k = 4

Let \(A=\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\) then
\(|A|=\left|\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right|=\cos ^{2} \theta+\sin ^{2} \theta=1 \neq 0 \)
\(\therefore A^{-1} \text {exists. } \)
\(\text { Also, } \operatorname{adj}(A)=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \) 
\(\left[\begin{array}{ll} \because \text { adj } \end{array}\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]\right]\) 
Now,\(A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=\left[\begin{array}{cr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\)

\(\left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right] ;\left[\begin{array}{ccc} 0 & -1 & 3 \\ 1 & 0 & 4 \\ -3 & -4 & 0 \end{array}\right]\)

±2√2

\(f^{\prime}(x)=\frac{1}{\cos x} \cdot(-\sin x)=-\tan x, \tan x>0 \text { for }\left(0, \frac{\pi}{2}\right)\)
f′(x) < 0. Hence, function is strictly decreasing

For reflexive : As (1, 1), (2, 2), (3, 3) ∈ R. Hence, reflexive
For Symmetric: (1, 2) ∈ R but (2, 1) ∉ R. Hence, not symmetric
For transitive: (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.Hence, not transitive
Hence, R is not an equivalence relation

Let \(\theta\) be the semi-vertical angle of the cone.

It is clear that \(\theta \in\left(0, \frac{\pi}{2}\right)\).
Let r, h and l be the radius, height and the slant height of the cone, respectively.
Since, slant height of the cone is given, so consider it as constant.
Now, in \(\triangle A B C, r=l \sin \theta\) and \(h=l \cos \theta\)
Let V be the volume of the cone.
Then, \(V=\frac{\pi}{3} r^2 h \Rightarrow V=\frac{1}{3} \pi\left(l^2 \sin ^2 \theta\right)(l \cos \theta)\)
\(\Rightarrow \quad V=\frac{1}{3} \pi l^3 \sin ^2 \theta \cos \theta\)
On differentiating both sides w.r.t. \(\theta\) two times, we get
\(\frac{d V}{d \theta} =\frac{l^3 \pi}{3}\left[\sin ^2 \theta(-\sin \theta)+\cos \theta(2 \sin \theta \cos \theta)\right] \)
\(=\frac{l^3 \pi}{3}\left(-\sin ^3 \theta+2 \sin \theta \cos ^2 \theta\right)\)
\( \text { and } \frac{d^2 V}{d \theta^2}=\frac{l^3 \pi}{3}\left(-3 \sin ^2 \theta \cos \theta+2 \cos ^3 \theta\right. - 4 \sin^2 \theta \cos \theta)\)
\(\Rightarrow \frac{d^2 V}{d \theta^2}=\frac{l^3 \pi}{3}\left(2 \cos ^3 \theta-7 \sin ^2 \theta \cos \theta\right)\)
For maxima or minima, put\(\frac{d V}{d \theta}=0\)
\(\Rightarrow \quad \sin ^3 \theta=2 \sin \theta \cos ^2 \theta \Rightarrow \tan ^2 \theta=2 \)
\( \Rightarrow \quad \tan \theta=\sqrt{2} \Rightarrow \theta=\tan ^{-1} \sqrt{2}\)
Now, when \(\theta=\tan ^{-1} \sqrt{2}\), then \(\tan ^2 \theta=2\)
\(\Rightarrow \quad \sin ^2 \theta=2 \cos ^2 \theta\)
Now, we have \(\frac{d^2 V}{d \theta^2} =\frac{l^3 \pi}{3}\left(2 \cos ^3 \theta-14 \cos ^3 \theta\right)\)
\(=-4 \pi l^3 \cos ^3 \theta<0, \text { for } \theta \in\left(0, \frac{\pi}{2}\right)\)
\(\therefore V\) is maximum, when \(\theta=\tan ^{-1} \sqrt{2}\) or
\(\theta  =\cos ^{-1} \frac{1}{\sqrt{3}} \)
\({[\because \cos \theta}  \left.=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+2}}=\frac{1}{\sqrt{3}}\right]\)
Hence, for given slant height, the semi-vertical angle of the cone of maximum volume is \(\cos ^{-1} \frac{1}{\sqrt{3}}\).

