We have,\(y=\sin ^{-1}(3 x+2)\) 
On differentiatingboth sides w.r.t.x, we get 
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-(3 x+2)^{2}}} \frac{d}{d x}(3 x+2)\) 
\(\left[\because \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}\right]\) 
\(=\frac{3}{\sqrt{1-(3 x+2)^{2}}}\)

\(LHS={ tan }^{ -1 }\frac { 1 }{ 4 } +{ tan }^{ -1 }\frac { 2 }{ 9 } \)
\(={ tan }^{ -1 }\left[ \frac { \frac { 1 }{ 4 } +\frac { 2 }{ 9 } }{ 1-\frac { 1\times 2 }{ 4\times 9 } } \right] \)
\(\left[ { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \frac { 9+8 }{ 36 } }{ \frac { 36-2 }{ 36 } } \right] \)
\(={ tan }^{ -1 }\left( \frac { 17 }{ 34 } \right) ={ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(=\frac { 1 }{ 2 } \left( 2{ tan }^{ -1 }\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 2\times \frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 4 } } \right) \)
\(\because \left[ 2{ tan }^{ -1 }={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right] \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 1 }{ 3/4 } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)

Equivalence relations could be the following:
{(1, 1), (2, 2), (3, 3), (1,2), (2, 1)} and 
{(1, I), (2, 2), (3, 3), (1, 2), (1, 3), (2,1), (2, 3), (3, 1), (3,2)} 
So, only two equivalence relations.

For x = 0,
f(x) is continuous and
at x = 1, LHL = 1 and RHL = 5,
hence discontinuous and
at x = 2;
f(x) = 5 which is continuous

9

\(y+2x=7,-x=-2\Rightarrow x=2,y=3\)

We have, \(y=e^{\sin ^2 x}\left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\}\)
On putting \(x=\cos 2 \theta\) in \(\left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\}\), we get
\(2 \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}} =2 \tan ^{-1} \sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}}\)
\( =2 \tan ^{-1}(\tan \theta)=2 \theta=\cos ^{-1} x\)
Then, from Eq. (i), we get
\(y=e^{\sin ^2 x} \cos ^{-1} x\)
On taking log both sides, we get
\(\log y=\sin ^2 x+\log \left(\cos ^{-1} x\right)\)
On differentiating both sides w.r.t. x, we get
\(\frac{1}{y} \times \frac{d y}{d x}=2 \sin x \cos x+\frac{1}{\cos ^{-1} x} \times \frac{-1}{\sqrt{1-x^2}} \)
\(=\sin 2 x-\frac{1}{\cos ^{-1} x \sqrt{1-x^2}} \\ \)
\(\therefore \frac{d y}{d x}=e^{\sin ^2 x} \cos ^{-1} x\left[\sin 2 x-\frac{1}{\cos ^{-1} x \sqrt{1-x^2}}\right]\)

We have, \(f(x)=[x], 0<x<2\)
\(\therefore \text { LHL(at }  x=1)=\lim _{x \rightarrow 1^{-}} f(x) \)
\(= \lim _{x \rightarrow 1^{-}}[x]=\lim _{h \rightarrow 0}[1-h] \)
\(=\lim _{h \rightarrow 0} 0=0\)
RHL (at x = 1) \(=\lim _{x \rightarrow 1^{+}} f(x)\)
\(=\lim _{x \rightarrow 1^{+}}[x]\)
\(=\lim _{h \rightarrow 0}[1+h]=\lim _{h \rightarrow 0} 1=1\)
and f(1) = [1] = 1
Since, LHL \(\neq \mathrm{RHL}\)
\(\therefore f(x)\) is discontinuous at x = 1
Hence, f(x) is not differentiable at x = 1.

