Function f(x)=cos ≤ft(log ≤ft(x+√(1+x^2))) is :

$f(-x)=\cos \left(\log \left(-x+\sqrt{1+x^2}\right)\right)$

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

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

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

$=\cos \left(\log \left(\sqrt{1+x^2}+x\right)\right)=f(x)$

Hence f(x) is even

Hence (1) is the correct answer.