Question
Download Solution PDFIf f(x) = ln (x + \(\sqrt{1+\text{x}^2}\)), then which one of the following is correct ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
- (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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