We have, maximise, Z = 60x + 40y             ...(i)
Subject to the constraints, x + 2y \(\leq 12\)     ...(ii)
\(\begin{aligned} 2 x+y & \leq 12 \end{aligned}\)                                   ...(iii)
\(\begin{aligned} 4 x+5 y & \geq 20 \end{aligned}\)                                 ...(iv)
and \(\begin{aligned} x, y & \geq 0 \end{aligned}\)                                     ...(v)
Table for  line x + 2y = 12 is

So, the line x + 2y = 12 is passing through the points (0, 6) and (12, 0).
On putting (0,0) in the inequality \(x+2 y \leq 12\) we get \(0+2(0) \leq 12 \Rightarrow 0 \leq 12\), which is true
So,the half plane is towards the origin.
Table for line 2x + y = 12 is 

So, the line 2x + y = 12 is passing through the points (0, 12) and (6, 0).
On putting (0, 0) in the inequality \(2 x+y \leq 12\) we get
\(2(0)+0 \leq 12 \Rightarrow 0 \leq 12\) , which is true.
So, the half plane is towards the origin.
Table for line 4x + 5y = 20 is

So, the line 4x + 5y = 20 is passing through the points (0,4) and (5, 0).
On putting (0, 0) in the inequality \(4 x+5 y \geq 20\), we get 4(0) + 5(0) \(\geq\) 20 \(\Rightarrow\) 0 > 20 which is not true.
So, the half plane is away from the origin.
Also, x, y \(\geq\) 0
So, the region lies in Ist quadrant.

On solving Eqs. x + 2y = 12 and 2x + y = 12, we get D(4, 4)
Clearly, the feasible region is ABCDEA.
The corner points of the feasible region are A(0, 4), B(5, 0), C(6, 0), D(4, 4) and E(0, 6).
The value of Z at corner points are given below.

The maximum value of Z is 400 at D(4, 4).

\(\left\{ \left( x+\frac { 1 }{ x } \right) ^{ x }\left[ \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } +log\left( x+\frac { 1 }{ x } \right) \right] +{ x }^{ 1+\frac { 1 }{ x } }\left( \frac { x+1-logx }{ { x }^{ 2 } } \right) \right\} \)

Given inequations are
\(2 x+4 y \leq 8 \text { or } \ x+2 y \leq 4 \)
\(3 x+y \leq 6, x+y \leq 4, \ x \geq 0, \ y \geq 0 \)
Maximise Z = 2x + 5y on plotting the graph of the inequations we notice shaded portion as feasible solution

Possible points for maximum Z are A(2, 0),\(B\left(\frac{8}{5}, \frac{6}{5}\right)\) C (0,2)

Z is maximum at qo, 2), i.e. x = 0, y = 2, maximum value = 10

\(\therefore \ \frac { dy }{ dx } =\frac { -y }{ x } \)

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.

Given, \(x=a \cos ^3 \theta\)and \(y=a \sin ^3 \theta\)
On differentiating both sides of x and y w.r.t. \(\theta\), we get
\(\frac{d x}{d \theta} =3 a \cos ^2 \theta \frac{d}{d \theta}(\cos \theta)=3 a \cos ^2 \theta \cdot(-\sin \theta)\)
\(=-3 a \cos ^2 \theta \cdot \sin \theta\)
and \(\frac{d y}{d \theta}=3 a \sin ^2 \theta \frac{d}{d \theta}(\sin \theta)\)
\(=3 a \sin ^2 \theta \cdot(\cos \theta)=3 a \sin ^2 \theta \cdot \cos \theta\)
Now, \(\frac{d y}{d x}=\left(\frac{d y / d \theta}{d x / d \theta}\right)\)
\(=\frac{3 a \sin ^2 \theta \cdot \cos \theta}{-3 a \cos ^2 \theta \cdot \sin \theta}=-\tan \theta\)
Again, on differentiating both sides w.r.t. x, we get
\(\frac{d^2 y}{d x^2}=\frac{d}{d x}(-\tan \theta)=-\frac{d}{d \theta}(\tan \theta) \frac{d \theta}{d x}\)
\(=-\sec ^2 \theta \cdot \frac{d \theta}{d x}\)
\(=-\sec ^2 \theta \cdot\left(\frac{-1}{3 a \cos ^2 \theta \cdot \sin \theta}\right) \)
\( \Rightarrow \quad\left[\because \frac{d \theta}{d x}=\frac{-1}{3 a \cos ^2 \theta \sin \theta}\right]\)
\(\therefore \quad \frac{d^2 y}{d x^2}=\frac{1}{3 a \cos ^4 \theta \cdot \sin \theta} \)
\(\therefore\left(\frac{d^2 y}{d x^2}\right)_{\text {at } \theta=\frac{\pi}{6}}=\frac{\frac{\pi}{6}}{3 a\left(\cos \frac{\pi}{6}\right)^4\left(\sin \frac{\pi}{6}\right)}\)
\(=\frac{1}{3 a\left(\frac{\sqrt{3}}{2}\right)^4\left(\frac{1}{2}\right)}\)
\(=\frac{1}{3 a\left(\frac{9}{16}\right)\left(\frac{1}{2}\right)}=\frac{32}{27 a}\)