Area of the triangle = \(\frac { 1 }{ 2 } \left| \begin{matrix} { at }_{ 1 }^{ 2 } & { 2at }_{ 1 } & 1 \\ { at }_{ 2 }^{ 2 } & { 2at }_{ 2 } & 1 \\ { at }_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \(\frac { 1 }{ 2 } (a)(2a)\left| \begin{matrix} { t }_{ 1 }^{ 2 } & { t }_{ 1 } & 1 \\ { t }_{ 2 }^{ 2 } & { t }_{ 2 } & 1 \\ { t }_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \({ a }^{ 2 }\left| \begin{matrix} { t }_{ 1 }^{ 2 }-{ t }_{ 2 }^{ 2 } & { t }_{ 1 }-{ t }_{ 2 } & 0 \\ { t }_{ 2 }^{ 2 }-{ t }_{ 3 }^{ 2 } & { t }_{ 2 }-{ t }_{ 3 } & 0 \\ { t }_{ 3 }^{ 2 } & { t }_{ 3 } & 0 \end{matrix} \right| \)
= \(a^2(t_2-t_2(t_2-t_3)\left| \begin{matrix} { t }_{ 1 }+{ t }_{ 2 } & 1 & 0 \\ { t }_{ 2 }+{ t }_{ 3 } & 1 & 0 \\ t_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \(a^2(t_1-t_2)(t_2-t_3)(1)\left| \begin{matrix} { t }_{ 1 }+{ t }_{ 2 } & 1 \\ { t }_{ 2 }+{ t }_{ 3 } & 1 \end{matrix} \right| \)
= \(a^2(t_1-t_2)(t_2-t_3)(t_1+t_2-t_2-t_3)\)
= \(a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)\)

The given system of equations can be expressed in the matrix form as AX=B
Where \(A=\begin{bmatrix}2&5\\3&2 \end{bmatrix}, X=\begin{bmatrix}x\\y \end{bmatrix}and \ B=\begin{bmatrix}1\\7 \end{bmatrix}\)
Now \(|A|=\begin{vmatrix} 2&5\\3&2\end{vmatrix}=4-15=-11\neq0\)
A is non-singular matrix \(\Rightarrow\) A^-1exists
From (1), A^-1(AX) = A^-1B \(\Rightarrow\)(A^-1A)X = A^-1B
\(\Rightarrow\)IX = A^-1 B \(\Rightarrow\)X = A^-1B
Now A₁₁ = 2; A₁₂=-3; A₂₁ = -5 and A₂₂ = 2
\(adj\ A=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22} \end{bmatrix}=\begin{bmatrix} 2&-3\\-5&2\end{bmatrix}\)
\(=\begin{bmatrix} 2&-5\\-3&2\end{bmatrix}\)
\(A^{-1}={adj\ A\over|A|}={1\over-11}\begin{bmatrix} 2&-5\\-3&2\end{bmatrix}\)
\(=\begin{bmatrix}-2/11&5/11\\3/11&-2/11 \end{bmatrix}\)
\(\therefore From(2), \begin{bmatrix}x\\y \end{bmatrix}=\begin{bmatrix} -2/11&5/11\\ 3/11&-2/11\end{bmatrix}\begin{bmatrix} 1\\7\end{bmatrix}\)
\(=\begin{bmatrix} {-2\over11}+{35\over11}\\{3\over11}-{14\over11}\end{bmatrix}=\begin{bmatrix}3\\-1 \end{bmatrix}\)
Hence, x = 3, y = -1

\(We\quad have:\quad x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]\)
\(\Rightarrow \left[ \begin{matrix} 2x \\ 3x \end{matrix} \right] +\left[ \begin{matrix} -y \\ y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]\)
\(\Rightarrow \left[ \begin{matrix} 2x-y \\ 3x+y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right] .\)
Equating corresponding elements:
2x - y = 10...(1)
and 3x + y = 5....(2)
Adding (1) and (2), 5x = 15 \(\Rightarrow \) x = 3.
Putting in (1), 2, (3) - y = 10 \(\Rightarrow\) y = 6-10 = -4.
Hence,  x = 3 and y = -4.

(i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 x 4.
(ii) Since the order of the matrix is 3 x 4, there are 3 x 4 = 12 elements in it.
(iii) \({ a }_{ 13 }=19,{ a }_{ 21 }=35,{ a }_{ 33 }=-5,{ a }_{ 24 }=12,{ a }_{ 23 }=\frac { 5 }{ 2 } .\)

Let \(y=(\sin x)^x+\sin ^{-1} \sqrt{x}\)
Let \(u=(\sin x)^x \ v=\sin ^{-1} \sqrt{x}\)
y=u+v
Differentiating both sides w.r.t. x.
\( \frac{d y}{d x}=\frac{d(u+v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)

