\(\frac{1}{\sqrt{6}},\) lines do not intersect
Hint: A pair of lines will intersect, if the shortest distance between them is zero.

As shown in the Figure, the line y = 3x + 2 meets x-axis at x\(=\frac{-2}{3}\) and its graph lies below x-axis for \(x \in\left(-1, \frac{-2}{3}\right)\) and above x-axis for \(x \in\left(\frac{-2}{3}, 1\right)\)
The required area = Area of the region ACBA + Area of the region ADEA
\(=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x\)
\(=\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}\)

It is given that A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2}.
Also, it is given that f, g : A → B are defined by f(x) = x² − x, x ∈ A and \(g(x)=2\left|x-\frac{1}{2}\right|-1, x \in A\)
It is observed that:
\(f(-1)=\left(1^{2}\right)-(-1)=1+1=2 \)
\(g(-1)=2\left|(-1)-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2 \)
⇒ f(-1) = g(-1)
f(0) = (0)^2 - 0 = 0
\(g(0)=2\left|0-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=1-1=0\)
⇒ f(0) = g(0)
f(1) = (1)^2 - 1 = 1 -   1 = 0
\(g(1)=2\left|a-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=1-1=0\)
⇒ f(1) = g(1)
f(2) = (2)^2 - 2 = 4 - 2 = 2
\(g(2)=2\left|2-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2\)
⇒ f(2) = g(2)
f(a) = g(a) ∀ a ∈ A
Hence, the functions f and g are equal.

\( \Delta=\left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right| \)
Expanding along \(\mathrm{C}_3\),
we have:
\( \Delta =-\sin \alpha\left(-\sin \alpha \sin ^2 \beta-\cos ^2 \beta \sin \alpha\right)+\cos \alpha\left(\cos \alpha \cos ^2 \beta+\cos \alpha \sin ^2 \beta\right) \)
\(=\sin ^2 \alpha\left(\sin ^2 \beta+\cos ^2 \beta\right)+\cos ^2 \alpha\left(\cos ^2 \beta+\sin ^2 \beta\right) \)
\(=\sin ^2 \alpha(1)+\cos ^2 \alpha(1)  =1 \)

\(k \in\left(0, \frac{1}{3}\right)\)

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.

\({(ax +b)^4\over 4a}+c\)​​​​​​​

\( \cot ^{-1}(\sqrt{3})=y \)
\(\Rightarrow \sqrt{3}=\cot y\left[\because \cot ^{-1} \mathbf{x}=\mathbf{y} \Rightarrow \mathbf{x}=\cot \mathbf{y}\right] \)
\(\Rightarrow \cot y=\sqrt{3} \)
\(\Rightarrow \cot y=\cot \frac{\pi}{6}\left[\because \cot \frac{\pi}{6}=\sqrt{\mathbf{3}}\right] \ \)
\(\Rightarrow y=\frac{\pi}{6}\)
So, the principal value of \(\cot ^{-1}(\sqrt{3}) \text { is } \frac{\pi}{6} \text {. }\)

Let us defme the events as
E₁ = selecting a two headed coin
E₂ = selecting a biased coin,
E ₃ = selecting an unbiased coin
A = head comes up.
Here, \(P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=1 / 3\) 
\([\because \) all coins have equal chances] 
P(A / E₁) = Probability that head comes up on a two headed coin = 1
P(A / E₂) = Probability that head comes up on a biased coin.
=  75% = 75/100
and P(A / E₃) = Probability that head comes up on an unbiased coin = 1/2
Now, by Baye's theorem, we get
\(P\left(E_{1} / A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)}{\left[\begin{array}{c} P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right) \\ +P\left(E_{3}\right) \cdot P\left(A / E_{3}\right) \end{array}\right]}\) 
\(\therefore P\left(E_{1} / A\right)=\frac{\frac{1}{3} \times 1}{\left(\frac{1}{3} \times 1\right)+\left(\frac{1}{3} \times \frac{75}{100}\right)+\left(\frac{1}{3} \times \frac{1}{2}\right)}\) 
\(=\frac{1}{1+\frac{75}{100}+\frac{1}{2}}=\frac{1}{\frac{100+75+50}{100}}=\frac{100}{225}=\frac{4}{9}\) 
Hence, the required probability is \(\frac{4}{9}\).

