We have
\(y=[x(x-2)]^{2}=\left[x^{2}-2 x\right]^{2} \)
\(\therefore \frac{d y}{d x}=y^{\prime}=2\left(x^{2}-2 x\right)(2 x-2)=4 x(x-2)(x-1) \)
\(\therefore \frac{d y}{d x}=0 \Rightarrow x=0, x=2, x=1 \)
\(\text {The points } x=0, x=1 \text { and } x=2 \) divide the real line into four disjoint intervals i.e.,
\((-\infty, 0),(0,1)(1,2), \text { and }(2, \infty) .\)
\(\text { In intervals }(-\infty, 0) \text { and }(1,2), \frac{d y}{d x}<0 \text { . }\)
∴y is strictly decreasing in intervals (−∞,0) and (1,2) . " id="MathJax-Element-580-Frame" role="presentation" style="position: relative;" tabindex="0">∴y is strictly decreasing in intervals  \( (-\infty, 0) \text { and }(1,2) \text { . }\)
However, in intervals \( (0,1) \text { and }(2, \infty), \frac{d y}{d x}>0 .\)
\(\therefore y \text { is strictly increasing in intervals }(0,1) \text { and }(2, \infty) .\)
\(\therefore y \text { is strictly increasing for } 02 \text { . }\)
Increasing:\(\left( 0,1 \right) \cup \left( 2,\infty \right) \) ;Decreasing:\(\left( -\infty ,0 \right) \cup \left( 1,2 \right) ;\)points are(0,0),(1,1),(2,0).

(i) f ′(c) = 0 if f attains its absolute maximum value at c.
(ii) f ′(c) = 0 if f attains its absolute minimum value at c.
In view of the above results, we have the following working rule for finding absolute maximum and/or absolute minimum values of a function in a given closed interval [a, b].
Working Rule
Step 1: Find all critical points of f in the interval, i.e., find points x where either f' (x) = 0 or f is not differentiable.
Step 2: Take the end points of the interval.
Step 3: At all these points (listed in Step 1 and 2), calculate the values of f .
Step 4: Identify the maximum and minimum values of f out of the values calculated in
Step 3. This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f .

 The equation of the curve is given as: " id="MathJax-Element-566-Frame" role="presentation" style="position: relative;" tabindex="0">The equation of the curve is given as: 
\(6 y=x^{3}+2\)
 The rate of change of the position of the particle with respect to time (t) is given by, " id="MathJax-Element-568-Frame" role="presentation" style="position: relative;" tabindex="0">The rate of change of the position of the particle with respect to time (t) is given by, 
\(6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0 \Rightarrow 2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}\)
 When the y -coordinate of the particle changes 8 times as fast as the " id="MathJax-Element-570-Frame" role="presentation" style="position: relative;" tabindex="0">When the y -coordinate of the particle changes 8 times as fast as the 
\(x \text { -coordinate i.e., }\left(\frac{d y}{d t}=8 \frac{d x}{d t}\right), \text { we have: }\)
\(2\left(8 \frac{d x}{d t}\right)=x^{2} \frac{d x}{d t} \)
\(\Rightarrow 16 \frac{d x}{d t}=x^{2} \frac{d x}{d t} \)
\(\Rightarrow\left(x^{2}-16\right) \frac{d x}{d t}=0 \)
\(\Rightarrow x^{2}=16 \)
\(\Rightarrow x=\pm 4 \)
\(\text { When } x=4, y=\frac{4^{3}+2}{6}=\frac{66}{6}=11 \text { . }\)
\(\text { When } x=-4, y=\frac{(-4)^{3}+2}{6}=-\frac{62}{6}=-\frac{31}{3} \text { . }\)
 Hence, the points required on the curve are (4,11) and (−4,−313) . " id="MathJax-Element-579-Frame" role="presentation" style="position: relative;" tabindex="0">Hence, the points required on the curve are (4,11) and 

Let x be the length of a side, V be the volume, and s be the surface area of the cube.
Then, V = x³ and S = 6x² where x is a function of time t.
 It is given that dVdt=8 cm3/s . " id="MathJax-Element-558-Frame" role="presentation" style="position: relative;" tabindex="0"> It is given that  \( \frac{d V}{d t}=8 \mathrm{~cm}^{3} / \mathrm{s} \text { . }\)
 Then, by using the chain rule, we have: " id="MathJax-Element-559-Frame" role="presentation" style="position: relative;" tabindex="0">Then, by using the chain rule, we have: 
\(\therefore 8=\frac{d V}{d t}=\frac{d}{d t}\left(x^{3}\right)=\frac{d}{d x}\left(x^{3}\right) \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t}\)
\(\Rightarrow \frac{d x}{d t}=\frac{8}{3 x^{2}} \ \text { (1) } \quad \text { [By chain rule] } \)
\(\text { Now, } \frac{d \mathrm{~S}}{d t}=\frac{d}{d t}\left(6 x^{2}\right)=\frac{d}{d x}\left(6 x^{2}\right) \cdot \frac{d x}{d t} \ 0 .\)
\(=12 x \cdot \frac{d x}{d t}=12 x .\left(\frac{8}{3 x^{2}}\right)=\frac{32}{x}\)
\(\text {Thus, when } x=12 \mathrm{~cm}, \frac{d S}{d t}=\frac{32}{12} \mathrm{~cm}^{2} / \mathrm{s}=\frac{8}{3} \mathrm{~cm}^{2} / \mathrm{s}\)
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \(\frac{8}{3}\) cm²/s

