JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 3)
Consider the function $$f:(0,2) \rightarrow \mathbf{R}$$ defined by $$f(x)=\frac{x}{2}+\frac{2}{x}$$ and the function $$g(x)$$ defined by
$$g(x)=\left\{\begin{array}{ll} \min \lfloor f(t)\}, & 0<\mathrm{t} \leq x \text { and } 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{array} .\right. \text { Then, }$$
$$g$$ is continuous but not differentiable at $$x=1$$
$$g$$ is continuous and differentiable for all $$x \in(0,2)$$
$$g$$ is not continuous for all $$x \in(0,2)$$
$$g$$ is neither continuous nor differentiable at $$x=1$$
Explanation
$$\begin{aligned} & f:(0,2) \rightarrow R ; f(x)=\frac{x}{2}+\frac{2}{x} \\ & f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^2} \end{aligned}$$
$$\therefore \mathrm{f}(\mathrm{x})$$ is decreasing in domain.
_27th_January_Evening_Shift_en_3_1.png)
$$g(x)= \begin{cases}\frac{x}{2}+\frac{2}{x} & 0 < x \leq 1 \\ \frac{3}{2}+x & 1 < x < 2\end{cases}$$
_27th_January_Evening_Shift_en_3_2.png)
Comments (0)
