Given \(a*b=\frac { a+b }{ 2 } \forall \in Q\)
i) Closure property:
Let a, b ∈ Q
∴ a*b = \(\frac{a+b}{2}\)∈Q
[∵ addition and division are closed on Q]
* is closed on Q.
(ii) Commutative property:
Let a, b ∈ Q
Then a+b \(=\frac { a+b }{ 2 } =\frac { b+a }{ 2 } =b*a\)
∴ a*b = b*a ∀a,b∈Q
∴ * is commutative on Q.
(iii) Associative property :
Let a, b, c ∈ Q
a*(b*c) = (a*b)*c
Let a = 2, b = 3, c-5
∴ a*(b*c) = 2*(3*-5)
\(=*\left( \frac { 3-5 }{ 2 } \right) \)
\(=2*(-1)=\frac { 2+(-1) }{ 2 } \)
\(=\frac { 1 }{ 2 } \quad \quad ...(1)\)
Now (a*b)*c = (2*3)*(-5)
\(=\left( \frac { 2+3 }{ 2 } \right) *(-5)\)
\(=\frac { 5 }{ 2 } *-5=\frac { \frac { 5 }{ 2 } +(-5) }{ 2 } \)
\(=\frac { 5-10 }{ 4 } =\frac { -5 }{ 4 } ...(2)\)
From (1) & (2), a*(b*c) ≠ (a*b)*c
∴ * is not associative on Q.

\(\frac { \partial z }{ \partial x } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } (y)+{ tan }^{ -1 }(xy)\)
\(\frac { \partial z }{ \partial y } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)(x)
\(\frac { \partial z }{ \partial z } =\frac { xy }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \) + tan^-1 (xy)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial x } =\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } +{ tan }^{ -1 }({ t }^{ 2 }{ se }^{ 2 })\)
\(\frac { \partial z }{ \partial y } =\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\frac { dx }{ dt } =2t;\frac { dy }{ dt } ={ s.e }^{ t }\)
Also, \(\frac { dx }{ ds } =0;\frac { dy }{ ds } ={ e }^{ t }\)
By chain rule
\(\frac { dw }{ ds } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial z }{ \partial y } .\frac { dy }{ ds } \)
= \(\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } (0)+\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } { (e }^{ t })=\frac { { e }^{ t }{ t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\therefore { \left( \frac { \partial z }{ \partial s } \right) }_{ s=t=1 }=\frac { e(1) }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
By chain rule
= \(\frac { dz }{ dt } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
\(\therefore { \left( \frac { \partial z }{ \partial t } \right) }_{ (s=t=1) }=\frac { e }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
= \(\frac { 2e+e }{ 1+{ e }^{ 2 } } =\frac { 3e }{ 1+{ e }^{ 2 } } \)+ 2 tan^-1(e)

Equation of the given curve is 2+x-x²+y = 0

x	0	2	-1
y	-2	0	0

\(\Rightarrow\)y = x²-x-2

\(\therefore\) Required area= \(\int _{ -3 }^{ -1 }{ ydx } +\int _{ -1 }^{ 2 }{ -y } dx+\int _{ 2 }^{ 3 }{ ydx } \)
\(\\ =\int _{ -3 }^{ -1 }{ \left( { x }^{ 2 }-x-2 \right) } dx\int _{ -1 }^{ 2 }{ (2+x-{ x }^{ 2 })dx } +\int _{ 2 }^{ 3 }{ ({ x }^{ 2 }-x-2) } dx\)
\(={ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ -3 }^{ -1 }+{ \left( 2x+\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 2 }+{ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ 2 }^{ 3 }\)
\(=\left( -\frac { 1 }{ 3 } -\frac { 1 }{ 2 } +2 \right) -\left( -9-\frac { 9 }{ 2 } +6 \right) +\left( 4+2-\frac { 8 }{ 3 } \right) -\left( -2+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } \right) +\left( 9-\frac { 9 }{ 2 } -6 \right) -\left( \frac { 8 }{ 3 } -2-4 \right) \)
\(\\ =\left( \frac { -2-3+6 }{ 6 } \right) -\left( \frac { -6-9 }{ 2 } \right) +\left( \frac { 18-8 }{ 3 } \right) -\left( \frac { -12+3+2 }{ 6 } \right) +\left( \frac { 6-9 }{ 2 } \right) -\left( \frac { 8-24 }{ 3 } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 15 }{ 2 } +\frac { 10 }{ 3 } +\frac { 7 }{ 6 } -\frac { 3 }{ 2 } +\frac { 16 }{ 3 } \)
\(=\frac { 1+45+20+7-9+32 }{ 6 } =\frac { 90 }{ 6 } =15 \)

Since the given function is a probability mass function, the total probability is one. That is \(\underset { x }{ \Sigma } f(x)=1\)
 From the given data k + 2k + 6k + 5k + 6k +10k+1
\(30k=1\Rightarrow k=\frac { 1 }{ 30 } \) 
Therefore the probability mass function is

x	1	2	3	4	5	6
f(x)	\(\cfrac { 1 }{ 30 } \)	\(\cfrac { 2 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 5 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 10 }{ 30 } \)

(i) P(2 < X < 6) = f(3)+ f(4)+ f(5) = \(\frac { 6 }{ 30 } +\frac { 5 }{ 30 } +\frac { 6 }{ 30 } =\frac { 17 }{ 30 } \) 
(ii) P(2≤X≤5) = f(2)+f(3)+f(4) = \(\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 13 }{ 30 } \) 
(iii) P(2≤4) = f(1)+f(2)+f(3)+f(4) = \(\frac { 1 }{ 30 } +\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 14 }{ 30 } \) 
(iv) P(3>X) = f(4)+f(5)+f(6) = \(\frac { 5 }{ 30 } +\frac { 6 }{ 30 } +\frac { 10 }{ 30 } =\frac { 21 }{ 30 } \)   

\(put\sqrt { t } =z\Rightarrow \frac { 1 }{ 2\sqrt { t } } dt=dz\)
\(\Rightarrow \frac { 1 }{ 2z } dt=dz\Rightarrow dt=2zdz\)

t	0	sin2x
z	0	sin x

t	0	cos2x
z	0	cos x

\(\therefore I=\int _{ 0 }^{ sin\quad x }{ 2z{ sin }^{ -1 }zdz } +\int _{ 0 }^{ cosx }{ 2z{ cos }^{ -1 }zdz } ...(1)\)
= I₁+ I₂ 
\(=2\int _{ 0 }^{ x }{ \theta sin\theta cos\theta d\theta +2\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta cos\theta (-sin\theta )d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta 2sin\theta cos\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta 1sin\theta cos\theta d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta sin2\theta d\theta } } \)
\([\because sin2\theta =sin\theta cos\theta ]\)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta \int _{ x }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } } \)
\(\left[ \because \int _{ a }^{ c }{ f(x)dx+\int _{ c }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(x)dx } } } \right] \)
\(={ \left[ -\frac { \theta cos2\theta }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos2\theta }{ 2 } d\theta } \)

