\(28\over45\)

We have : \(R=\{ (a_{ 1 },b_{ 1 }),(a_{ 2 },b_{ 1 }),(a_{ 3 },b_{ 2 }),(a_{ 4 },b_{ 2 }),(a_{ 5 },b_{ 2 })\} \)
R is not one-are
\(\left[ \because (a_{ 1 },b_{ 1 }),(a_{ 2 },b_{ 1 })\in R \right] \)
R is not onto 
\(\left[ \because b_{ 3 }\ and\ b_{ 4 }\ have\ no\ pre-images \right] \)
\(\Rightarrow \) R is not bijective.
Yes, true friend -ship makes life easier.

(i) Let Rs. x and Rs. y by charges for typing 1  English and 1 Hindi page respectively.
By the question 10x + 3y = 145
and  3x + 10y = 180
These can be written AX = B
where \(A=\begin{bmatrix}10&3\\3&10 \end{bmatrix},X=\begin{bmatrix} x\\y\end{bmatrix}, B=\begin{bmatrix} 145\\180\end{bmatrix}\)
Now \(|A|=\begin{bmatrix} 10&3\\3&10\end{bmatrix}=100-9=91\)
= A^-1exists
Now adj A = \(\begin{bmatrix}10&-3\\-3&10 \end{bmatrix}=\begin{bmatrix} 10&-3\\-3&10\end{bmatrix}\)
\(\therefore\ A^{-1}={1\over 91}\begin{bmatrix} 10&-3\\-3&10\end{bmatrix}\)
From(1) \(X=A^{-1}B\Rightarrow X={1\over91}\begin{bmatrix}10&-3\\-3&10 \end{bmatrix}\begin{bmatrix}145\\180 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix}x\\y \end{bmatrix}={1\over91}\begin{bmatrix}1450-540\\-435+1800 \end{bmatrix}\)
\(={1\over91}\begin{bmatrix}910\over1365 \end{bmatrix}=\begin{bmatrix} 10\\15\end{bmatrix}\)
⇒ x = 10 and y = 15
Hence charges for one English and one Hindi pages are Rs. 10 and Rs. 15 respectively.
(ii) Shyam is to pay = 5x2 = Rs. 10
Less charged = Rs. (5x15-10) = Rs. 65
Value: Poor should be charged subsidised rates.

(i) Here x, y, z refer to honesty, punctuality and obedience respectively.
By the question,
\(\left( \begin{matrix} 5 & 4 & 3 \\ 4 & 3 & 5 \\ 1 & 1 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 11000 \\ 10700 \\ 2700 \end{matrix} \right) \)
\(\left( \begin{matrix} 5x+4y+3z \\ 4x+3y+5z \\ x+y+z \end{matrix} \right) =\left( \begin{matrix} 11000 \\ 10700 \\ 2700 \end{matrix} \right) \)
The 5x+6y+3z = 11000
      4x+3y+5z = 10700
           x+y+z = 2700
(ii) Let \(A=\left( \begin{matrix} 5 & 4 & 3 \\ 4 & 3 & 5 \\ 1 & 1 & 1 \end{matrix} \right) \)
\(\therefore\ |A|=\left| \begin{matrix} 5 & 4 & 3 \\ 4 & 3 & 5 \\ 1 & 1 & 1 \end{matrix} \right| \)
= 5(3-5)-4(4-4)+3(4-3)
= 5(-2)-4(-1)+3(1)|
= -10+4+3 = -3 ≠ 0
A^-1 exists ⇒ equations have a unique solution
Yes, it is possible to solve the system of equations, using matrix multiplication
(iii) We prefer to be awarded most the value of honesty (x) because it has highest value

The given system of equations is:
x + 2y + z = 7
x - y + z = 4
x + 3y + 2z = 10
Solving usual we get : x = 3, y = 1, z = 2
Food, taken at home is always the best way.

