Verified Solution • 2025–26 Session moderate

QJEEM24Ma6a3e: JEE Main 2024 Mathematics - Class 12 JEEM24Ma6a3e Solution: Consider the function defined by and the function defined by

Consider the function f:(0,2)→Rf:(0,2) \rightarrow \mathbf{R} defined by f(x)=x2+2xf(x)=\frac{x}{2}+\frac{2}{x} and the function g(x)g(x) defined by

g(x)={min⁡⌊f(t)},amp;0lt;t≤x and 0lt;x≤132+x,amp;1lt;xlt;2. Then, 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, }

Multiple Choice Options

  • A gg is continuous but not differentiable at x=1x=1
  • B gg is continuous and differentiable for all x∈(0,2)x \in(0,2)
  • C gg is not continuous for all x∈(0,2)x \in(0,2)
  • D gg is neither continuous nor differentiable at x=1x=1

Step-by-Step Verified Solution

amp;f:(0,2)→R;f(x)=x2+2xamp;f′(x)=12−2x2\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}

∴f(x)\therefore \mathrm{f}(\mathrm{x}) is decreasing in domain.

JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Limits, Continuity and Differentiability Question 35 English Explanation 1

g(x)={x2+2xamp;0lt;x≤132+xamp;1lt;xlt;2g(x)= \begin{cases}\frac{x}{2}+\frac{2}{x} & 0 < x \leq 1 \\ \frac{3}{2}+x & 1 < x < 2\end{cases}

JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Limits, Continuity and Differentiability Question 35 English Explanation 2

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