The given function is f (x) = | x – 1|, x R
It is known that a function f is differentiable at a point x = c in its domain if both are finite and equal
To check the differentiability of the given function at x = 1,
consider the left hand limit of f at x = 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

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)

We have \(F(\alpha)=\begin{vmatrix}cos\alpha&-sin\alpha&0\\sin\alpha&cos\alpha&0\\0&0&1 \end{vmatrix}\)
\(F(-\alpha)=\begin{bmatrix}cos(-\alpha) &-sin(-\alpha)&0\\sin(-\alpha)&cos{-\alpha}&0\\0&0&1\end{bmatrix}\)
\(=\begin{bmatrix} cos\alpha&sin\alpha&0\\-sin\alpha&cos\alpha&0\\0&0&1\end{bmatrix}\)
\(\therefore F(\alpha).F(-\alpha)=\begin{bmatrix}cos\alpha&-sin\alpha&0\\sin\alpha&cos\alpha&0\\0&0&1 \end{bmatrix}\begin{bmatrix}cos\alpha&sin\alpha&0\\-sin\alpha &cos\alpha &0\\0&0&1\end{bmatrix}\)
\(=\begin{bmatrix} cos^2\alpha+sin^2\alpha&cos\alpha sin\alpha-sin\alpha cos\alpha&0+0+0\\sin\alpha cos\alpha-sin\alpha cos\alpha&sin^2\alpha+cos^2\alpha&0+0+0\\0+0+0&0+0+0&0+0+0\end{bmatrix}\)
\(=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3\)
\(\Rightarrow [F(\alpha)]^{-1}=F(\alpha)...(1)\)
Similarly \([G(\beta)]^{-1}=G(-\beta).....(2)\)
Now \([F(\alpha)G(\beta)]^{-1}=[G(\beta)]^{-1}[F(\alpha)]^{-1}=G(\beta)F(-\beta).\)
Hence the result

\(\frac { 2{ cos }^{ 3 }t+3\ \ cos\ t }{ 2{ sin }^{ 3 }t+3 \ \ sin \ t } \)

Here, \(f(x)=\left\{\begin{array}{cl} \lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0 \end{array}\right.\) 
At \(x=0, \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right)\) 
\(\therefore \mathrm{LHL}=\lim _{h \rightarrow 0} \lambda\left[(0-h)^{2}-2(0-h)\right]=\lim _{h \rightarrow 0}\left[\lambda\left(h^{2}+2 h\right)\right]=0\) 
\(\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(4 x+1)\) 
\(\therefore \mathrm{RHL}=\lim _{h \rightarrow 0}[4(0+h)+1]=\lim _{h \rightarrow 0}[4 h+1]=0+1=1\)
\(\text { [put } x=0+h \text { ; when } x \rightarrow 0^{+} \text {, then } \left.h \rightarrow 0\right] \) 
\(\therefore \mathrm{LHL} \neq \mathrm{RHL}\) 
Thus, f(x) is not continuous at x = 0 for any value of λ.
At x = 1,
 \( \mathrm{LHL} =\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(4 x+1) \)
\(\therefore \mathrm{LHL} =\lim _{h \rightarrow 0}[4(1-h)+1]=\lim _{h \rightarrow 0}[5-4 h]=5-0=5 \)
[put x=1−h; when x→1−,then h→0]   
\( \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x+1) \)
\(\therefore \lim _{h \rightarrow 0}[4(1+h)+1]=\lim _{h \rightarrow 0}(5+4 h)=5+0=5 \)
[put x=1+h; when x→1, then h→0]   
Also, f(1)=4×1+1=5
\([\because f(x)=4 x+1]\)
Thus f(x) is continuous at x=1 for all values of λ.

A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
We know that any number (x) is divisible by itself.
⇒ (x, x) ∈R
∴ R is reflexive.
Now,
(2, 4) ∈R [as 4 is divisible by 2]
But,
(4, 2) ∉ R. [as 2 is not divisible by 4]
∴ R is not symmetric.
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.
∴ z is divisible by x.
⇒ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.

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.

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
x + 3 = 0, z + 4 = 6, 2y – 7 = 3y – 2
a – 1 = – 3, 0 = 2c + 2 b – 3 = 2b + 4,
Simplifying, we get
a = – 2, b = – 7, c = – 1, x = – 3, y = –5, z = 2

\(\frac{d y}{d x}=e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}\)

For x = -3 function is not defined. Hence, not continuous for x ∈ R.

\(We\quad have\quad :\quad x={ 2at }^{ 2 },y={ at }^{ 4 }\)
\(\frac { dx }{ dt } =4at,\frac { dy }{ dt } ={ 4at }^{ 3 }\)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { { 4at }^{ 3 } }{ 4at } ={ t }^{ 2 }\)

We have
\(f(x)=\log \sin x \)
\(\therefore f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x \)
In interval \( \left(0, \frac{\pi}{2}\right), f^{\prime}(x)=\cot x>0\)
\(\therefore f \text { is strictly increasing in }\left(0, \frac{\pi}{2}\right) \text { . }\)
In interval \( \left(\frac{\pi}{2}, \pi\right), f^{\prime}(x)=\cot x<0\)
\(\therefore f \text { is strictly decreasing in }\left(\frac{\pi}{2}, \pi\right) \text { . }\)

\(We\quad have:\quad y=3{ e }^{ 2x }+{ 2e }^{ 3x }\)
\(\frac { dy }{ dx } =6{ e }^{ 2x }+6{ e }^{ 3x }=6({ e }^{ 2x }+{ e }^{ 3x })\)
\(and\quad \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =12{ e }^{ 2x }+18{ e }^{ 3x }=6(2{ e }^{ 2x }+3{ e }^{ 3x })\)
\(Now\quad \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -5\frac { dy }{ { dx }^{ 2 } } +6y\)
\(=6(2{ e }^{ 2x }+3{ e }^{ 3x })-5.6({ e }^{ 2x }+{ e }^{ 3x })+6(3{ e }^{ 2x }+{ 2e }^{ 3x })\)

We differentiate the relationship directly with respect to x, i.e.,
\(\frac{d y}{d x}+\frac{d}{d x}(\sin y)=\frac{d}{d x}(\cos x)\)
which implies using chain rule
\(\frac{d y}{d x}+\cos y \cdot \frac{d y}{d x}=-\sin x\)
This gives, \( \frac { dy }{ dx } =-\frac { sin\quad x }{ 1+cos\quad y } \)
\(where\ y\neq (2n+1)\pi ,\ n\epsilon I.\)

Area of \(\Delta ABC={1\over 2}\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|\)
\(={1\over2}\begin{vmatrix} a&a+b&1\\b&c+a&1\\c&a+b&1 \end{vmatrix}\)
\(={1\over2}\begin{vmatrix}a+b+C&b+c&1\\a+b+c&c+a&1\\a+b+c&a+b&1 \end{vmatrix}\)
\({1\over2}(a+b+c)\begin{vmatrix}1&b+c&1\\1&c+a&1\\1&a+b&1 \end{vmatrix}\)
\(={1\over2}(a+b+c)(0)=0\)

Let  \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =y\), Then \(\cot y=\frac{-1}{\sqrt{3}}=-\cot \left(\frac{\pi}{3}\right)=\cot \left(\pi-\frac{\pi}{3}\right)=\cot \left(\frac{2 \pi}{3}\right)\)
We know that the range of principal value branch of cot^–1 is (0, π) and \(\cot \left(\frac{2 \pi}{3}\right)=\frac{-1}{\sqrt{3}}\)
Hence, principal value of \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =\frac { 2\pi }{ 3 } .\)