(i)Let 'x', 'y' and 'z' be the amount invested in three investments.
Then
x + y + z = 65000  ......(1)
\({6x\over100}+{8y\over100}+{9z\over100}=4800\)
\(\Rightarrow\) 6x + 8y + 9z = 480000  ...(2)
\({9z\over100}=600+{8y\over100}\)
\(\Rightarrow\) 0x - 8y + 9z = 60000
These can be written as AX = B where:
\(A=\begin{bmatrix} 1&1&1\\6&8&9\\0&-8&9\end{bmatrix},X=\begin{bmatrix} x\\y\\z\end{bmatrix}\)
and \(B=\begin{bmatrix} 650000\\480000\\60000\end{bmatrix}\)
(ii) Now \(|A|=\begin{bmatrix}1&1&1\\6&8&9\\0&-8&9 \end{bmatrix}\)
= 1.(72 + 72) - 6(9+8)
= 144 - 102 = 42 ≠ 0
Hence, the equations have a unique solution.
(iii) No.We are not fools because most of such companies are frauds.

Let Rs. x, Rs. y and Rs. z be the cost of reach carry bag.
By the question,
20x + 30y + 40z = 250
30x + 40y + 20z = 270
40x + 20y + 30z = 200
i.e. 2x + 3y + 4z = 25
3x + 4y + 2z = 27
4x + 2y + 3z = 20
Solving a susual, we will get : x = 1, y = 5, z = 2.
Shopkeeper (A) is better for environmental conditions because he is using least number of polythene.
Shopkeeper (B) is better for social conditions because he using more made bags, prepared by prisoners.

(i) Let x be the number of people belive in morning walk.
Let y be the number of people belive in yoga.
Let z be the number of people joining gym
x + y + z = 70
20x + 30y + 40z = 2,100
\(\Rightarrow\)2x + 3y + 4z = 210 and
500y + 400z = 23,000
\(\Rightarrow\) 5y + 4z = 230
This system of equation can be written in the matrix form as
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 0 & 5 & 4 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 70 \\ 210 \\ 230 \end{matrix} \right] \)
Let AX = B
where A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 0 & 5 & 4 \end{matrix} \right] \)
X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and B = \(\left[ \begin{matrix} 70 \\ 210 \\ 230 \end{matrix} \right] \)
X = A^-1B
|A| = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 0 & 5 & 4 \end{matrix} \right] \)
= 1(12-20)-1(8-0)+1(10-0)
= -8-8+10 = -6 \(\neq \) 0
\(\therefore\) A^-1 exists
adj A= \(\left[ \begin{matrix} -8 & -8 & 10 \\ 1 & 4 & -5 \\ 1 & -2 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -8 & 1 & 1 \\ -8 & 4 & -2 \\ 10 & -5 & 1 \end{matrix} \right] \)
\(A^{ -1 }=\frac { 1 }{ |A| } adjA\)
\(\Rightarrow\) A^-1 = \(\frac { 1 }{ -6 } \left[ \begin{matrix} -8 & 1 & 1 \\ -8 & 4 & -2 \\ 10 & -5 & 1 \end{matrix} \right] \)
X = \(-\frac { 1 }{ 6 } =\left[ \begin{matrix} -8 & 1 & 1 \\ -8 & 4 & -2 \\ 10 & -5 & 1 \end{matrix} \right] \left[ \begin{matrix} 70 \\ 210 \\ 230 \end{matrix} \right] \)
= \(-\frac { 1 }{ 6 } \left[ \begin{matrix} -560+210+230 \\ -560+840-460 \\ 700-1050+230 \end{matrix} \right] \)
= \(-\frac { 1 }{ 6 } \left[ \begin{matrix} -120 \\ -180 \\ -120 \end{matrix} \right] \)
\(\Rightarrow\) \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 20 \\ 30 \\ 20 \end{matrix} \right] \)
The number of people believe in morning walk = 20
The number of people believe in yoga = 30
The number of people joining gym = 20
(iii) Exercise keeps a person fit and healthy.

Let Rs. x crores, Rs. y crores and Rs. z crores be spent on three Schemes.
By the question.
x + y + z = 600 .....(1)
x + 2z = 500 ......(2)
3x + y + z = 1200 .....(3)
Solving (1), (277444) and (3) as usual we get:
x = Rs. 300crores, y = Rs. 200 crores and z = Rs.100 crores.
(i) In our country, male population is more than female population
(ii) It is essential for a human being to save the life of all.

