{a_1, a_2, ....., a_n, .....} is a progressio | vedclass

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

Question

(B) Let $f(x) = \frac{x^2}{x^3 + 200}$. To find the maximum,we differentiate with respect to $x$:$f'(x) = \frac{(x^3 + 200)(2x) - x^2(3x^2)}{(x^3 + 20

Vedclass · Accepted Answer

$a_7$

Vedclass · Answer

(B) Let $f(x) = \frac{x^2}{x^3 + 200}$. To find the maximum,we differentiate with respect to $x$:$f'(x) = \frac{(x^3 + 200)(2x) - x^2(3x^2)}{(x^3 + 200)^2} = \frac{2x^4 + 400x - 3x^4}{(x^3 + 200)^2} = \frac{x(400 - x^3)}{(x^3 + 200)^2}$.For $f(x)$ to be increasing,$f'(x) > 0$,which implies $400 - x^3 > 0$,or $x Since $7^3 = 343$ and $8^3 = 512$,we have $7 Thus,the function increases until $x = (400)^{1/3}$ and decreases thereafter.We compare $a_7$ and $a_8$:$a_7 = \frac{7^2}{7^3 + 200} = \frac{49}{343 + 200} = \frac{49}{543} \approx 0.0902$.$a_8 = \frac{8^2}{8^3 + 200} = \frac{64}{512 + 200} = \frac{64}{712} = \frac{8}{89} \approx 0.0898$.Since $a_7 > a_8$,the largest term is $a_7$.

${a_1, a_2, ....., a_n, .....}$ is a progression where $a_n = \frac{n^2}{n^3 + 200}$. The largest term of this progression is

Similar Questions

$A$ square piece of tin of side $30\,cm$ is to be made into a box without a top by cutting a square of side $x$ from each corner and folding up the flaps to form a box. If the volume of the box is maximum,then its surface area (in $cm^2$) is equal to $............$.

The difference between the greatest and the least values of the function $f(x) = \sin(2x) - x$ on the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ is:

If $x+y=60$,$x>0$,$y>0$,then the maximum value of $x y^3$ is

Divide $10$ into two parts such that the sum of twice the first part and the square of the second part is minimum. The two parts are:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module