\(LHL=\lim _{ x\rightarrow { 0 }^{ - } }{ (2x=3) } =\lim _{ h\rightarrow { 0 } }{ [2(0-h) } +3]=3\)
\(RHL=\lim _{ x\rightarrow { { 0 }^{ + } } }{ 3(x+1) } =\lim _{ h\rightarrow { 0 } }{ [3(0+h+1)]=3 } \)
Ans.3

Let x denotes the rupees per litre and y denotes the quantity of milk in litre.Then, we have the following ordered pairs in the cartesian plane. (14,980) and (16,1220)
Now, the equation of line joining these two points is 
 \(y-980=\frac { 1220-980 }{ 16-14 } (x-14)\Longrightarrow 120x-y=700\)
He will sell weekly 1340 L milk at the rate of Rs.17 per litre

Let the length of shortest side be x cm
Then, according to question, we have 
Length of third side = (x+2) cm
Since the perimeter of the triangle is more than 166 cm.
2x+x+(x+2)>166
\(\Rightarrow 4x+2>166\)
\( \Rightarrow 4x>164\quad [subtracting\quad 2\quad from\quad both\quad sides]\)
\(\Rightarrow x>41\quad [dividing\quad both\quad sides\quad by\quad 4]\)
Hence,the length of the shortest side should be greater than 41cm.

\(cosA=\frac { \sin { B } }{ 2\sin { C } } \Rightarrow \frac { \quad { a }^{ 2 }+b^{ 2 }-{ c }^{ 2 } }{ 2ab } =\frac { b }{ 2c } \)
\(\Rightarrow { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 }={ b }^{ 2 }\Rightarrow { c }^{ 2 }={ a }^{ 2 }\Rightarrow c=a\)
\(\therefore \triangle ABC\quad is\quad isosceles\)

A' = {2, 5, 6, 7, 8, 9, 10}
B' = {1, 2, 3, 4, 7, 8, 9, 10}
A \(\cap\) B' ={1, 3, 4}
and B' - A' ={1, 3, 4}

(i) 4320
(ii) 36000

282240

x²+ y² + 6x + 2y - 90=0

Here f(x) = 2x - 5
putting x = - 3
\(\therefore\) f(-3) = 2 x -3 - 5 = - 11

L(3,0,0), M(0,4,0) and N(0,0,5)

First, flag can be selected in 5 ways and the second flag can be selected in 4 ways.
Ans. 20 

Also, IQ = \(\frac { MA }{ CA } \) \(\times \) 100
\(\therefore \) 80 \(\le \) \(\frac { MA }{ CA } \)\(\times \) 100 \(\le \)140
\(\Rightarrow \) 80 \(\le \)\(\frac { MA }{ 12 } \) \(\times \) 100 \(\le \) 140
9.6 \(\le \) MA \(\le \) 16.8
range of mental age is[9.6,16.8]

\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1,\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

Let P(n) = (xⁿ-yⁿ) is divisible by (x - y) for all n ∊ N.
For n =1
P(1) = (x¹-y¹) is divisible by (x-y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k)=(x^k-y^k) is divisible by (x-y)
⇒ (x^k-y^k) =m(x-y) for some m ∊ Z     ..(i)
For n = k+1
∴  P(k+1)=x^k+1-y^k+1 is divisible by (x-y)
x^k+1-y^k+1 = x^k+1-x^ky+x^ky-y^k+1
= x^k(x-y)+y(x^k-y^k)
= x^k(x -y)+y.m(x-y)
= (x-y)[x^k+my]
which is divisible by (x-y)
∴  P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, pen) is true for all n∊N.