We have: A = Set of all students of class XII in a school and \(R=\{ (S_{ 1 },S_{ 2 }):\)Now  and have same sex}.
Now \(R=\{ (L_{ 1 },L_{ 2 })\in R\)    \(\left[ \because L_{ 1 }\quad and\quad L_{ 2 }\quad have\quad same\quad sex \right] \)
Thus R is reflexive.
Again \((L_{ 1 },L_{ 2 })\in R\)   \(\Rightarrow \quad L_{ 1 }\quad and\quad L_{ 2 }\quad have\quad same\quad sex\)
\(\Rightarrow L_{ 2 }\ and\ L_{ 1 }\ have\ same\ sex\)
\( \Rightarrow (L_{ 2 },L_{ 1 })\in R\)
Thus R is symmetric.
Last by \((L_{ 1 },L_{ 2 })\in R\) and \((L_{ 2 },L_{ 3 })\in R\)
\(\Rightarrow \) \( L_{ 1 }\) and \(L_{ 2 }\) have same sex
and \(L_{ 2 }\) and \(L_{ 3 }\)  have same sex
\(\Rightarrow \) \( L_{ 1 }\) and \(L_{ 3 }\) have same sex \(\Rightarrow (L_{ 2 },L_{ 3 })\in R.\)
Thus R is transitive.
Hence, R is an equivalence relation.
Co-education is very useful because it leads to the balanced development of the children so that they become good citizens in future.

\(\lim _{ x\rightarrow 1 }{ f\left( x \right) } =\lim _{ x\rightarrow 1 }{ \frac { 1-{ x }^{ m } }{ 1-x } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { 1-{ \left( 1+h \right) }^{ m } }{ 1-(1+h) } } =\lim _{ h\rightarrow 0 }{ \frac { { \left( 1+h \right) }^{ m }-1 }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { \left( 1+mh+\frac { m(m-1) }{ 2! } { h }^{ 2 }+.... \right) -1 }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { mh+\frac { m(m-1) }{ 2! } { h }^{ 2 }+.... }{ h } } \)
\( =\lim _{ h\rightarrow 0 }{ \left( mh+\frac { m(m-1) }{ 2! } { h }^{ 2 }+.... \right) } \)
And f(1) = m-1
Thus, \(\lim _{ x\rightarrow 1 }{ f\left( x \right) } \neq f(1)\)
f(x) is discontinuous at x = 1.
No, because life is precious OR drives carefully.

We have: \(\frac { dy }{ dx } +\frac { y }{ x } =0\)
\(\Rightarrow\) \(\frac { dy }{ y } +\frac { dx }{ x } =0\)
|Variables Separable
Integrating, \(log|y|+log|x|=log|c|\)
\(\Rightarrow\)  \(log|xy|=log|c|\)
\(\Rightarrow\)  \(xy=c\)  ..(1)
When \(x=100,y=1,\) then \((100)(1)=c\Rightarrow=c=100.\)
Putting in (1), \(xy=100\)
which is the required particular solution.
Yes,  as the population increases, area for living decreases, which is very harmful to us

Let 'x' be the population of a colony at any time t.
By the question, \(\frac { dx }{ dt } =kx\)
\(\Rightarrow \)   \(\frac { dx }{ x } =k\quad dt.\) 
|Variables Separable 
Integrating, \(log|x| =kt\ +c\) ...(1)
When  \(t=0,x={ x }_{ 0 },\)
\( \therefore log|{ x }_{ 0 }|=c\) ...(2)
When  \(t=50,x=2{ x }_{ 0 },log|2{ x }_{ 0 }|+50k+c\) ...(3)
Subtracting (2) from (3), \(log|2{ x }_{ 0 }|-log|{ x }_{ 0 }|=50k\)
\(\Rightarrow log\quad 2=50k\)  
\(\Rightarrow k=\frac { 1 }{ 50 } \ log2.\)
Putting in (1),  \(log|x|=\frac { 1 }{ 50 } \quad log2.t+log|{ x }_{ 0 }|.\)
When  \(x=3{ x }_{ 0 },\quad log|3{ x }_{ 0 }|=\frac { 1 }{ 50 } log\quad 2.t+|{ x }_{ 0 }|\)
\(\Rightarrow log\ 3=\frac { 1 }{ 50 } log\quad 2.t\Rightarrow t=50\frac { log\quad 3 }{ log\quad 2 } \)
Hence, the population becomes triple in months \(50\frac { log\quad 3 }{ log\quad 2 } \).

