Ans.Yes.

Total number of persons  = 12, which can be arranged in a row in 12! ways.
Now, if 6 girls sit together, then we have 7 persons (Considering 6 girls as one person), which can be in a row in 7! ways. But 6 girls can be arrange in 6! ways. 
\(\therefore\)  Required probability
\(\frac { 7!\times 6! }{ 12! } =\frac { 1 }{ 132 } \)

The first n natural numbers are 1,2,3...,n.
\(\because\) Standard deviation,
\(SD=\sqrt { \frac { \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } }{ n } -({ \frac { \sum _{ i=1 }^{ n }{ { x }_{ i } } }{ n } ) }^{ 2 } } \)
\(\therefore \ SD=\sqrt { \frac { (n(n+1)(2n+1) }{ 6n } -({ \frac { n(n+1) }{ 2n } ) }^{ 2 } } \)
\([\because \ \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } =\frac { n(n+1)(2n+1) }{ 6 } and\ \sum _{ i=1 }^{ n }{ { x }_{ i } } =\frac { n(n+1) }{ 2 } ]\)
\(=\sqrt { (n+1)(\frac { 2n+1 }{ 6 } -\frac { n+1 }{ 4 } ) } \)
\(=\sqrt { (n+1)(\frac { 4n+2-3n-3 }{ 12 } ) } \)
\(=\sqrt { \frac { (n+1)(n-1) }{ 12 } } =\sqrt { \frac { { n }^{ 2 }-1 }{ 12 } } \)
\(\therefore \ Variance={ (SD) }^{ 2 }=\frac { { n }^{ 2 }-1 }{ 12 } \)

\(\lim_{ h\rightarrow 0 } \frac { (a+h)^{ 2 }sin(a+h)-a^{ 2 }sin\quad a }{ h } \)
\(=\lim_ { lim }{ h\rightarrow 0 } \frac { (a^{ 2 }+h^{ 2 }+2ah)[sin \ a\ cos \ h+cos\ a\ sin \ h)-a^{ 2 }sin\quad a }{ h } \)
\(\left[ \because \ sin(C+D)=sinCcosD+CosCsinD \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sin \ a(cos \ h-1) }{ h } +\frac { a^{ 2 }cos \ a \ sin \ h) }{ h } +(h+2a)(sin \ a \ cos \ h+cos\ a \ sin\ h) \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sina(-2sin^{ 2 }\frac { h }{ 2 } ) }{ \frac { h^{ 2 } }{ 2 } } .\frac { h }{ 2 } \right] +\overset { lim }{ h\rightarrow 0 } \frac { a^{ 2 }cosasinh }{ h } +\lim_{ h\rightarrow 0 } (h+2a)sin(a+h)\)
\(\left[ \because cosm \ h-1=-2sin^{ 2 } h/2\quad and\quad sinacos\quad h+cosasin\quad h=sin(a+h) \right] \)
\(=a^{ 2 }sin\quad a\times 0+a^{ 2 }\quad cosa(1)+2asina \quad \left[ \because \lim_{ x\rightarrow 0 } \frac { sin\quad x }{ x } =1 \right]\)
\( =a^{ 2 }cos \ a+2asin \ a\)

We know that, AM \(\ge \) GM
\(\therefore \frac { { 4 }^{ x }+\frac { 4 }{ { 4 }^{ x } } }{ 2 } >\ge \sqrt { { 4 }^{ x } } \times \frac { 4 }{ { 4 }^{ x } } \)
\(\Rightarrow { 4 }^{ x }+\frac { 4 }{ { 4 }^{ x } } \ge \sqrt [ 2 ]{ 4 } \)
\(\Rightarrow { 4 }^{ x } +\frac { 4 }{ { 4 }^{ x } } \ge 4\)
Hence, the minimum value of given expression is 4.
 

The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 0 cannot be included. Therefore, the number of ways in which thousands place can be filled is 9.The hundreds, tens, and units place can be filled by any of the digits from 0 to 9. However, the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and unitsplace is to be filled by the remaining 9 digits.Therefore, there will be as many such 3-digit numbers as there are permutations of 9 different digits taken 3 at a time. Number of such 3-digit numbers  
\(={ }^{9} \mathrm{P}_{3}=\frac{9 !}{(9-3) !}=\frac{9 !}{6 !} \)
\(=\frac{9 \times 8 \times 7 \times 6 !}{6 !}=9 \times 8 \times 7=504\)
Thus, by multiplication principle, the required number of 4-digit numbers is 9 × 504 = 4536

