The given function is f(x) = sin x + cos x.
\(\therefore f^{\prime}(x)=\cos x-\sin x\)
\(\text { Now, }f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}\)
Then, we evaluate the value of f at critical point and at the end points\(x=\frac{\pi}{4}\) of the interval [0, π].
\(f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \)
\(f(0)=\sin 0+\cos 0=0+1=1 \)
\(f(\pi)=\sin \pi+\cos \pi=0-1=-1 \)
Hence, we can conclude that the absolute maximum value of f on [0, π] is \(\sqrt{2}\) occurring at \(x=\frac{\pi}{4}\)and the absolute minimum value of f on [0, π] is −1 occurring at x = π.

Let r and h be the radius and height of the cylinder respectively.
Then, the surface area (S) of the cylinder is given by
\(S= 2 \pi r^{2}+2 \pi r h \)
\(\Rightarrow h =\frac{S-2 \pi r^{2}}{2 \pi r} \)
\(=\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r \)
Let V be the volume of the cylinder. Then,
\(V=\pi r^{2} h=\pi r^{2}\left[\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\right]=\frac{S r}{2}-\pi r^{3} \)
\(\text { Then, } \frac{d V}{d r}=\frac{S}{2}-3 \pi r^{2}, \frac{d^{2} V}{d r^{2}}=-6 \pi r \)
\(\text { Now, } \frac{d V}{d r}=0 \Rightarrow \frac{S}{2}=3 \pi r^{2} \Rightarrow r^{2}=\frac{S}{6 \pi} \)
\(\text { When } r^{2}=\frac{S}{6 \pi}, \text { then } \frac{d^{2} V}{d r^{2}}=-6 \pi\left(\sqrt{\frac{S}{6 \pi}}\right)<0 \)
∴ By second derivative test, the volume is the maximum when \(r^{2}=\frac{S}{6 \pi}\)
\(\text { Now, when } r^{2}=\frac{S}{6 \pi} \text { , then } h=\frac{6 \pi r^{2}}{2 \pi}\left(\frac{1}{r}\right)-r=3 r-r=2 r \text { . }\)
Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter

We have
\( f(x) =\frac{3}{10} x^4-\frac{4}{5} x^3-3 x^2+\frac{36}{5} x+11 \)
\(f^{\prime}(x) =\frac{3}{10}\left(4 x^3\right)-\frac{4}{5}\left(3 x^2\right)-3(2 x)+\frac{36}{5} \)
\(=\frac{6}{5}(x-1)(x+2)(x-3) \)(on simplification)

Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3. The points x = 1, – 2, and 3 divide the real line into four disjoint intervals namely, (– ∞, – 2), (– 2, 1), (1, 3) and (3, ∞)
Consider the interval (– ∞, – 2), i.e., when – ∞ < x < – 2. In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0.
(In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1) (– 6) < 0)
Therefore, f ′(x) < 0 when – ∞ < x < – 2. Thus, the function f is decreasing in (– ∞, – 2).
Consider the interval (– 2, 1), i.e., when – 2 < x < 1. In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0
(In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3) = 6 > 0)
So f ′(x) > 0 when – 2 < x < 1.
Thus, f is increasing in (– 2, 1).
Now consider the interval (1, 3), i.e., when 1 < x < 3. In this case, we have x – 1 > 0, x + 2 > 0 and x – 3 < 0.
So, f ′(x) < 0 when 1 < x < 3.
Thus, f is decreasing in (1, 3).
Finally, consider the interval (3, ∞), i.e., when x > 3. In this case, we have x – 1 > 0, x + 2 > 0 and x – 3 > 0. So f ′(x) > 0 when x > 3.
Thus, f is increasing in the interval (3, ∞).

We have
\(y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta \)
\(\therefore \frac{d y}{d x} =\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}-1 \)
\(\text { Now, } \frac{d y}{d x}=0 .\)
\(\Rightarrow \frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}=1\)
\(\Rightarrow 8 \cos \theta+4=4+\cos ^{2} \theta+4 \cos \theta \)
\(\Rightarrow \cos ^{2} \theta-4 \cos \theta=0\)
\(\Rightarrow \cos \theta(\cos \theta-4)=0 \)
\(\Rightarrow \cos \theta=0 \text { or } \cos \theta=4\)
\(\text { Since } \cos \theta \neq 4, \cos \theta=0\)
\(\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}\)
\(\text { Now, }\frac{d y}{d x}=\frac{8 \cos \theta+4-\left(4+\cos ^{2} \theta+4 \cos \theta\right)}{(2+\cos \theta)^{2}}=\frac{4 \cos \theta-\cos ^{2} \theta}{(2+\cos \theta)^{2}}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}\)
\(\text { In interval }\left(0, \frac{\pi}{2}\right), \text { we have } \cos \theta>0 . \text { Also, } 4>\cos \theta \Rightarrow 4-\cos \theta>0 \text { . }\)
\(\therefore \cos \theta(4-\cos \theta)>0 \text { and also }(2+\cos \theta)^{2}>0 \)
\(\Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}>0 \)
\(\Rightarrow \frac{d y}{d x}>0 \)
 Therefore, y is strictly increasing in interval \( \left(0, \frac{\pi}{2}\right) \text { . }\)
Also, the given function is continuous at  \( x=0 \text { and } x=\frac{\pi}{2} \text { . }\)
Hence, y is increasing in interval  \( \left[0, \frac{\pi}{2}\right] \text { . }\)

Maximum value = 2; Minimum value = -10

\(\frac{1}{2 \pi} \mathrm{cm} / \mathrm{s}\)

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.

a=-2

f'(x) 3x²- 6x + 3 = 3(x² - 2x + 1)
= 3 (x - 1)² > 0
Hence, function is increasing in R.

