JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 13)
The number of real roots of the equation $${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$$ is :
2
4
6
1
Explanation
$${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$$
$$ \Rightarrow {\left( {{e^{3x}} - 1} \right)^2} - {e^x}\left( {{e^{3x}} - 1} \right) = 12{e^{2x}}$$
$${\left( {{e^{3x}} - 1} \right)^2}\left( {{e^x} - {e^{ - x}} - {e^{ - 2x}}} \right) = 12$$
$$ \Rightarrow \underbrace {{e^x} - {e^{ - x}} - {e^{ - 2x}}}_{increa{\mathop{\rm sing}\nolimits} \,(let\,f(x))} = {{12} \over {\underbrace {{e^{3x}} - 1}_{decrea{\mathop{\rm sing}\nolimits} \,(let\,g(x))}}}$$
_25th_July_Morning_Shift_en_13_1.png)
$$\Rightarrow$$ No. of real roots = 2
$$ \Rightarrow {\left( {{e^{3x}} - 1} \right)^2} - {e^x}\left( {{e^{3x}} - 1} \right) = 12{e^{2x}}$$
$${\left( {{e^{3x}} - 1} \right)^2}\left( {{e^x} - {e^{ - x}} - {e^{ - 2x}}} \right) = 12$$
$$ \Rightarrow \underbrace {{e^x} - {e^{ - x}} - {e^{ - 2x}}}_{increa{\mathop{\rm sing}\nolimits} \,(let\,f(x))} = {{12} \over {\underbrace {{e^{3x}} - 1}_{decrea{\mathop{\rm sing}\nolimits} \,(let\,g(x))}}}$$
$$\Rightarrow$$ No. of real roots = 2
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