If x 1 and x 2 are the roots of the equation e 2 x ln x - x 3 with x 1- x 2- thenA- 2 x 1- x 2B- x 12- x 23C- x 1-2 x 2D - x 1- x 22

The correct option is D x_1=x_2^2Clearly, x 0 e^2x^ln x=x^3 ⇒ln e^2+ln x·ln x=3ln x ⇒ 2+(ln x)^2=3ln x Let ln x=t Then t^2 3t+2=0 ⇒ t=1,2 ∴ x_1=e^2, x_2=e ⇒ x ...

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Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

If x 1 and x ...

Question

If $x_{1}$ and $x_{2}$ are the roots of the equation $e^{2} x^{ln x} = x^{3}$ with $x_{1} > x_{2}$ , then

x_{1} = 2 x_{2}

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x_{1} = x_{2}^{2}

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2 x_{1} = x_{2}

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x_{1}^{2} = x_{2}^{3}

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Solution

The correct option is B $x_{1} = x_{2}^{2}$
Clearly, $x > 0$
$e^{2} x^{ln x} = x^{3}$
$\Rightarrow ln e^{2} + ln x \cdot ln x = 3 ln x \Rightarrow 2 + (ln x)^{2} = 3 ln x$
Let $ln x = t$
Then $t^{2} - 3 t + 2 = 0$
$\Rightarrow t = 1, 2 ∴ x_{1} = e^{2}, x_{2} = e \Rightarrow x_{1} = x_{2}^{2}$

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If x1 and x2 are the roots of the equation e2xlnx=x3 with x1>x2, then

The correct option is B x1=x22Clearly, x>0 e2xlnx=x3 ⇒lne2+lnx⋅lnx=3lnx⇒2+(lnx)2=3lnx Let lnx=t Then t2−3t+2=0 ⇒t=1,2∴x1=e2, x2=e⇒x1=x22

If $x_{1}$ and $x_{2}$ are the roots of the equation $e^{2} x^{ln x} = x^{3}$ with $x_{1} > x_{2}$ , then

The correct option is B $x_{1} = x_{2}^{2}$
Clearly, $x > 0$
$e^{2} x^{ln x} = x^{3}$
$\Rightarrow ln e^{2} + ln x \cdot ln x = 3 ln x \Rightarrow 2 + (ln x)^{2} = 3 ln x$
Let $ln x = t$
Then $t^{2} - 3 t + 2 = 0$
$\Rightarrow t = 1, 2 ∴ x_{1} = e^{2}, x_{2} = e \Rightarrow x_{1} = x_{2}^{2}$