Let wire is cut at x m and made into a rectangle,
whose breadth is a and length 2a.
For rectangle,
\(\therefore 2(a+2 a)=x \Rightarrow 6 a=x \)
\(\Rightarrow a=\frac{x}{6} \mathrm{~m} \)
\(\therefore \text { Area of rectangle }=a(2 a)=2 a^{2}=2\left(\frac{x}{6}\right)^{2}=\frac{x^{2}}{18} \)
For square,
\(4 \times \text { side }=(34-x) \Rightarrow \text { side }=\frac{34-x}{4} \mathrm{~m} \)
\(\therefore \text { Area of square }=\frac{1}{16}(34-x)^{2}\)
\(\text { Combined area }(A)=\frac{x^{2}}{18}+\frac{1}{16}(34-x)^{2} \)
\(\frac{d A}{d x} =\frac{2 x}{18}+\frac{2}{16}(34-x) \cdot(-1) \)
\(=\frac{x}{9}-\left(\frac{34-x}{8}\right)=\frac{8 x-306+9 x}{72} \)
\(\frac{d A}{d x} =\frac{17 x-306}{72}\)
\(\text { For minimum area, } \frac{d A}{d x}=0\)
\(\Rightarrow 17 x-306=0 \)
\(\Rightarrow x=\frac{306}{17}=18 \)
\(\frac{d^{2} A}{d x^{2}} =\frac{17}{72} \)
\(\left.\frac{d^{2} A}{d x^{2}}\right|_{x=18} =\frac{17}{72}>0\)
For x = 18, A is minimum. Hence, wire must be cut at 18 m to made into a rectangle and remaining 16 m piece to made into a square for a minimum combined area.

\(\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2} \)
\(\Rightarrow \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{2}-\tan ^{-1} z \)
\(\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{2}-\tan ^{-1} z \)
\(\Rightarrow \frac{x+y}{1-x y}=\tan \left(\frac{\pi}{2}-\tan ^{-1} z\right) \)
\(\Rightarrow \frac{x+y}{1-x y}=\cot \left(\tan ^{-1} z\right)=\cot \left(\cot ^{-1} \frac{1}{z}\right) \)
\(\Rightarrow \frac{x+y}{1-x y}=\frac{1}{z} \)
\(\Rightarrow x z+y z=1-x y \)
\(\Rightarrow x y+y z+z x=1 \)

We have, f(x) = 3x + 5
Let \(x_{1}, x_{2} \in R\) ,such that \(x_{1} 
Then, \(x_{1} 
\(\Rightarrow 3 x_{1}<3 x_{2}\) [multiplying both sides by 3]
\(\Rightarrow 3 x_{1}+5<3 x_{2}+5\) [adding 5 on both sides]
\(\Rightarrow \quad f\left(x_{1}\right) 
\(\Rightarrow \quad f\left(x_{1}\right) 
Hence, f (x) is strictly increasing on R.

Given \(A=\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \) 
|A| = 1(1+3)-2(-1-1)+1(3-1)
= 4 + 4 + 2 = 10
Co-factor Matrix is :
\(=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right] \)
\(\therefore\) adj A = transpose of above matrix
\(=\left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\therefore { A }^{ -1 }=\frac { adj\quad A }{ |A| } \)
\(=\frac { 1 }{ 10 } \left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\Rightarrow { A }^{ -1 }=\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \)
Given set of equations are:
x + 2y + z = 4
-x + y + z = 0
x - 3y + z = 4
\(\Rightarrow \quad \left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \quad A.X=B\)
Multiplying both sides by A^-1 , we get
\({ A }^{ -1 }AX={ A }^{ -1 }B\)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} \frac { 8 }{ 5 } +\frac { 2 }{ 5 } \\ \frac { 4 }{ 5 } -\frac { 4 }{ 5 } \\ \frac { 4 }{ 5 } +\frac { 6 }{ 5 } \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 0 \\ 2 \end{matrix} \right] \)
\(\therefore \ x=2,y=0,z=2\)

