You visited us 1 times! Enjoying our articles? Unlock Full Access!

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?

$n < 11$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$n = 10$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$n = 11$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$n > 11$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$n < - 11$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

p

q

x^{2} + p x + q \neq 0

x

n

X^{2} + p X + q I_{n} \neq 0_{n}

n \times n

\frac{m x^{2} + 3 x + 4}{x^{2} + 2 x + 2} < 5

x \in R,

x

\frac{x^{2} + 2 x + c}{x^{2} + 4 x + 3 c}

(x^{2} - 2 x - 35) < 0

\frac{x^{2} + x + d}{x^{2} + 2 x + d}

[\frac{5}{6}, \frac{3}{2}]

Join BYJU'S Learning Program