Our problem is to minirnise
Z = 5x + 10y   ...(i)
subject to constraints
\(x+2 y \leq 120\)   ...(ii)
\(x+y \geq 60\)   ...(iii)
\(x-2 y \geq 0\)   ...(iv)
and \(x \geq 0, y \geq 0\)   ...(v)
Firstly, draw the graph of the line x + 2y = 120.
\(\begin{array}{c|c|c} \hline x & 0 & 120 \\ \hline y & 60 & 0 \\ \hline \end{array}\) 
Put (0, 0) in the inequality \(x+2 y \leq 12 \overrightarrow{0}\) , we get
\(0+2 \times 0 \leq 120 \Rightarrow 0 \leq 120\) , which is true
So, the half plane is towards the origin. ,Secondly, draw the graph of the line x + y = 60.
\(\begin{array}{c|c|c} \hline x & 0 & 60 \\ \hline y & 60 & 0 \\ \hline \end{array}\) 
Put (0, 0) in the inequality \(x+y \geq 60\) , we get
\(0+0 \geq 60 \Rightarrow 0 \geq 60\) ,which is false
So, the half plane is away from the origin.
Thirdly, draw the graph of the line x - 2y = 0.
\(\begin{array}{c|c|c} \hline x & 0 & 10 \\ \hline y & 0 & 5 \\ \hline \end{array}\) 
Put (5, 0) in the inequality \(x-2 y \geq 0\)  we get
\(5-2 \times 0 \geq 0=5 \geq 0\) which is true
So, the half plane is towards the x-axis. Since \(x, y \geq 0\) the feasible region lies in the first quadrant.
 
On solving equations x - 2y = 0 and x + y = 60, we get
D(40, 20) and solving equations x - 2y = 0 and
x + 2y = 120, we get C (60, 30)
So, the feasible region is ABCDA. The corner points of the feasible region are A (60, 0), B (120, 0), C (60, 30) and D (40, 20).
The values of Z at these points are as follows.
\(\begin{array}{c|c} \hline \text { Corner point } & z=5 x+10 y \\ \hline A(60,0) & 300 \text { (minimum) } \\ B(120,0) & 600 \\ C(60 \mid 30) & 600 \\ D(40,20) & 400 \\ \hline \end{array}\) 
So, the minimum value of Z is 300 at the point (60, 0).

\(\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ - } }{ \left( 3ax+b \right) } \)
\(=\lim _{ h\rightarrow 0 }{ [3a(1-h)+b] } \)
\(=3a(1-0)+b\)
\(=3a+b\)
\(\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ \left( 5ax-2b \right) } \)
\(=\lim _{ h\rightarrow 0 }{ [5a(1+0)-2b] } \)
\(=5a-2b\)
\(f(1)=11\)
Also
Since'f' is continuous at x = 1
\(\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =f(1)\)
From first and third 3a + b = 11 ....(1)
From last two 5a - 2b = 11 .....(2)
Multiplying (1) by 2, 6a + 2b = 22 .....(3)
Adding (2) and (3), 11a = 33 = a = 3
Putting in (1), 3(3) + b = 11
b = 11 - 9 = 2
Hence a = 3 and b = 2.

