4(nC₂) = (n + 2)C₃
\(4 \frac{n(n-1)}{1 \times 2}=\frac{(n+2)(n+1)(n)}{1 \times 2 \times 3}\)
\(12\left( n-1 \right) =\left( n+2 \right) \left( n+1 \right) \)
\(12\left( n-1 \right) =\left( { n }^{ 2 }+3n+2 \right) \)
\({ n }^{ 2 }-9n+14=0\)
\( \left( n-2 \right) \left( n-7 \right) =0\Rightarrow n=2,n=7\)

The given system can be written as
\(\left[ \begin{matrix} 2 & 5 \\ 3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \end{matrix} \right] \)
i.e., AX = B
X = A^–1B
where \(A=\left[ \begin{matrix} 2 & 5 \\ 3 & 2 \end {matrix} \right] ,X=\left [ \begin{matrix} x \\ y \end {matrix} \right] and\ B=\left [ \begin{matrix} 1 \\ 7 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 2 & 5 \\ 3 & 2 \end{matrix} \right] \)
= –11 \(\neq \) 0
A^–1 exists.
\(adj\ A=\left[ \begin{matrix} 2 & -5 \\ -3 & 2 \end{matrix} \right] \)
\(A^{-1}=\frac{1}{|A|}adjA\)
\(=\frac{1}{-11}\left[ \begin{matrix} 2 & -5 \\ -3 & 2 \end{matrix} \right] \)
\(X=A^{-1} B\)
\(=\frac{1}{-11}\left[ \begin{matrix} 2 & -5 \\ -3 & 2 \end{matrix} \right] \ \left[ \begin{matrix} 1 \\ 7 \end{matrix} \right] \)
\(=\frac{1}{-11}\left[ \begin{matrix} -33 \\ 11 \end{matrix} \right] = \left[ \begin{matrix} 3 \\ -1 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -1 \end{matrix} \right] \)
x = 3, and y = -1.

\(A=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
=16 ≠ 0
Since A is a nonsingular matrix, A ^-1 exists
Now  adj \(A=\left[ \begin{matrix} 2 & -4 \\3 & 2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
\(=\frac { 1 }{ 16 } \left[ \begin{matrix} 2 & -4 \\ 3 & 2 \end{matrix} \right] \)

\(y=16-x^2\)

\(\tan ^{-1} 2 x+\tan ^{-1} 3 x=\pi / 4\)
\(\tan^{-1}\left(\frac{2x+3x}{1-(2x)(3x)}\right)=\frac{\pi}{4}\)
\(\frac{5 x}{1-6 x^2}=\tan \pi / 4=1 \text { if } 6 x^2<1 \)
\(5 x=1-6 x^2 \quad x^2<1 / 6 \)
\(6 x^2+5 x-1=0 \quad \frac{-1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}} \)
\(6 x^2+6 x-x-1=0 \)
\(x=-1, \frac{1}{6} \text { and } \frac{-1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}} \)
\(x=1 / 6\)

u = sin² x and v = x²
\(\frac { du }{ dx } =2 \sin x \cos x;\quad \frac { dv }{ dx } =2x\)
= 2 sinx 
\(\therefore \frac { du }{ dv } =\frac { \frac { du }{ dx } }{ \frac { dv }{ dx } } =\frac { sin2x }{ 2x } \)

\(A^{-1}=\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\)
\(A^2=\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\)
\(=\left(\begin{array}{ll} 1-2 & -1-3 \\ 2+6 & -2+9 \end{array}\right)=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)\)
\(\mathrm{LHS}=A^2-4 A+5 l_2\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)-4\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)+5\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)-\left(\begin{array}{cc} 4 & -4 \\ 8 & 12 \end{array}\right)+\left(\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right)\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)+\left(\begin{array}{cc} 1 & 4 \\ -8 & -7 \end{array}\right)=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right)\)
\(=O=R H S\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A\)
\(|A|=3+2=5 \neq 0\)
\(A^{-1}=\frac{1}{5}\left(\begin{array}{cc} 3 & 1 \\ -2 & 1 \end{array}\right)\)

Expanding along R₁ we get,
\(\left|\begin{array}{lll} 2 & x & 3 \\ 4 & 1 & 6 \\ 1 & 2 & 7 \end{array}\right|=0\)
⇒ 2 (7 -12) -x (28- 6) + 3 (8 - 1) = 0
⇒ -10 - 22x + 21 = 0
⇒ 11 = 22x
⇒ \(x={11\over 22}={1\over 2}\)

