Let the radius of spherical ball of the salt be r.
\(\therefore \text { Volume of the ball, } V=\frac{4}{3} \pi r^{3}\) 
and surface area, \(S=4 \pi r^{2}\) 
\(\therefore \frac{d V}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=-\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}=-4 \pi r^{2} \frac{d r}{d t}\) 
[here, we take negative sign because salt is dissolving]
According to the given condition,
\(\frac{d V}{d t} \propto S \Rightarrow \frac{d V}{d t}=k S\) wehre - k )ISproportio.na II' t)' constant.
\(\Rightarrow -4 \pi r^{2} \frac{d r}{d t}=k \cdot 4 \pi r^{2} \Rightarrow \frac{d r}{d t}=-k\) 
Hence, the radius of ball is decreasing at constant rate.

Given is y \(y=5 x-2 x^{3} \text { and } \frac{d x}{d t}=2 \text { units } / \mathrm{s}\) 
Now, slope of the curve \(\frac{d y}{d x}=5-6 x^{2}=M(\text { say })\) 
Rate of change of the slope
\( \frac{d M}{d t} =-6 \frac{d}{d t}\left(x^{2}\right) \)
\( \Rightarrow \frac{d M}{d t} =-12 x \frac{d x}{d t} \) 
When x = 3 , then
\(\frac{d M}{d t}=-12 \times 3 \times 2=-72 \text { unit/s }\) 
Thus, the slope of decreasing at a rate of 72 units/s.

Let r be the radius, A be the area and P be the perimeter.
Then, we have \(\frac{d A}{d t}=\text { constant }=k(\text { say })\)

Let \(\theta\) denote the angle at instant t
\(\frac{d\theta}{dt}=2\frac{d}{dt}(\sin \theta)\)
\(\frac{d\theta}{dt}=2\cos\theta.(\frac{d\theta}{dt})\)
\(1=2\cos\ \theta\)
\(2\cos\theta=1 cos\theta=\frac{1}{2}\)
\(\Rightarrow \theta=\cos^{-1}(\frac{1}{2})\)
Hence required angles is \(\frac{\pi}{3}\)

\(f^{\prime}(x)=\frac{1+2 x^{2}}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}}=\frac{1+2 x^{2}-\sqrt{1+x^{2}}}{1+x^{2}}\) 
Now,\(f^{\prime}(x) \geq 0\) 
\( \Rightarrow \frac{1+2 x^{2}-\sqrt{1+x^{2}}}{1+x^{2}} \geq 0\)
\(\Rightarrow 1+2 x^{2} \geq \sqrt{1+x^{2}}\)
\( \Rightarrow 1+4 x^{4}+4 x^{2} \geq 1+x^{2} \) 
\(\Rightarrow 4 x^{4}+3 x^{2} \geq 0\) which is true for all \(x \in R\) .
So,\(f^{\prime}(x) \geq 0, \forall x \in R\)