ExamPlay Light Logo
Oturum aç

JEE Advance - Mathematics (1983 - No. 38)

Show that $$1+x$$ $$In\left( {x + \sqrt {{x^2} + 1} } \right) \ge \sqrt {1 + {x^2}} $$ for all $$x \ge 0$$
The inequality is false for x = 0.
The inequality is true for all x ≥ 0.
We cannot determine if the inequality is true or false without further analysis.
The inequality is only true for specific values of x.
The inequality can be solved directly for x.

Yorumlar (0)

Yorum yapmak için giriş yapın
Reklamcılık
BrainBehindX Inc Logo
©2026; Tarafından desteklenmektedir BrainBehindX Inc