The solution set contained in R of the inequa | vedclass

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(B) Given the inequation: $3^x+3^{1-x}-4 < 0$ Substitute $3^{1-x}$ as $\frac{3}{3^x}$: $3^x+\frac{3}{3^x}-4 < 0$ Multiply the entire inequality by $3^

Vedclass · Accepted Answer

$(0,1)$

Vedclass · Answer

(B) Given the inequation: $3^x+3^{1-x}-4 Substitute $3^{1-x}$ as $\frac{3}{3^x}$: $3^x+\frac{3}{3^x}-4 Multiply the entire inequality by $3^x$ (since $3^x > 0$): $(3^x)^2 - 4(3^x) + 3 Let $y = 3^x$. The inequality becomes $y^2 - 4y + 3 Factor the quadratic expression: $(y-1)(y-3) This implies $1 Substituting back $y = 3^x$: $1 Since $3^0 = 1$ and $3^1 = 3$,we have $3^0 Comparing the exponents,we get $0 Thus,the solution set is $(0,1)$.

The solution set contained in $R$ of the inequation $3^x+3^{1-x}-4 < 0$ is:

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