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 \(y=\left[x+\sqrt{x^{2}+a^{2}}\right]^{n}\) 
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=n\left[x+\sqrt{x^{2}+a^{2}}\right]^{n-1} \cdot \frac{d}{d x}\left[x+\sqrt{x^{2}+a^{2}}\right]\) 
[by chain rule of derivative]
\(=n\left[x+\sqrt{x^{2}+a^{2}}\right]^{n-1}\left[1+\frac{1}{2 \sqrt{x^{2}+a^{2}}} \frac{d}{d x}\left(x^{2}+a^{2}\right)\right]\) 
[by chain nile of derivative]
\(=n\left[x+\sqrt{x^{2}+a^{2}}\right]^{n-1}\left[1+\frac{1}{2 \sqrt{x^{2}+a^{2}}} \cdot 2 x\right]\) 
\(=n\left[x+\sqrt{x^{2}+a^{2}}\right]^{n-1}\left[\frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}}}\right]\) 
\(=\frac{n\left[x+\sqrt{x^{2}+a^{2}}\right]^{n}}{\sqrt{x^{2}+a^{2}}}\) 
\(\Rightarrow \frac{d y}{d x}=\frac{n y}{\sqrt{x^{2}+a^{2}}} \quad\left[\because y=\left(x+\sqrt{x^{2}+a^{2}}\right)^{n}\right][1]\) 
Hence proved.

\({ \left( { tan }^{ -1 }x \right) }^{ 2 }+{ \left( { cot }^{ -1 }x \right) }^{ 2 }=\frac { 5{ \pi }^{ 2 } }{ 8 } \)
\(\Rightarrow { \left( { tan }^{ -1 }x \right) }^{ 2 }+{ \left( \frac { \pi }{ 2 } -{ tan }^{ -1 }x \right) }^{ 2 }=\frac { 5\pi ^{ 2 } }{ 8 } \)
\(\Rightarrow 2{ \left( { tan }^{ -1 }x \right) }^{ 2 }-\pi { tan }^{ -1 }x+\frac { { \pi }^{ 2 } }{ 4 } -\frac { { 5\pi }^{ 2 } }{ 8 } =0\)
\(\Rightarrow 2{ \left( { tan }^{ -1 }x \right) }^{ 2 }-\pi { tan }^{ -1 }x-\frac { 3{ \pi }^{ 2 } }{ 8 } =0\)
\(\Rightarrow { tan }^{ -1 }x=\frac { \pi \pm \sqrt { { \pi }^{ 2 }+3{ \pi }^{ 2 } } }{ 4 } \)
\(\Rightarrow { tan }^{ -1 }x=\frac { 3\pi }{ 4 } ,\frac { -\pi }{ 4 } \)
\(\Rightarrow x=-1\)

\(LHS=2{ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(={ tan }^{ -1 }\frac { 2\times \frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 4 } } +tan^{ -1 }\frac { 1 }{ 7 } \)
\(={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(={ tan }^{ -1 }\frac { \frac { 4 }{ 3 } +\frac { 1 }{ 7 } }{ 1-\frac { 4 }{ 3 } .\frac { 1 }{ 7 } } \)
\(={ tan }^{ -1 }\frac { 31 }{ 17 } \)
Using the above triangle, we get
\({ tan }^{ -1 }\frac { 31 }{ 17 } ={ sin }^{ -1 }\left( \frac { 31 }{ 25\sqrt { 2 } } \right) \)
=RHS
Hence   LHS = RHS

Let Rs. 3x and Rs. 4x bee the month;y income of Aryan and Babban respectively.
Let Rs. 5y7 and Rs. 7y be the monthly expenditure of Aryan and Babban respectively.
By the equation 3x-5y = 15000
and  4x-7y = 15000
where \(A=\begin{bmatrix}3&-5\\4&-7 \end{bmatrix}, X=\begin{bmatrix} x\\y\end{bmatrix},B=\begin{bmatrix} 15000\\15000\end{bmatrix}\)
Now \(|A|=\begin{bmatrix}3&-5\\4&-7 \end{bmatrix}=-21+20=-1\neq0\)
⇒ A^-1exists.
\(\therefore\ adj\ A=\begin{bmatrix} -7&-4\\5&3\end{bmatrix}=\begin{bmatrix} -7&5\\-4&3\end{bmatrix}\)
\(\therefore\ A^{-1}={1\over-1}\begin{bmatrix}-7&5\\-4&3 \end{bmatrix}=\begin{bmatrix}7&-5\\4&-3 \end{bmatrix}\)
From (1), \(X=A^{-1}B\Rightarrow X=\begin{bmatrix}7&-5\\4&-3 \end{bmatrix}=\begin{bmatrix}15000\\15000\end{bmatrix}\)
\(\Rightarrow\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 105000-75000\\60000-450000\end{bmatrix}=\begin{bmatrix}30000\\15000 \end{bmatrix}\)
⇒ x = 30,000 and y = 15,000
Hence, monthly income of Aryan = 3 x 30000 = Rs. 90,000
and of Babban = 4 x 30000 = Rs. 1,20,000

