Positive $\times$ Positive = Positive.

Algebra Level 5

$\dfrac{\log(1+\lambda^2) \times |1-\lambda^2|}{\sqrt{1+\lambda^2} \times e^{1-\lambda^2}}>0$

Find the set of all real values of $\lambda$ for which the above inequality holds.

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\mathbb{R}-\{0\}

None of the given choices.

(0,\infty)

\mathbb{R}

\mathbb{R}-\{0,1\}

1 solution

Harshit Singhania
Jun 4, 2015

The absolute value function will always be positive, e raised to anything will also be a positive value, the square root if is referring to the principal root (the positive root)(assumption) will also obviously be positive. Now the logarithmic function for base e or 10 will only be negative for (1+x^2)<1 which is not possible for real values of x. So the answer should be all reals but the expression may be equal to 0 for values of x like for x=1 or for the log being equal to 0 so none of the choices is the answer as the expression has to be greater than 0 and not greater or equal to 0.

I can't really understand what you are saying, so I am posting my own comments.

It's clear that $\sqrt{1 + \lambda^2} > 0$ and $e^{1 - \lambda^2} > 0$ for all $\lambda$ . Also, $\log (1 + \lambda^2) \ge 0$ for all $\lambda$ , with equality if only if $\lambda = 0$ , and $|1 - \lambda^2| \ge 0$ for all $\lambda$ , with equality if only if $\lambda = -1$ or $\lambda = 1$ . Therefore, the solution to the inequality is $\mathbb{R} \setminus \{-1, 0, 1\}$ .

Jon Haussmann - 6 years ago

that kind of is exactly what i was saying :-)

Harshit Singhania - 6 years ago

Positive × \times × Positive = Positive.

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Positive $\times$ Positive = Positive.