Let, r be radius and s be area of surface of sphere at time t.
By question, dr/dt = 1/2π
Now, s = 4πr²;
Thus, ds/dt = 4π * 2r(dr/dt) = 8?r(dr/dt)
When r = 5cm and dr/dt = 1/2π
Then ds/dt = 8π*5*(1/2π) = 20
Thus, correct answer is 20 sq cm.

The circumference of circle is given by C=2πr, where r is radius of circle.
&re4;\(\frac{dC}{dt}\)=2π.\(\frac{dr}{dt}\)=5 cm/s
\(\frac{dr}{dt}\)=5/2π cm/s
\(\frac{dA}{dt}\)|_r=6=2πr.\(\frac{dr}{dt}\)=2πr.\(\frac{5}{2 \pi}\)=5r=5(6)=30 cm²/s.

We have, x = t³ + 6t² ? 15t + 18
Let, ?v? be velocity of particle and ?a? be acceleration of particle at end of t seconds. Then,
v = dx/dt = d/dt(t³ + 6t² ? 15t + 18)
So, v = 3t² + 12t ? 15
Therefore, a = dv/dt = d/dt(3t² + 12t ? 15)
So, a = 6t + 12
Thus, acceleration of particle at end of 2 seconds is,
[dv/dt]_{t = 2} = 6(2) + 12 = 24cm/sec².

We have, x = t³ + 6t² ? 15t + 18
Let, v be velocity of particle at end of t seconds. Then,
v = dx/dt = d/dt(t³ + 6t² ? 15t + 18)
So, v = 3t² + 12t ? 15
Thus, velocity of particle at end of 2 seconds is,
[dx/dt]_{t = 2} = 3(2)² + 12(2) ? 15 = 21cm/sec.

The relation between f(x) and ℓ when minimum value or least value function f is f(x) ≥ (ℓ) ∀ x ∈ A where function is defined on a set A and ℓ ∈ f(A).

f(x) = sinx+cosx.
f?(x) = cosx ? sinx. Now f?(x) = 0 gives sinx = cosx which gives that x=?/4, 5?/4 as 0 ? x ? 2?.
Therefore on checking values we get f is decreasing in (?/4, 5?/4).

Differentiating 2x²+3y²=3 w.r.t x, we get
4x+6y \(\frac{dy}{dx}\)=0
\(\frac{dy}{dx} = -\frac{2x}{3y}\)
\(\frac{dy}{dx}\)]_(3,4)=-\(\frac{2(3)}{3(4)}=-\frac{1}{2}\)
Therefore, equation of tangent at (3,4) is
y-y₀=m(x-x₀)
y-4=-\(\frac{1}{2}\) (x-3)
2(y-4)=-x+3
2y-8=-x+3
x+2y=11

There are two types of decreasing functions in maxima and minima. They are strictly decreasing and monotonically decreasing. The mamatical expression for strictly decreasing function is x₁ < x₂ ⇒ f(x₁) > f(x₂) ∀ x₁, x₂ ∈ (a,b).

One of properties of a function is to be constant. A function is said to be constant when it satisfies condition f?(x) = 0 ∀ x₁, x₂ ∈ (a,b) where function ?f? should be continuous and differentiable on (a,b).

f(x) = x³ ? 3x² + 4x
f?(x) = 3x² ? 6x + 4.
f?(x) = 3(x² ? 2x + 1) + 1.
=> 3(x-1)² + 1>0, in every interval of R. Therefore function f is increasing on R.