If function is continuous at \(x=\frac{\pi}{2}, \text { then }\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\mathrm{RHL}=f\left(\frac{\pi}{2}\right)\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \)
\(=\lim _{h \rightarrow 0} \frac{1-\sin ^{3}\left(\frac{\pi}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}=\lim _{h-0} \frac{1-\cos ^{3} h}{3 \sin ^{2} h} \)
\(=\lim _{h \rightarrow 0} \frac{(1-\cos h)\left(1+\cos ^{2} h+\cos h\right)}{3(1-\cos h)(1+\cos h)} \)
\(=\lim _{h \rightarrow 0} \frac{1+\cos ^{2} h+\cos h}{3(1+\cos h)}=\frac{1+1+1}{3(1+1)}=\frac{1}{2} \ldots .(i i) \)
\(\operatorname{RHL} =\lim _{h=\frac{\pi}{2}} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0} \frac{b\left\{1-\sin \left(\frac{\pi}{2}+h\right)\right\}}{\left\{\pi-2\left(\frac{\pi}{2}+h\right)\right\}^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{(\pi-\pi-2 h)^{2}}=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{4 h^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b \cdot 2 \sin ^{2} \frac{h}{2}}{4 h^{2}}=\lim _{h \rightarrow 0} \frac{b}{8}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2} \)
\(=\frac{b}{8} \times 1=\frac{b}{8} \)
Substituting from (ii) and (iii) in (i), we get
\(\frac{1}{2}=\frac{b}{8}=a \Rightarrow a=\frac{1}{2}, b=4\)
Hence, for \(a=\frac { 1 }{ 2 } ,\quad b=4 \text { function is continuous at} x=\frac { \pi }{ 2 } .\)

\(A_{12}=(-1)^{1+2} \sin \theta=-\sin \theta \)
\(A_{21}=(-1)^{1+2}(-\sin \theta)=\sin \theta \)
\(A_{12}+A_{21}=-\sin \theta+\sin \theta=0 \)

Given \(x=a \cos \theta+b \sin \theta\) ...(i)
and \(y=a \sin \theta-b \cos \theta\)  ...(ii)
Here,\(\theta\) is the parameter 
On differentiating Eqs. (i) and (ii), respectively w.r.t. θ, we get
\( \frac{d x}{d \theta}=\frac{d}{d \theta}(a \cos \theta+b \sin \theta) \)
\(\Rightarrow \frac{d x}{d \theta}=-a \sin \theta+b \cos \theta \) 
and \(\frac{d y}{d \theta}=\frac{d}{d \theta}(a \sin \theta-b \cos \theta)\) 
\(\Rightarrow \frac{d y}{d \theta}=a \cos \theta+b \sin \theta\) 
Now,\(\frac{d y}{d x}=\frac{(d y / d \theta)}{(d x / d \theta)}=\frac{a \cos \theta+b \sin \theta}{-(a \sin \theta-b \cos \theta)}=-\frac{x}{y}\) 
Again, differentiating both sides w.r.t. x, we get
\( \frac{d^{2} y}{d x^{2}} =-\left[\frac{y \frac{d}{d x}(x)-x \frac{d y}{d x}}{y^{2}}\right] \)
\(\Rightarrow y^{2} \frac{d^{2} y}{d x^{2}} =-y+x \frac{d y}{d x} \Rightarrow y^{2} \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+y=0 \)