\(={ \left[ -\frac { \theta cos2\theta }{ 2 } +\frac { sin2\theta }{ 4 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=\frac { -\frac { \pi }{ 2 } cos\pi }{ 2 } +\frac { sin\pi }{ 4 } -(0+0)\)
\(=\frac { -\frac { \pi }{ 2 } (-1) }{ 2 } +\frac { 0 }{ 4 } =\frac { \pi }{ 4 } \)

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

By definition, we have |x + 3| =  \(\begin{cases} \begin{matrix} x+3 & x\ge -3 \\ -x-3 & x<-3 \end{matrix}\end{cases}\)
for the graph of y =| x + 3 | in −4 \(\le\) x \(\le\) 4.
∴ \(\int ^4_{-4}\)|x+3| dx =  \(\int ^{-3}_{-4}\) |x+3| dx + \(\int ^{4}_{-3}\) |x+3| dx = \(\int ^{-3}_{-4}\)|-x-3| dx+ \(\int ^{4}_{-3}\)(x+3)dx
= \([- \frac {x^2}{2}-3x]^{-3}_{-4}\) +  \([ \frac {x^2}{2}+3x]^{4}_{-3}\)
= \(( -\frac {9}{2}+9)\) - \(( -\frac {16}{2}+12)\) + \(( \frac {16}{2}+12)\) - \(( \frac {9}{2}-9)\) = \(( \frac {9}{2})\) - 4+20 + \(( \frac {9}{2})\) = 25

Let the price of low-grade steel be Rs. p per tonne. Then the price of high-grade steel is Rs. 2p per tonne.
The total receipt per day is given by \(R=px+py=px+2p\left( \frac { 40-5x }{ 10-x } \right) \). Hence the problem is to maximise R . Now, simplifying and differentiating R with respect to x , we get
\(R=p\left( \frac { 80-{ x }^{ 2 } }{ 10-x } \right) \)
\(\frac { dR }{ dx } =p\left( \frac { { x }^{ 2 }-20x+80 }{ (10-x)^{ 2 } } \right) \)
\(\frac { dR }{ dx } =-\frac { 40P }{ \left( 10-x \right) ^{ 3 } } \)
Now, \(\frac { dR }{ dx } =0\Rightarrow { x }^{ 2 }-20x+80=0\) and hence \(x=10\pm 2\sqrt { 5 } \) 
At \(x=10-2\sqrt { 5 } ,\frac { { d }^{ 2 }R }{ { dx }^{ 2 } } <0\) and hence R will be maximum. If x \(x=10-2\sqrt { 5 } \) then \(y=5-\sqrt { 5 } \)  
Therefore the steel plant must produce low-grade and high-grade steels respectively in tonnes per day are \(10-2\sqrt { 5 } \)  and \(5-5\sqrt { 5 } \)

We have,
\(f(x)=log(1+x)-\frac{x}{1+x}\)
Therefore, \(f'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\).
Hence, f′(x) is \(\begin{cases} <0 \ when-10 \ whenx>0 \end{cases}\)
Therefore f (x) is strictly increasing for x > 0 and strictly decreasing for x < 0. Since f′(x) changes from negative to positive when passing through x = 0, the first derivative test tells us there is a local minimum at x = 0 which is f (0) = 0. Further, for x > 0, f(x) > f (0) = 0 gives
\(log(1+x)-\frac{x}{1+x}>0 \Rightarrow log(1+x)>\frac{x}{1+x}\).

Let T be the temperature of the coffee at time t
and T_m' the temperature of the kitchen.
By Newton's law of cooling
\(\frac { dT }{ dt } =K(T-{ T }_{ m })\)
\(\Rightarrow \frac { dT }{ dt } =K(T-70)\)
\(\Rightarrow \int { \frac { dT }{ T-70 } =K\int { dt } } \)
\(\Rightarrow log(T-70)=kt+logC\)
\(\Rightarrow log(T-70)-logC=Kt\)
\(\Rightarrow log\left( \frac { T-70 }{ C } \right) =Kt\)
\(\Rightarrow \frac { T-70 }{ C } ={ e }^{ Kt }\)
\(\Rightarrow T-70={ Ce }^{ Kt }...(1)\)
\(\\ When\ t=0,\ T={ 180 }^{ o }F\)
\(\therefore { 180 }^{ o }-{ 70 }^{ o }={ Ce }^{ 0 }\)
\(\Rightarrow C={ 11 }0^{ 0 }\)
\(\\ \therefore (1)\Rightarrow T-70=110{ e }^{ Kt } ..(2)\)
\(When\ t=0,T=160\)
\(\therefore 160-70=110{ e }^{ 10K }\)
\(90=110{ e }^{ 10K }\)
\(\Rightarrow { e }^{ 10K }=\frac { 9 }{ 11 } \)
\(\Rightarrow { e }^{ K }={ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } }...(3)\)
(i) when t = 15, (2) becomes,
\(\Rightarrow T-70=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } \times 15 }\)
\(=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 3 }{ 2 } }\)
\(=110\times { \left( \frac { 9 }{ 11 } \right) }\left( \sqrt { \frac { 9 }{ 11 } } \right) \)
\(=110\times \frac { 9 }{ 11 } \times \frac { 3 }{ \sqrt { 11 } } \)
\(=\frac { 270 }{ \sqrt { 11 } } =\frac { 270 }{ 3.32 } =81.33\)
\(\Rightarrow\) T=81.33+70=151.3F
\(\therefore\) T = 151.3F
\(\therefore\) The temperature of the coffee at 10.15 am is 151.3F
(ii) when T = 130F, (2) becomes
T-70 = 110e^kt  ...(2)
\(\Rightarrow\) 130-70 = 110e^kt 
60 = 110e^kt 
e^kt  = \(\frac{6}{11}\)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 6 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 6 }{ 11 } \right) }{ log\left( \frac { 9 }{ 11 } \right) } \)
\(=\frac { log(0.545) }{ log(0.818) } =\frac { -0.264 }{ -0.087 } \)
= 3.34
t = 30.34min
T = 140F (2)becomes
140-70 = 110e^kt ...(2)
\(\Rightarrow 70={ 110e }^{ kt }\)
\({ e }^{ kt }=\frac { 7 }{ 11 } \)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 7 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 7 }{ 11 } \right) }{ log\left( \frac { 7 }{ 11 } \right) } =\frac { -0.197 }{ -0.087 } \)
= 2.26
t = 22.6min
\(\therefore\) Between 10.22 min to 10.30 min, the woman should have drunk the coffee.

