JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 9)
The sum of the roots of the equation
$$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$, is :
$$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$, is :
log2 14
log2 11
log2 12
log2 13
Explanation
$$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$
$${\log _2}({2^{x + 1}}) - {\log _2}{(3 + {2^x})^2} + {\log _2}(10 - {2^{ - x}}) = 0$$
$$lo{g_2}\left( {{{{2^{x + 1}}.(10 - {2^{ - x}})} \over {{{(3 + {2^x})}^2}}}} \right) = 0$$
$${{2({{10.2}^{ - x}} - 1)} \over {{{(3 + {2^x})}^2}}} = 1$$
$$ \Rightarrow {20.2^x} - 2 = 9 + {2^{2x}} + {6.2^x}$$
$$\therefore$$ $${({2^x})^2} - 14({2^x}) + 11 = 0$$
Roots are 2x1 & 2x2
$$\therefore$$ 2x1 . 2x2 = 11
x1 + x2 = log2(11)
$${\log _2}({2^{x + 1}}) - {\log _2}{(3 + {2^x})^2} + {\log _2}(10 - {2^{ - x}}) = 0$$
$$lo{g_2}\left( {{{{2^{x + 1}}.(10 - {2^{ - x}})} \over {{{(3 + {2^x})}^2}}}} \right) = 0$$
$${{2({{10.2}^{ - x}} - 1)} \over {{{(3 + {2^x})}^2}}} = 1$$
$$ \Rightarrow {20.2^x} - 2 = 9 + {2^{2x}} + {6.2^x}$$
$$\therefore$$ $${({2^x})^2} - 14({2^x}) + 11 = 0$$
Roots are 2x1 & 2x2
$$\therefore$$ 2x1 . 2x2 = 11
x1 + x2 = log2(11)
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