Number of real roots of equation, mathrm{e}^{4 mathrm{x}}+mathrm{e}^{3 mathrm{x}}-4 mathrm{e}^{2 mat

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4-2.Quadratic Equations and Inequations

hard

The number of real roots of the equation, $\mathrm{e}^{4 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}-4 \mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{\mathrm{x}}+1=0$ is

$4$

$2$

$3$

$1$

(JEE MAIN-2020)

Solution

$\mathrm{e}^{4 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}-4 \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{x}}+1=0$

Divide by e $2 x$

$\Rightarrow \quad \mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{\mathrm{x}}-4+\frac{1}{\mathrm{e}^{\mathrm{x}}}+\frac{1}{\mathrm{e}^{2 \mathrm{x}}}=0$

$\Rightarrow\left(\mathrm{e}^{2 \mathrm{x}}+\frac{1}{\mathrm{e}^{2 \mathrm{x}}}\right)+\left(\mathrm{e}^{\mathrm{x}}+\frac{1}{\mathrm{e}^{\mathrm{x}}}\right)-4=0$

$\Rightarrow\left(\mathrm{e}^{\mathrm{x}}+\frac{1}{\mathrm{e}^{\mathrm{x}}}\right)^{2}-2+\left(\mathrm{e}^{\mathrm{x}}+\frac{1}{\mathrm{e}^{\mathrm{x}}}\right)-4=0$

Let $\mathrm{e}^{\mathrm{x}}+\frac{1}{\mathrm{e}^{\mathrm{x}}}=\mathrm{t} \Rightarrow(\mathrm{e^x}-1)^{2}=0 \Rightarrow \mathrm{x}=0$

Number of real roots $=1$

Standard 11

Mathematics

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normal

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hard

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hard

(IIT-1980)

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hard

(JEE MAIN-2017)

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normal

The number of real roots of the equation, $\mathrm{e}^{4 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}-4 \mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{\mathrm{x}}+1=0$ is

Solution

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