(i) A² = AA
= \(\begin{vmatrix} 0&4&3\\1&-3&^-3\\-1&4&4\end{vmatrix}\begin{vmatrix}0&4&3\\1&-3&-3\\-1&4&4 \end{vmatrix}\)
= \(\begin{vmatrix}0+4-3&0-12+12&0-12+12\\0-3+3&4+9-12&3+9-12\\-0+4-4&-4-12+16&-3-12+16 \end{vmatrix}\)
= \(\begin{vmatrix}1&0&0\\0&1&0\\0&0&1 \end{vmatrix}=I\)
(ii) \(A^2=I\Rightarrow A^{-1}A^2=A^{-1}I\)
\(\Rightarrow A^{-1}(A)=A^{-1}I\)
\(\Rightarrow (A^{-1}A)A=A^{-1}I\Rightarrow IA=A^{-1}I\)
\(\Rightarrow A=A^{-1}\)
Hence, \(A^{-1}=A\)

Let r, l and h denote respectively the radius, slant height and height of the cone at any time t. Then,
\(\begin{aligned} l^{2} &=r^{2}+h^{2} \\ \Rightarrow & 2 l \frac{d l}{d t}=2 r \frac{d r}{d t}+2 h \frac{d h}{d t} \\ \Rightarrow & l \frac{d l}{d t}=r \frac{d r}{d t}+h \frac{d h}{d t} \end{aligned}\) 
\(\begin{array}{ll} \Rightarrow & l \frac{d l}{d t}=7 \times 3+24 \times(-4) \quad\left[\because \frac{d h}{d t}=-4 \text { and } \frac{d r}{d t}=3\right] \\ \Rightarrow & l \frac{d l}{d t}=-75 \end{array}\) 
When r = 7 and h = 24, then we have
\( l^{2} =7^{2}+24^{2} \\ \Rightarrow \quad l =25 \\ \therefore \quad l \frac{d l}{d t} =-75 \Rightarrow \frac{d l}{d t}=-3 \) 
Let S denote the lateral surface area, then 
\( S=\pi r l \\ \Rightarrow \frac{d S}{d t}=\pi\left(\frac{d r}{d t} l+r \frac{d l}{d t}\right)=\pi(3 \times 25+7 \times(-3)) \\ \Rightarrow 54 \pi \mathrm{cm}^{2} / \mathrm{min} \)

Let the side of an equilateral triangle be x cm.
\(\therefore\) Area of equilateral triangle \(A=\frac{\sqrt{3}}{4} x^{2}\) 
Also,\(\frac{d x}{d t}=4 \mathrm{~cm} / \mathrm{s}\) 
On differentiating Eq. (i) w.r.t. t, we get
\(\frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \cdot \frac{d x}{d t}\) 
\( =\frac{\sqrt{3}}{4} \cdot 2 \cdot 5 \cdot 4 {\left[\because x=5 \text { and } \frac{d x}{d t}=4\right]} \\ = 10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{s} \)

Given that \(x=e^{x / y}\) 
Taking log on both sides, we get
\( \log x =\frac{x}{y} \cdot \log e=\frac{x}{y} \)
\(\Rightarrow=y \log x \) 
Now, differentiating w.r.t. x, we get
\(\Rightarrow 1=y \cdot \frac{1}{x}+\log x \cdot \frac{d y}{d x}\)
\( \Rightarrow \frac{d y}{d x}=\frac{x-y}{x \log x} \) 

We have \(f(x)=\left\{\begin{array}{cl} \frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\ 3, & \text { if } x=\frac{\pi}{2} \end{array}\right.\) 
f(x) is continuous at \(x=\frac{\pi}{2}\) 
\(\therefore \ \lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{k \cos x}{\pi-2 x}=3 \)
\(\Rightarrow \lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=3 \)
\(\Rightarrow \lim _{h \rightarrow 0} \frac{k \sin h}{2 h}=3 \)
\(\therefore \frac{k}{2}=3 \Rightarrow k=6 \)