 Let △ABC be isosceles where BC is the base of fixed length b_ . " id="MathJax-Element-542-Frame" role="presentation" style="position: relative;" tabindex="0">Let △ABC be isosceles where BC is the base of fixed length  \( \underline{b} \)
Let the length of the two equal sides of  \( \triangle \mathrm{ABC} \text { be } a \text { . }\)
\(\text {Draw } \mathrm{AD} \perp \mathrm{BC}\)
\(\text {Now, in } \triangle A D C \) by applying the Pythagoras theorem, we have: 
\(\mathrm{AD}=\sqrt{a^{2}-\frac{b^{2}}{4}}\)
\(\therefore \text { Area of triangle }(A)=\frac{1}{2} b \sqrt{a^{2}-\frac{b^{2}}{4}}\)
 The rate of change of the area with respect to time (t) is given by " id="MathJax-Element-548-Frame" role="presentation" style="position: relative;" tabindex="0"> The rate of change of the area with respect to time (t) is given by 
\(\frac{d A}{d t}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^{2}-\frac{b^{2}}{4}}} \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^{2}-b^{2}}} \frac{d a}{d t}\)
 It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. " id="MathJax-Element-550-Frame" role="presentation" style="position: relative;" tabindex="0">It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. 
\(\therefore \frac{d a}{d t}=-3 \mathrm{~cm} / \mathrm{s} \)
\(\therefore \frac{d A}{d t}=\frac{-3 a b}{\sqrt{4 a^{2}-b^{2}}} \)
 Then, when a=b , we have: " id="MathJax-Element-553-Frame" role="presentation" style="position: relative;" tabindex="0">Then, when a=b , we have: 
\(\frac{d A}{d t}=\frac{-3 b^{2}}{\sqrt{4 b^{2}-b^{2}}}=\frac{-3 b^{2}}{\sqrt{3 b^{2}}}=-\sqrt{3} b\)
 Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the " id="MathJax-Element-555-Frame" role="presentation" style="position: relative;" tabindex="0">Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of  \( \sqrt{3} b \mathrm{~cm}^{2} / \mathrm{s}\) Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the " role="presentation" style="position: relative;" tabindex="0"> 
 

Let f be continuous at a critical point c in I. Then
(i) If f ′(x) changes sign from positive to negative as x increases through c, i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.
(ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.
(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection

Let r be the radius, h be the height and V be the volume of the sand cone
Also given that, \(\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}, h=\frac{1}{6} r\) 
\(\Rightarrow r=6 h \text { and } h=4 \mathrm{~cm}\) 
Volume of sand cone,
\(V=\frac{1}{3} \pi r^{2} h\) 
\( \Rightarrow V=\frac{1}{3} \pi(6 h)^{2} h \)
\(\Rightarrow V=\frac{1}{3} \pi \times 36 h^{2} \times h=12 \pi h^{3} \) 
On differentiating both sides w.r.t. t, we get
\( \frac{d V}{d t}=12 \pi \times 3 h^{2} \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t} \)
\(\Rightarrow 12=36 \pi(4)^{2} \frac{d h}{d t} \)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi \times 16}=\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \) 
Hence, the height of the sand cone is increasing at the rate of \(\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}\). when the height is 4 cm,

Let P be the perimeter and A be the area of the rectangle of length x and width y.
Given \(\frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}\) 
[negative (-) sign for decreasing rate]
\(x=8 \mathrm{~cm}, \frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min} \text { and } y=6 \mathrm{~cm}\) 
(ii) Perimeter of the rectangle.\(P=2(x+y)\) 
On differentiating both sides w.r.t. t, we get
\( \frac{d P}{d t} =2\left(\frac{d x}{d t}+\frac{d y}{d t}\right) \)
\(=2(-5+4)=-2 \mathrm{~cm} / \mathrm{min}\) 
So, perimeter decreases at the rate of 2 cm/min.
(ii) Area of the rectangle, A = xy
On differentiating both sides w.r.t. t, we get
\( \frac{d A}{d t} =x \frac{d y}{d t}+y \frac{d x}{d t} \)
\(=8 \times 4+6 \times(-5)=32-30=2 \mathrm{~cm}^{2} / \mathrm{min} \) 
Hence, area increases at the rate of 2 cm ² /min.

