The function fx=logx+√(x^2+1) is

Correct answer: (A) An odd function. f-x=log-x+√(x^2+1) =log(-x+√(x^2+1))/(x+√(x^2+1))×x+√(x^2+1) =log(-x^2+x^2+1)/(x+√(x^2+1)) =log1/(x+√(x^2+1)) =-logx+√(x^2+1) =-fx ⇒f-x+fx=0 Hence, fx is an odd function.

Q42 / 160 · TS EAMCET 2021 (04 Aug Shift 2) · Mathematics

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

Correct answer
A
An odd function
B
A periodic function
C
Neither an even nor an odd function
D
An even function

Solution

f (- x) = \log (- x + \sqrt{x^{2} + 1})

$= \log (\frac{- x + \sqrt{x^{2} + 1}}{x + \sqrt{x^{2} + 1}} \times (x + \sqrt{x^{2} + 1}))$

$= \log (\frac{- x^{2} + x^{2} + 1}{x + \sqrt{x^{2} + 1}})$

$= \log (\frac{1}{x + \sqrt{x^{2} + 1}})$

$= - \log (x + \sqrt{x^{2} + 1})$

$= - f (x)$

$\Rightarrow f (- x) + f (x) = 0$

Hence, $f (x)$ is an odd function.

Hence, the correct answer is option: 'A'.

The function fx=logx+x2+1 is

Solution

More Functions Questions

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