\(\int _{ 0 }^{ 2 }{ \frac { 6x+3 }{ { x }^{ 2 }+4 } } dx\)
\(=\int _{ 0 }^{ 2 }{ \frac { 6x }{ { x }^{ 2 }+4 } } dx+\int _{ 0 }^{ 2 }{ \frac { 3 }{ { x }^{ 2 }+4 } } dx\)
\(=3\int _{ 0 }^{ 2 }{ \frac { 2x }{ { x }^{ 2 }+4 } } dx+3\int _{ 0 }^{ 2 }{ \frac { 1 }{ { 2 }^{ 2 }+{ x }^{ 2 } } } dx\)
\(=3{ \left[ \log { \left| { x }^{ 2 }+4 \right| } \right] }_{ 0 }^{ 2 }+\frac { 3 }{ 2 } { \left[ \tan ^{ -1 }{ \frac { x }{ 2 } } \right] }_{ 0 }^{ 2 }\)
\(=3\left[ \log { \left| 8 \right| -\log { \left| 4 \right| } +\frac { 3 }{ 2 } } \left[ \tan ^{ -1 }{ (1)-\tan ^{ -1 }{ (0) } } \right] \right] \)
\(=3\left[ \log { 8 } -\log { 4 } \right] +\frac { 3 }{ 2 } \left[ \frac { \pi }{ 4 } -0 \right] \)
\(=3\log { \frac { 8 }{ 4 } } +\frac { 3\pi }{ 8 } =3\log { 2 } +\frac { 3\pi }{ 8 } .\)

The feasible region determined by the system of constraints, x + 3y ≥ 3, x + y ≥ 2, and x, y ≥ 0, is as follows.

It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (3, 0) B \(\left( \frac { 3 }{ 2 } ,\frac { 1 }{ 2 } \right) \) and C (0, 2).
The values of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 7 may or may not be the minimum value of Z.
For this, we draw the graph of the inequality, 3x + 5y < 7, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 5y < 7. Therefore, the minimum value of Z is 7 at (3/2,1/2).

\(\text { Let } \log x=t \)
\(\therefore \frac{1}{x} d x=d t \)
\(\Rightarrow \int \frac{1}{x(\log x)^{m}} d x=\int \frac{d t}{(t)^{m}} \)
\(=\left(\frac{t^{-m+1}}{1-m}\right)+\mathrm{C} \)
\(=\frac{(\log x)^{1-m}}{(1-m)}+\mathrm{C}\)

Let 'P' be the principal at any time t.
By the question, \(\frac { dP }{ dt } =\frac { r }{ 100 } \times P\)
\(\Rightarrow\) \(\frac { dP }{ P } =\frac { r }{ 100 } dt\) 
Variables Separable
Integrating, \(\int { \frac { dP }{ P } =\frac { r }{ 100 } \int { 1.dt+C } } \)
\(\Rightarrow\) \(log\quad P=\frac { rt }{ 100 } +C\) ...... (1) [ \(\because \)P > 0]
When t = 0, P = 100, \(\therefore\)log 100 = C.
Putting in (1),\(log\quad P=\frac { rt }{ 100 } +log\quad 100\)... (2)
When  \(t=10,P=200,\)
\(log\quad 200=\frac { 10r }{ 100 } +log\quad 100\)
\(\Rightarrow\) \(log\frac { 200 }{ 100 } =\frac { r }{ 10 } \Rightarrow r=10\quad log\quad 2\)
\(=10\left( .6931 \right) =6.931\)
Hence,  \(r=6.93%\)        

\((i)\quad { e }^{ -x }\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } ={ e }^{ -x }.\frac { d }{ dx } (-x)={ e }^{ -x }\quad (-1)=-{ e }^{ -x }\)
\((ii)\quad Let\quad y=sin(log\quad x)\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } =cos(log\quad x).\frac { d }{ dx } (log\quad x)\)
\(=cos(log\quad x).\frac { 1 }{ x } =\frac { cos(log\quad x) }{ x } ,x>0\)
\((iii)\quad Let\quad y={ cos }^{ -1 }({ e }^{ x })\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } =\frac { -1 }{ \sqrt { 1-{ \left( { e }^{ x } \right) }^{ 2 } } } .\frac { d }{ dx } ({ e }^{ x })\)
\(=-\frac { 1 }{ \sqrt { 1-{ e }^{ 2x } } } .{ e }^{ x }=\frac { -{ e }^{ x } }{ \sqrt { 1-{ e }^{ 2x } } } \)
\((iv)\quad Let\quad y={ e }^{ cos\quad x }\)
\(By\quad Chain\quad Rule,\quad \frac { dy }{ dx } ={ e }^{ cos\quad x }\frac { d }{ dx } (cos\quad x)\)
\(={ e }^{ cos\quad x }(-sin\quad x)=-(sin\quad x){ e }^{ cos\quad x }\)

