alleryThe function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:
A$x = 0$
B$x = 1$ (max value $1/2$)
C$x = \sqrt 3$
D$x \to \infty$
Answer & Solution
Correct answer: B. $x = 1$ (max value $1/2$)
We have $f'(x) = ((1+x^2) - x(2x))/(1+x^2)^2$.
That is $f'(x) = (1 - x^2)/(1+x^2)^2$.
So $f'(x) = 0$.
⇒ $x = 1$ (in [0,∞)). $f(1) = 1/2$.
**Check:** $f(0) = 0$, $f \to 0$ as $x \to \infty$.
We have **So max = 1/2 at.
So $x = 1$.**.
Keep practising Application of Derivatives
Browse all Application of Derivatives questions for AP Intermediate 2nd Year, or explore more Mathematics topics.
Related questions
If the tangent to the curve $y = e^{2x}$ at point $(a, e^{2a})$ passes through origin, theThe volume of a sphere is increasing at $8\pi$ cm³/s. The rate of change of surface area wIf $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:If $y = a \ln x + b x^2 + x$ has extreme values at $x = 1$ and $x = 2$, then $(a, b)$ is:Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):A rectangle is inscribed in a semicircle of radius $R$ with base on the diameter. The **maFind the **maximum value** of $f(x) = in x + \cos x$ for $x \in [0, 2\pi]$.The function $f(x) = x^3 - 3x + 1$ has critical points at: