ExamPlay Dark Logo
登入

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.

评论 (0)

登录发表评论
广告
BrainBehindX Inc Logo
©2026; 供电 BrainBehindX Inc