For x > 0, the minimum value of x/(_e x) is Option 1: 1/e Option 2: 2e Option 3: e Option 4: 2/e

Answer: e. Solution: Let f(x) = x/(_e x) where _e x = x To find the minimum, take the derivative using the quotient rule. For u = x and v = x: u' = 1 v' = 1/x f'(x) = ((1)( x) - (x)(1/x))/(( x)^2) f'(x) = ( x - 1)/(( x)^2) Setting the derivative equal to zero: ( x - 1)/(( x)^2) = 0 The numerator must equal zero: x - 1 = 0 x = 1 x = e Checking the sign of f'(x) around x = e: When x e: x > 1, so f'(x) > 0 (increasing) The function has a minimum at x = e. The minimum value occurs at x = e: f(e) = e/( e) f(e) = e/1 f(e) = e Therefore, the minimum value of x/(_e x) is e.

Calculus

Application of Derivatives

Medium

common

For $x > 0$ , the minimum value of $\frac{x}{\log_e x}$ is

$\frac{1}{e}$

$2e$

$e$

$\frac{2}{e}$

✅ Correct Option: 3

Let $f(x) = \frac{x}{\log_e x}$ where $\log_e x = \ln x$

To find the minimum, take the derivative using the quotient rule.

For $u = x$ and $v = \ln x$ :

$u' = 1$

$v' = \frac{1}{x}$

$f'(x) = \frac{(1)(\ln x) - (x)\left(\frac{1}{x}\right)}{(\ln x)^2}$

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

Setting the derivative equal to zero:

$\frac{\ln x - 1}{(\ln x)^2} = 0$

The numerator must equal zero:

$\ln x - 1 = 0$

$\ln x = 1$

$x = e$

Checking the sign of $f'(x)$ around $x = e$ :

When $x < e$ : $\ln x < 1$ , so $f'(x) < 0$ (decreasing)

When $x > e$ : $\ln x > 1$ , so $f'(x) > 0$ (increasing)

The function has a minimum at $x = e$ .

The minimum value occurs at $x = e$ :

$f(e) = \frac{e}{\ln e}$

$f(e) = \frac{e}{1}$

$f(e) = e$

Therefore, the minimum value of $\frac{x}{\log_e x}$ is $e$ .

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Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

For $x > 0$ , the minimum value of $\frac{x}{\log_e x}$ is