Consider the function f:(0,2)→R defined by f(x)=x/2+2/x and the function g(x) defined by gx= min{f(t)}, 0<t≤x and 0<x≤1 3/2+x, 1<x<2 . Then

Correct answer: (A) g is continuous but not differentiable at x=1. Given, f:(0,2)→R;f(x)=x/2+2/x ⇒f^(')x=1/2-(2x^2⇒f^(')x=x^2-42x^2∴fx is decreasing in domain 0,2.⇒gx=x2+2)/x 0<x≤1 3/2+x 1<x<2 ⇒g1=1/2+2/1=5/2 ⇒g1^(+)=3/2+1=5/2 ⇒gx is continuous in 0,2. ⇒g^(')x= f^(')x, 0<x≤1 1, 1<x<2 ⇒g^(')1=f^(')1=(-3)/2 ⇒g^(')1≠g^(')1^(+) So, gx is not differentiable at x=1.

Q48 / 90 · JEE Main 2024 (27 Jan Shift 2) · Mathematics

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = \{\begin{matrix} min {f (t)}, & 0 < t \leq x and 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{matrix}$ . Then

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

Solution

Given,

$f : (0, 2) \to R; f (x) = \frac{x}{2} + \frac{2}{x}$

$\Rightarrow f^{'} (x) = \frac{1}{2} - \frac{2}{x^{2}}$

$\Rightarrow f^{'} (x) = \frac{x^{2} - 4}{2 x^{2}}$

$∴$ $f (x)$ is decreasing in domain $(0, 2)$ .

$\Rightarrow g (x) = \{\begin{cases} \frac{x}{2} + \frac{2}{x} 0 < x \leq 1 \\ \frac{3}{2} + x 1 < x < 2 \end{cases}$

$\Rightarrow g (1) = \frac{1}{2} + \frac{2}{1} = \frac{5}{2}$

$\Rightarrow g (1^{+}) = \frac{3}{2} + 1 = \frac{5}{2}$

$\Rightarrow g (x)$ is continuous in $(0, 2)$ .

$\Rightarrow g^{'} (x) = \{\begin{cases} f^{'} (x), 0 < x \leq 1 \\ 1, 1 < x < 2 \end{cases}$

$\Rightarrow g^{'} (1) = f^{'} (1) = \frac{- 3}{2}$

$\Rightarrow g^{'} (1) \neq g^{'} (1^{+})$

So, $g (x)$ is not differentiable at $x = 1$ .

Hence, the correct answer is option: 'A'.

Consider the function f:(0,2)→R defined by f(x)=x2+2x and the function g(x) defined by gx=min{f(t)},0<t≤x and 0<x≤132+x,1<x<2. Then

Solution

Relevant Questions

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = \{\begin{matrix} min {f (t)}, & 0 < t \leq x and 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{matrix}$ . Then