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}

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

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.

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

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

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

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]

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

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

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

x²+ y² + 6x + 2y - 90=0

282240

(i) 4320
(ii) 36000

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.

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

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

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
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)

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]