Show that $$1+xloge$$⁡$$x+x2+1$$≥$$1+x2$$ for all x≥$$0$$

Correct option is 11+x 1n (x+x2+1))≥(1+x2) for x≥0

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$Show that 1 + x \log_{e} (x + \sqrt{(x^{2} + 1))} \geq \sqrt{(1 + x^{2}}) for all x \geq 0$

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1+x 1n (x+x2+1))≥(1+x2) for x≥0

1+x 1n (x+x2+1))≤(1+x2) for x≥0

1+x 1n (x+x2+1))≠(1+x2) for x≥0

1+x 1n (x+x2+1)) #(1+x2) for x≥0

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detailed solution

Correct option is A

Let f(x)=1+xloge⁡x+x2+1−1+x2⇒ f′(x)=ln⁡x+x2+1+xx+x2+1 1+2x2x2+1−2x21+x2⇒ f′(x)=1nx+x2+1≥0∀x≥0so, f(x) increases when x≥0hence x≥0⇒f(x)≥f(0) Thus, 1+xln⁡x+x2+1−1+x2≥0 for x≥0 So, 1+xln⁡x+x2+1≥1+x2 for x≥0

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Show that 1+xloge⁡x+x2+1≥1+x2 for all x≥0

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$Show that 1 + x \log_{e} (x + \sqrt{(x^{2} + 1))} \geq \sqrt{(1 + x^{2}}) for all x \geq 0$