\(​​​​​​\underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }\)
This is an indeterminate of the form 0⁰
Let g(x) x^x
Taking logarithm, we get
\(log \ g(x)=log({ x }^{ 2 })=xlogx=\frac { log\quad x }{ \frac { 1 }{ x } } \)
\(\therefore ​​​​​​\underset { x\rightarrow { 0 }^{ + } }{ lim } log \ g(x)={ \left[ \frac { logx }{ \frac { 1 }{ x } } \right] }=\frac { \infty }{ \infty } \)
\(=\underset { x\rightarrow { 0 }^{ + } }{ lim } \left( \frac { \frac { 1 }{ x } }{ -\frac { 1 }{ { x }^{ 2 } } } \right) \) [by L' Hopital rule]
= \(\underset { x\rightarrow { 0 }^{ + } }{ lim } \frac { 1 }{ x } \times \frac { { x }^{ 2 } }{ 1 } =\underset { x\rightarrow { 0 }^{ + } }{ lim } -x\)
= 0
But \(\underset { x\rightarrow { 0 }^{ + } }{ lim } (log(g(x))=log(\underset { x\rightarrow { 0 }^{ + } }{ lim } (g(x))\)
ஃ \(\underset { x\rightarrow { 0 }^{ + } }{ lim } (log(g(x))=0\)
\(\Rightarrow { e }^{ log }(\underset { x\rightarrow { 0 }^{ + } }{ lim } log(g(x))={ e }^{ 0 }\)
\(\Rightarrow \underset { x\rightarrow { 0 }^{ + } }{ lim } g(x)=1\)
\(\therefore \underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }=1\)

Let A be the number of bacteria at any time t
Given \(\frac { dA }{ A } \alpha A\)
\(\Rightarrow \frac { dA }{ dt } =kA\)
\(\Rightarrow \frac { dA }{ dt } =kdt\)
\(\Rightarrow \int { \frac { dA }{ dt } } =k\int { \frac { dt }{ t } } \)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow \frac { A }{ c } ={ e }^{ kt }\)
\(\Rightarrow A=c.{ e }^{ kt }\) ...(1)
Initially when t = 0, assume that A = A₀
\(\therefore\) becomes, = ce₀ \(\Rightarrow\) c = A₀ 
\(\therefore\) A = A0ekt  ...(2)
Given when t = 5, A = 3A₀
3A₀ = A₀e^5k \(\Rightarrow\) 3 = 35k 
when t = 10, 2 becomes,
A = A₀e^10k
= A₀(e^5k)2 = A₀(3)²
\(\Rightarrow\) A = 9A₀
Hence, the amount of bacteria will be present after 10 hours is 9 times the initial amount.

Let (x) = cos2 x
f^I(x) = 2cos x (- sin x)
= - sin 2x ⇒ f^l(0) = 0
f^Il(x) = - 2 cos 2x ⇒ f^Il(0) = -2
f^III(x) = + 4 sin 2 x ⇒ f^III(0) = 0
f^IV(x) = 8 cos 2 x ⇒ f^IV(0) = 8
f^V(x) = -16 sin 2x ⇒ f^V(0) = 0
f^VI(x) = - 32 cos 2x ⇒ f^VI(0) = -32
∴ Maclaurin's series
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\) .......
∴ cos x = 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { 8{ x }^{ 4 } }{ 4! } +\frac { { 32x }^{ 6 } }{ 6! } \) + ...
= 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { { 2 }^{ 3 }{ x }^{ 4 } }{ 4! } +\frac { { { 2 }^{ 5 }x }^{ 6 } }{ 6! } \)+ ...

The given linear differential euation is of the form
\(\frac{d y}{d x}+P y=Q  \)
 where \(\mathrm{P}=\frac{1}{(1-x) \sqrt{x}} ; \mathrm{Q}=1-\sqrt{x}  \)
Take \(\sqrt{x}=\mathrm{t} \Rightarrow \mathrm{t}^2=x \)
\( x^{1 / 2}=\mathrm{t}  \)
\( \frac{1}{2} x^{1 / 2-1} \mathrm{~d} x=\mathrm{dt} \Rightarrow \frac{1}{2} x^{-1 / 2} d x=\mathrm{dt} \)
\( \frac{1}{2 x^{1 / 2}} d x=\mathrm{dt} \Rightarrow \frac{1}{2 \sqrt{x}} d x=\mathrm{dt}  \)
\( \frac{d x}{\sqrt{x}}=2 \mathrm{dt} \)
\( \text { I.F }=e^{\int P d x}=e^{\int \frac{1}{(1-x) \sqrt{x}} d x}  \)
\( =e^{\int \frac{1}{1-t^{t^2} 2 d t}}=e^{\int \frac{2 d t}{1-1^2}} \)
\(\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 x} \log \left|\frac{a+x}{a-x}\right|  \) 
\(\text { Here } a=1 ; x=t\)
\(=e^{\log \left(\frac{1+1}{1-t}\right)} \)
\(\mathrm{I} . \mathrm{F}=\frac{1+t}{1-t} \)
\({ I.F }=\frac{1+\sqrt{x}}{1-\sqrt{x}} \)
The solution is y\(\times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times \mathrm{I}.Fd x+c\)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=\int(1-\sqrt{x}) \frac{1+\sqrt{x}}{1-\sqrt{x}} d x+c  \)
\( =\int(1+\sqrt{x}) d x+c  \)
\(  y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{x^{3 / 2}}{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x^{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x \sqrt{x}+c \quad \because x^{3 / 2}=x \sqrt{x} \)
Which is the required solution

The given equation can be rewritten as \(\frac { dy }{ dx } +\frac { (x\quad tan\quad x-1) }{ x } y=xcosx\)
This is a linear differential equation. Here \(P=\frac { (x\quad tan\quad x-1) }{ x } ;Q=xcosx\)
\(\int { Pdx } =\int { \frac { (xtanx-1) }{ x } } dx=-log|cosx|-log|x|=-log|xcosx|=log\frac { 1 }{ |xcosx| } \)
Thus, \(I.F.={ e }^{ \int { pdx } }={ e }^{ log\frac { 1 }{ |xcosx| } }=\frac { 1 }{ xcosx } \)
Hence the solution is \({ ye }^{ \int { Pdf } }=\int { Q{ e }^{ \int { Pdx } }dx+C } \)
i.e., \(y\frac { 1 }{ xcosx } =\int { (xcosx)\frac { 1 }{ xcosx } dx+C } \)
or \(\\ \\ \\ \\ y\frac { 1 }{ xcosx } =x+C\)
or y = x² cos x + Cx cos x is the required solution.

f (x) is defined and differentiable for all x∈(-∞, ∞)
\(f'\left( x \right) =\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \)
\(f''(x)=\frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) \)
f"(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) =0\Rightarrow { e }^{ x }-{ e }^{ -x }=0\) 
\(\Rightarrow { e }^{ x }={ e }^{ -x }\Rightarrow { e }^{ x }=\frac { 1 }{ { e }^{ x } } \) 
\(\Rightarrow { e }^{ 2x }=1\Rightarrow { e }^{ 2x }={ e }^{ 0 }\) 
\(\Rightarrow 2x=0\Rightarrow x=0\) 
The possible intervals are (-∞,0) and (0,∞)

Intervel	(-∞, 0)	(0, ∞)
Sign of f"(x)	Say x = -1 \(\cfrac { 1 }{ 2 } \left( { e }^{ -1 }-{ e }^{ 1 } \right) =-ve\)	Say x = 1 \(\cfrac { 1 }{ 2 } \left( { e }^{ -1 }-{ e }^{ 1 } \right) =+ve\)
Concavity	Concave down	Concave up

ஃ f(x) is concave up in (0, ∞) and concave down in (∞, 0).
Since f"(x) changes its position from negative to positive, when it passes through x = 0 the points of inflection is (0,1(0))
\(f(0)=\frac { 1 }{ 2 } \left( { e }^{ o }-{ e }^{ o } \right) =\frac { 1 }{ 2 } \left( 1-1 \right) =0\) 
ஃ (0, 0) is the point of inflection.

