(D) $3^x+3^{1-x}-4 0$.$\Rightarrow t+\frac{3}{t}-4 < 0$$\Rightarrow t^2-4t+3 < 0$$\Ri

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(D) $3^x+3^{1-x}-4 $\Rightarrow 3^x+\frac{3}{3^x}-4 Let $3^x=t$, where $t > 0$.$\Rightarrow t+\frac{3}{t}-4 $\Rightarrow t^2-4t+3 $\Rightarrow (t-1)(t-3) This inequality holds for $1 Substituting $t=3^x$ back, we get $1 $\Rightarrow 3^0 Since the base $3 > 1$, the inequality sign remains the same:$0 Thus, the solution set is $x \in (0,1)$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic inequations and Newton Formula

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

TS EAMCET 2024 All TS EAMCET

A
$(1,2)$
B
$(1,3)$
C
$(0,2)$
D
$(0,1)$

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