Number of integral values of x satisfying the inequality ((3/4))6x+10-x2

Question

Number of integral values of x satisfying the inequality ((3/4))6x+10-x2 < (27/64) is (A) 5 (B) 6 (C) 7 (D) 8

Tardigrade Admin · Accepted Answer

((3/4))6x+10-x2 < (27/64) ⇒ ((3/4))6x+10-x2 < ((3/4))3 ⇒ 6x+10-x2 >3 (as base (3/4) < 1) ∴ x2-6x-7 <0 ∴ (x+1)(x-7) <0 -1 < x < 7 ∴ number of integral values is = 7

Q. Number of integral values of x satisfying the inequality $(\frac{3}{4})^{6 x + 10 - x^{2}} < \frac{27}{64}$ is

$5$

$6$

$7$

$8$

$(\frac{3}{4})^{6 x + 10 - x^{2}} < \frac{27}{64}$
$\Rightarrow (\frac{3}{4})^{6 x + 10 - x^{2}} < (\frac{3}{4})^{3}$
$\Rightarrow 6 x + 10 - x^{2} > 3$ (as base $(3/4) < 1$ )
$∴ x^{2} - 6 x - 7 < 0$
$∴ (x + 1) (x - 7) < 0$
$- 1 < x < 7$
$∴$ number of integral values is = 7

Q. Number of integral values of x satisfying the inequality (43​)6x+10−x2<6427​ is

5

6

7

8

(43​)6x+10−x2<6427​ ⇒(43​)6x+10−x2<(43​)3 ⇒6x+10−x2>3 (as base (3/4)<1) ∴x2−6x−7<0 ∴(x+1)(x−7)<0 −1<x<7 ∴ number of integral values is = 7

Q. Number of integral values of x satisfying the inequality $(\frac{3}{4})^{6 x + 10 - x^{2}} < \frac{27}{64}$ is

$5$

$6$

$7$

$8$

$(\frac{3}{4})^{6 x + 10 - x^{2}} < \frac{27}{64}$
$\Rightarrow (\frac{3}{4})^{6 x + 10 - x^{2}} < (\frac{3}{4})^{3}$
$\Rightarrow 6 x + 10 - x^{2} > 3$ (as base $(3/4) < 1$ )
$∴ x^{2} - 6 x - 7 < 0$
$∴ (x + 1) (x - 7) < 0$
$- 1 < x < 7$
$∴$ number of integral values is = 7