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Permutation

If x2+2x+n>10...

Question

If $x^{2} + 2 x + n > 10$ for all real numbers $x$ , then which of the following conditions is correct?

$n < 11$

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$n = 10$

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$n = 11$

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$n > 11$

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$n < - 11$

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Solution

The correct option is D
$n > 11$

Explanation for the correct option.
As $x^{2} + 2 x + n > 10$ , so $D < 0$ .
$x^{2} + 2 x + n > 10$ can be written as $x^{2} + 2 x + n - 10 > 0$ .
$\begin{array}{rcl} 4 - 4 n + 40 & < & 0 (by D = b^{2} - 4 a c) \\ \Rightarrow 44 & < & 4 n \\ \Rightarrow n & > & 11 \end{array}$
Hence, option D is correct.

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