Question

plz give its solution

Answer 1

there is one fact that.... for an equation to have positive value, discriminant of the equation should be negative and the coefficient of x square should be positive ....here
$a {9}^{x} + 4(a - 1) {3}^{x} + a - 1 \\ let \: {3}^{x} = t \\ so \: \\ {9}^{x} = {t}^{2} \\ eqn \: becomes \\ a {t}^{2} + 4(a - 1)t + (a - 1) = 0 \\ first \: of \: all \: coeff \: of \: {t}^{2} > 0 \\ i.e \: \: a > 0 \\ \\ now \: discriminant \: < 0 \\ 16(a - 1) {}^{2} - 4a(a - 1) < 0 \\ (a - 1)(16(a - 1) - 4a) < 0 \\ (a - 1)(12a - 16) < 0 \\ 4(a - 1)(3a - 4) < 0 \\ (a - 1)(3a - 4) < 0$


now we have 2 conditions for "a"

we have to take intersection of values

conditions are

(1) a. > 0

(2) (a-1)(3a-4)<0

I'm making a number line.... see figure...

so ans is

$a \: belongs \: to \: (1 \: to \: \frac{4}{3} )$


a. € (1,4/3)


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