Differentiation and its Application: Consider the function f:(0,2)tomathbb R defined by f(x)=frac… | JEE Main 2024

JEE Main 2024 (Online) 27th January Evening ShiftQuestion 1

Consider the function $f:(0,2)\to\mathbb R$ defined by $f(x)=\frac x2+\frac2x$ and the function $g(x)$ defined by $g(x)=\begin{cases}\min\{f(t)\},&0<t\le x\text{ and }0<x\le1\\\frac32+x,&1<x<2\end{cases}$. Then :

Differentiation and its Application

$g$ is continuous but not differentiable at $x=1$✓ Correct answer$g$ is not continuous for all $x\in(0,2)$$g$ is neither continuous nor differentiable at $x=1$$g$ is continuous and differentiable for all $x\in(0,2)$

Solution

Concept. $f$ is decreasing on $(0,2)$, so the minimum of $f$ on $(0,x]$ is $f(x)$. Formulas used. (1) $f'(x)=\frac12-\frac{2}{x^2}<0$ for $0<x<2$Step 1 - $g$ on $(0,1]$. By (1), $g(x)=f(x)=\frac x2+\frac2x$. Step 2 - Continuity at $1$. $g(1)=\frac52$ and $\lim_{x\to1^+}\left(\frac32+x\right)=\frac52$. Continuous. Step 3 - Slopes at $1$. Left: $\frac12-2=-\frac32$. Right: $1$. Not differentiable. Answer: (A)

Differentiation and its Application - JEE Mathematics question (JEE Main 2024 (Online) 27th January Evening Shift)

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