The set of values of p for which the equation | vedclass

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(A) The equation is $|\ln x| = px$. Let $f(x) = |\ln x|$ and $g(x) = px$. For $x \ge 1$,$f(x) = \ln x$. The line $y = px$ is tangent to $y = \ln x$ wh

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$\left( 0, \frac{1}{e} \right)$

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(A) The equation is $|\ln x| = px$. Let $f(x) = |\ln x|$ and $g(x) = px$. For $x \ge 1$,$f(x) = \ln x$. The line $y = px$ is tangent to $y = \ln x$ when the derivative of $\ln x$ equals the slope $p$,i.e.,$1/x = p$. Since the point of tangency $(x_1, y_1)$ lies on both curves,$y_1 = \ln x_1$ and $y_1 = px_1$. Thus,$px_1 = \ln x_1$. Substituting $p = 1/x_1$,we get $(1/x_1)x_1 = \ln x_1$,which implies $\ln x_1 = 1$,so $x_1 = e$. Then $p = 1/e$. For $x \in (0, 1)$,$f(x) = -\ln x$. The line $y = px$ always intersects $y = -\ln x$ at exactly one point for any $p > 0$. For $x \ge 1$,the line $y = px$ intersects $y = \ln x$ at two points if $0  1/e$. Therefore,for three distinct roots (one in $(0, 1)$ and two in $(1, \infty)$),we must have $0 < p < 1/e$.

The set of values of $p$ for which the equation $|\ln x| - px = 0$ possesses three distinct roots is

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