\(We\quad have\quad :\quad x=a(cos\theta +sin\theta ),\quad y=a(sin\theta -\theta cos\theta )\)
\(\frac { dx }{ d\theta } =\alpha \left( -sin\theta +\theta cos\theta +sin\theta \right) =a\theta cos\theta \)
\(and\quad \frac { dy }{ d\theta } =\alpha (cos\theta +\theta sin\theta -cos\theta )=a\theta sin\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { \alpha \theta sin\theta }{ \alpha \theta cos\theta } =tan\theta \)

The shaded region is the feasible region ABC determined by the system of constraints (2) to (4), which is bounded. The coordinates of corner points

A, B and C are (0,5), (4,3) and (0,6) respectively. Now we evaluate Z = 200x + 500y at these points.
Hence, minimum value of Z is 2300 attained at the point (4, 3)

\(we\ have\quad :\ { x }^{ 2 }+xy+{ y }^{ 2 }=100\)
\(Diff\ w.r.t.x,\ 2x+\left( x.\frac { dy }{ dx } +y.1 \right) +2y\frac { dy }{ dx } =0\)
\(\Rightarrow (x+2y)\frac { dy }{ dx } =\quad -2x-y\)
\(Hence,\ \frac { dy }{ dx } =-\frac { 2x+y }{ x+2y } \)

Let sin^-1 x = \(\theta\) then  sin^-1 x = \(\theta\). we have
sin^-1 \((2x\sqrt { 1-{ x }^{ 2 } } )\) = sin^-1\(2\sin { \theta } \sqrt { 1-{ sin }^{ 2 } } \)
= sin^–1 (2sinθ cosθ) = sin^–1 (sin2θ) = 2θ = 2 sin^–1 x

\( \left.\begin{array}{l} \text {We have, } \tan ^{-1}\left[\sqrt{\frac{1-\cos x}{1+\cos x}}\right. \end{array}\right]=\tan ^{-1}\left[\sqrt{\frac{2 \sin ^{2} x / 2}{2 \cos ^{2} x / 2}}\right] \text { [1] } \)
\(\left[\because 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2}, 1+\cos \theta=2 \cos ^{2} \frac{\theta}{2}\right]\)
\( \left[ta n^{-1}\left[\sqrt{\tan ^{2} x / 2}\right]=\tan ^{-1}\left[\tan \frac{x}{2}\right]=\frac{x}{2}\right. \)

\(\sin ^{-i}\left[\cos \left(\frac{33 \pi}{5}\right)\right]=\sin ^{-1}\left[\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right] \)
\(=\sin ^{-1}\left[\cos \frac{3 \pi}{5}\right] \quad[\because \cos (2 n \pi+\theta)=\cos \theta] \)
\(=\sin ^{-1}\left[\cos \left(\frac{\pi}{2}+\frac{\pi}{10}\right)\right]=\sin ^{-1}\left[-\sin \left(\frac{\pi}{10}\right)\right] \)
\(=-\sin ^{-1}\left(\sin \frac{\pi}{10}\right) \quad[1]\)
\(=-\frac{\pi}{10} \quad\left[\because \sin ^{-1}(\sin x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right][1] \)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

\(\frac { dx }{ d\theta } =\theta cos\theta +sin\theta \)
\(\frac { dy }{ d\theta } =-\theta sin\theta +cos\theta \)
\(\frac { dy }{ dx } =\frac { cos\theta -\theta sin\theta }{ \theta cos\theta +sin\theta } \)
dy/dx at \(\theta\)= \(\pi/4\)
\(=\frac { cos\frac { \pi }{ 4 } -\frac { \pi }{ 4 } sin\frac { \pi }{ 4 } }{ \frac { \pi }{ 4 } cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } } \)
\(=\frac { \frac { 1 }{ \sqrt { 2 } } -\frac { \pi }{ 4 } \times \frac { 1 }{ \sqrt { 2 } } }{ \frac { \pi }{ 4 } \times \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } } \)
=\(\frac { 1-\frac { \pi }{ 4 } }{ \frac { \pi }{ 4 } +1 } \)
\(\Rightarrow\)\(\frac { dy }{ dx } =\frac { 4-\pi }{ 4+\pi } \)

We have y = log(sin x)
dy/dx = d/dx \(\left| log(sinx) \right| \)
= \(\frac { 1 }{ sinx } \times cosx\)
\(\Rightarrow\) dy/dx = cot x
\(\therefore\)  \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } =-{ cosec }^{ 2 }x\)

\(Let\quad { sec }^{ -1 }x=\theta ,\quad then\quad x=sec\theta \quad and\quad for\quad x<-1,\)
\(\frac { \pi }{ 2 } <\theta <\pi \)
Given expression = cot^-1(-cot \(\theta \))
\(={ cot }^{ -1 }\left[ cot\left( \pi -\theta \right) \right] =\pi -{ sec }^{ -1 }x\quad as\quad 0<\pi -\theta <\frac { \pi }{ 2 } \)

As sin (\(-\frac{\pi}{2}\))= -1, and tan^-1(-1) = \(-\frac{\pi}{4}\).

