CUET UG 2025 — Mathematics Calculus

Let $f(x) = \frac{x}{\log_e x}$f(x)=loge​xx​ where $\log_e x = \ln x$loge​x=lnx

---

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

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

$$u' = 1$$u′=1

$$v' = \frac{1}{x}$$v′=x1​

$$f'(x) = \frac{(1)(\ln x) - (x)\left(\frac{1}{x}\right)}{(\ln x)^2}$$f′(x)=(lnx)2(1)(lnx)−(x)(x1​)​

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

---

Setting the derivative equal to zero:

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

The numerator must equal zero:

$$\ln x - 1 = 0$$lnx−1=0

$$\ln x = 1$$lnx=1

$$x = e$$x=e

---

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

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

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

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

---

The minimum value occurs at $x = e$x=e:

$$f(e) = \frac{e}{\ln e}$$f(e)=lnee​

$$f(e) = \frac{e}{1}$$f(e)=1e​

$$f(e) = e$$f(e)=e

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