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  • The function f is defined by <img alt="y=f(x), where x=3 t-|t|, y=2 t^2+t|t|" src="https://entrancecorner.oncodecogs.com/gif.latex?y%3Df%28x%29%2C%20where%20x%3D3%20t-%7Ct%7C%2C%20y%3D2%2

The function f is defined by y=f(x), where x=3 t-|t|, y=2 t^2+t|t|, then which of the following options are correct?

Option: 1

 f(x) is continuous at x=0 but not differentiable at x=0.


Option: 2

 f(x) is continuous at x=0.


Option: 3

 f(x) is continuous at x=0 and is also differentiable at x=0
 


Option: 4

 f(x) is differentiable at x=0.


Answers (1)

best_answer

(b, c, d) Given function is x=3 t-|t|, y=2 t^2+t|t|.

\begin{array}{ll} \therefore & x=3 t-|t|= \begin{cases}4 t, & \text { if } t<0 \\ 2 t, & \text { if } t \geq 0\end{cases} \\ \text { and } & y=2 t^2+t|t|= \begin{cases}3 t^2, & \text { if } t<0 \\ t^2, & \text { if } t \geq 0\end{cases} \end{array}

Case I When t<0 \Rightarrow x=4 t and y=t^2

\Rightarrow \quad y=\frac{x^2}{16}

Case II When t \geq 0 \Rightarrow x=2 t and y=3 t^2

\Rightarrow \quad y=\frac{3 x^2}{4}

\therefore On combining the cases I and II, we get

y=f(x)= \begin{cases}\frac{x^2}{16}, & x<0 \\ \frac{3 x^2}{4}, & x \geq 0\end{cases}

The above function is both continuous and differentiable at x=0.

Posted by

Ritika Harsh

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