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f-x-ln-x-1-x-2-is-even-or-odd-give-reasion-

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Question Number 11167 by agni5 last updated on 15/Mar/17
f(x)=ln∣x+(√((1+x^2 )))∣ is even or odd? give reasion.
f(x)=ln∣x+(1+x2)∣isevenorodd?givereasion.
Answered by mrW1 last updated on 15/Mar/17
f(x)=ln∣x+(√((1+x^2 )))∣=ln (x+(√(1+x^2 ))) f(−x)=ln[−x+(√(1+(−x)^2 ))] =ln(−x+(√(1+x^2 ))) =ln[(−x+(√(1+x^2 )))×((x+(√(1+x^2 )))/(x+(√(1+x^2 ))))] =ln ((1+x^2 −x^2 )/(x+(√(1+x^2 )))) =ln (1/(x+(√(1+x^2 )))) =−ln (x+(√(1+x^2 ))) =−f(x) ⇒f(x) is odd.
f(x)=ln∣x+(1+x2)∣=ln(x+1+x2)f(−x)=ln[−x+1+(−x)2]=ln(−x+1+x2)=ln[(−x+1+x2)×x+1+x2x+1+x2]=ln1+x2−x2x+1+x2=ln1x+1+x2=−ln(x+1+x2)=−f(x)⇒f(x)isodd.
Answered by ajfour last updated on 15/Mar/17
f(−x)=ln ∣−x+(√(x^2 +1)) ∣ =ln ∣(1/(x+(√(x^2 +1))))∣=−ln∣x+(√(x^2 +1)) ∣ so it appears odd; should it not so be?
f(−x)=ln∣−x+x2+1∣=ln∣1x+x2+1∣=−ln∣x+x2+1∣soitappearsodd;shoulditnotsobe?

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