\( y e^{\left(\frac{1}{r}\right)} \cdot d x   =\left(x e^{\frac{1}{y}}+y\right) d y \)
\(\frac{d x}{d y}   =\frac{x \cdot e^{\left(\frac{6}{y}\right)}+y}{y e^{\left(\frac{x}{y}\right)}} \)
\(\frac{d x}{d y}   =\left(\frac{x}{y}\right)+\frac{1}{e^{\left(\frac{6}{y}\right)}}\)
Put x = \( \mathrm{vy}\) \( \Rightarrow\left(\frac{x}{y}\right)=\mathrm{v}\) and \(\frac{d x}{d y}=\cdot v+y \cdot \frac{d v}{d y} \)
\((1) \Rightarrow \ v+y \cdot \frac{d v}{d y}=v+\frac{1}{e^y} \)
\(\mathrm{e}^v \cdot \mathrm{dv}=\frac{d y}{y}\)
Integrating on both sides,
ie) \(\int e^v \cdot d v   =\int \frac{d y}{y} \)
\(e^v   =\log |y|+\log |c|  \) 
\(e^{\left(\frac{x}{y}\right)}   =\log |c y|\)

\(\\ \\ \\ \frac { dy }{ dx } ={ tan }^{ 2 }(x+y)...(1)\)
Take x + y = t
\(\Rightarrow 1+\frac { dy }{ dx } =\frac { dt }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { dt }{ dx } -1\)
∴ (1) becomes,
\(\frac { dt }{ dx } -1={ tan }^{ 2 }t\)
\(\Rightarrow \frac { dt }{ dx } ={ tan }^{ 2 }t1\)
\(\Rightarrow \frac { dt }{ dx } ={ sec }^{ 2 }(t)\)
\(\Rightarrow \frac { dt }{ { sec }^{ 2 }t } =dx\)
\(\Rightarrow { cos }^{ 2 }t\quad dt=dx\)
\(\left(\frac{1+\cos 2 t^{\circ}}{2}\right) d t=\mathrm{d} x \quad\left(\because \cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\right)\)
\(\left[ cos\quad 2x=2{ cos }^{ 2 }x-1{ cos }^{ 2 }x=\frac { 1+cos2x }{ 2 } \right] \)
Taking integration on both sides, we get
\(\Rightarrow \left( \frac { 1+cos2\quad t }{ 2 } \right) dt=dx\)
\(\Rightarrow \frac { 1 }{ 2 } \int { (1+cos2t)dt=\int { dx } } \)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { sin2t }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { 2sintcost }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } [t+sin\ t\ cost]=x+c\ [\because t=x+y]\)
\(\Rightarrow \frac { 1 }{ 2 } [x+y+sin(x+y)cos(x+y)=x+c\)

Given h = 12 m
r = 5 m
∴ tan α = \(\frac { OA }{ VO } =\frac { 5 }{ 12 } =\frac { r }{ h } \)
⇒ 12r = 5h 
⇒ r = \(\frac{5h}{12}\) ...(1)
Given rate of changeof volume = \(\frac{dv}{dt}=10\)
Volume of cone V = \(\frac13\) πr²h
= \(\frac { 1 }{ 3 } \)π\({ \left( \frac { 5h }{ 12 } \right) }^{ 2 }.h\) [From (1)]
\(v=\frac { 1 }{ 3 } \pi { \left( \frac { 25h }{ 144 } \right) }^{ 2 }=\frac { { 25\pi h }^{ 3 } }{ 3\times 144 } \)
Differentiating with respect to 't' we get,
\(\frac { dv }{ dt } =\frac { 25\pi }{ 3\times 144 } 3{ h }^{ 2 }\frac { dh }{ dt } \)
= \(\frac { 25\pi }{ 144 } { h }^{ 2 }\frac { dh }{ dt } \)
[∵ Given h = 8, \(\frac { dh }{ dt } \) = 10m³/min]
\(\Rightarrow \frac { 10\times 144 }{ 25\pi \times 64 } =\frac { dh }{ dt } \)
\(\Rightarrow \frac { dh }{ dt } =\frac { 9 }{ 10\pi } \)
The depth of the water increasing at the rate of \(=\frac { 9 }{ 10\pi } \)m / min.

The volume of the baloon of radius r is \(V =\frac{4}{3}\pi r ^{3} \)
We are given \(\frac{dV }{dt }=1000 \) and we need to find \(\frac{dr }{dt} \) when r = 7. Now,
\(\frac{dV}{dt}=3\times\frac{4}{3}\pi r^{2}\times \frac{dr}{dt}\)
Substituting r = 7 an \(\frac{dV}{dt}\) = 1000, we get 1000 \(= 4\pi \times 49 \times \frac{dr}{dt}\)
Hence, \(\frac{dr}{dt}=\frac{1000}{4\times49\times \pi}=\frac{250}{49\pi}\)

The surface area S of the baloon is S = 4ㅠr². Therefore, \(\frac{dS}{dt}=8\pi \times r \times \frac{dr}{dt}\)
Substituting\(\frac{dr}{dt}=\frac{250}{49\pi}\) and r = 7, we get \(\frac{dS}{dt}=8\pi\times7\times \frac{250}{49\pi}=\frac{2000}{7}\)
Therefore, the rate of change of radius is \(\frac{250}{49 \pi}\) cm/sec and the change of surface area is \(\frac{2000}{7}\) cm² / sec

3x + 3y − z = 11, 2x − y + 2z = 9, 4x + 3y + 2z = 25.
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-2-6)-3(4-8)-1(6+4)·
= 3(- 8) - 3(- 4) - 1(10)
= - 24 + 12 - 10 = - 22
Δ₂ = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= \(11\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -1\left| \begin{matrix} 9 & -1 \\ 25 & 3 \end{matrix} \right| \)
= 11(-2-6)-3(18-50)-1(27+25)
= 11(-8)-3(-32)-1(52)
= -88+96-52 = -44
Δ₂ = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -11\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| 2\begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| \)
= 3(18 - 50) -11(4 - 8) - 1(50- 36)
= 3(- 32) - 11(- 4) - 1(14)
= -96+44-14 = - 66
Δ₃ = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
\(3\left| \begin{matrix} -1 & 9 \\ 3 & 25 \end{matrix} \right| -3\left| \begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| -11\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-25-27)-3(50-36)+ 11(6+4)
= 3(- 52) - 3(14) + 11(10)
= -156 - 42 + 110= - 88
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -44 }{ -22 } \) = 2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -66 }{ -22 } \) = 3
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { -88 }{ -22 } \) = 4
∴ Solution set is {2, 3, 4}