We have
f(x) = sin x + cos x,
or f'(x) = cos x – sin x

Now f'(x) = 0 gives sin x = cos x which gives that \(x=\frac{\pi}{4}, \frac{5 \pi}{4} \text { as } 0 \leq x \leq 2 \pi\)
The points \(x=\frac{\pi}{4} \text { and } x=\frac{5 \pi}{4}\) divide the interval [0, 2p] into three disjoint intervals namely,
\(\left[0, \frac{\pi}{4}\right),\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \text { and }\left(\frac{5 \pi}{4}, 2 \pi\right]\)
\(\text {Note that } f^{\prime}(x)>0 \text {if } x \in\left[0, \frac{\pi}{4}\right) \cup\left(\frac{5 \pi}{4}, 2 \pi\right]\)
or f is increasing in the intervals \(\left[0, \frac{\pi}{4}\right) \text { and }\left(\frac{5 \pi}{4}, 2 \pi\right]\)
\(\text {Also }f^{\prime}(x)<0 \text { if } x \in\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)
or is plecreasing in \( \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)

Let r be the radius, h be the height, α be semi-vertical
Then \(\tan \alpha=\frac{r}{h}\)
Given, \( \alpha =\tan ^{-1}(0.5) \)
\(\frac{r}{h} =0.5 \\ r =\frac{h}{2} \)
 
Let V be the volume of the cone. Then
\(\mathrm{V}=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi\left(\frac{h}{2}\right)^2 h=\frac{\pi h^3}{12}\)
Therefore \( \frac{d \mathrm{~V}}{d t} =\frac{d}{d h}\left(\frac{\pi h^3}{12}\right) \cdot \frac{d h}{d t} \)
\(=\frac{\pi}{4} h^2 \frac{d h}{d t} \) 
Now rate of change of volume, i.e., \(\frac{d \mathrm{~V}}{d t}=5 \mathrm{~m}^3 / \mathrm{h} \text { and } h=4 \mathrm{~m}\) 
\(5=\frac{\pi}{4}(4)^2 \cdot \frac{d h}{d t}\) 
\(\frac{d h}{d t}=\frac{5}{4 \pi}=\frac{35}{88} \mathrm{~m} / \mathrm{h}\left(\pi=\frac{22}{7}\right)\)
Thus, the rate of change of water level is \(\frac{35}{88} \mathrm{~m} / \mathrm{h}\)
\(V=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi\left(\frac{h}{2}\right) h \quad \text { Ans. } \left.\frac{35}{88} \mathrm{~m} / \mathrm{h}\right]\)

Let radius of cone be x and its height be h.
\(\therefore\) OD = (h - r)

Volume of cone (V)
\(=\frac { 1 }{ 3 } \pi { x }^{ 2 }h\) ...(i)
In \(\Delta OCD,\quad { x }^{ 2 }+({ h-r) }^{ 2 }={ r }^{ 2 }or\quad { x }^{ 2 }={ r }^{ 2 }-{ (h-r) }^{ 2 }\)
\(\therefore V=\frac { 1 }{ 3 } \pi h\{ { r }^{ 2 }-(h-r{ ) }^{ 2 }\} \)
\(=\frac { 1 }{ 3 } \pi (-{ h }^{ 3 }+{ 2h }^{ 2 }r)\)
\(\Rightarrow \frac { dV }{ dh } =\frac { \pi }{ 3 } (-3{ h }^{ 2 }+4hr)\)
\(\therefore \quad \frac { dV }{ dh } =0\Rightarrow h=\frac { 4r }{ 3 } \)
\(\frac { { d }^{ 2 }V }{ { dh }^{ 2 } } =\frac { \pi }{ 3 } (-6h+4r)\)
\(=\frac { \pi }{ 3 } \left( -6\left( \frac { 4r }{ 3 } \right) +4r \right) \)
\(=-\frac { 4\pi r }{ 3 } <0\)
\(\therefore \ at\quad h=\frac { 4r }{ 3 } \), Volume is maximum
Maximum volume
\(=\frac { 1 }{ 3 } \pi .\left\{ -{ \left( \frac { 4r }{ 3 } \right) }^{ 3 }+2{ \left( \frac { 4r }{ 3 } \right) }^{ 2 }r \right\} \)
\(=\frac { 8 }{ 27 } .\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)
\(=\frac { 8 }{ 27 } \) (volume of sphere)

As \(\frac{dy}{dt}\) = 2x.\(\frac{dx}{dt}\)
⇒ 6.\(\frac{dx}{dt}\) = 2x.\(\frac{dx}{dt}\) ⇒ x = 3
From curve, y = 9. Point is (3, 9)

The area of a circle (A) with radius (r) is given by.
Therefore, the rate of change of area (A) with respect to time (t) is given by,
[By chain rule]
It is given that.
Thus, when r = 8 cm,

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80 π cm²/s.

Note that
f '(x) = 3x² – 6x + 4
= 3(x² – 2x + 1) + 1
= 3(x – 1)² + 1 > 0, in every interval of R
Therefore, the function f is increasing on R.