If f(x) = ln (x + \(\sqrt{1+\text{x}^2}\) ), then which one of the following is correct ?
f(x) + f(−x) = 0
The question asks us to find the correct relationship between \(f(x)\) and \(f(-x)\) for the given function \(f(x) = \ln (x + \sqrt{1+x^2})\). This involves evaluating the function at \(-x\) and comparing the result to the original function \(f(x)\).
We substitute \(-x\) into the expression for \(f(x)\):
\[f(-x) = \ln ((-x) + \sqrt{1+(-x)^2})\]Since \( (-x)^2 = x^2 \), this simplifies to:
\[f(-x) = \ln (-x + \sqrt{1+x^2})\]The options suggest relationships between \(f(x)\) and \(f(-x)\), such as sums or differences equaling zero, or one being a multiple of the other. We will test these relationships using the expressions for \(f(x)\) and \(f(-x)\).
Let's add \(f(x)\) and \(f(-x)\):
\[f(x) + f(-x) = \ln (x + \sqrt{1+x^2}) + \ln (-x + \sqrt{1+x^2})\]Using the logarithm property \(\ln a + \ln b = \ln (ab)\), we combine the terms:
\[f(x) + f(-x) = \ln \left[ (x + \sqrt{1+x^2}) (-x + \sqrt{1+x^2}) \right]\] \[f(x) + f(-x) = \ln \left[ (\sqrt{1+x^2} + x) (\sqrt{1+x^2} - x) \right]\]This expression is in the form \((a+b)(a-b) = a^2 - b^2\), where \(a = \sqrt{1+x^2}\) and \(b = x\). Applying this identity:
\[f(x) + f(-x) = \ln \left[ (\sqrt{1+x^2})^2 - x^2 \right]\] \[f(x) + f(-x) = \ln \left[ (1+x^2) - x^2 \right]\] \[f(x) + f(-x) = \ln \left[ 1+x^2 - x^2 \right]\] \[f(x) + f(-x) = \ln (1)\]Since \(\ln(1) = 0\), we have:
\[f(x) + f(-x) = 0\]This confirms that the relationship in Option 1 is correct.
Since \(f(x) + f(-x) = 0\), it means \(f(-x) = -f(x)\).
Therefore, the only correct relationship is \(f(x) + f(-x) = 0\).
A function \(f(x)\) is called an odd function if \(f(-x) = -f(x)\) for all \(x\) in its domain. Our finding that \(f(x) + f(-x) = 0\) is equivalent to \(f(-x) = -f(x)\). Thus, the function \(f(x) = \ln (x + \sqrt{1+x^2})\) is an odd function.
A function \(f(x)\) is called an even function if \(f(-x) = f(x)\) for all \(x\) in its domain.
| Concept | Definition | Example Property |
|---|---|---|
| Odd Function | \(f(-x) = -f(x)\) for all \(x\) | Graph is symmetric about the origin |
| Even Function | \(f(-x) = f(x)\) for all \(x\) | Graph is symmetric about the y-axis |
| Logarithm Property | \(\ln a + \ln b = \ln (ab)\) | Used to combine log terms |
| Difference of Squares | \((a+b)(a-b) = a^2 - b^2\) | Used to simplify algebraic expressions |
The natural logarithm function, \(\ln(y)\), is defined for \(y > 0\). For \(f(x) = \ln (x + \sqrt{1+x^2})\), the argument \(x + \sqrt{1+x^2}\) must be greater than 0. We know that \(\sqrt{1+x^2} > \sqrt{x^2} = |x|\). So, \(\sqrt{1+x^2} > -x\). This implies \(x + \sqrt{1+x^2} > 0\) for all real \(x\). Thus, the domain of \(f(x)\) is all real numbers, \((-\infty, \infty)\).
The fact that \(f(x)\) is an odd function means its graph has rotational symmetry about the origin. If a point \((a, b)\) is on the graph, then the point \( (-a, -b) \) is also on the graph.
Consider the following statements:
1. The relation f defined by \(f(x)= \begin{cases}x^3, & 0 \leq x \leq 2 \\ 4 x, & 2 \leq x \leq 8\end{cases}\) is a function.
2. The relation g defined by \(g(x)= \begin{cases}x^2, & 0 \leq x \leq 4 \\ 3 x, & 4 \leq x \leq 8\end{cases}\) is a function.
Which of the statements given above is/are correct?
A function satisfies \(f(x-y)=\frac{f(x)}{f(y)}\), where f(y) ≠ 0. If f(1) = 0.5, then what is f(2) + f(3) + f(4) + f(5) + f(6) equal to ?
If f(x) = x(4x2 - 3), then what is f(sinθ) equal to ?
Let R be a relation on the set N of natural numbers defined by ‘nRm ⟺ n is a factor of m’. Then which one of the following is correct?
f(xy) = f(x) + f(y) is true for all