Least integral value of $x$ satisfying the in equation $(x^{2} + 1) < {(x + 2)}^{2} < 2 x^{2} + 4 x - 12$ is

- 2

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

- 5

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

2

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

5

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

Open in App

Solution

The correct option is D $5$

Given, $(x^{2} + 1) < {(x + 2)}^{2} < (2 x^{2} + 4 x - 12)$

Put x=5

$(x^{2} + 1) < {(x + 2)}^{2} < (2 x^{2} + 4 x - 12)$

$(5^{2} + 1) < {(5 + 2)}^{2} < (2 \times 5^{2} + 4 \times 5 - 12)$

$26 < 49 < 58$
Inequality is satisfied

Which is correct hence option D is correct

Suggest Corrections

Similar questions

x

(x^{2} + 1) < {(x + 2)}^{2} < 2 x^{2} + 4 x - 12

x

\frac{x^{2} - 7 x + 12}{2 x^{2} + 4 x + 5} > 0

| x^{4} {.3}^{| x - 2 |} {.5}^{x - 1} | = - x^{4} {.3}^{| x - 2 |} {.5}^{x - 1}

x

\frac{x^{2} - 7 x + 12}{x^{2} - 13 x + 22} \leq 1

x

{log}_{1 / 2} (x^{2} - 6 x + 12) \geq 2

Join BYJU'S Learning Program