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Algebra Level 3

$x$ is a real number and $x+\frac{1}{x}=-\dfrac{7}{\sqrt{6}}$ .

Read the following statements.

$[1]$ . $\sin^{-1} (x+\frac{1}{x})$ is a real number.

$[2]$ . $x^2+\frac{1}{x^2}=\frac{37}{6}$

$[3]$ The given information can not be true because by the AM-GM inequality, $x+\frac{1}{x}$ has to be greater than or equal to $2$ . And $-\dfrac{7}{\sqrt{6}}$ is clearly less than $2$ .

Which of these statements are correct?

This problem is from the set "MCQ Is Not As Easy As 1-2-3". You can see the rest of the problems here .

All of them are correct.

[1]

and

[2]

Only

[2]

[2]

and

[3]

3 solutions

John Ashley Capellan
May 12, 2014

For (1), not all ranges of f(x) = x + 1/x gives the sine inverse a real number. If f(x) lies in the interval [-1 , 1], it has its sine inverse, otherwise, imaginary solutions which are obviously not real. By algebraic manipulations (squaring and subtracting both sides), statement (2) holds. For statement (3), AM - GM Inequality holds for positive reals, the given expression has negative real solution, hence, statement (3) is wrong.

Debtanu Bhuiya
Jun 10, 2014

The 1st option is wrong as range of sin x is [-1,1] the second option is correct as u solve u'll get 37/6 and as far as the third option is concerned x must be > or = 0 else u can't apply am-gm method and here by no means x is positive so 3rd option Can't be provided here

Ramesh Goenka
May 12, 2014

am gm inequality only holds for x>=0 not for x<0

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