The function $f\left( x \right) = {\rm{sec}}\left[ {\log (x + \sqrt {1 + {x^2})} } \right]$ is

JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Analyzing +4 | -1

The function

f\left( x \right) = \mathrm{sec}\left[ {\log (x + \sqrt {1 + {x^2})} } \right]

View answer and explanation

Correct option: B

Even

Option B (Even) is correct because the function can be shown to satisfy the condition f(x) = f(-x), indicating even symmetry.

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}