\( \sqrt{1+\sin x}=\cos \frac{x}{2}+\sin \frac{x}{2} \text { and }\\ \sqrt{1-\sin x}=\cos \frac{x}{2}-\sin \frac{x}{2} \)

The given system of equations will have no solution, if \(|A|=0\) 
\(\Rightarrow \left|\begin{array}{ccc} 2 & -1 & 2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{array}\right|=0\)
\(\Rightarrow 2(-2 \lambda-1)+(\lambda-1)+2(1+2)=0\)
\(\Rightarrow-3 \lambda+3=0 \Rightarrow \lambda=1 \)

As distance = \(\left| \frac { 5-\frac { 2 }{ 3 } }{ \sqrt { 4+1+4 } } \right| =\frac { 13 }{ 9 } \) units

As direction cosines of a line whose direction ratio are 2,3, -6 are
\(\frac { -2 }{ 7 } ,\frac { 3 }{ 7 } ,\frac { 6 }{ 7 } \)
As angle with the y-axis is obtuse,
∴ cos β < 0,
Therefore direction ratios are \(\frac { -2 }{ 7 } ,\frac { -3 }{ 7 } ,\frac { 6 }{ 7 } \)

Let r be the radius and h be the height of conical part.
Here, \(\tan 45^{\circ}=\frac{r}{h} \Rightarrow 1=\frac{r}{h} \Rightarrow h=r\)
The volume of water in the tank
\(V=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^3 \mathrm{~cm}^3 \quad[\because h=r]\)
(ii) \( \frac{d V}{d t}=\frac{3}{3} \pi r^2 \frac{d r}{d t} \)
\(\Rightarrow \quad-2=\pi r^2 \frac{d r}{d t} \quad\left[\because \frac{d V}{d t}=-2\right]\)
\( \Rightarrow \quad \frac{d r}{d t}=\frac{-2}{\pi r^2}=\frac{-2}{\pi(2 \sqrt{2})^2}=\frac{-1}{4 \pi}\)
\({[r=2 \sqrt{2} \text { (given) }]}\)
Hence, the rate of decrease of radius is \(\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\).
(iii) (b) \(\therefore S=\pi r l \Rightarrow S=\pi h l\) \([\because r=h]\)
\(\Rightarrow \quad S=\pi h \sqrt{h^2+r^2}\)
(a) \(\therefore S=\pi r l \)
\(=\pi r \sqrt{h^2+r^2}=\pi r \sqrt{2 r^2} \)
\({\left[\because l=\sqrt{h^2+r^2} ; \ \because h=r\right]}\)
\(=\sqrt{2} \pi r^2\)
\(\Rightarrow \frac{d S}{d t}=\sqrt{2} \pi(2 r) \frac{d r}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi r \frac{d r}{d t}\)
\(\Rightarrow\left(\frac{d S}{d t}\right)_{r=2 \sqrt{2}}=2 \sqrt{2} \pi(2 \sqrt{2})\left(\frac{-1}{4 \pi}\right)\)
\(=-2 \mathrm{~cm}^2 / \mathrm{s} \)
Or \(\text { (b) } \therefore S=\pi r l \Rightarrow S=\pi h l\) \({[\because r=h]}\)
\( \Rightarrow S=\pi h \sqrt{h^2+r^2} \)
\(\Rightarrow \quad S=\sqrt{2} \pi h^2 \)
\(\Rightarrow \quad \frac{d S}{d t}=\sqrt{2} \pi(2 h) \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi h \frac{d h}{d t}\)
\( \Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi(2 \sqrt{2}) \frac{d h}{d t}\)
\( \text { [since, } l^2=r^2+h^2 \Rightarrow l^2=2 h^2\)
\(\left.\Rightarrow 16=2 h^2 \Rightarrow h=2 \sqrt{2}\right]\)
\( \Rightarrow \quad-2=8 \pi \frac{d h}{d t}\) \({[\because \text { from Eq. (i) }]}\)
\( \Rightarrow \quad \frac{d h}{d t}=-\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\)
\(\text { (b) } \therefore S=\pi r l \Rightarrow S=\pi h l\)
\([\because r=h]\)
\(\Rightarrow S=\pi h \sqrt{h^2+r^2}\)
\(\Rightarrow \quad S=\sqrt{2} \pi h^2\)
\(\Rightarrow \quad \frac{d S}{d t}=\sqrt{2} \pi(2 h) \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi h \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi(2 \sqrt{2}) \frac{d h}{d t}\)
\(\text { [since, } l^2=r^2+h^2 \Rightarrow l^2=2 h^2\)
\(\left.\Rightarrow 16=2 h^2 \Rightarrow h=2 \sqrt{2}\right]\)
\(\Rightarrow \quad-2=8 \pi \frac{d h}{d t}\)
\([\because \text { from Eq. (i) }]\)
\(\Rightarrow \quad \frac{d h}{d t}=-\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\)

