f(x) = sin3x - cos3x, 0 < x < π 
\(\Rightarrow f'(x)=3\cos3x+3\sin3x\)
\(=3(\cos3x+\sin3x)\)
Put \(f'(x)=0\)
\(\Rightarrow \cos3x+\sin3x=0\)
\(\Rightarrow \sin3x=-\cos3x\)
\(\Rightarrow -\tan3x=1\)
\(\Rightarrow \tan3x=-1\)
As 0\(\pi\), 0 < 3x > 3\(\pi\)
ஃ tan 3x is negative for the following values:
\(3x=\frac{3\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{4}\)
\(3x=\pi+\frac{3\pi}{4}=\frac{7\pi}{4}\)
\(\Rightarrow x=\frac{7\pi}{12}\)
\(3x=\frac{7\pi}{4}+\pi=\frac{11\pi}{4}\)
\(\Rightarrow x=\frac{11\pi}{12}\)
Hence we have intervals:
Hence, \(f(x)-\sin3x-\cos3x\) is strictly increasing in the intervals \((0,\frac{\pi}{4})\cup(\frac{7\pi}{12},\frac{11\pi}{12}) \) and strictly decreasing in intervals \((\frac{\pi}{4},\frac{7\pi}{12})\cup(\frac{11\pi}{12},\pi)\)

Given \(f(x)=\sin x-\cos x, \ \ \ \ 0
\(f'(x)=\cos x+\sin x,\ \ \ \ \ 0
\(\therefore\ f'(x)=0\)
\(\Rightarrow \cos x+\sin x=0\ \ or\ \tan x=-1\)
\(\therefore\ x=\frac{3\pi}{4},\frac{7\pi}{4}\)
Getting   \(f'(x)=\cos x+\sin x\)
i.e., \(f''(\frac{3\pi}{4})\) negative, so at x = \(\frac{3\pi}{4},f(x)\) is local maxima
Hence, local maximum value = \(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}=\sqrt2\) and \(f''(\frac{7\pi}{2})=\cos\frac{7\pi}{4}-\sin\frac{7\pi}{4}=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\)
i.e., \(f''(\frac{7\pi}{2})\) is positive so \(x=\frac{7\pi}{4}\) is local minima
Hence, local minimum value \(=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}=-\sqrt2\)

Let l, b, h be the length, breadth and depth of the tank, respectively.
\(\therefore\ l\times b\times 3=75\)
\(\Rightarrow l\times b=25\)
Let C be the cost, then
C = 100(1x b) + 100[h(b + I)] 
\(=100(l\times \frac{25}{l})+300(\frac{25}{l}+l)\)
\(=2500+300(\frac{25}{l}+l)\)
Differentiating w.r.t. l,
\(\therefore\ \frac{dC}{dl}=0+300(\frac{-25}{l^2}+1)\)
Putting \(\frac{dC}{dl}=0\)
\(\Rightarrow 300(-\frac{25}{l^2}+1)=0\)
\(\Rightarrow l^2=25 \ or\ l=5\)
Getting \(\frac{d62C}{dl^2}=300(\frac{50}{l^3})\)
 \(\Rightarrow (\frac{d62C}{dl^2})_{at\ l=5}=\frac{15000}{125}>0\)
i.e., C is minimum when l = 5
\(\Rightarrow b=5\)
ஃ C = 100(25) + 300(10)
= 2,500 + 3,000
= 5,500
Hence the minimum cost is Rs. 5,500.