3x + 2y + 7z = 0, 4x − 3y − 2z = 0, 5x + 9y + 23z = 0
Reducing the augmented matrix to row-echelon form we,
[A|0] =\(\left[ \begin{matrix} 3 & 2 & 7 \\ 4 & -3 & -2 \\ 5 & 9 & 23 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-\frac { 4 }{ 4 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 3 & 2 & 7 \\ 0 & -\frac { 17 }{ 3 } & \frac { -34 }{ 3 } \\ 5 & 9 & 23 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 5 }{ 3 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 2 & 7 \\ 0 & -\frac { 17 }{ 3 } & \frac { -34 }{ 3 } \\ 5 & \frac { 17 }{ 3 } & \frac { 34 }{ 3 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & 2 & 7 \\ 0 & -\frac { 17 }{ 3 } & \frac { -34 }{ 3 } \\ 5 & 0 & 0 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times \frac { -3 }{ 7 } }{ \longrightarrow } \left[ \begin{matrix} 3 & 2 & 7 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|0] = 2
So, \(\rho \)(A) = \(\rho \)(A|0]) = 2<3 = number of unknowns 
So put z = t where t \(\in \) R
writing the equations using the echelon form, we get
3x+2y+7z = 0 ............(1)
y+2z = 0 .............(2)
put z = t, (2) becomes
y+2t = 0
⇒ y = 2t
∴ (1) becomes, 3x+2(-2)+7t = 0
⇒ 3x-4t+7t = 0
⇒ 3x+3t = 0
⇒ 3x = -3t
⇒ x = -t
∴ Solution set is {-t, -2t, t} where t \(\in \) R

\(\vec { r } =(5\hat { i } +7\hat { j } -3\hat { k } )+s(-4\hat { i } +4\hat { j } -5\hat { k } )\) and \(\vec { r } =(8\hat { i } +4\hat { j } +5\hat { k } )+t(7\hat { i } +\hat { j } +3\hat { k } )\)
Let \(\vec { a } =5\hat { i } +7\hat { j } -3\hat { k } ,\vec { b } =4\hat { i } +4\hat { j } -5\hat { k } \)
\(\vec { c } =8\hat { i } +4\hat { j } +5\hat { k } \ and\ \vec { d } =7\hat { i } +\hat { j } +3\hat { k } \)
We know that the two given lines are co-planar
if \(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{matrix} \right| \)
= \(\hat { i } \left( 12+5 \right) -\hat { j } \left( 12+35 \right) +\hat { k } \left( 4-28 \right) \)
= \(17\hat { i } -47\hat { j } -24\hat { k } \)
\(\left( \vec { c } -\vec { a } \right) =\left( 8-5 \right) \hat { i } +\left( 4-7 \right) \hat { j } +\left( 5+3 \right) \hat { k } \)
= \(3\hat { i } -3\hat { j } +8\hat { k } \)
Now,\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =\left( 3\hat { i } -3\hat { j } +8\hat { k } \right) \)
\(\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \)
= \(51+141-192=192-192=0\)
\(\therefore\) The two given lines are co-planar.
The plane containing the two given co-planar lines is
\(\left( \vec { r } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\Rightarrow \left( \vec { r } -5\hat { i } +7\hat { j } -3\hat { k } \right) \times \left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) -\left[ \left( 5\hat { i } +7\hat { j } -3\hat { k } \right) .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \right] =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =\left[ 85-329+72 \right] \)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =-172 \)
which is the required vector equation of the plane.

We have to find \((\sqrt{3}+1)^{\frac{1}{3}}\). Let \(z=(\sqrt{3}+i)^{\frac{1}{3}}\). Then \({ z }^{ 3 }=\sqrt { 3 } +i=r\left( cos\theta +isin\theta \right) \)
Then, \(r=\sqrt { 3+1 } =2\) and \(\alpha =\theta =\frac { \pi }{ 6 } \) (\(\because \sqrt{3}+i\) lies in the first quadrant)
Therefore, \({ z }^{ 3 }=\sqrt { 3 } +i=2\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow z=\sqrt [ 3 ]{ 2 } \left( cos\left( \frac { \pi +12k\pi }{ 18 } \right) +isin\left( \frac { \pi +12k\pi }{ 18 } \right) \right) \), k = 0, 1, 2.
Taking k = 0, 1, 2, we get
k = 0, z \(={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 1, \(z={ z }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 2, \(z={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { 25\pi }{ 18 } +sin\frac { 25\pi }{ 18 } \right) ={ 2 }^{ \frac { 1 }{ 3 } }\left( -cos\frac { 7\pi }{ 18 } -sin\frac { 7\pi }{ 18 } \right) \)

As per the given information, we can take the parabola as open downward.

∴ Its equation is x² = -4ay ...(1)
Let P be a point on the flow paths, 2.5 m below the line of the pipe and 3m beyond the vertical line through the end of the pipe.

∴ Pis (3, -2.5)
∴ (1) becomes 3² = -4a(-2.5)
⇒  \(\frac { -9 }{ 2.5 } \) = 4a
∴ (1) becomes, x₁² = \(\frac { -9 }{ 2.5 } \)y .....(2)
Let x₁ be the distance between the bottom of the verticallineon the ground from the pipeendand the point on which this water touches the ground.
But the height of the pipe from the ground is 7.5 m.
∴ (x₁, -7.5) lies on ....(2)
⇒ x₁² = \(\frac { -9 }{ 2.5 } \)(-7.5)
⇒ x₁ = \(\sqrt { 9\times 3 } =3\sqrt { 3 } \) m
The water strikes the ground \(3\sqrt { 3 } \) m beyond the vertical line.

kx-2y+z = 1, -2ky+z = -2, x-2y+k = 1 
The matrix form of the system is AX = B where
\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
Applying elementary row operation on the augment matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k & 1 \\ k & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k+2 & k \\ 0 & -2+2k & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ - \\ 1-k \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & k \\ 0 & 0 & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ -3 \\ 1-k \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & { k }^{ 2 }-k+2 \end{matrix}\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & (k+2)(1-k) \end{matrix}|\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \).........(1)
Case (i): when k = 1
\([A|B]\rightarrow \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -3 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B] ⇒ The system has no solution
Case (ii): when k ≠ 2, k ≠ -2 
\(\left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 1 \\ -3 \\ not\quad zero \end{matrix} \right] \)
⇒ \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
so, \(\rho \)(A) =\(\rho \)[A|B] = 3 = the number of unknowns Hence, the system has unique solution.
Case (iii): when k = -2
\(\rho [A|B]\rightarrow \left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} -2 \\ 6 \\ 0 \end{matrix}\begin{matrix} -2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \) (A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2<3 the number of unknowns so the system is consistent with infinitely many solutions.