We have, \(x=a(\theta+\sin \theta)\) 
and \(y=a(1-\cos \theta)\) 
Here, \(\theta\) is the parameter
On differentiating Eqs. (i) and (ii), respectively w.r.t. \(\theta\) we get
\(\frac{d x}{d \theta}=\frac{d}{d \theta}[a(\theta+\sin \theta)]=a(1+\cos \theta)\) .. (iii)
and \(\frac{d y}{d \theta}=\frac{d}{d \theta}[a(1-\cos \theta)]=a(0+\sin \theta)=a \sin \theta\)   ...(iv)
On putting the values from Eqs. (iii) and (iv) in the formula \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\) we get
\( \frac{d y}{d x}=\frac{a \sin \theta}{a(1+\cos \theta)} \)
\(\Rightarrow \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}=\frac{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)}=\tan \frac{\theta}{2} \) 
Now, differentiating both sides w.r.t. x, we get 
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\tan \frac{\theta}{2}\right)=\frac{d}{d \theta}\left(\tan \frac{\theta}{2}\right) \cdot \frac{d \theta}{d x}\) 
\(=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{1}{a(1+\cos \theta)}\)  \(=\frac{\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)}\) 
Now, at \(\theta=\frac{\pi}{2}\) 
\(\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{2} \frac{\pi}{4}}{2 a\left(1+\cos \frac{\pi}{2}\right)}=\frac{2}{2 a(1+0)}=\frac{1}{a}\)

 For principal value of \(sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
Let \(y=\sin ^{-1} \frac{\sqrt{3}}{2} \Rightarrow \sin y=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin y=\sin \frac{\pi}{3}\)
\(\left[\because \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right]\)
\(\Rightarrow y=\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
\(\left[\because\right.\) principal value branch of \(\sin ^{-1}\) is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
\(\therefore \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\)
For principal value of \(\tan ^{-1}(\sqrt{3})\)
Again, let \(x=\tan ^{-1}(\sqrt{3}) \Rightarrow \tan x=\sqrt{3}\)
\(\Rightarrow \tan x=\tan \left(\frac{\pi}{3}\right) \quad\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]\) 
\( \Rightarrow x=\frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
\( \Rightarrow\left[\quad \because \tan ^{-1} \text {is }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\right] \)
\(\therefore \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3} \)

Let AB be the lamp post and a man CD be at a distance x
from the lamp post and let CE = y be his shadow

Given that 
\( \frac{d x}{d t}=6 \mathrm{~km} / \mathrm{h}, A B=6 \mathrm{~m}=\frac{6}{1000} \mathrm{~km} \text { and }\)
\( C D=2 \mathrm{~m}=\frac{2}{1000} \mathrm{~km}[\because 1 \mathrm{~km}=1000 \mathrm{~m}] \) 
Here,\(\Delta B A E \sim \Delta D C E\) 
\( \Rightarrow \frac{6}{\frac{1000}{2}}{\frac{1000}{y}}=\frac{x+y}{y} \Rightarrow 3=\frac{x+y}{y}\)
\( \Rightarrow 3 y =x+y \Rightarrow 3 y-y=x \Rightarrow 2 y=x \) 
On differentiating both sides w.r.t. t, we get
\(2 \frac{d y}{d t}=\frac{d x}{d t} \)
\( \Rightarrow 2 \frac{d y}{d t}=6 \quad\left[\because \frac{d x}{d t}=6\right] \) 
\(\Rightarrow \frac{d y}{d t}=\frac{6}{2}=3 \mathrm{~km} / \mathrm{h}\) 

The given curve is : x = 1 \(cos\theta ,y-=\theta -sin\theta \)
\(\therefore \frac { dx }{ d\theta } =sin\theta ,\frac { dy }{ d\theta } =1-cos\theta \)
\(\therefore \frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { 1-cos\theta }{ sin\theta } =\frac { 2sin^{ 2 }\frac { \theta }{ 2 } }{ 2sin\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } } \)
\(=\frac { sin\frac { \theta }{ 2 } }{ cos\frac { \theta }{ 2 } } =tan\frac { \theta }{ 2 } \)
At \(\theta \) \(=\frac { \pi }{ 4 } \)    \(x=1-cos\frac { \pi }{ 4 } =1-\frac { 1 }{ \sqrt { 2 } } \)
\(y=\frac { \pi }{ 4 } -sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
and  \(\frac { dy }{ dx } =tan\frac { \pi }{ 8 } \)