[Newton's Law of Cooling: The temperature of body changes at a rate, which is proportional to the difference in temperature between that of the surrounding medium and that of the body itself.]
Let 'T' be the temperature of the body at time t.
By Newton's Law of cooling,
\(\frac { dT }{ dt } =k\ \left( T-70 \right) \)
\(\Rightarrow \frac { dT }{ T-70 } k\quad dt.\)   
|Variables Separable 
Integrating, \(\int { \frac { dT }{ T-70 } } =k\int { 1.dt+log|c| } \)
\(\Rightarrow log|T-70|-logc|=kt\)
\(\Rightarrow log|\frac { T-70 }{ c } |=\quad kt\quad \Rightarrow \frac { T-70 }{ c } ={ e }^{ k\quad t }\)
\(\Rightarrow T=70+c { e }^{ k\ t }\)
When  \(t=0,T=94.6,\ 94.6=70+c { e }^{ 0 } \Rightarrow c=24.6\)
Putting in(1),  \(t=0,T=94.6,\quad 94.6=70+c{ e }^{ 0 } \Rightarrow c=24.6\)
When \(=60,T=93.4,\quad 93.4=70+24.6\quad { e }^{ 60kt }\) 
\(\Rightarrow 24.6 { e }^{ 60kt } =23.4 \Rightarrow { e }^{ 60kt }=\frac { 23.4 }{ 24.6 } =\frac { 117 }{ 123 } \)
\(\Rightarrow 60\ k=log\frac { 117 }{ 123 } \Rightarrow k=\frac { 1 }{ 60 } log\frac { 117 }{ 123 } <0\) ...(2)
Thus the time t, that has elapsed after death is given by:
\(98.6=70+(24.6){ e }^{ kt }\)
\(\Rightarrow \left( 24.6 \right) { e }^{ kt }=28.6\Rightarrow { e }^{ kt }=\frac { 28.6 }{ 24.6 } =\frac { 143 }{ 123 } \)
\(\Rightarrow \)  \(t=\frac { 1 }{ k } log\frac { 143 }{ 123 } \)
\(=-3.01\)
Hence, the estimated time of death  \(=11.30-3.01\)\(=8.30\) P.M. approx.

We have: \(m\frac { dv }{ dt } +V\frac { dm }{ dt } =0\)
\(\Rightarrow m\frac { dv }{ dt } +V\frac { dm }{ dt } =0\) ...(1)
Integrating,  \(\int { dv+V\int { \frac { dm }{ m } =C } } \)
\(\Rightarrow \) \(v+V\quad logm=C\)..(2)\(\left[ \because m>0 \right] \)
When \(v=0,m={ m }_{ 0 }.\)
\(\therefore\)   \(0+\ V\ log{ m }_{ 0 }=C\Rightarrow C=V\ log{ m }_{ 0 }\)
Putting In (2),  \(v+Vlogm=V\ log{ m }_{ 0 }\)
\(\Rightarrow\) \(v=V(log{ m }_{ 0 }-log\ m).\)
Hence,  \(v=V\log\left( \frac { { m }_{ 0 } }{ m } \right) \)
We should not encourage rocket technology because they can be used nuclear warheads to attack other countries.

Let 'x' be the bacteria at any time 't'
By the question,\(\frac { dx }{ dt } =kx\)
\(\Rightarrow \)  \(\frac { dx }{ x } =k\quad dt\) 
|Variables Separable
Integrating, \(log|x|=kt+c\) ...(1)
When  \(t=0,x={ x }_{ 0 }\)
\(\therefore\)   \(t=0,x={ x }_{ 0 }\) ...(2)
When\(t=2,x=2{ x }_{ 0 }\)
\(\therefore\)  \(t=2,x=2{ x }_{ 0 }\) ...(3)
Subtracting (2) from (3), \(log|2{ x }_{ 0 }|-log|{ x }_{ 0 }|=2k\)
\(\Rightarrow \)  \(log2=2k\Rightarrow k=\frac { 1 }{ 2 } log\quad 2\)
Putting in (1), \(log|x|=\frac { 1 }{ 2 } 2.t+log|{ x }_{ 0 }|\)
When \(x=5{ x }_{ 0 },log|5{ x }_{ 0 }|=2.t+log|{ x }_{ 0 }|\)
\(\Rightarrow \)  \(log5=\frac { 1 }{ 2 } log\quad 2.t\)
\(\Rightarrow \)   \(t=2\frac { log5 }{ log2 } \)
Hence, the number of bacteria will be five times after 2\(\frac { log5 }{ log2 } \) hours.