Let \(\theta\) = 18° then 5\(\theta\) = 90° \(\Rightarrow\) 3\(\theta\) + 2\(\theta\) = 90°
\(\Rightarrow\)2\(\theta\)=90°-3\(\theta\)
\(\Rightarrow\)sin 2\(\theta\) = sin (90°-3\(\theta\)) = cos 3\(\theta\)
\(\Rightarrow\) 2 sin \(\theta\) cos \(\theta\) = 4 cos³\(\theta\) - 3 cos \(\theta\)
\(\Rightarrow\) 2 sin \(\theta\) cos \(\theta\) = cos \(\theta\) (4 cos²\(\theta\)- 3)
\(\Rightarrow\)cos \(\theta\) (4 cos²\(\theta\) - 3) - 2 sin \(\theta\) cos \(\theta\) = 0
\(\Rightarrow\) cos \(\theta\) (4 cos²\(\theta\) - 3 - 2 sin \(\theta\)) = 0
\(\Rightarrow\) 4cos²\(\theta\)-2 sin\(\theta\)-3=0 [\(\because\)cos 18°\(\neq\)0]
\(\Rightarrow\) 4 (1- sin²\(\theta\)) - 2 sin \(\theta\)- 3 =0
\(\Rightarrow\) 4 - 4 sin²\(\theta\) - 2 sin \(\theta\) - 3 = 0
\(\Rightarrow\) 4 sin²\(\theta\) + 2 sin \(\theta\) - 1 = 0
\(\Rightarrow sin \theta ={-2\pm\sqrt{(2)^2-4\times4(-1)}\over 2\times4}\)
\(={-2\pm \sqrt{20}\over 8}={-2\pm2\sqrt{5}\over8}\)
\(\therefore sin 18^o={\sqrt{15}-1\over 4}\)
[\(\because\) 9 lies in I quadrant]

Here the sample space S = {I, 2, 3,4,5,6}
\(\therefore \) n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5} \(\Rightarrow \) n(A) = 3
\(Thus\ P(A)=\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event of getting a number greater than or equal to 3
B = {3, 4, 5, 6} \(\Rightarrow \) n(B) = 4
\(Thus\ P(B)=\frac { n(B) }{ n(S) } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Let C be the event of getting a number less than or equal to 1
C = {I} \(\Rightarrow \) n(C) = 1
\(Thus\ P(C)=\frac { n(C) }{ n(S) } =\frac { 1 }{ 6 } \)
(iv) Let D be the event of getting a number more than 6
\(D=\phi \Rightarrow n(D)=0\)
\(Thus\ P(D)=\frac { n(D) }{ n(S) } \frac { 0 }{ 6 } =0\)
Let E be the event of getting a number less than 6
E = {I, 2, 3, 4, 5} \(\Rightarrow \) n(E) = 5
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)

It is given that 
\(f(x)=\left\{\begin{array}{cc} m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^{3}+m, & x>1 \end{array}\right.\)
\(\lim _{x \rightarrow 0^{-}} f(x) =\lim _{x \rightarrow 0}\left(m x^{2}+n\right) \)
\(=m(0)^{2}+n \)
\(=n \)
\(\lim _{x \rightarrow 0^{+}} f(x) =\lim _{x \rightarrow 0}(n x+m) \)
\(=n(0)+m \)
\(=m .\)
\(\text { Thus, } \lim _{x \rightarrow 0} f(x) \text { exists if } m=n \text { . }\)
\(\lim _{x \rightarrow 1^{\prime}} f(x) =\lim _{x \rightarrow 1}(m x+m) \)
\(=n(1)+m \)
\(=m+n \)
\(\lim _{x \rightarrow 1^{+}} f(x) =\lim _{x \rightarrow 3}\left(n x^{3}+m\right) \)
\(=n(1)^{3}+m \)
\(=m+n \)
\(\therefore \lim _{x \rightarrow 1^{-}} f(x) =\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) .\)
\(\text { Thus, } \lim _{x \rightarrow 1} f(x) \text { exists for any integral value of } m \text { and } n \text { . }\)

\(\frac{N}{2}=\frac{50}{2}=25\)∴ Median class is 20 - 30
∴ Median=\(20+\frac { 25-14 }{ 14 } \times 10=20+7.86=27.86\)
M.D. about median =\(\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-M \right| } =\frac { 1 }{ 50 } \times 517.16=10.34\)