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  • If <img alt="f(x)=\left\{\begin{array}{ll} \left|1-4 x^2\right|, & 0 \leq x<1 \\ {\left[x^2-2 x\right],} & 1 \leq x<2 \end{array},\right." src="https://entrancecorner.oncodecogs.

If f(x)=\left\{\begin{array}{ll} \left|1-4 x^2\right|, & 0 \leq x<1 \\ {\left[x^2-2 x\right],} & 1 \leq x<2 \end{array},\right. where [.] denotes the greatest integer function, then \mathrm{f(x)} is

Option: 1

differentiable for all x


Option: 2

continuous at x=1


Option: 3

f(x) is non-differentiable at x=1


Option: 4

none of the above 


Answers (1)

best_answer

Since \mathrm{1 \leq x<2 \Rightarrow 0 \leq x-1<1}

\mathrm{\begin{aligned} & \Rightarrow \quad\left[x^2-2 x\right]=\left[(x-1)^2-1\right]=-1 \\ & \therefore \quad f(x)=\left\{\begin{array}{cc} 1-4 x^2, & 0 \leq x<\frac{1}{2} \\ 4 x^2-1, & \frac{1}{2} \leq x<1 \\ -1, & 1 \leq x<2 \end{array}\right. \end{aligned}}

Graph of f(x) is given by 

It is clear from graph that f(x) is discontinuous at x=1 and not differentiable at x=1 and x=\frac{1}{2}

Posted by

manish painkra

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