Number of value(s) of $x$ which disobey(s) the condition ${log}_{3} (2 x^{2} + 6 x - 5) > 1, x \in N$ is

0

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

1

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

4

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!

Open in App

Solution

The correct option is B $1$
$2 x^{2} + 6 x - 5 > 0 \forall x \in N$

Now, ${log}_{3} (2 x^{2} + 6 x - 5) > 1$
$\Rightarrow 2 x^{2} + 6 x - 5 > 3$
$\Rightarrow x^{2} + 3 x - 4 > 0$
$\Rightarrow (x + 4) (x - 1) > 0$
$\Rightarrow x < - 4$ or $x > 1$
Since $x \in N$ , we have $x = 1$ which disobeys the inequality.

Suggest Corrections

Similar questions

x

{log}_{3} (2 x^{2} + 6 x - 5) > 1, x \in N

x

{log}_{3} (2 x^{2} + 6 x - 5) > 1

{log}_{3} (2 x^{2} + 6 x - 5) > 1

x

{log}_{3} (2 x^{2} + 6 x - 5) > 1

x

{log}_{3} (2 x^{2} + 6 x - 5) > 1

Join BYJU'S Learning Program