Let V₁ be the vertex of the parabola and
V₂ be the vertex of the hyperbola.
\(\overset { \_ \_ \_ \_ \_ \_ }{ { F }_{ 1 }{ F }_{ 2 } } \) = 14−2 = 12m, 2c = 12, c = 6
The distance of centre to the vertex of the hyperbola is a = 6−1 = 5
b² = c² - a²
= 36−25 = 11.
Therefore the equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 11 } =1\)

Giveny = ax² + bx + c .............(1)
(-6, 8) lies on (1)
⇒ 8 = a(-6)²+b(-6)+c
⇒ 8 = 36z-6b+c ..........(2)
(-2,12) lies on (1)
⇒ -12 = a(-2)²+b(-2)+c
⇒ -12 = 4a-2b+c ...........(3)
Also (3, 8) lies on (1)
⇒ 8 = a(3)²+b(3)+c
⇒ 8 = 9a+3b+c ............(4)
Reducing the augment matrix to an equivalent row-echelon form by using elementary. row operations, we get,
\(\left[ \begin{matrix} 36 & -6 & 1 \\ 4 & -2 & 1 \\ 0 & 3 & 1 \end{matrix}|\begin{matrix} 8 \\ -12 \\ 8 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { 9R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow 4{ R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -12 & 8 \\ 0 & 18 & 3 \end{matrix}|\begin{matrix} 8 \\ -116 \\ 24 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div 4\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 3 }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -8 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -50 \end{matrix} \right] \)
Writing the equivalent equation from the row echelon matrix, we get
36a - 6b + c = 8 ........(1)
-3b+2c = -29 ....(2)
5c = -50
⇒ c = \(\frac{-50}{5}\) = -10
Substituting c = -10 in (2) we get,
-3b+2(-10)= -29
⇒ -3b+2(-10) = -29
⇒ -3b-20 = -29
⇒ -3b = -9
⇒ b = \(\frac{-9}{-3}\) = 3
Substituting b = 3 and c = -10 in (1) we get,
36a-6(3)-10 = 8
⇒ 36a-18-10 = 8
⇒ 36a-28 = 8
⇒ 6a = 8+28 = 36
⇒ a = \(\frac{36}{36}\) = 1
∴ a = 1, b = 3, c = -10
Hence the path of the boy is
y = 1(x²)+3(x)-10
⇒ y = x²+3x-10
Since his friend is at P(7, 60),
60 = (7)²+3(7)-10
⇒ 60 = 49+21-10
⇒ 60 = 70-10 = 60
⇒ 60 = 60
Since (7, 60) satisfies his path, he can meet his friend who is at P(7, 60)

Now, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } =tan^{ -1 }{ a }_{ 2 }-tan^{ -1 }{ a }_{ 1 }\)
Similarly, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) =tan^{ -1 }{ a }_{ 3 }-tan^{ -1 }{ a }_{ 2 }\)
Continuing inductively, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ n-1 } }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }{ a }_{ n }-tan^{ -1 }{ a }_{ n-1 }\)
Adding vertically, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)

Given z = x + iy and arg\(\left( \frac { z-i }{ z+2 } \right) =\frac { \pi }{ 4 } \)
⇒ arg(z-i) - arg(z+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x + iy-i) - arg(x+iy+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x+i(y-1)-arg((x+2)+iy) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { y-1 }{ x } \right) -tan^{ -1 }\left( \frac { y }{ x+2 } \right) \) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { \frac { y-1 }{ x } -\frac { y }{ x+2 } }{ 1+\frac { y-1 }{ x } .\frac { y }{ x+2 } } \right) \)
= \(\frac { \pi }{ 4 } \)\(\left[ \because tan^{ -1 }x-tan^{ -1 }y=tan^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(\Rightarrow \frac{\left(\frac{(x+2)(y-1)- x y}{\not {x (\not x+\not2)}}\right)}{\left(\frac{x(x+2)+y(y-1)}{\not x(\not x+\not 2)}\right)}=\tan \frac{\pi}{4}=1\)
⇒ \(\frac { (x+2)(y-1)-xy }{ x(x+2)+y(y-1) } \) = 1
⇒ -x + 2y-2 = x²+ 2x + y²-y
⇒ x² + 2x + y²-y + x-2y + 2 = 0
⇒ x² + y²+3x-3y + 2 = 0
Hence proved.

Let the prices per unit for the commodities A, B and C be Rs. x, Rs. y and Rs. z.
By the given data,
2x - 4y + 5z = 15000
3x + y - 2z = 1000
-x + 3y + z = 4000
The matrix form of the system of equations is
\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 12 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and B =\(\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| +4\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| +5\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \)
= 2 (1 + 6) + 4 (3 - 2) + 5 (9 + 1)
= 2 (7) + 4 (1) + 5(10) = 14 + 4 + 50 = 68.
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 5 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 5 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -1 & 3 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 5 \\ 1 & -2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 5 \\ 3 & -2 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ 3 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
= \(\left[ \begin{matrix} +(1+6) & -(3-2) & +(9+1) \\ -(-4-15) & +(2+5) & -(6-4) \\ +(8-5) & -(4-15) & +(2+12) \end{matrix} \right] \)
= \(\left[ \begin{matrix} 7 & -1 & 10 \\ 19 & 7 & -2 \\ 3 & 19 & 14 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adj=\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 105000+19000+12000 \\ -15000+7000+76000 \\ 150000-2000+56000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 136000 \\ 68000 \\ 204000 \end{matrix} \right] =\left[ \begin{matrix} 2000 \\ 1000 \\ 3000 \end{matrix} \right] \)
∴ x = 2000, y = 1000, z = 3000
Hence the prices per unit of the commodities A, B and C are Rs. 2000, Rs. 1000 and Rs. 3000 respectively.

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

By the definition of a conic  \(\frac{SP}{PM}\)= e or SP² = e²PM²
Then, (x−2)² + (y−3)² = \(\frac { 1 }{ 4 } \) (x-7)² 
3x²+ 4y²−2x − 24y + 3 = 0
\({ 3\left( x-\frac { 1 }{ 3 } \right) }^{ 2 }+4(y-3)^{ 2 }=3\left( \frac { 1 }{ 9 } \right) +4\times 9-3=\frac { 100 }{ 3 } \)
\(\frac { { \left( x-\frac { 1 }{ 3 } \right) }^{ 2 } }{ \frac { 100 }{ 9 } } +\frac { (y-3{ ) }^{ 2 } }{ \frac { 100 }{ 12 } } \) = 1 which is in the standard form.
Therefore, the length of major axis = 2a = 2\(\sqrt { \frac { 100 }{ 9 } = } \frac { 20 }{ 3 } \) and
the length of minor axis = 2b = 2\(\sqrt { \frac { 100 }{ 12 } = } \frac { 10 }{ \sqrt { 3 } } \).

