The greatest negative integer satisfying $x^{2} - 4 x - 77 < 0$ & $x^{2} > 4$ is

- 7

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

- 9

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

- 6

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

None of the above

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

Open in App

Solution

The correct option is D None of the above
$x^{2} - 4 x - 77 < 0$
or, $x^{2} - 11 x + 7 x - 77 < 0$
or, $x (x - 11) + 7 (x - 11) < 0$
or, $(x - 11) (x + 7) < 0$
either $x - 11 < 0$ or, $x + 7 > 0$
$\Rightarrow x < 11$ or, $x > - 7$
and $x^{2} > 4 \Rightarrow x > \pm 2$
So, there is no such integer

Suggest Corrections

Similar questions

x^{2} - 4 x - 77 < 0

x^{2} > 4

x^{2} + 4 x - 77 < 0

x^{2} > 4

x

{[x^{2} - 2]}^{2} - 9 [x^{2} - 2] + 14 = 0

x \in [0, 10]

x

x^{2} + 4 x + [x] + 6 = 0

I = \int_{1}^{3} ({tan}^{- 1} (\frac{x^{2} - 5 x + 6}{x^{3} - 6 x^{2} + 12 x - 7}) + {tan}^{- 1} (\frac{1}{x^{2} - 4 x + 4}))

[I]

Join BYJU'S Learning Program