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 \( \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)

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 \((-\infty,-2) \cup(1,3)\)

Let ABCD be a rectangle inscribed in a given circle with centre at 0 and radius a.
Let AB = 2x and BC = 2y

Then, OA² = OM² + AM² 
\(\Rightarrow a^2=y^2+x^2\)
\(\Rightarrow y=\sqrt{a^2-x^2}\)
Let A be the area of the rectangle.
\(\therefore A=4xy=4x\sqrt{x^2-x^2}\)
\(\Rightarrow \ \ \frac{dA}{dx}=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}\)
For maximum or minimum value of A,
\(\frac{dA}{dx}=0\)
\(=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}=0\Rightarrow\ \ x=\frac{a}{\sqrt{2}}\)
Now, \(\frac{d^2A}{dx^2}=4\frac{d}{dx}\{(a^2-2x^2)(a^2-x^2)^{-1/2}\}\)
\(\Rightarrow \frac{d^2A}{dx^2}=4[-4x(a^2-x^2)^{-1/2}+(a^2-2x^2)\times(-1/2)(a^2-x^2)^(-3/2)(-2x)]\)
\(=[\frac{-4x}{\sqrt{a^2-x^2}}+\frac{x(a^2-2x^2)}{(a^2-x^2)^{3/2}}]\)
\(\therefore\ (\frac{d^2A}{dx^2})_{x=\frac{a}{\sqrt2}}=-16<0\)
Thus A is maximum when \(x=\frac{a}{\sqrt2}\)
putting \(x=\frac{a}{\sqrt2}\) in (i) \(y=\frac{a}{\sqrt2}\)
Therefore \(x=y=\frac{a}{\sqrt2}\)
Hence area is maximum when x = y ⇒ 2x = 2y
i.e., the rectangle is a square.

We have,
\( f(x)=4 \sin ^{3} x-6 \sin ^{2} x+12 \sin x+100 \\ \therefore f^{\prime}(x) =12 \sin ^{2} x \cdot \cos x-12 \sin x \cdot \cos x+12 \cos x \\ =12\left[\sin ^{2} x \cdot \cos x-\sin x \cdot \cos x+\cos x\right] \\ =12 \cos x\left[\sin ^{2} x-\sin x+1\right] \\ \Rightarrow f^{\prime}(x) =12 \cos x\left[\sin ^{2} x+(1-\sin x)\right] \\ \because \quad 1-\sin x \geq 0 \text { and } \sin ^{2} x \geq 0 \\ \therefore \sin ^{2} x +1+\sin x \geq 0 \) 
Hence,\(f^{\prime}(x)>0\) when \(x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text { and } f^{\prime}(x)<0\) 
when \(\cos x<0 \text { i.e., } x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\) 
Hence, f(x) is decreasing when \(x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\) 
Since,\(\left(\frac{\pi}{2}, \pi\right) \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\) 
Hence, f(x) is decreasing in \(\left(\frac{\pi}{2}, \pi\right)\) 

(i) (b) : Since, side of square is of length 20 cm therefore \(\begin{equation} x \in(0,10) \end{equation}\) .
(ii) (a) : Clearly, height of open box = x cm 
Length of open box = 20 - 2x
and width of open box = 20 - 2x
\(\therefore\) Volume (V) of the open box 
= x x (20 - 2x) x (20 - 2x
(iii) (d) : We have, V = x(20 - 2X)²
\(\begin{equation} \therefore \frac{d V}{d x}=x \cdot 2(20-2 x)(-2)+(20-2 x)^{2} \end{equation}\) 
= (20 - 2x)( -4x + 20 - 2x) = (20 - 2x)(20 - 6x)
Now, \(\begin{equation} \frac{d V}{d x}=0 \Rightarrow 20-2 x=0 \text { or } 20-6 x=0 \end{equation}\) 
\(\begin{equation} \Rightarrow x=10 \text { or } \frac{10}{3} \end{equation}\) 
(iv) (c) : We have, V = x(20 - 2X)²
and \(\begin{equation} \frac{d V}{d x}=(20-2 x)(20-6 x) \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d^{2} V}{d x^{2}}=(20-2 x)(-6)+(20-6 x)(-2) \end{equation}\) 
= (-2)[60 - 6x + 20 - 6x] = (-2)[80 - 12x] = 24x - 160
For \(\begin{equation} x=\frac{10}{3}, \frac{d^{2} V}{d x^{2}}<0 \end{equation}\) 
and for \(\begin{equation} x=10, \frac{d^{2} V}{d x^{2}}>0 \end{equation}\) 
So, volume will be maximum when \(\begin{equation} x=\frac{10}{3} \end{equation}\) .
(v) (d) : We have, V = x(20 - 2x)²,which will be maximum when \(\begin{equation} x=\frac{10}{3} \end{equation}\) .
\(\begin{equation} \therefore \ \text { Maximum volume }=\frac{10}{3}\left(20-2 \times \frac{10}{3}\right)^{2} \end{equation}\) 
\(\begin{equation} =\frac{10}{3} \times \frac{40}{3} \times \frac{40}{3}=\frac{16000}{27} \mathrm{~cm}^{3} \end{equation}\)

(c) Assertion is correct, Reason is incorrect

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.