The true set of values of ‘a’ for | vedclass

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Question

$\int\limits_a^0 {{3^{ - x}}\,({3^{ - x}}\, - \,2)\,\,dx\,\, \ge \,0} $ 

put $3^{-x} = t 3^{-x} ln3\, dx = -dt$
$\ln 3\,\int\limits_1^{{3^{ -

Vedclass · Accepted Answer

$(-\infty , - 1] \cup [0, \infty )$

Vedclass · Answer

$\int\limits_a^0 {{3^{ - x}}\,({3^{ - x}}\, - \,2)\,\,dx\,\, \ge \,0} $ put $3^{-x} = t 3^{-x} ln3\, dx = -dt$ $\ln 3\,\int\limits_1^{{3^{ - a}}} {(t - 2)\,\,\,dt\,\,\, \ge \,\,0} $ $ \Rightarrow $ $\left. {\frac{{{t^2}}}{2}\,\,\, - \,\,2t} \right]_1^{{3^{ - a}}}\,\,\, \ge \,\,\,0$ $\left( {\frac{{{3^{ - 2a}}}}{2}\, - \,2\,.\,{3^{ - a}}} \right)\,\, - \,\,\left( {\frac{1}{2}\,\, - \,2} \right)\,\,\, \ge \,\,\,0$ $3^{-2a} - 4×3^{-a} + 3 > 0$ $(3^{-a} - 3) (3^{-a} - 1) > 0$ $3^{-a} > 3^1 a < 1$ or $3^{-a} < 3^0 a > 0$ Hence $a\,\, \in \,\,( - \infty , - 1)\,\, \cup \,[0,\infty )$

The true set of values of $‘a’$ for which the inequality $\int\limits_a^0 {} (3^{ -2x} - 2. 3^{-x})\, dx \geq 0$ is true is:

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