Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(0≤X≤10)=P(X=0)+P(X=10)
\(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

Given
The difference table is

To findy when x = x ⇒ x₀+ nh = x ⇒ 1 + n (1) = x ⇒ n = x-1
Newton's forward interpolation formula is
y_(x=x) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ .....
y_(x=x) = 1 + (x-1)(-2) + \(\frac { (x-1)(x-2) }{ 2 } (4)+\frac { (x-1)(x-2)(x-3) }{ 6 } (-8)+\frac { (x-1)(x-2)(x-3)(x-4) }{ 24 } (16)\)
⇒ y = 1 - 2x + 2(x² - 3x + 2) - \(\frac{4}{3}\) (x - 1) (x - 2) (x - 3) + \(\frac{2}{3}\) (x - 1) (x - 2) (x - 3) (x -4)
⇒ y = 3 - 2x + 2x² - 6x + 4 - \(\frac{4}{3}\) [(x² - 3x + 2) (x- 3)] + \(\frac{2}{3}\) [(x² - 3x + 2)(x² - 7x + 12)]
⇒ y = 2x² - 8x+ 7- \(\frac{4}{3}\) [x³- 3x²+ 2x- 3x² + 9x- 6] + \(\frac{2}{3}\) [x⁴ - 3x³ + 2x² - 7x³ + 21x² - 14x + 12x² - 36x + 24]
⇒ y = 2x²-8x+7- \(\frac{4}{3}\) x³ + 4x² - \(\frac{8}{3}\) x + 4x² - 12x + 8 + \(\frac { { 12x }^{ 4 } }{ 3 } -\frac { 20 }{ 3 } { x }^{ 3 }+\frac { 70 }{ 3 } { x }^{ 2 }-\frac { 100x }{ 3 } +\frac { 48 }{ 3 } \)
⇒ y = \(\frac{2}{3}\) x⁴ + x³ \(\left( \frac { -4 }{ 3 } \frac { -20 }{ 3 } \right) \) + x²\(\left( 2+4+4+\frac { 70 }{ 3 } \right) \) + x \(\left( -8-\frac { 8 }{ 3 } -12-\frac { 100 }{ 3 } \right) \) + 31
⇒ y =  \(\frac{2}{3}\) x⁴ - 8x³ + \(\frac{100}{3}\) x² - 56x + 31 which is the required polynomial.

Here the intervals are unequal. By Lagrange’s interpolation formula we have
x₀ = 5, x₁ = 6, x₂ = 9, x₃ = 11
y₀ = 12, y₁ = 13, y₂ = 14, y₃ = 16
\(y=f(x)=\frac { \left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 0 }-{ x }_{ 1 } \right) \left( x_{ 0 }-{ x }_{ 2 } \right) \left( { x }_{ 0 }-{ x }_{ 3 } \right) } \times { y }_{ 0 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 1 }-{ x }_{ 0 } \right) \left( x_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) } \times { y }_{ 1 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 2 }-{ x }_{ 0 } \right) \left( x_{ 2 }-{ x }_{ 1 } \right) \left( { x }_{ 2 }-{ x }_{ 3 } \right) } \times { y }_{ 2 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 3 }-{ x }_{ 0 } \right) \left( x_{ 3 }-{ x }_{ 1 } \right) \left( { x }_{ 3 }-{ x }_{ 2 } \right) } \times { y }_{ 3 }\)
\(=\frac { (x-6)(x-9)(x-11) }{ (5-6)(5-6)(5-11) } (12)+\frac { (x-5)(x-9)(x-11) }{ (6-5)(6-9)(6-9) } (13)+\frac { (x-5)(x-6)(x-11) }{ (9-5)(9-6)(9-11) } (14)+\frac { (x-5)(x-6)(x-9) }{ (11-5)(11-6)(11-9) } (16)\)
Put x = 10
\({ y }_{ (10) }=f\left( 10 \right) =\frac { 4(1)(-1) }{ (-1)(-4)(-6) } (12)+\frac { (5)(1)(-1) }{ (1)(-3)(-5) } (13)+\frac { 5(4)(-1) }{ 4(3)(-2) } (14)+\frac { (5)(4)(1) }{ 6(5)(2) } (16)\)}
= \(\frac { 1 }{ 6 } \left( 12 \right) -\frac { 13 }{ 3 } +\frac { 5\left( 14 \right) }{ 3\times 2 } +\frac { 4\times 16 }{ 12 } \)
= 14.6663

