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  • The function <img alt="\mathrm{f} \text { is defined by } \mathrm{y}=\mathrm{f}(\mathrm{x}) \text { where } \mathrm{x}=3 \mathrm{t}-|\mathrm{t}|, \quad \mathrm{y}=2 \mathrm{t}^2+\mathrm{t}|\ma

The function \mathrm{f} \text { is defined by } \mathrm{y}=\mathrm{f}(\mathrm{x}) \text { where } \mathrm{x}=3 \mathrm{t}-|\mathrm{t}|, \quad \mathrm{y}=2 \mathrm{t}^2+\mathrm{t}|\mathrm{t}| \text { then }

Option: 1

\mathrm{f(x) \text { is continuous at } x=0 \text { but not differentiable at } x=0}


Option: 2

\mathrm{f(x) \text { is not continuous at } x=0}


Option: 3

\mathrm{f(x) \text { is continuous at } x=0 \text { and is also differentiable at } x=0}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

Case I: when  \mathrm{t<0 \Rightarrow x=3 t+t=4 t}

\mathrm{ \mathrm{y}=2 \mathrm{t}^2-\mathrm{t}^2=\mathrm{t}^2 \Rightarrow \mathrm{y}=\frac{\mathrm{x}^2}{16} \quad \forall \mathrm{x}<0 }

Case II: when \mathrm{\mathrm{t} \geq 0 \Rightarrow \mathrm{x}=3 \mathrm{t}-\mathrm{t}=2 \mathrm{t}}

\mathrm{ \begin{aligned} & y=2 t^2+t^2=3 t^2 \Rightarrow y=\frac{3 x^2}{4} \forall x \geq 0 \\ & \begin{aligned} \therefore y & =\frac{x^2}{16}, x<0 \\ & =\frac{3 x^2}{4}, x \geq 0 \end{aligned} \end{aligned} }

which is continuous and differentiable at  \mathrm{x=0 \text {. }}

Posted by

Ritika Kankaria

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