Let  α, β and γ  be the roots of x³+ ax²+ bx + c = 0
Then, we get
Σ₁ = α + β + γ = -a ....(1)
Σ₂ = αβ + βγ + γα = b ...(2)
Σ₃ = αβγ = -c ...(3)
We have to form the equation whose roots are  α², β² and γ².
Using (1), (2) and (3), we find the following
Σ₁ = α² + β² + γ² = (α + β + γ )² - 2( αβ + βγ + γα) = (-a)² -2(b) = a²-2b,
Σ₂ = α²β² +  β²γ² + γ²α² = (αβ + βγ + γα)² - 2((αβ)( βγ)(γα) + (γα)(αβ))
= (αβ + βγ + γα)² - 2αβγ (β + γ + α) = (b)² - 2(-c)(-a) = b²-2ca
Σ₃ = α²β²γ² = (αβγ)² = (-c)² = c²
Hence, the required equation is
x³-(α² + β² + γ²)x² + (α²β² + β²γ² + γ²α²)x -  α²β²γ² = 0
That is, x³-(a²-2b)x² + (b²-2ca)x-c² = 0

(i) Since f (x) is a probability mass function, f (x) ≥ 0 for all x , and d \(\sum_{x} f(x)=1\)
Thus, \(\sum_{x} f(x)=1\)
\(c^{2}+2 c^{2}+3 c^{2}+4 c^{2}+c+2 c=0\)
\(c=\frac{1}{5} \text { or }-\frac{1}{2}\)
Since f x( ) ≥ 0 for all x , the possible value of c is \(\frac{1}{5}\)
Hence, the probability mass function is

x	1	2	3	4	5	6
f(x)	\( \frac{1}{25} \)	\( \frac{2}{25} \)	\( \frac{3}{25} \)	\(\frac{4}{25} \)	\(\frac{1}{5}\)	\( \frac{2}{5}\)

 (ii) To find mean and variance, let us use the following table

x	f(x)	xf(x)	x2f(x)
1	\(\cfrac { 1 }{ 25 } \)	\(\cfrac { 2 }{ 25 } \)	\(\cfrac { 1 }{ 25 } \)
2	\(\cfrac { 2 }{ 25 } \)	\(\cfrac { 4 }{ 25 } \)	\(\cfrac { 8 }{ 25 } \)
3.	\(\cfrac { 3 }{ 25 } \)	\(\cfrac { 9 }{ 25 } \)	\(\cfrac { 27 }{ 25 } \)
4.	\(\cfrac { 4 }{ 25 } \)	\(\cfrac { 16 }{ 25 } \)	\(\cfrac { 64 }{ 25 } \)
5.	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 5 }{ 5 } \)	\(\cfrac { 25 }{ 5 } \)
6.	\(\cfrac { 2 }{ 25 } \)	\(\cfrac { 12 }{ 5 } \)	\(\cfrac { 72 }{ 5 } \)
	\(\Sigma f(x)=1\)	\(\Sigma xf(x)=\cfrac { 115 }{ 25 } \)	\({ \Sigma x }^{ 2 }f(x)=\cfrac { 585 }{ 25 } \)

Mean :  \(E(X)=\Sigma xf(x)=\frac { 115 }{ 25 } =4.6\) 
Variance : \(V(x)=E\left( x \right) ^{ 2 }=\Sigma { x }^{ 2 }f(x)-\left( \Sigma xf(x) \right) ^{ 2 }\) 
= \(\frac { 585 }{ 25 } -\left( \frac { 115 }{ 25 } \right) ^{ 2 }=23.40-21.16=2.24\)
Therefore the mean and variance are 4.6 and 2.24 respectively.

Let p be the probability that indicates the defective rate of an electronic device
n = 10
\(P=5\%=0.05 \) 
q = 1 - p
n = 10, p = 0.05, X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Atleast 1 defective item
P(X ≥ 1) = 1 - P(X < 1)
= 1-P(X = 0)
= 1-¹⁰C₀ (0.05)⁰ (0.95)¹⁰
P(X ≥1) = 1 - (0.95)¹⁰
(ii) Exactly two defective items  
P(X = 2) =¹⁰C₂(0.05)² (0.95)⁸

\(Let\ u={ x }^{ 2 }\ v=cos 2x\ dx\)
\(u'=2x\quad { v }_{ 1 }=\frac { sin2x }{ 2 } \)
\(u"=2\quad { v }_{ 2 }=-\frac { cos2x }{ 4 } \)
\(\\ { v }_{ 3 }=-\frac { sinx }{ 8 } \)
Bernoulli's' formula:
\(\int { uvdx } =u{ v }_{ 1 }-u'{ v }_{ 2 }+u''{ v }_{ 3 }\)
\(\therefore \int _{ 0 }^{ \pi /2 }{ { x }^{ 2 }cos2x\quad dx={ \left[ \frac { { x }^{ 2 }sin2x }{ 2 } +\frac { 2xcos2x }{ 4 } -\frac { 2sin2x }{ 8 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } } } \)
\(=\left[ \frac { { \pi }^{ 2 } }{ 8 } sin\pi +\frac { 2\pi }{ 8 } cos\pi -\frac { sin\pi }{ 8 } \right] -[0+0-0]\)
\(=\frac {- 2\pi }{ 8 } (-1)=\frac { \pi }{ 4 }\quad [\because sin\pi =0,cos\pi =-1]\)

Equation of the given curve is x = cos t; y = 2 sin² t
\(\frac { dx }{ dt } \) = - sin t;
\(\frac { dy}{ dt } \) = 4 sin t cos t
\(\therefore \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \) = \(\frac{4 \ sin \ t \ cos \ t}{-sin \ t}\)
= 4 cost
∴ Slope = m = \(\left( \frac { dy }{ dx } \right) \)t = \(\frac { \pi }{ 3 } \)
= -4 cos \(\frac { \pi }{ 3 } \) = -4 \(\left( \frac { 1 }{ 2 } \right) \)
Equation of the tangent is y - y₁ = m (x - x₁)
When t = \(\frac { \pi }{ 3 } \), x cos \(\frac { \pi }{ 3 } \) = \(\frac12\)
when t =  \(\frac { \pi }{ 3 } \), y = 2 \({ \left( sin\ \frac { \pi }{ 3 } \right) }^{ 2 }\)
= 2 \({ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }\) = 2 \(\frac34\) = \(\frac32\)
∴ Equating of the tangent is y - \(\frac32\) = -2 (x-\(\frac12\))
\(\Rightarrow \frac{2 y-3}{\not 2}=\frac{1}{2}\left(\frac{2 x-1}{\not 2}\right)\)
⇒ 2y - 3 = -4x + 2
⇒ 4x + 2y = 5
Equating of the normal is y - y₁ = \(\frac{-1}{m}\) (x-x1)
⇒ y - \(\frac32\) = \(\frac12\)(x - \(\frac12\))
\(\Rightarrow \frac{2 y-3}{\not 2}=\frac{1}{2}\left(\frac{2 x-1}{\not 2}\right)\)
⇒ 4y-6 = 2x-1
⇒ 2x - 4y +5 = 0

The equation of the family of  parabolas with latus rectum (4a) and whose axes are parallel to the x-axis is shown in sketch
Let vertex 'V' be (h, k) and focus at 'F" and let
L-L' be latus rectum = 4a
Hence, FL = 2a and VF = a.
Thus 'F' is at (h+a, k).

