The function f:R→R.given by fx =ln|x+√(1+x^2)| is neither even nor odd.true or false
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f ( x ) = ln | x +√(1 +x ^2)|
put x = -x
f( -x ) = ln | -x +√(1+x^2 ) |
add f(x ) and f(- x )
f(x) + f(-x )=ln |x +√(1+x^2) +ln|-x +√(1+x^2)
we know ,
logx + logy = logxy
so,
f(x ) + f(-x) =ln|{x+√(1+x^2}{-x+√(1+x^2)}
use (a-b)(a+b)=a^2-b^2
=ln{(√(1+x^2))^2-x^2}
=ln(1+x^2 -x^2) =0
hence,
f(x )+ f(-x ) =0
so, f(x ) is an odd function .
statement is false that f(x ) is neither odd nor even .
put x = -x
f( -x ) = ln | -x +√(1+x^2 ) |
add f(x ) and f(- x )
f(x) + f(-x )=ln |x +√(1+x^2) +ln|-x +√(1+x^2)
we know ,
logx + logy = logxy
so,
f(x ) + f(-x) =ln|{x+√(1+x^2}{-x+√(1+x^2)}
use (a-b)(a+b)=a^2-b^2
=ln{(√(1+x^2))^2-x^2}
=ln(1+x^2 -x^2) =0
hence,
f(x )+ f(-x ) =0
so, f(x ) is an odd function .
statement is false that f(x ) is neither odd nor even .
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