Consider the function defined by and the function defined by [2024], Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question-1

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Solution

Q.
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, [2024]

1 $g$ is neither continuous nor differentiable at $x = 1$

2 $g$ is continuous and differentiable for all $x \in (0, 2)$

3 $g$ is continuous but not differentiable at $x = 1$

4 $g$ is not continuous for all $x \in (0, 2)$

Ans.
(3)

Given, $f : (0, 2) \to R$

$f (x) = \frac{x}{2} + \frac{2}{x}$

and $g (x) =$ ${\begin{matrix} \min {f (t)}, & 0 < t \leq x and 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{matrix}$

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

$\Rightarrow$ $f (x)$ is decreasing in (0, 2).

$\Rightarrow \min (f (t)) = f (x), 0 < t \leq x$

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

From the graph, we see that $g (x)$ is continuous but not differentiable at $x = 1 .$

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