Since the required value is at the end of the table, apply backward interpolation formula

.\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 7.5
\(\therefore\) xn + nh = 7.5, xn = 8, h = 1 \(\Rightarrow\) n = –0.5
\({ y }_{ (x=7.5) }=512+\frac { -0.5 }{ 1! } 169+\frac { 0.5(-0.5+10 }{ 2! } 42+\frac { -0.5(-0.5+1)(-0.5+2) }{ 3! } 6\)
= 421.88

To find the population for the year 1905 (i.e) the value of y at x = 1905
Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\cfrac { n }{ n! } \Delta { y }_{ 0 }+\cfrac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\cfrac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 } + ...\)
To find y at x = 1905
\(\therefore\) x₀+nh = 1905 , x₀ = 1891, h = 10
1891+n(10) = 1905 \(\Rightarrow\) n = 1.4

y_(x=1905) = \(98,752+(1.4)(33533)+\frac { (1.4)(0.4) }{ 2 } (2258)+\frac { (1.4)(0.4)(-0.6) }{ 6 } (-10435)+\frac { (1.4)(0.6)(-0.6)(-1.6) }{ 24 } (41358)\)
=  98,752 + 46946.2 + 632.4 + 584.36 + 1389.63
= 1,48,304.43
= 1,48,304

The given equation can be written as \(\frac { dy }{ dx } +\frac { 1 }{ { cos }^{ 2 }x } y=\frac { tanx }{ { cos }^{ 2 }x } \)
\(\frac { dy }{ dx } \) + y sec²x = tan x sec²x
It is of the form \(\frac{dy}{dx}\) + Py + Q
Here P = sec²x,Q = tanx sec²x
ഽPdx = ഽsec² x dx = tanx
I.F = eഽ^pdx = e^{tan x}
The required solution is y(I.F) = ഽQ(I.F)dx + c
ye^{tan x} = ഽtan x sec²xe^tan xdx + c
Put tan x = t
Then sec² xdx = dt
∴ ye^{tan x} = ഽte^t dt + c
= ഽtd(e^t) + c
= te^t − e^t + c
= tanx e^{tan x}−e^{tan x}+ c
ye^{tan x} = e^{tan x}(tan x − 1) + c

Given (y³−2yx²)dx + (2xy²−x³)dy = 0
(y³−2yx²)dx = - (2xy²−x³)dy
(y³−2yx²)dx = - (x³-2xy²)dy
⇒ \(\frac { dy }{ dx } =\frac { { y }^{ 3 }-2yx^{ 2 } }{ { x }^{ 3 }-2xy^{ 2 } } \)
Since the numerator and denominator are homogeneous functions of degree 3,
put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

