The function f(x) = frac{x}{1 + x tan x} atta | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $f(x) = \frac{x}{1 + x 	an x}$. To find the maxima of $f(x)$,we can find the minima of its reciprocal $g(x) = \frac{1}{f(x)} = \frac{1 + x

Vedclass · Accepted Answer

$x = \cos x$

Vedclass · Answer

(B) Let $f(x) = \frac{x}{1 + x 	an x}$. To find the maxima of $f(x)$,we can find the minima of its reciprocal $g(x) = \frac{1}{f(x)} = \frac{1 + x 	an x}{x} = \frac{1}{x} + 	an x$.Now,differentiate $g(x)$ with respect to $x$:$g'(x) = -\frac{1}{x^2} + \sec^2 x$.For critical points,set $g'(x) = 0$:$-\frac{1}{x^2} + \sec^2 x = 0 \implies \sec^2 x = \frac{1}{x^2} \implies \cos^2 x = x^2 \implies x = \cos x$ (considering the domain where $x > 0$).Now,find the second derivative $g''(x)$:$g''(x) = \frac{2}{x^3} + 2 \sec x (\sec x 	an x) = \frac{2}{x^3} + 2 \sec^2 x 	an x$.At $x = \cos x$,$g''(x) = \frac{2}{\cos^3 x} + 2 \sec^2 x 	an x = 2 \sec^3 x + 2 \sec^2 x 	an x = 2 \sec^2 x (\sec x + 	an x)$.Since $x = \cos x$ lies in the first quadrant,$\sec x > 0$ and $	an x > 0$,so $g''(x) > 0$.Thus,$g(x)$ has a minimum at $x = \cos x$,which implies $f(x)$ has a maximum at $x = \cos x$.

The function $f(x) = \frac{x}{1 + x \tan x}$ attains its maximum value at:

Similar Questions

If $x$ and $y$ are two positive real numbers such that $xy = 4$, then the minimum value of $\left(\sqrt{x} + \frac{y^2}{2}\right)$ is

The curve $y = 2x^3 + ax^2 + bx + c$ passes through the origin,and the tangents at $x = -1$ and $x = 2$ are parallel to the $X$-axis. Then the values of $a, b,$ and $c$ are respectively:

If the sum of two numbers is $3$,then the maximum value of the product of the first and the square of the second is:

Let $f(x) = \alpha x^2 - 2 + \frac{1}{x}$,where $\alpha$ is a real constant. The smallest $\alpha$ for which $f(x) \geq 0$ for all $x > 0$ is

$A$ cylindrical tank without a top lid is being manufactured to hold a fixed volume of $125\pi \text{ cm}^3$. The minimum surface area required to construct this tank is ............ $\text{cm}^2$. (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module