Mathematics · 2024
JEE Main · 27 January 2024, Shift 2 · Q9
Consider the function f:(0,2) → R defined by f(x)=x/2+2/x and the function g(x) defined by g(x)= {min {f(t)}, 0< t ≤ x and 0<x ≤ 1; 3/2+x, 1<x<2}.…
Consider the function $\displaystyle f:(0,2) \rightarrow \mathbf{R}$ defined by $\displaystyle f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $\displaystyle g(x)$ defined by $\displaystyle g(x)=\left\{\begin{array}{lll}\min \{f(t)\}, & 0<\mathrm{t} \leq x \text { and } & 0<x \leq 1 \\ \frac{3}{2}+x, & 1<x<2\end{array}\right.$. Then,
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle g$ is continuous but not differentiable at $\displaystyle x=1$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.