LHS = \(tan\quad \theta \quad tan\quad ({ 60 }^{ o }-\theta )tan\quad ({ 60 }^{ o }+\theta )\)
\(=\frac { sin\quad \theta }{ cos\quad \theta } .\frac { sin\quad \left( { 60 }^{ o }-\theta \right) }{ cos\quad \left( { 60 }^{ o }-\theta \right) } .\frac { sin\quad \left( { 60 }^{ o }+\theta \right) }{ cos\quad \left( { 60 }^{ o }+\theta \right) } \)
\(=\frac { sin\quad \theta [2sin\quad \left( { 60 }^{ o }-\theta \right) sin\quad \left( { 60 }^{ o }+\theta \right) ] }{ cos\quad \theta [2cos\quad \left( { 60 }^{ o }-\theta \right) cos\quad \left( { 60 }^{ o }+\theta \right) ] } \)
\(=\frac { sin\quad \theta (cos\quad 2\theta \quad -\quad cos\quad { 120 }^{ o }) }{ cos\quad \theta (cos\quad { 120 }^{ o }+cos\quad 2\theta ) } \)
\(\because\)  2 sinA sinB = cos(A−B)−cos(A+B) and 2 cosA cosB = cos(A+B)+cos(A−B)]
\(=\frac { sin\quad \theta \left( cos\quad 2\theta +\frac { 1 }{ 2 } \right) }{ cos\quad \theta \quad \left( cos\quad 2\theta -\frac { 1 }{ 2 } \right) } \left[ \because cos { 120 }^{ o }= cos \left( { 180 }^{ o }- { 60 }^{ o } \right) = -cos\quad { 60 }^{ o }=-\frac { 1 }{ 2 } \right] \)
\(=\frac { sin\quad \theta cos\quad 2\theta +\frac { 1 }{ 2 } sin\theta }{ cos\quad \theta \quad cos\quad 2\theta -\frac { 1 }{ 2 } cos\theta } =\frac { 2sin\quad \theta cos\quad 2\theta +sin\theta }{ 2cos\quad \theta \quad cos\quad 2\theta -cos\quad \theta }\)
\(=\frac { sin\quad \left( \theta +2\theta \right) +sin\quad (-\theta )+sin\theta }{ cos\quad (\theta +2\theta )+cos\quad (-\theta )-cos\quad \theta } \)
\(\left[ \therefore \quad 2sin\quad A\quad cos\quad B\quad =\quad sin\quad (A+B)+sin\quad (A-B)\quad and\quad 2cos\quad A\quad cos\quad B\quad =\quad cos\quad (A+B)+cos\quad (A-B) \right] \)
\(=\frac { sin\quad 3\theta -sin\quad \theta +sin\quad \theta }{ cos\quad 3\theta +cos\quad \theta -cos\quad \theta } \quad \quad \left[ \because sin\quad (-\theta )=-sin\theta \quad and\quad cos\quad (-\theta )=cos\quad \theta \right] \)
\(=\frac { sin\quad 3\theta }{ cos\quad 3\theta } =tan\quad 3\theta =RHS\)
Hence proved.

Given A={1,2,3,4}, B = {1,2,3} And C={2,4}
Now, P(A)={ \(\phi\), {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}} ....(i)
P(B) ={\(\phi\){1},{2},{3},{1,2},{1,3},{1,3},{2,3},{1,2,3}} ...(ii)
and P(c) ={\(\phi\)),{2},{4},{2,4}}  ...(iii)
Given condition is X \(\subset\) A, X \(\subset\) B  and X \(\subset\) C.
\(\Rightarrow\) X \(\in\) p(A), X \(\in\) P(B), X \(\in\) P(c)and X \(\neq\)A,B,C
\(\Rightarrow\) X is proper subset of A,B and C
\(\Rightarrow\)   X=\(\phi\),{2}

The value of the function for every rational number is 1 and for every irrational number is -1.
\(\therefore \quad \frac { 1 }{ 2 } \in Q\Rightarrow f\left( \frac { 1 }{ 2 } \right) =1\)
\(and \quad \pi \notin Q\Rightarrow f\left( \pi \right) =-1\)

Let A, B and C denote the set of students who study English, Economics and Mathematics, respectively.
Then, we have,
Total number of students, n (A \(\cup \) B \(\cup \) C) = 40
Number of students who study English, n (A) = 16
Number of students who study Eoconomics, n (B) = 22
Number of students who study Mathematics, n (C) = 26
Number of students who study English and Economics, n (A  \(\cap \) B) = 5
Number of students who study Mathematics and Economics, n( B \(\cap \) C) = 14
and number students who study all subjects,
n (A \(\cup \) B \(\cup \) C) = 2
Clearly, n (A \(\cup \) B \(\cup \) C)  = n(A) + n(B) - n(C) - n (A \(\cap \) B) - n( B \(\cap \) C)  - n( A \(\cap \) C) -  n (A \(\cap \) B \(\cap \) C)                       
40 = 16 +22+ 26 - 5- 14 n ( C \(\cap \) A) + 2
40 = 66 - 19 - n ( C \(\cap \) A)
n ( C \(\cap \) A) = 47 - 70
=7
Hence the number of students who study English and Mathematics are 7.