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  • <img alt="\mathrm{\text { If } x+2|y|=3 y \text {, where } y=f(x) \text {, then } f(x) \text { is }}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext%20%7B%20If%20%7D%20xa

\mathrm{\text { If } x+2|y|=3 y \text {, where } y=f(x) \text {, then } f(x) \text { is }}

Option: 1

continuous everywhere
 


Option: 2

differentiable everywhere
 


Option: 3

discontinuous at x=0
 


Option: 4

none of the above


Answers (1)

best_answer

\mathrm{\left.\begin{array}{ll} x+2 y=3 y & (y \geq 0) \\ x-2 y=3 y & (y<0) \end{array}\right\} \Rightarrow y= \begin{cases}x & (x \geq 0) \\ \frac{x}{5} & (x<0)\end{cases}}

function is continuous everywhere, but not differentiable at  x=0 \text {. }

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