Let A be the area and \(\theta\) be the sector angle. Then 
\( A=\frac{1}{2} \times 30^{2} \times \theta=450 \theta \\ \frac{d A}{d \theta}=450 \) 
Let \(\Delta \theta\) be an error in \(\theta \text { and } \Delta A\) be the corresponding error in A.
Then,\(\Delta A=\frac{d A}{d \theta} \Delta \theta\) 
\(\begin{array}{ll} \Rightarrow & \Delta A=450 \times \frac{\pi}{180} \quad\left[\because \Delta \theta=1^{\circ}=\frac{\pi}{180} \text { radians }\right] \\ \Rightarrow & \Delta A=2.5 \pi \mathrm{cm}^{2} \end{array}\)

We have
\(f(x)=x^{2}+2 x-8, x \in[-4,2]\) 
Rolle'stheorem is satisfied
\(\therefore f^{\prime}(c)=0 \)
\(\Rightarrow f^{\prime}(c)=2 c+2=0 \)
\(c+1=0 \Rightarrow c=-1 \)

\(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)

 Consider \(((f \cdot g) o h)(x)\)
\(=(f \cdot g)(h(x))=f(h(x)) \cdot g(h(x))\)
\(=(f o h)(x) \cdot(g o h)(x)\) 
\(=\{(f o h) \cdot(g o h)\}(x)\) 
\(\text { Hence, }(f \cdot g) o h=(f o h) \cdot(g o h)\) 
Now, consider \(((f+g) o h)(x)=(f+g)(h(x))\)
\(=f(h(x))+g(h(x))=(\text { foh })(x)+(\text { goh })(x)\) 
\(=\{(f o h)+(g o h)\}(x) \)
Hence \((f+g) o h=f o h+g o h\)

(i) (a) : Let \(\begin{equation} y=\cos \sqrt{x} \end{equation}\) 
\(\begin{equation} \therefore \quad \frac{d y}{d x}=\frac{d}{d x}(\cos \sqrt{x})=-\sin \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x}) \end{equation}\) 
\(\begin{equation} =-\sin \sqrt{x} \times \frac{1}{2 \sqrt{x}}=\frac{-\sin \sqrt{x}}{2 \sqrt{x}} \end{equation}\) 
(ii) (a) : Let \(\begin{equation} y=7^{x+\frac{1}{x}} \quad \therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(7^{x+\frac{1}{x}}\right) \end{equation}\) 
\(\begin{equation} =7^{x+\frac{1}{x}} \cdot \log 7 \cdot \frac{d}{d x}\left(x+\frac{1}{x}\right)=7^{x+\frac{1}{x}} \cdot \log 7 \cdot\left(1-\frac{1}{x^{2}}\right) \end{equation}\) 
\(\begin{equation} =\left(\frac{x^{2}-1}{x^{2}}\right) \cdot 7^{x+\frac{1}{x}} \cdot \log 7 \end{equation}\) 
(iii) (a) : Let \(\begin{equation} y=\sqrt{\frac{1-\cos x}{1+\cos x}}=\sqrt{\frac{1-1+2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}-1+1}}=\tan \left(\frac{x}{2}\right) \end{equation}\) 
\(\begin{equation} \therefore \frac{d y}{d x}=\sec ^{2} \frac{x}{2} \cdot \frac{1}{2}=\frac{1}{2} \sec ^{2} \frac{x}{2} \end{equation}\) 
(iv) (b) : Let \(\begin{equation} y=\frac{1}{b} \tan ^{-1}\left(\frac{x}{b}\right)+\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right) \end{equation}\) 
\(\begin{equation} \therefore \quad \frac{d y}{d x}=\frac{1}{b} \times \frac{1}{1+\frac{x^{2}}{b^{2}}} \times \frac{1}{b}+\frac{1}{a} \times \frac{1}{1+\frac{x^{2}}{a^{2}}} \times \frac{1}{a} \end{equation}\) 
\(\begin{equation} =\frac{1}{b^{2}+x^{2}}+\frac{1}{a^{2}+x^{2}} \end{equation}\) 
(v) (d) : Let \(\begin{equation} y=\sec ^{-1} x+\operatorname{cosec}^{-1} \frac{x}{\sqrt{x^{2}-1}} \end{equation}\) 
Put  \(\begin{equation} x=\sec \theta \Rightarrow \theta=\sec ^{-1} x \end{equation}\) 
\(\begin{equation} \therefore \quad y=\sec ^{-1}(\sec \theta)+\operatorname{cosec}^{-1}\left(\frac{\sec \theta}{\sqrt{\sec ^{2} \theta-1}}\right) \end{equation}\) 
\(\begin{equation} =\theta+\sin ^{-1}\left[\sqrt{1-\cos ^{2} \theta}\right] \end{equation}\) 
\(\begin{equation} =\theta+\sin ^{-1}(\sin \theta)=\theta+\theta=2 \theta=2 \sec ^{-1} x \end{equation}\) 
\(\begin{equation} \therefore \quad \frac{d y}{d x}=2 \frac{d}{d x}\left(\sec ^{-1} x\right)=2 \times \frac{1}{|x| \sqrt{x^{2}-1}}=\frac{2}{|x| \sqrt{x^{2}-1}} \end{equation}\)