Given, volume of the box = 1024 cm³.Let length of the side ofsquare base be x cm and height ofthe box be y cm.
= x cm,
height of box = h cm.
Volume of box = 1024 cm³
\(\Rightarrow x.x.h=1024\)
\(\therefore h=\frac{1024}{x^2}\)
C (cost of box) = 5(2x²) + 2.5(4xh)
= 10x² + 10hx
\(=10x^2+\frac{10,240}{x}\)
\(\Rightarrow \frac{dC}{dx}=20x-\frac{10,240}{x^2}\)
Solving \(\frac{dC}{dx}=0,\)  we get x³ = 512   ஃ x = 8
\(=20+\frac{1(10240)}{x^3}>0\)
Thus, cost of box is least at x = 8 and least cost of
box is:
\(C(8)=10(8)^2+\frac{10240}{8}\)
= Rs. 1,920

\(f(x)=x^5-5x^4+5x^3-1\)
\(\Rightarrow f'(x)=5x^4-20x^3+15x^2\)
\(\Rightarrow f'(x)=5x^2(x-1)(x-3)\)
\(f'(x)=0\Rightarrow x=0,x=1,x=3\)
\(f''(x)=20x^3-60x^2+30x\)
\(=10x[2x^2-6x+3]\)
\(f'(0)=0,f'(1)=-ve,f'(3)=+ve\)
also , \(f'(<0)=-ve\)
\(f'(>0)=-ve\)
x = 0 is a point of inflexion
x = 1 is a point of local maxima
x = 3 is a point of local minima
∴ Local max. value = f(1) = 0
and Local min. value = f(3) = -28

Let length and breadth of rectangle be x and y.
\(\therefore\ P=2y+x+\pi\frac{x}{2}\)
\(\therefore\ A=xy+\frac{1}{2}\pi\frac{x^2}{4}\)
\(=\frac{x}{2}[P-x-\frac{\pi x}{2}]+\frac{\pi x^2}{8}\)
\(=P\frac{x}{2}-\frac{x^2}{2}-\pi\frac{x^2}{4}+\pi\frac{x^2}{8}\)
​​​​​​
\(\frac{dA}{dx}=\frac{P}{2}-x-\frac{\pi x}{2}+\frac{\pi x}{4}\)
\(=\frac{P}{2}-x-\frac{\pi x}{4}\)
\(\frac{dA}{dx}=0\)
\(\Rightarrow\ x=\frac{2P}{4+\pi}\ and\ y=\frac{P}{4+\pi}\)
\(\frac{d^2A}{dx^2}=-1-\pi / 4<0\)
⇒Area is maximum when length
\(=\frac{2P}{4+\pi}\)
breath \(=\frac{P}{4+\pi}\)

Let the sides of rt. ΔABC be x and y.
\(\therefore\ x^2+y^2=4r^2\)
and A = area of \(\Delta=\frac{1}{2}xy\)
Let, \(S=A^2 =\frac{1}{4}x^2y^2\)
\(=\frac{1}{4}x^2(4r^2-x^2)\)
\(=\frac{1}{4}(4r^2x^2-x^4)\)
\(\therefore\ \frac{dS}{dx}=\frac{1}{4}[8r^2x-4x^3]\)
\(\Rightarrow \ \frac{dS}{dx}=0\Rightarrow x^2=2r^2 \ or\ x=\sqrt{2r}\)
\(y^2=4r^2-2r^2=2r^2\Rightarrow y=\sqrt2r\)
i.e. x = y and  \(\frac{d^2S}{dx^2}=(2r^2-3x^2)=2r^2-6r^2<0\)
⇒ Area is maximum when Δ is isosceles 

let radius, height and slant height of cone be r, h and l respectively.
\(\therefore r^2+h^2=l^2\)