Let 'y' be the number of bacteria at any time t.
By the question,  \(\frac { dy }{ dt } =ky\)
\(\Rightarrow \)  \(\frac { dy }{ y } =k\quad dt\)             
|Variables Separable
Integrating,  \(\int { \frac { dy }{ y } } =k\int { 1. } dt+c\)      
\(\Rightarrow \)  \(log|y|=kt+c\)              
\(\Rightarrow \)  \(log\quad y\quad =kt+c\) ...(1) \(\left[ \because \quad y>0 \right] \)
Let   \(y={ y }_{ 0 }\left( =1,00,000 \right) \) when \(t=0\)
\(\therefore \)  \(log\quad { y }_{ 0 }=0+c\Rightarrow c=log\quad { y }_{ 0 }\)
Putting in(1),\(log\quad { y }_{ 0 }=0+c\Rightarrow c=log\quad { y }_{ 0 }\)
\(\Rightarrow \)  \(log\frac { y }{ { y }_{ 0 } } =kt\)...(2)
By the question, when \(log\frac { y }{ { y }_{ 0 } } =kt\)
\(\therefore \)   \(log\frac { \frac { 11{ y }_{ 0 } }{ 10 } }{ { y }_{ 0 } } =2k\Rightarrow k=\frac { 1 }{ 1 } log\frac { 11 }{ 10 } .\)
Putting in (2), \(log \frac { y }{ { y }_{ 0 } } =\frac { 1 }{ 2 } \left( log\frac { 11 }{ 10 } \right) t.\)
When \(y=2{ y }_{ 0 }\left( 2,00,000 \right) ,\)  
then  \(log\frac { 2{ y }_{ 0 } }{ { y }_{ 0 } } =\frac { 1 }{ 2 } log\frac { 11 }{ 10 } t\)
\(\Rightarrow \)  \(log2=\frac { 1 }{ 2 } log\left( \frac { 11 }{ 10 } t \right) \)
\(\Rightarrow \) \(t=\frac { 2log2 }{ log\frac { 11 }{ 10 } } \)
Hence, the bacteria count will reach \(2,00,000\) after \(\frac { 2log2 }{ log\frac { 11 }{ 10 } } \)hours.

Let 'x' be the number of bacteria present at time 't'.
By the question, \(\frac { dx }{ dt } =kx\)
\(\Rightarrow\)  \(\frac { dx }{ x } =k\quad dt\)                
|Variables Separable 
Integrating, \(log|x|=kt+c\)..(1)
When \(t=0,x={ x }_{ 0 }\) (say)
\(\therefore \)  \(log|x|=kt+log|{ x }_{ 0 }|\)
Putting in(1),   \(log|x|=kt+log|{ x }_{ 0 }|\) ...(2)
When \(t=5,x=3{ x }_{ 0 }.\)
\(\therefore \)  \(log|3{ x }_{ 0 }|\quad =5kt+log|{ x }_{ 0 }|\)
\(\Rightarrow\)  \(log3=5k\Rightarrow k=\frac { 1 }{ 5 } log3.\)
Putting in (2), \(log|x|=\frac { 1 }{ 5 } log\quad 3t+log|{ x }_{ 0 }|\)...(3)
(i) When \(t=10,log|x|=2log\quad 3+log|{ x }_{ 0 }|=log|9{ x }_{ 0 }|\)
\(\Rightarrow\)\(x=9{ x }_{ 0 }.\)
Hence, the bacteria will be 9 times after 10 hours.
(ii) When \(x=10{ x }_{ 0 }\)
\(\therefore\) From(3),\(log|10{ x }_{ 0 }|=\frac { 1 }{ 5 } log\quad 3.t+log|{ x }_{ 0 }|\)
\(\Rightarrow \)  \(log\quad 10=\frac { 1 }{ 5 } log\quad 3.t\)
\(\Rightarrow \)  \(t=5\frac { log10 }{ log3 } \)
Hence, the number of bacteria will be 10 times after \(5\frac { log10 }{ log3 } \) hours.