\(\left(x+\frac{1}{y}\right)^7 =7 C_0 x^7+7 C_1 x^6\left(\frac{1}{y}\right)+7 C_2 x^5\left(\frac{1}{y}\right)^2 +7 C_4 x^3\left(\frac{1}{y}\right)^4+7 C_5 x^2\left(\frac{1}{y}\right)^5 +7 C_6 x\left(\frac{1}{y}\right)^6+7 C_7\left(\frac{1}{y}\right)^7\)
\(=x^7+\frac{7 x^6}{y}+\frac{21 x^5}{y^2}+\frac{35 x^4}{y^3} +\frac{35 x^3}{y^4}+\frac{21 x^2}{y^5}+\frac{7 x}{y^6}+\frac{1}{y^7}\)

Investment = 200 x 100 x \(\frac{8}{100}\)= Rs. 1600

Since there is no common area between the lines x₁ + x₂ ≤ 1 and 5x₁ + 5x₂ ≥ 0

\(\frac{-1}{\operatorname{cosec} 45^{\circ}}=\frac{-1}{\sqrt{2}}\)

\(\frac{\pi}{8}=\frac{180^{\circ}}{8}=22 \frac{1}{2}^{\circ}=22^{\circ} 30^{\prime}\)

\(m=\frac{-a}{b}=\frac{-7}{5}\)

nP₂ = 20
n(n - 1) = 5 x 4
n = 5

(Since \(|k A|=k^n|A|,\) n is the order of matrix A

\(\left|\begin{array}{lll} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{array}\right|=0 \Rightarrow-1\left[x^2-x\right]=0\)
\(\Rightarrow x(x-1)=0 \Rightarrow x=0,1\)

\(AM=\frac { \sum { fx } }{ N } =\frac { 2357 }{ 140 } =16.84\)
GM = Anti \(\log { \left( \frac { \sum { flogX } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 167.209 }{ 140 } \right) } \)
= Anti log(1.1944) = 15.65
\(HM=\frac { N }{ \sum { \left( \frac { f }{ X } \right) } } =\frac { 140 }{ 9.6852 } =14.46\)
i.e.,16.84 >15.65 >14.46
\(\therefore\) AM > GM > HM

(i) First convert the given equations Y on X and X on Y in standard form and find their regression coefficients respectively.
Given regression lines are
3X–2Y = 5      ... (1)
X–4Y = 7        ... (2)
Let the line of regression of X on Y is
3X–2Y = 5
3X = 2Y+5
X = \(\frac{1}{3}\)(2Y+5)
X = \(\frac{1}{3}\)(2Y+5)
X = \(\frac { 2 }{ 3 } Y+\frac { 5 }{ 3 } \)
∴ Regression coefficient of X on Y is
b_xy =\(\frac{2}{3}\)(<1)
Let the line of regression of Y on X is
X–4Y = 7
–4Y = –X+7
4Y = X–7
Y = \(\frac{1}{4}\)(X-7)
Y = \(\frac{1}{4}\)X-\(\frac{7}{4}\)
∴ Regression coefficient of Y on X is
b_yx = \(\frac{1}{4}\)(<1)
(ii) Coefficient of correlation
Since the two regression coefficients are positive then the correlation coefficient is also positive and it is given by
r = \(\sqrt { { b }_{ yx }.{ b }_{ xy } } \)
= \(\sqrt { \frac { 2 }{ 3 } .\frac { 1 }{ 4 } } \)
= \(\sqrt { \frac { 1 }{ 6 } } \)
= 0.4082
∴ r = 0.4082

Given
Ordering cost per order : C₃ = Rs. 150 per order.
Number of units per order: q = 500 units
Annual demand = 500 × 4 = 2000 units
∴ Demand rate : R = 2000 per year
Carrying cost : C₁ = 20% of unit value
C₁ = \(\frac { 20 }{ 100 } \times 125=Rs25\)
(i) Total annual cost of due existing inventory policy
= \(\frac { R }{ q } \times { C }_{ 3 }+\frac { q }{ 2 } { C }_{ 1 }\) = \(\frac { 2000 }{ 500 } \times 150+\frac { 500 }{ 2 } \times 25\)
= Rs. 6850
(ii) EOQ =  \(\sqrt { \frac { 2Rc_{ 3 } }{ { c }_{ 1 } } } \)
= \(\sqrt { \frac { 2\times 2000\times 150 }{ 25 } } \)
= \(\sqrt { 12\times 2000 } \)
= 155 units (app.)
(iii) Minimum annual cost \(= \sqrt { { 2Rc_{ 3 } }{ { c }_{ 1 } } } \)
= \(\sqrt { 2\times 2000\times 150\times 25 } \)
= Rs. 3873.
By applying the economic order quantity, money saved by a company = 6850 – 3873  = Rs. 2977.

