\rm{b}} = 0 \]

Thus, the value of the expression \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\) is 0.

Final Answer Summary

The simplification process using algebraic manipulation, specifically the Componendo and Dividendo rule, leads directly to the relationship \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}} = 0\). This confirms that the value of the given expression is 0.

The correct option is the one that equals 0.

Was this answer helpful?

Similar Questions

  1. What is \(\rm \frac{x^2+a b}{x^2+m^2 a b}\) equal to? 

  2. What is x equal to ?

  3. For \(x = \frac{{4\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }},\)what is the value of \(\frac{{x + 2\sqrt 2 }}{{x - 2\sqrt 2 }} + \frac{{x\; + \;2\sqrt 3 }}{{x - 2\sqrt 3 }}?\)

  4. If \(\rm \frac{a+b}{b+c}=\frac{c+d}{d+a}\) a ≠ c, then which one of the following is correct ?

  5. If (3a + 6b + c + 2d) × (3a - 6b - c + 2d) = (3a - 6b + c - 2d) × (3a + 6b - c - 2d), then which one of the following is correct?

  6. Consider the following statements:

    1. If (a + b) is directly proportional to (a - b), then (a2 + b2) is is directly proportional to ab.

    2. If a is directly proportional to b, then (a2 - b2) is directly proportional to ab.

    Which of the statements given above is/are correct?

  7. If \(\rm\frac{\sqrt{x+20}+\sqrt{x-1}}{\sqrt{x+20}-\sqrt{x-1}}=\frac{7}{3}\) , then what is the value of  \(\rm \sqrt{(x + 20)(x-1)}\)  ?


Important Questions from Componendo or Dividendo

  1. If \(\frac{x}{y} = \frac{5}{3}\), then \(\frac{x + y}{x - y}\) is equal to

  2. What is \(\rm \frac{x^2+a b}{x^2+m^2 a b}\) equal to? 

  3. What is x equal to ?

  4. If \(\frac b a = 0.7,\)  find the value of  \(\frac {a-b}{a+b} + \frac {11}{34}.\)

  5. For \(x = \frac{{4\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }},\)what is the value of \(\frac{{x + 2\sqrt 2 }}{{x - 2\sqrt 2 }} + \frac{{x\; + \;2\sqrt 3 }}{{x - 2\sqrt 3 }}?\)

Need Expert Advice?
Test Series
CDS img
Defence
UPSC CDS 2026 Mock Test Series
536 Tests 4 Tests Free
1647 Attempts
4.3(174)
English, Hindi
More Questions from CDS

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App