V(volume) \(=\frac{\pi}{3}r^2h\)
\(\Rightarrow V^2=\frac{\pi^2r^4h^2}{9}\)
\(\Rightarrow h^2=\frac{9V^2}{\pi^2r^4}\)
\(A=\pi r l,z=A^2=\pi^2r^2l^2\)
\(=\pi^2r^2(r^2+h^2)\)
\(let \ \ A^2=z\)
\(\therefore\ z=\pi^2r^2[r^2+\frac{9V^2}{\pi^2r^4}]\)
\( z=\pi^2[r^4+\frac{9V^2}{\pi^2r^2}]\)
\( \frac{dz}{dr}=\pi^2[4r^3-\frac{18V^2}{\pi^2r^3}]\)
\(\therefore \ \ \ \frac{dz}{dr}=0\)
\(\Rightarrow\ r=6\sqrt{\frac{9V^2}{2\pi^2}}\)
at \( r=6\sqrt{\frac{9V^2}{2\pi^2}}\)
\(\frac{d^2z}{dr^2}=\pi^2(12r^2+\frac{54V^2}{\pi^2r^4 })>0\)
\(\therefore\) curved surface area is minimum iff
\(2\pi^2r^6=9V^2\)
i.e., \(2\pi^2r^6=\pi^2r^4h^2\)

let, and h be the radius and height of the cylinder, then Volume of cylinder (V)
\(\pi r^2h=128\pi\)
\(\therefore \ h=\frac{128\pi}{\pi r^2}=\frac{128}{r^2}\)
Surface area of cylinder
\(2\pi r^2+2\pi rh\)
\(=2\pi(r^2+rh)\)
\(\therefore\ s=2\pi(r^2+\frac{128}{r})\)
\(\therefore \frac{dS}{dr}=2\pi(2r-\frac{128}{r^2})\)
\(\frac{dS}{dr}=0\)
\(\Rightarrow\ r^3=64\)
r = 4
at r = 4: \(\frac{d^2S}{dr^2}=2\pi(2+\frac{256}{r^3})\)
\(=2\pi(2+\frac{256}{64})\)
\(=12\pi>0\)
ஃ  Surface area is minimum at r = 4 cm
h = 8 cm

Given, \(f(x)=\frac{3}{10} x^{4}-\frac{4}{5} x^{3}-3 x^{2}+\frac{36 x}{5}+11\) 
On differentiating both sides w.r.t. x, we get
\(f^{\prime}(x)=\frac{12 x^{3}}{10}-\frac{12 x^{2}}{5}-6 x+\frac{36}{5}+0\) 
Put \(f^{\prime}(x)=0\) 
\(\Rightarrow \frac{12 x^{3}}{10}-\frac{12 x^{2}}{5}-6 x+\frac{36}{5}=0\)
\(\Rightarrow \frac{6 x^{3}-12 x^{2}-30 x+36}{5}=0\)
\(\Rightarrow\)\(x^{3}-2 x^{2}-5 x+6=0\left[\right. divide by \left.\frac{6}{5}\right]\)
\(\Rightarrow (x-1)\left(x^{2}-x-6\right)=0\)
\(\Rightarrow (x-1)(x+2)(x-3)=0\)
\(\Rightarrow\) x - 1 = 0
x + 2 = 0 or x - 3 = 0
x = -2, 1, 3
Now, we find the intervals in which f(x) is strictly increasing or strictly decreasing.
\(\begin{array}{ccc} \hline \text { Interval } & \begin{array}{c} \text { Sign of } f^{\prime}(x) \\ f^{\prime}(x)=(x-1)(x+2)(x-3) \end{array} & \begin{array}{c} \text { Nature of } \\ \text { function } \end{array} \\ \hline(-\infty,-2] & (-)(-)(-)=(-)<0 & \begin{array}{l} \text { Strictly } \\ \text { decreasing } \end{array} \\ \hline[-2,1] & (-)(+)(-)=(+)>0 & \text { Strictly increasing } \\ \hline[1,3] & (+)(+)(-)=(-)<0 & \begin{array}{l} \text { Sthictly } \\ \text { decreasing } \end{array} \\ \hline[3, \infty) & (+)(+)(+)=(+)>0 & \text { Strictly increasing } \\ \hline \end{array}\) 
(a) f(x) is strictly increasing in the interval \((-2,1) \cup(3, \infty)\) 
(b) f(x) is strictly decreasing in the interval \((-\infty,-2) \cup(1,3)\)