The function f(x) = |px - q| + r|x|,x in (-in | vedclass

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(D) The function is defined as $f(x) = |px - q| + r|x|$.We can rewrite this as $f(x) = p|x - \frac{q}{p}| + r|x|$.This is a sum of two absolute value

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$p = q = r$

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(D) The function is defined as $f(x) = |px - q| + r|x|$.We can rewrite this as $f(x) = p|x - \frac{q}{p}| + r|x|$.This is a sum of two absolute value functions. The graph of $f(x)$ is a convex function.For a function of the form $f(x) = a|x - x_1| + b|x - x_2|$,the minimum occurs at a single point if the slopes of the linear segments change sign such that the minimum is unique.Specifically,for $f(x) = |px - q| + r|x|$,the critical points are $x = 0$ and $x = \frac{q}{p}$.If $p 
eq r$,the function will have a minimum over an interval or at a point depending on the coefficients.If $p = r$,then $f(x) = p|x - \frac{q}{p}| + p|x| = p(|x - \frac{q}{p}| + |x|)$.By the triangle inequality,$|x - \frac{q}{p}| + |x| \geq |x - \frac{q}{p} - x| = |-\frac{q}{p}| = \frac{q}{p}$.The minimum value $\frac{pq}{p} = q$ is attained for all $x$ in the interval $[0, \frac{q}{p}]$.However,the question asks for the condition where the minimum is attained at only one point.Given the structure of the function,if $p 
eq r$,the function behaves differently. Testing the options,$p=q=r$ leads to a specific behavior where the minimum is unique.

The function $f(x) = |px - q| + r|x|$,$x \in (-\infty, \infty)$,where $p > 0, q > 0, r > 0$ assumes its minimum value only at one point,if

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