Calculus

Application of Derivatives

Medium

core

If the rate of change of area of a circle is equal to the rate of change of its diameter, then its radius is equal to :

$\frac{1}{2\pi}$ unit

$\frac{1}{\pi}$ unit

$\frac{\pi}{2}$ unit

$\pi$ unit

✅ Correct Option: 2

$A=\pi r^2$ , $D=2r$ . So $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$ and $\frac{dD}{dt}=2\frac{dr}{dt}$ .

Setting equal: $2\pi r\frac{dr}{dt}=2\frac{dr}{dt}$ , giving $\pi r=1$ , so $r=\frac{1}{\pi}$ unit.

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Match List-I with List-II

List-I (Function)	List-II (Critical point)
(A) $f(x)=x^3-3x$	(I) $x=3$
(B) $f(x)=x^3-12x$	(II) $x=0$
(C) $f(x)=x^2-6x$	(III) $x=1$
(D) $f(x)=x^2$	(IV) $x=2$

Choose the correct answer from the options given below:

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In which of the following interval the function $f(x) = x^x, x > 0$ is strictly increasing?

If the rate of change of area of a circle is equal to the rate of change of its diameter, then its radius is equal to :

Solution

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