Equation of parabola with vertex at (h, k) and focal length 'a', latus rectum 4a' is
(y - k)² = 4a(x - h) .......(1)
Differentiating with respect to 'x' we got
2(y - k) \(\frac{dy}{dx}\) = 4a 
⇒ (y - k). \(\frac{dy}{dx}\) = 2a ....... (2)
Again differentiating with respect to x,
(y-k) y''+y'\(\times\) y' = 0  ............(3)
From(2),(y-k) = From(2),(y-k) = \(\frac{2a}{y'}\)
Putting in (3), we get
\((\frac{2a}{y'})y''+(y')^2=0 (or) 2ay'' +(y')^3 = 0\)
This is the required differential equation.

The Cartesian equations of the straight line \(\vec { r } =(\hat { i } +\hat { 3j } -\hat { k } )+t(2\hat { i } +3\hat { j } +2\hat { k } )\) is
\(\frac { x-1 }{ 2 } =\frac { y-3 }{ 3 } =\frac { z+1 }{ 2 } \) = s(say)
Then any point on this line is of the form (2s + 1, 3s + 3, 2s -1) ...............(1)
The Cartesian equation of the second line is \(\frac { x-2 }{ 1 } =\frac { y-4 }{ 2 } =\frac { z+3 }{ 4 } =t\) (say)
Then any point on this line is of the form (t + 2, 2t + 4, 4t - 3)....(2)
If the given lines intersect, then there must be a common point. Therefore, for some s, t ∈ R
we have (2s + 1, 3s + 3, 2s −1 ) = (t + 2, 2t + 4, 4t − 3)
Equating the coordinates of x, y and z we get
2s − t = 1, 3s − 2t = 1 and s − 2t =  −1.
Solving the first two of the above three equations, we get s = 1 and t = 1. These values of s and t satisfy the third equation. So, the lines are intersecting.
Now, using the value of s in (1) or the value of t in (2), the point of intersection (3,6,1) of these two straight lines is obtained.
If we take \(\vec{b}=(2\hat { i } +3\hat { j } +2\hat { k } )\) and \(\vec{d}=\hat { i } +\hat { 3j } -\hat { k } \),
then \(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 2 \\ 1 & 2 & 4 \end{matrix} \right| =8\hat { i } -6\hat { j } +\hat { k } \) is a vector perpendicular to both the given straight lines.
Therefore, the required straight line passing through (3,6,1)
and perpendicular to both the given straight lines is the same as the straight line passing through (3,6,1) and parallel to \(8\hat { i } -6\hat { j } +\hat { k } \). Thus, the equation of the required straight line is
\(\vec { r } =(\hat { 3i } +\hat { 6j } -\hat { k } )+m(8\hat { i } -6\hat { j } +\hat { k } )\), k ∈ R.

With usual notations in triangle, ABC let \(\vec { BC } =\vec { a } \), \(\vec { CA } =\vec { b } \) and \(\vec { AB } =\vec { c} \). Then \(|\vec { BC }| =\vec { a } \), \(|\vec { CA } |=\vec { b } \) and \(|\vec { AB }| =\vec { c } \)
Since in ΔABC, \(\vec { BC }+\vec { CA }+\vec { AB}=0\) we have \(\vec { BC }\times(\vec { BC }+\vec { CA }+\vec { AB })=\vec { 0 }\)
Simplifying, we get,
\(\vec { BC }\times\vec { CA }=\vec { AB }\times\vec { BC }\) ....(1)

Similarly, since \(\vec { BC }+\vec { CA }+\vec { AB}=\vec 0\), we have
\(\vec {CA} \times (\vec { BC }+\vec { CA }+\vec { AB})=\vec 0\) .... (2)
On Simplification, we obtain \(\vec { BC }\times\vec { CA }=\vec {CA }\times\vec {AB }\)
From equations (1) and (2), we get 
\(\vec { AB }\times\vec {BC }\) = \(\vec { CA }\times\vec {AB }\)=\(\vec { BC}\times\vec {CA}\)
So, \(\left| \overrightarrow { AB} \times \overrightarrow { BC } \right| =\left| \overrightarrow { CA } \times \overrightarrow { AB } \right| =\left| \overrightarrow { BC } \times \overrightarrow { CA } \right| \). Then, we get
ca sin(π − B) = bc sin(π - A) = ab sin (π - C)
That is, ca sin B = bc sin A = absinC . Dividing by abc, we get
\(\frac { sinA }{ A } =\frac { sinB }{ b } =\frac { sinC }{ c } \) or \(\frac { a }{ sinA }= \frac { b }{ sinB } =\frac { c }{ sinC } \)

Given A =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \)
A² = \(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 25-3 & 15-6 \\ -5+2 & -3+4 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22 & 9 \\ -2 & 1 \end{matrix} \right] \)
∴ A²- 3A - 7I₂
=\(\left[ \begin{matrix} 22 & 9 \\ -3 & 1 \end{matrix} \right] -3\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -7\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22-15-7 & 9-9+0 \\ -3+3+0 & 1+6-7 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)= O₂
Hence proved.
∴ A²-3A-7I₂ = O₂
Postmultiplying by A^-1 we get,
A²-A^-1-3AA^-1-7I₂A^-1 = 0.A^-1
⇒ A(AA^-1)-3(AA^-1)-7(A^-1) = 0
[∵ I₂A^-1 = A^-1 and | (0)A^-1= 0]
⇒ AI-3I-7A-1 = 0  [∵ AA^-1= 1]
⇒ AI-3I = 7A^-1
⇒ A^-1 = \(\frac { 1 }{ 7 } \)[A - 3I] [∴ AI = A]
⇒ A^-1 = \(\frac { 1 }{ 7 } =\left[ \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \right] \)
⇒ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 5-3 & 3-0 \\ -1-0 & -2-3 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \).

(1) Mount Everest is the highest mountain of the world.
(2) 3+ 4 = 8 .
(3) 7 + 5 >10 .
(4) Give me that book.
(5) (10 − x) = 7.
(6) How beautiful this flower is!
(7) Where are you going?
(8) Wish you all success.
(9) This is the beginning of the end.
The truth value of the sentences (1) and (3) are T, while that of (2) is F. Hence they are statements.
The sentence (5) is true for x = 3 and false for x ≠ 3 and hence it may be true or false but not both. So it is also a statement.
The sentences (4), (6), (7), (8) are not statements, because (4) is a command, (6) is an exclamatory, (7) is a question while (8) is a sentence expressing one’s wishes and (9) is a paradox.