(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.

(i) (a) : Now, \(\begin{equation} \frac{d f(\tan x)}{d g(\sec x)}=\frac{f^{\prime}(\tan x) \sec ^{2} x}{g^{\prime}(\sec x) \sec x \tan x} \end{equation}\) 
\(\begin{equation} =\frac{f^{\prime}(\tan x) \sec x}{g^{\prime}(\sec x) \tan x} \end{equation}\) 
\(\begin{equation} \therefore\left[\frac{d f(\tan x)}{d g(\sec x)}\right]_{x=\pi / 4}=\frac{f^{\prime}(1) \sqrt{2}}{g^{\prime}(\sqrt{2}) \cdot 1}=\frac{2 \sqrt{2}}{4 \cdot 1}=\frac{1}{\sqrt{2}} \end{equation}\) 
(ii) (b)
(iii) (c) : Let \(\begin{equation} y=e^{x^{3}}, z=\log x \end{equation}\) 
Differentiating w.r.t. x, we get
\(\begin{equation} \frac{d y}{d x}=e^{x^{3}}\left(3 x^{2}\right)=3 x^{2} e^{x^{3}} \end{equation}\) and \(\begin{equation} \frac{d z}{d x}=\frac{1}{x} \end{equation}\) 
\(\begin{equation} \therefore \ \frac{d y}{d z}=\frac{\frac{d x}{d x}}{\frac{d z}{d x}}=\frac{3 x^{2} e^{x^{3}}}{\left(\frac{1}{x}\right)}=3 x^{3} e^{x^{3}} \end{equation}\) 
(iv) (a): Let y = cos^-1(2x² - 1) = 2cos^-1x
Differentiating w.r.t. cos'" x, we get
\(\begin{equation} \frac{d y}{d\left(\cos ^{-1} x\right)}=\frac{2 d\left(\cos ^{-1} x\right)}{d\left(\cos ^{-1} x\right)}=2 \end{equation}\) 
(v) (a) : We have \(\begin{equation} y=\frac{1}{4} u^{4} \Rightarrow \frac{d y}{d u}=\frac{1}{4} \cdot 4 u^{3}=u^{3} \end{equation}\) 
and \(\begin{equation} u=\frac{2}{3} x^{3}+5 \Rightarrow \frac{d u}{d x}=\frac{2}{3} \cdot 3 x^{2}=2 x^{2} \end{equation}\) 
\(\begin{equation} \therefore \ \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=u^{3} \cdot 2 x^{2}=\left(\frac{2}{3} x^{3}+5\right)^{3}\left(2 x^{2}\right) \end{equation}\) 
\(\begin{equation} =\frac{2}{27} x^{2}\left(2 x^{3}+15\right)^{3} \end{equation}\)

we have,\(\begin{equation} f(x)=\left\{\begin{array}{ll} x-3 & , x \geq 3 \\ 3-x & , 1 \leq x<3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} & , x<1 \end{array}\right. \end{equation}\) 
(i) (b) : \(\begin{equation} \mathrm{R} f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{3-(1+h)-2}{h}=\lim _{h \rightarrow 0}-\frac{h}{h}=-1 \end{equation}\) 
(ii) (b) : \(\begin{equation} \mathrm{L}_{\mathrm{s}}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{-1}{h}\left[\frac{(1-h)^{2}}{4}-\frac{3(1-h)}{2}+\frac{13}{4}-2\right] \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{1+h^{2}-2 h-6+6 h+13-8}{-4 h}\right) \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{h^{2}+4 h}{-4 h}\right)=-1 \end{equation}\) 
(iii) (c) : Since, R.H.D. at x = 3 is 1 and L.H.D. at x = 3 is-1
\(\therefore\)  f(x) is non-differentiable at x = 3.
(iv) (d)
(v) (c) : From above, we have
\(\begin{equation} f^{\prime}(x)=\frac{x}{2}-\frac{3}{2}, x<1 \end{equation}\) 
\(\begin{equation} \therefore f^{\prime}(-1)=\frac{-1}{2}-\frac{3}{2}=-2 \end{equation}\)
 

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

(d) Assertion is incorrect, Reason is correct.