Q₁ = Size of \(\left( \frac { N }{ 4 } \right) ^{ th }\)value =\(\left( \frac { 67 }{ 4 } \right) ^{ th }\)= 16.75^th value Thus Q₁ lies in the class 20 – 30; and the corresponding values are
L = 20,\(\frac { N }{ 4 } \) = 16.75 ; pcf = 12, f = 19, c = 10
\({ Q }_{ 1 }=L+\left( \frac { \frac { N }{ 4 } -pcf }{ f } \right) \times c\)
\({ Q }_{ 1 }=20+\left( \frac { 16.75-12 }{ 19 } \right) \times 10\)= 20 + 2.5 = 22.5
Q₃ = Size of \(\left( \frac { 3N }{ 4 } \right) ^{th}\) value = 50.25^th value
Thus Q₃ lies in the class 50 – 60 and corresponding values are 
L = 50 ; \(\frac { 3N }{ 4 } \)= 50.25 ; pcf = 46, f = 9, c = 10
\({ Q }_{ 3 }=L+\left( \frac { \frac { 3N }{ 4 } -pcf }{ f } \right) \times c\)
\({ Q }_{ 3 }=20+\left[ \frac { 50.25-46 }{ 9 } \right] \times 10=54.72\)
\(QD=\frac{1}{2}(Q_3-Q_1)\)
\(=\frac { 54.72-22.5 }{ 2 } =16.11\)
\(\therefore\)  QD = 16.11

Face value of a share = Rs. 10
Face value of 20 shares = Rs. 200
Dividend = \(\frac{9}{100}\times 200\) = Rs. 18 
\([S.I=\frac{PNR}{100},N=1]\)
Investment = \(\frac{18\times 100}{1\times 12} = Rs. 150\)
\([P=\frac{100\times SI}{NR},(N=1)]\)
A person purchased 20 shares = Rs. 150
The market value of one share = Rs. \(\frac{150}{20}\) = Rs. 7.50

\(\rho =1-\frac{6 \sum d^2}{n\left(n^2-1\right)} \)
\(=1-\frac{6(22)}{11(120)}=1-\frac{1}{10}=1-0.1=0.9\)

u = log\(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \)
e^u = \(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \) = f(x, y) ... (1)
Consider f(x, y)  = \(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \)
f(tx, ty) = \(\frac { { t }^{ 4 }{ x }^{ 4 }+{ t }^{ 4 }{ y }^{ 4 } }{ tx+ty } ={ t }^{ 3 }\left( \frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \right) ={ t }^{ 3 }f(x,y)\)
\(\therefore\)  f is a homogeneous function of degree 3.
Using Euler’s theorem we get
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3f\)
Consider f(x, y) = e^u
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3e\)^u
\(∴ { e }^{ u }x.\frac { \partial u }{ \partial u } +{ e }^{ u }y.\frac { \partial u }{ \partial y } =3{ e }^{ u }\)
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3\)

We know that average cost [AC] is minimum when average cost [AC] = marginal cost [MC].
Cost: \(C={1\over3}x^3-3x^2+9x\)
AC = \({1\over3}-3x^2+9\) and MC = x² - 6x + 9
Now, AC = MC ⇒ \({1\over 3}x^2-3x+9-x^2-6x+9\)
⇒ 2x² - 9x = 0 ⇒ \(x={9\over 2}\) unit (∵ x > 0)

\(\text { LHS }=\sin \theta \cdot \cos \theta\left\{\sin \left(\frac{\pi}{2}-\theta\right) \cdot \operatorname{cosec} \theta\right. \left.+\cos \left(\frac{\pi}{2}-\theta\right) \cdot \sec \theta\right\} \)
\(=\sin \theta \cos \theta\left\{\cos \theta\left(\frac{1}{\sin \theta}\right)+\sin \theta\left(\frac{1}{\cos \theta}\right)\right\} \)
\(=\cos ^2 \theta+\sin ^2 \theta=1=\text { RHS } \)
Hence proved.

Given f(2) = 0
= 2 + 2
\(=4 \neq 0\)
\(\therefore \lim _{x \rightarrow 2} f(x) \neq f(2)\)
The function is not continuous at x = 2

Compare the equation
12x² - 10xy + 2y² + 14x - 5y + 2 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = 12, 2h = -10, b = 2, 2g = 14, 2f = -5
\(h=-5\quad g=7\quad f=-\frac { 5 }{ 2 } ,c=2\)
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=\left|\begin{array}{ccc} 12 & -5 & 7 \\ -5 & 2 & \frac{-5}{2} \\ 7 & \frac{-5}{2} & 2 \end{array}\right|\)
\(=12\left(4-\frac{25}{4}\right)+5\left(-10+\frac{35}{2}\right)+7\left(\frac{25}{2}-14\right)\)
\(=48-75-50+\frac{175}{2}+\frac{175}{2}-98\)
= -175 + 175 = 0
Hence the given equations represent a pair of straight lines.
To find separate equation
\(12 x^2-10 x y+2 y^2 =12 x^2-6 x y-4 x y+2 y^2 \)
\(=6 x(2 x-y)-2 y(2 x-y) \)
\(=(6 x-2 y)(2 x-y)\)
\(12 x^2-10 x y+2 y^2+ 14 x-5 y+2 =(6 x-2 y+l)(2 x-y+\mathrm{m})\)
Comparing the coefficient of x and y
14 = 6m + 2l
divided by 2
7 = 3m + l .........(1)
-5 = -2m - l .......(2)
Solving (1) and (2) we get m = 2, 1 = 1
The separate equations are
6x - 2y + 1 = 0
2x - y + 2 = 0

LHS =  \(\begin{vmatrix} {1\over a}&bc&b+c\\{1\over b}&ca&c+a\\{1\over c}&ab&a+b \end{vmatrix}\)
Multiplying R₁ by a, R₂ by b, and R₃ by c respectively and dividing the determinant by abc we get.
\(LHS={1\over abc}\begin{vmatrix} 1&abc&ab+ac\\1&abc&bc+ab\\{1}&abc&ac+bc \end{vmatrix}\)
\(={1\over abc}(abc)\begin{vmatrix} 1&1&ab+ac\\1&1&bc+ab\\{1}&1&ac+bc \end{vmatrix}\) = 0  \([\because C_1\equiv C_2]\) = RHS.
Hence Proved.

\(\frac { 1 }{ { x }^{ 2 }-1 } =\frac { 1 }{ (x+1)(x-1) } =\frac { A }{ x+1 } +\frac { B }{ x-1 } \)
⇒ \(\frac{1}{(x+1)(x-1)}=\frac{A(x-1)+B(x+1)}{(x+1)(x-1)}\)
⇒ 1 = A (x - 1) + B (x + 1)       ...(1)
If x = -1 in (1) we get
1 = A(-2) + 0 ⇒ A = \(\frac { -1 }{ 2 } \)
If x = 1 in (1) we get,
1 = 0 + B(1 + 1) ⇒ 1 = 2B ⇒ B = \(\frac { 1 }{ 2 } \)
\(\frac{1}{x^2-1}=\frac{-1}{2(x+1)}+\frac{1}{2(x-1)}\).