If f(x) = ln (x + \(\sqrt{1+\text{x}^2}\)), then which one of the following is correct ?

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NDA 02/2022 Mathematics Official Paper (Held On 04 Sep 2022)
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  1. f(x) + f(−x) = 0
  2. f(x)−f(−x) = 0
  3. 2f(x) = f(−x)
  4. f(x) = 2f(−x)

Answer (Detailed Solution Below)

Option 1 : f(x) + f(−x) = 0
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NDA 01/2025: English Subject Test
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Detailed Solution

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Concept:

  • (x + y)(x - y) = x2 - y2
  • \(\ln { 1\over a} = \ln 1 - \ln a = 0 - \ln a = -\ln a\)

Calculation:

Given: f(x) = ln (x + \(\sqrt{1+\text{x}^2}\)),__(i)

Replace x by (-x) in (i),

⇒  f(-x) = ln (-x + \(\sqrt{1+(\text{-x})^2}\)),

⇒  f(-x) = ln (-x + \(\sqrt{1+\text{x}^2}\)),

Multiply and divide by (x + \(\sqrt{1+\text{x}^2}\)) inside the ln function,

⇒  f(-x) =  \(\ln (-x + \sqrt{1+\text{x}^2}.{ x + \sqrt{1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({- x^2 + {1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({1 \over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(-\ln ({ x + \sqrt{1+\text{x}^2}})\),

From (i),

⇒  f(-x) =  - f(x)\

⇒  f(-x) + f(x) = 0

∴ The correct answer is option (1).

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