Given, P(E₁) = 0.65, P(E/E₁) = 0.35 and \(P\left(\frac{E}{E_2}\right)=0.80\)
(i) P(E₂) = 1 - P(E₁) = 1 - 0.65 = 0.35
(ii) \(\begin{aligned} P(E) & =P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right) \end{aligned}\)
\(\begin{aligned} =0.65 \times 0.35+0.35 \times 0.80 \end{aligned}\)
\(\approx 0.23+0.28=0.51\)
(iii) (a) \(\begin{aligned} P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) P\left(E / E_2\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.65 \times 0.35}{0.65 \times 0.35+0.35 \times 0.80}=0.45 \end{aligned}\)

or (b) \(\begin{aligned} P\left(\frac{E_2}{E}\right) & =\frac{P\left(E_2\right) P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) P\left(E / E_2\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.35 \times 0.80}{0.65 \times 0.35+0.35 \times 0.80} \end{aligned}\)
= 0.55

(i) (c) : Let OAB be a triangle such that

\(\overrightarrow{A O}=-\vec{p}, \overrightarrow{A B}=\vec{q}, \)\(\overrightarrow{B O}=\vec{r}\)
Now, \(\vec{q}+\vec{r} =\overrightarrow{A B}+\overrightarrow{B O}\)
\(=\overrightarrow{A O} =-\vec{p}\)
(ii) (c) : From triangle law of vector addition,
\(\overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{B C}+\overrightarrow{C D}\)

\( =\overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C D}\)
\(=\overrightarrow{A B}+2 \overrightarrow{B C}-\overrightarrow{A B}=2 \overrightarrow{B C} \)
 (iii) (b) : \(\operatorname{In} \Delta A B C, \overrightarrow{A C}=2 \vec{a}+2 \vec{b}\)

and in \(\Delta A B D, 2 \vec{b}=2 \vec{a}+\overrightarrow{B D}\)..(ii)
[By triangle law of addition]
Adding (i) and (ii), we have
\( \overrightarrow{A C}+2 \vec{b}=4 \vec{a}+\overrightarrow{B D}+2 \vec{b} \)
⇒\( \overrightarrow{A C}-\overrightarrow{B D}=4 \vec{a} \)
(iv) (d) :  \(\text { In } \Delta A B C, \overrightarrow{B A}+\overrightarrow{A C}=\overrightarrow{B C}\)... (i)
[By triangle law]
In \(\Delta B C D, \overrightarrow{B C}+\overrightarrow{C D}=\overrightarrow{B D}\) ..(i)
From (i) and (ii), \(\overrightarrow{B A}+\overrightarrow{A C}=\overrightarrow{B D}-\overrightarrow{C D}\)
\(\Rightarrow \overrightarrow{B A}+\overrightarrow{C D}=\overrightarrow{B D}-\overrightarrow{A C}=\overrightarrow{B D}+\overrightarrow{C A}\)
(v) (b): Since T is the mid point of YZ.
So,\(\overrightarrow{Y T}=\overrightarrow{T Z}\)
Now,\(\overrightarrow{X Y}+\overrightarrow{X Z}=(\overrightarrow{X T}+\overrightarrow{T Y})+(\overrightarrow{X T}+\overrightarrow{T Z})\)
[By triangle law]
\(=2 \overrightarrow{X T}+\overrightarrow{T Y}+\overrightarrow{T Z}=2 \overrightarrow{X T} \quad[\because \overrightarrow{T Y}=-\overrightarrow{Y T}]\)