⇒ \(v+x\frac { dv }{ dx } =\frac { { v }^{ 3 }-2v }{ 1-2v^{ 2 } } \)
⇒ \(x\frac { dv }{ dx } =\frac { { v }^{ 3 }-2v }{ 1-2v^{ 2 } } -v=\frac { { v }^{ 3 }-2v-v(1-2v^{ 2 }) }{ 1-2v^{ 2 } } \)
⇒ \(x\frac { dv }{ dx } =\frac { { v }^{ 3 }-2v-v+2v^{ 3 } }{ 1-2v^{ 2 } } =\frac { { v }^{ 3 }+2v^{ 3 }-3v }{ 1-2v^{ 2 } } \)
= \(\frac { 3v^{ 3 }-3v }{ 1-2v^{ 2 } } \)
Separating the variables we get,
\(\frac { 1-2v^{ 2 } }{ 3v^{ 3- }3v } dv=\frac { dx }{ x } \)
⇒ \(\frac { 1-2v^{ 2 } }{ { v }^{ 3 }-v } dv=3\frac { dx }{ x } \)
\(\frac { 1-2v^{ 2 } }{ { v }^{ 3 }-v } =\frac { 1-2v^{ 2 } }{ v({ v }^{ 3 }-1) } \)
= \(\frac { 1-2v^{ 2 } }{ v(v+1)(v-1) } \)
= \(\frac { A }{ v } +\frac { B }{ v+1 } +\frac { C }{ v-1 } \)
1-2v² = A(v+1)(v-1)+Bv(v-1)+Cv(v+1)
put v = 1 ⇒ -1 = 2c ⇒ c = -\(\frac { 1 }{ 2 } \)
put v = -1 ⇒ -1 = -B = -\(\frac { 1 }{ 2 } \)
put v = -1 ⇒ -1 = -B(-2)
⇒ -1 = 2B ⇒ B = -\(\frac { 1 }{ 2 } \)
put v = 0 ⇒ 1 = -A + 0 + 0
⇒ A = -1
⇒ \(\int { \left( \frac { -1 }{ v } \frac { -\frac { 1 }{ 2 } }{ v+1 } \frac { -\frac { 1 }{ 2 } }{ v-1 } \right) } dv=3\int { \frac { dx }{ x } } \)
⇒ -log v -\(\frac { 1 }{ 2 } \)log(v+1) -\(\frac { 1 }{ 2 } \)log(v-1)
= 3log x + log c
⇒ -\(\frac { 1 }{ 2 } \)log(v-1) = 3log x + log c
⇒ log v+\(\frac { 1 }{ 2 } \)log(v+1)+\(\frac { 1 }{ 2 } \)log(v-1)
= -3log x + log c
⇒ log v.\(\sqrt { v+1 } \sqrt { v-1 } =log\left( \frac { 1 }{ x^{ 3 } } \right) \).c
⇒ v\(\sqrt { v^{ 2 }-1 } =\frac { c }{ { x }^{ 3 } } \)
Replace v by \(\frac { y }{ x } \) we get,
\(\frac { y }{ x } \sqrt { \frac { { y }^{ 2 } }{ x^{ 2 } } -1 } =\frac { c }{ { x }^{ 3 } } \)
⇒ \(\frac { y }{ x } \sqrt { \frac { { y }^{ 2 }-{ x }^{ 2 } }{ x } } =\frac { c }{ { x }^{ 3 } } \)
⇒ \(\sqrt { { y }^{ 2 }-x^{ 2 } } =\frac { c }{ x } \Rightarrow \sqrt { { y }^{ 2 }-x^{ 2 } } \)= c        (1)
Since the curve passes through (1, 2) we get
(1) (2) \(\sqrt { 4-1 } =c\Rightarrow c=2\sqrt { 3 } \)
Substituting c = \(2\sqrt { 3 } \) in (1) we get
\(xy\sqrt { { y }^{ 2 }-x^{ 2 } } =2\sqrt { 3 } \).

Cost of Living Index Number
= \(\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 1000 }{ 758 } \times 100=131.926 \)

Plot the variable X = 50 on the left side and
X = 86 on the right side of the curve.
Given P( -∞< Z₁ < -Z₁) = 0.3
⇒ P(-Z₁ < Z < Z₁) = 0.2       [∵ 0.5 - 0.3 = 0.2]
⇒ P(0 < Z < Z₁) = 0.2 [By symmetry]
⇒ Z₁ = -0.52 [From the normal distribution table 'and it lies on the negative side]
⇒ -0.52 = \(\frac { X-\mu }{ \sigma } \) ⇒ -0.52 σ = 50-μ
⇒ 50-μ = -0.52 σ ....(1)
Also given P(Z₂ ∴ P(Z₂2) = 0.4    [∵ 0.5-0.1 = 0.4]
⇒ Z₂ = 1.28 (from the table)
⇒ 1.28 =\(\frac { 86-\mu }{ \sigma } \)
⇒ 86-μ = 1.28 σ ...(2)

Substituting σ = 20 in (2) we get,
86-μ = (1.28)(20)
86-μ = 25.6
μ = 86-25.6
μ = 60.4

Hence, the mean is 60.4 and standard deviation is 20.

