The function f(x) = sin (log (x + sqrt {x^2 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \sin (\log (x + \sqrt {x^2 + 1}))$. To check if the function is even or odd,we find $f(-x)$: $f(-x) = \sin (\log (-x + \sqrt {(-x)^2

Vedclass · Accepted Answer

Odd function

Vedclass · Answer

(B) Given $f(x) = \sin (\log (x + \sqrt {x^2 + 1}))$. To check if the function is even or odd,we find $f(-x)$: $f(-x) = \sin (\log (-x + \sqrt {(-x)^2 + 1})) = \sin (\log (\sqrt {x^2 + 1} - x))$. Multiply and divide the argument of the logarithm by $(\sqrt {x^2 + 1} + x)$: $f(-x) = \sin \left( \log \left( (\sqrt {x^2 + 1} - x) \cdot \frac{\sqrt {x^2 + 1} + x}{\sqrt {x^2 + 1} + x} ight) ight)$. Using $(a-b)(a+b) = a^2 - b^2$: $f(-x) = \sin \left( \log \left( \frac{x^2 + 1 - x^2}{\sqrt {x^2 + 1} + x} ight) ight) = \sin \left( \log \left( \frac{1}{x + \sqrt {x^2 + 1}} ight) ight)$. Using the property $\log(1/a) = -\log(a)$: $f(-x) = \sin (-\log (x + \sqrt {x^2 + 1}))$. Since $\sin(-	heta) = -\sin(	heta)$: $f(-x) = -\sin (\log (x + \sqrt {x^2 + 1})) = -f(x)$. Therefore,$f(x)$ is an odd function.

The function $f(x) = \sin (\log (x + \sqrt {x^2 + 1}))$ is

Similar Questions

Let $f(x) = \cos(\sqrt{P}x),$ where $P = [\lambda]$ and $[.]$ denotes the Greatest Integer Function. If the period of $f(x)$ is $\pi$,then:

The function $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right],$ defined as $f(x) = \frac{x}{1 + x^2}$ is

The number of functions $f: \{1, 2, 3, 4\} \to \{a, b, c\}$, which are not onto, is:

The function $f(x) = \log (x + \sqrt{x^2 + 1})$ is

Find the number of all onto functions from the set $\{1, 2, 3, \ldots, n\}$ to itself.

Vedclass Test Series

Exam Paper Generator

Online Exam Module