Let P(n) denote the statement 5²ⁿ - 1 is divisible by 24.
Step-1: Put n = 1
5²(1) - 1 = 25 - 1 = 24 is divisible by 24
∴ P(1) is true.
Step-2:
Let us assume that P(k) is true
∴ 5^2k - 1 is divisible by 24.
⇒ 5^2k - 1 = 24m
⇒ 5^2k = 24m + 1    ...(1)
To prove that P(k + 1) is true 
P(k + 1) = 5^2(k+1) - 1
= 5^2k+2 - 1 = 5^2k.5² - 1
= (24m + 1)25 - 1
= 24m. 25 + 25 - 1
= 24m + 25 - 24
= 24(25m + 1)
Which is divisible by 24
∴ P (k + 1) is true whenever P(k) is true.
∴ p(n) is true for all \(n\in N\).

Let A \(=\begin{vmatrix} 1 & a&a^2&-bc \\1 &b&{b}^{2}&-ca\\1&c&c^2&-ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -b c \\ 1 & b & -c a \\ 1 & c & -a b \end{array}\right|\)
\(A=\begin{vmatrix} 1 & a&{a}^{2} \\ 1 &b&b^2\\1&c&c^2 \end{vmatrix}-\begin{vmatrix} 1 & a&bc \\1 &b&ca\\1&c&ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{ccc} a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c \end{array}\right|\)
(Multiplying R₁, R₂ and R₃ of II det by a, b, c respectively)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)

\(\underset { x\rightarrow 1 }{ lim } \frac { { x }^{ 3 }-1 }{ x-1 } \) is of the type \(\frac { 0 }{ 0 } \)
\(\underset { x\rightarrow 1 }{ lim } \cfrac { { x }^{ 3 }-1 }{ x-1 } =\underset { x\rightarrow 1 }{ lim } \cfrac { \left( x-1 \right) \left( { x }^{ 2 }+x+1 \right) }{ \left( x-1 \right) } \)
\(=\underset { x\rightarrow 1 }{ lim } \left( { x }^{ 2 }+x+1 \right) \)
\(=\left( 1 \right) ^{ 2 }+1+1\) 
= 3

  \(\cfrac { 1 }{ 9! } +\cfrac { 1 }{ 10! } =\cfrac { n }{ 11! } \)
\(\cfrac { 1 }{ 9! } +\cfrac { 1 }{ 10\times 9! } =\cfrac { n }{ 11! }\)
\( \cfrac { 1 }{ 9! } \left[ 1+\cfrac { 1 }{ 10 } \right] =\cfrac { n }{ 11! } \)
\( \cfrac { 1 }{ 9! } \times \cfrac { 11 }{ 10 } =\cfrac { n }{ 11! } \)
\(n=\cfrac { 11!\times 11 }{ 9!\times 10 } \)
\(=\cfrac { 11!\times 11 }{ 10! } \)
\( =\cfrac { 11\times 10!\times 11 }{ 10! } \)
\(n=121\)

Given C = 10- 4x³ + 3x⁴
(i) Average Cost (AC) = \({C\over x}=\frac { 10 }{ x } \) - 4x² + 3x³
(ii) Marginal Cost (MC) = \({dC\over dx}\)= -12x² + 12x³
(iii) Marginal Average Cost (MAC) = \({d\over dx }(AC)=-\frac { 10 }{ { x }^{ 2 } } \) - 8x + 9x²

Let x be the number of passengers.
\(P=136 -\frac{40}{100}(x-100) \text { if } x \geq 100 \)
\(R=p x =136 x-\frac{2 x^2}{5}+40 x \)
\(=176 x-\frac{2 x^2}{5} \)
\(\frac{d R}{d x} =176-\frac{4 x}{5}\)
For max. revenue \(\frac{d R}{d x}=0\)
\(176 =\frac{4 x}{5} \)
\(x =\frac{176 \times 5}{4}=220 \)
\(\frac{d^2 R}{d x^2} =-\frac{4}{5}<0\)
∴ At  x = 220,
Revenue is maximum.

Given that f(x) = x² + 2x – 5 … (1)
f '(x) = 2x + 2
At stationary points, f'(x) = 0
\(\Rightarrow\)  2x + 2 = 0 \(\Rightarrow\)  x = –1
f(x) has stationary value at x = –1
When x = –1, from (1)
f(–1) = (–1)² + 2(–1) – 5 = – 6
Stationary value of f (x) is – 6
Hence stationary point is (–1,–6)

\(\sin A + \sin 2 A=2 \sin \frac{A + 2 A}{2} \cos \frac{A-2 A}{2}\)
\(=2 \sin \frac{3 A}{2} \cos \frac{A}{2}\)

\({ 60 }^{ o }=60\times \frac { \pi }{ 180 } = \frac { \pi }{ 3 } \)