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  4. Number of integral values of x satisfying the inequality ((3/4))6x+10-x2 < (27/64) is

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

Linear Inequalities

Solution:

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