\(\sin ^{ -1 }{ (\sin { x } ) } =x\)
\(\Rightarrow \quad \sin ^{ -1 }{ \left( \sin { \frac { 4\pi }{ 5 } } \right) } =\frac { 4\pi }{ 5 } \)
\(But\quad \frac { 4\pi }{ 5 } \notin \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(\sin ^{ -1 }{ \left( \sin { \frac { 4\pi }{ 5 } } \right) } =\sin ^{ -1 }{ \left( \sin { \left( \pi -\frac { \pi }{ 5 } \right) } \right) } \)
\(=\sin ^{ -1 }{ \left[ \sin { \left( \frac { \pi }{ 5 } \right) } \right] } \)
\(\left[ \because \sin ^{ -1 }{ (\sin { \theta } ) } =\theta \forall \theta \epsilon \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \quad \right] \)
\(=\frac { \pi }{ 5 } \epsilon \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
All that we see or feel from senses is not always true. We must verify it by other means also, before reaching a conclusion.

\(\tan ^{ -1 }{ 3x } +\tan ^{ -1 }{ 2x } =\frac { \pi }{ 4 } \)
\(\Rightarrow \tan ^{ -1 }{ \left( \frac { 3x+2x }{ 1-6{ x }^{ 2 } } \right) } =\frac { \pi }{ 4 } \)
\(\Rightarrow \frac { 5x }{ 1-6{ x }^{ 2 } } =1\)
\(\Rightarrow 6{ x }^{ 2 }+5x-1=0\)
\(\Rightarrow (6x-1)(x+1)=0\)
\(\Rightarrow x=\frac { 1 }{ 6 } ,x=-1\)
But on putting x = -1
The value of \(\tan ^{ -1 }{ 3x } +\tan ^{ -1 }{ 2x } \)  becomes-ve. Hence x = -1 is not possible.
Thus \(x=\frac {1}{6}\)
In real life, both type of situations occurred +ve and -ve, but it must be verified and then final result should be accepted. We should always be prepared to handle both the situations.

It is given that * P(X) x P(X) -7 P(X) is defined as
\(A*B=A\cap B\forall A,B\in \ P(X)\)
we know that
\(A\cap X=A=X\cap A\forall \in \quad P(X)\)
\(\Rightarrow A*X=A=X*A\forall A\in P(X)\)
Thus, X is the identity element for the given binary operation * .
Now, an element A  P(X) is invertible if there exists B P(X) such that, A*B = X = B*A.
(AS X is the identity element)
\(A\cap B=A=B\cap A\)
Thus, X is the only invertible element in P(X) with
respect to the given operation * .
Hence, the given result is proved. 
Values:
Respect for elders,
Concern for the aged,
Lending a helping hand.

\(f(x)=\{2x-1, \ if\ x<0 \ 2x+1,\ if\ x\ge0\)
L.H.L = \(\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ 2x-1 } c\) = -1
R.H.L = \(\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ 2x-1 } c\) = 1
L.H.L. ≠ R.H.L
ஃ f(x) = is not continues at x = 0
ஃ The dangerous point on the path is at x = 0
The rider should not pass through the point at x = 0.

\(​​\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1 }{ \frac { (x-1)(x+1) }{ (x-1) } } =\lim _{ x\rightarrow 1 }{ (x+1) } =2\)
No, a person cannot be successful without any failure. if he wins his failure, he can reach his goal.
Value: Patience with rigorous efforts.

Given total revenue
\(R(x)=3x^2+36x+5\)
Now, marginal revenue \(=\frac{d(R)}{dx}\)
\(=\frac{d}{dx}(3x^2+36x+5)\)
\(=6x+36\)
Marginal revenue at x = 5
= 6 x 5 + 35 = Rs. 66.

Projection of a vector \(\overrightarrow { x } \) on \(\overrightarrow { y } \)= \(\frac { \overrightarrow { x } .\hat { y } }{ \left| \overrightarrow { y } \right| } \)
Projection of a \(\overrightarrow { x } \) on \(\overrightarrow { y } \)
\(=\frac { \left( \hat { i } +2\hat { j } +\hat { k } \right) \left( 2\hat { i } +3\hat { j } +2\hat { k } \right) }{ \left| { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 2 }^{ 2 } \right| } \)
\(=\frac { 10 }{ \sqrt { 17 } } \)