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  • The function defined by <img alt="\mathrm{ f(x)=\left\{\begin{array}{cc} |x-3| & ; \quad x \geq 1 \\ \frac{1}{4} x^2-\frac{3}{2} x+\frac{13}{4} & ; \quad x<1 \end{array}\right. }" s

The function defined by \mathrm{ f(x)=\left\{\begin{array}{cc} |x-3| & ; \quad x \geq 1 \\ \frac{1}{4} x^2-\frac{3}{2} x+\frac{13}{4} & ; \quad x<1 \end{array}\right. } is

Option: 1

differentiable nowhere
 


Option: 2

differentiable everywhere
 


Option: 3

differentiable at x=1
 


Option: 4

none of these


Answers (1)

best_answer

Since \mathrm{|x-3|=x-3, \, \, if \, \, x \geq 3=-x+3, \, \, if \, \, x<3}

\mathrm{ \therefore \quad f(x)=\left\{\begin{array}{ccc} \frac{1}{4} x^2-\frac{3}{2} x+\frac{13}{4} & ; & x<1 \\ 3-x & ; & 1 \leq x<3 \\ x-3 & ; & x \geq 3 \end{array}\right. }

Now proceed to check the continuity and differentiability at \mathrm{x=1}

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