(i) (d) : Let be the area of the triangle then,

\(\begin{equation} =\frac{1}{2}|-2(-6-5)-6(3-1)+1(15+6)| \end{equation}\) [Expanding along R_1]
\(\begin{equation} \Rightarrow \quad \Delta=\frac{1}{2}|43-12|=15.5 \mathrm{sq} . \text { units } \end{equation}\) 
(ii) (d) : The given points are collinear.
\(\begin{equation} \therefore \frac{1}{2}\left|\begin{array}{ccc} 2 & -3 & 1 \\ k & -1 & 1 \\ 0 & 4 & 1 \end{array}\right|=0 \end{equation}\) 
Expanding along R₁',we get
\(\begin{equation} 2(-1-4)+3(k)+1(4 k)=0 \end{equation}\) 
\(\begin{equation} \Rightarrow 7 k-10=0 \Rightarrow k=\frac{10}{7} \Rightarrow 4 k=\frac{40}{7} \end{equation}\) 
(iii) (a) : Area of \(\begin{equation} \Delta A B C=3 \text { sq. units } \end{equation}\) [Given]
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 3 \Rightarrow\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 6 \end{equation}\) 
\(\begin{equation} \Rightarrow 1(0-0)-3(0-k)+1(0-0)=\pm 6 \end{equation}\) 
\(\begin{equation} \Rightarrow 3 k=\pm 6 \Rightarrow k=\pm 2 . \end{equation}\) 
(iv) (a) : Let Q(x, y) be any point on the line joining
A(1, 2) and B(3, 6). Then, area of \(\begin{equation} \Delta A B Q=0 \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{array}\right|=0 \end{equation}\) 
\(\begin{equation} \Rightarrow 1(6-y)-2(3-x)+1(3 y-6 x)=0 \end{equation}\) 
\(\begin{equation} \Rightarrow 6-y-6+2 x+3 y-6 x=0 \end{equation}\) 
\(\begin{equation} \Rightarrow-4 x=-2 y \Rightarrow 2 x=y \end{equation}\)
(v) (c) : Area of ΔABC is given by
\(\begin{equation} \frac{1}{2}\left|\begin{array}{rrr} 11 & 7 & 1 \\ 5 & 5 & 1 \\ -1 & 3 & 1 \end{array}\right|=\frac{1}{2}[11(5-3)-7(5+1)+1(15+5)] \end{equation}\) 
\(\begin{equation} =\frac{1}{2}[22-42+20]=0 \end{equation}\) 
\(\therefore\) Points are collinear