Sets, Relations & Functions
Functions
star_batch_jee_advanced_2025
Grade 11
$f(x)$ has horizontal tangent at $x = e$
$f(x)$ cuts the $x$-axis at only one point
$f(x)$ is many-one function
$f(x)$ has one vertical tangent
Step-by-Step Solution
Key Concept: A horizontal tangent occurs where the derivative equals zero, many-one functions fail the horizontal line test, and vertical tangents require the derivative to approach infinity.
For $f(x) = \frac{\log x}{x}$, we find $f'(x) = \frac{1-\log x}{x^2}$. Setting $f'(x) = 0$ gives $\log x = 1$, so $x = e$ with a horizontal tangent, confirming option 1. The function $f(x) = 0$ only when $\log x = 0$, i.e., $x = 1$, confirming option 2. Since $f'(x) > 0$ for $0 < x < e$ and $f'(x) < 0$ for $x > e$, the function increases then decreases, making it many-one (option 3 correct). For option 4, no vertical tangent exists since the denominator $x^2$ never causes $f'(x) = \infty$.
Correct Answer: 1,2,3