Clearly,
\( f(a) =\left|\begin{array}{ccc} 0 & 0 & a-b \\ 2 a & 0 & a-c \\ a+b & a+c & 0 \end{array}\right| \)
\(=[(a-b)\{2 a \cdot(a+c)\}] \neq 0 \) 
\( \therefore \ f(b) =\left|\begin{array}{ccc} 0 & b-a & 0 \\ b+a & 0 & b-c \\ 2 b & b+c & 0 \end{array}\right|\)
\( =-(b-a)[2 b(b-c)] \)
\( =-2 b(b-a)(b-c) \neq 0 \) 
 

By definition of invertible matrix

Total number of functions = (n(B)^n(A) = 2ⁿ. Clearly a function will not be onto if all elements of A map to either a or b.

A function f : A ➝ B is said to be onto, if for every b ∈ B, there exists an element a in A such that f(a) = b

As x^o = \(\frac { \pi }{ 180 } { x }^{ c }\)
\(\therefore \frac { d }{ dx } (cot{ x }^{ o })=\)\(\frac { d }{ dx } \left( cot\frac { \pi }{ 180 } x \right) \)
\(=-\frac { \pi }{ 180 } { cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }{ x }^{ 0 }\)

(i) (b) \(\ln \triangle A B D\)
\(\tan \alpha  =\frac{B D}{A D}\)
\( =\frac{30}{30 \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow  \alpha  =30^{\circ}\)
\(\therefore  \sin \alpha =\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow  \alpha  =\sin ^{-1}\left(\frac{1}{2}\right)\)
(ii) (c) \(\cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \alpha =\cos ^{-1}\left(\frac{\sqrt{3}}{2})\right.\)
(iii) (d) \(\ln \triangle B DC\)
\(\tan \beta  =\frac{(BD)}{(C D)}\)
\(=\frac{30}{10 \sqrt{3}}=\sqrt{3} \)
\(\Rightarrow \beta =\tan ^{-1}(\sqrt{3})\)
(iv) \( \text { (c) Since, } \alpha=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \text {, } \)
\(\beta=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)
\(\therefore \quad \angle A B C=\pi-\left(\frac{\pi}{6}+\frac{\pi}{3}\right)\)
\( =\pi-\frac{\pi}{2}=\frac{\pi}{2}\)
(v) (c) We know that domain and range of \(\cos ^{-1} x\) are [-1,1] and \([0, \pi]\) respectively.

Combined sales in September and October for each farmer in each variety is given by

(i) (c) : Combined sales of Masoor in September and October for farmer Balwan Singh = Rs. 40000
(ii) (d) : Combined sales of Urad in September and October for farmer Shyam = Rs. 15000
(iii) (a) : Change in sales from September to October is given by
 
\(\therefore\) Decrease in sales of Mung from September to October for farmer Shyam = Rs. 24000.
(iv) (b) : Required profit is given by
\(2 \% \text { of } B=\frac{2}{100} \times B=0.02 \times B\) 

 
thus,in October Shyam receives Rs. 100, Rs. 200 and Rs. 120 as profit in the sale of each variety of pulses, respectively and Balwan Singh receives a profit of Rs. 400, Rs. 200 and Rs. 200 in the sale of each variety of pulses respectively. 

(i) (a) : For f(x) to be defined \(x-2 \neq 0\) i.e.,\(x \neq 2\) 
\(\therefore\) Domain of f = R - {2}
(ii) (b) : Let y =J(x), then \(y=\frac{x-1}{x-2}\) 
\(\Rightarrow x y-2 y=x-1 \Rightarrow x y-x=2 y-1 \Rightarrow x=\frac{2 y-1}{y-1}\) 
Since, \(x \in R-\{2\}\),therefore \(y \neq 1\)
Hence, range of f = R-{1} 
(iii) (d): We have,g(x) = 2f(x) - 1
\(=2\left(\frac{x-1}{x-2}\right)-1=\frac{2 x-2-x+2}{x-2}=\frac{x}{x-2}\)
(iv) (a) : We have, \(g(x)=\frac{x}{x-2}\) ,
Let  \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{x_{1}}{x_{1}-2}=\frac{x_{2}}{x_{2}-2}\) 
 \(\Rightarrow x_{1} x_{2}-2 x_{1}=x_{1} x_{2}-2 x_{2} \Rightarrow 2 x_{1}=2 x_{2} \Rightarrow x_{1}=x_{2}\) 
Thus, \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\) 
Hence, g(x) is one-one.
(v) (c)