JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 18)
If the sum of all the roots of the equation
$${e^{2x}} - 11{e^x} - 45{e^{ - x}} + {{81} \over 2} = 0$$ is $${\log _e}p$$, then p is equal to ____________.
$${e^{2x}} - 11{e^x} - 45{e^{ - x}} + {{81} \over 2} = 0$$ is $${\log _e}p$$, then p is equal to ____________.
Answer
45
Explanation
Given that
$$e^{2 x}-11 e^x-45 e^{-x}+\frac{81}{2}=0 $$
$$\Rightarrow 2 e^{3 x}-22 e^{2 x}-90+81 e^x=0 $$
$$\Rightarrow 2\left(e^x\right)^3-22\left(e^x\right)^2+81 e^x-90=0$$
Let $ e^x=y$
$$ \Rightarrow 2 y^3-22 y^2+81 y-90=0 $$
Product of roots $\left(y_1, y_2, y_3\right)$
$$ y_1 \cdot y_2 \cdot y_3=\frac{-(-90)}{2}=45 $$
Let $x_1, x_2$, and $x_3$ be roots of given equation
$$\Rightarrow e^{x_1} \cdot e^{x_2} \cdot e^{x_3} = 45 $$
$$\Rightarrow e^{x_1+x_2+x_3} =45$$
$$\Rightarrow x_1+x_2+x_3 =\log _e 45=\log _e p $$
$$\Rightarrow p = 45$$
$$e^{2 x}-11 e^x-45 e^{-x}+\frac{81}{2}=0 $$
$$\Rightarrow 2 e^{3 x}-22 e^{2 x}-90+81 e^x=0 $$
$$\Rightarrow 2\left(e^x\right)^3-22\left(e^x\right)^2+81 e^x-90=0$$
Let $ e^x=y$
$$ \Rightarrow 2 y^3-22 y^2+81 y-90=0 $$
Product of roots $\left(y_1, y_2, y_3\right)$
$$ y_1 \cdot y_2 \cdot y_3=\frac{-(-90)}{2}=45 $$
Let $x_1, x_2$, and $x_3$ be roots of given equation
$$\Rightarrow e^{x_1} \cdot e^{x_2} \cdot e^{x_3} = 45 $$
$$\Rightarrow e^{x_1+x_2+x_3} =45$$
$$\Rightarrow x_1+x_2+x_3 =\log _e 45=\log _e p $$
$$\Rightarrow p = 45$$
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