The mean of the binomial distribution
Ex = \(\overset { n }{ \underset { x=0 }{ \Sigma } } x.p(x)=\overset { n }{ \underset { x=0 }{ \Sigma } } \left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
= \(p.\overset { n }{ \underset { x=0 }{ \Sigma } } x.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x-1 }q^{ n-x }\)
[Take p common]

=  \(np.\overset { n }{ \underset { x=1 }{ \Sigma } } \left( \begin{matrix} n-1 \\ x-1 \end{matrix} \right) { p }^{ x-1 }q^{ n-x }\)
= np (q +p)^n-1 [using binomial theorem]
(x+a)ⁿ = \({ x }^{ n }+{ n }_{ { C }_{ 1 } }{ x }^{ n-1 }{ a }^{ 1 }+...+{ a }^{ n }\)
= np(1)^n-1 [∵ p+q=1]
= np
∴ Mean = E(x) = np..(1)
Now, E(X²) =\(\overset { n }{ \underset { x=0 }{ \Sigma } } { x }^{ 2 }.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=0 }{ \Sigma } } \{ x(x-1)+x\} \left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=0 }{ \Sigma } } x(x-1).\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }+{ \Sigma }_{ x }.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=2 }{ \Sigma } } x(x-1)\frac { n(n-1) }{ x(x-1) } \left( \begin{matrix} n-2 \\ n-2 \end{matrix} \right) { p }^{ x-2 }{ q }^{ n-x }\)+\(\Sigma x\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
= \(n(n-1){ p }^{ 2 }\left\{ \Sigma \left( \begin{matrix} n-2 \\ n-2 \end{matrix} \right) { p }^{ x-2 }{ q }^{ n-x } \right\} \) +np (using (1))
= n(n-1)p² (q+p)^n-2 + np
[using binomial theory]
= n(n-1)p² (1) + np  ∴ p+ q = 1
= n(n-1)p²+ np .... (2)
Variance = E(X²) - [E(X)]²
= n(n-1)p²+np-(np)²
[From (1) & (2)]

= np (1-p) = npq       [ ∵ p+q = 1⇒ q = 1-p]
∴ Mean = np and variance = npq

We know that, the expected random variable 
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\)
\(=\int _{ 100 }^{ \infty }{ { xe }^{ -\frac { x }{ 100 } }dx } \)
\(=\left\{ { \left[ x\left( \frac { { e }^{ \frac { x }{ 100 } } }{ -\frac { 1 }{ 100 } } \right) \right] }_{ 100 }^{ \infty }-\int _{ 100 }^{ \infty }{ \left( \frac { { e }^{ \frac { x }{ 100 } } }{ -\frac { 1 }{ 100 } } \right) dx } \right\} (\because \int { udv=uv-\int { vdu } } )\)
\(=\left[ \left( 10000 \right) \left( { e }^{ -1 } \right) +\left( 10000 \right) \left( { e }^{ -1 } \right) \right] \)
\(=\left[ \left( 10000 \right) \left( 0.3679 \right) +\left( 10000 \right) (0.3679) \right] \)
= 7358 hours 
Therefore, the mean life of a radio tube is 7,358 hours.

The present value of payment for t year = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
Present value of 12 years = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
= 72000\({ \left[ \frac { { e }^{ -0.09t } }{ { -0.09 } } \right] }_{ 0 }^{ \\ 12 }\)
= \(\frac { 72000 }{ -0.09 } \left[ { e }^{ -0.09(12) }-{ e }^{ 0 } \right] \)
= \(-8,00,000[{ e }^{ -1.08 }-{ e }^{ 0 }]\)
= −8,00,000 [0.3396 −1]
= 5,28,320
Cost of the machine = 5,00,000 − 30,000
= 4,70,000
Hence it not advisable to rent the machine
It is better to buy the machine.

Since y = x² is symmetric
about Y-axis, the required
Area = \(2\int _{ 0 }^{ 4 }{ x\quad dy } \)
When \(y={ x }^{ 2 }\Rightarrow x=\sqrt { y } \)
∴ Area \(=2\int _{ 0 }^{ 4 }{ \sqrt { y } dy } \)
\(=2\int _{ 0 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } \)
\(=2\times \frac { 2 }{ 2 } { \left[ { y }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 4 }\)
\(=\frac { 4 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } }-{ 0 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 4 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 4 }{ 3 } \times { ({ 2 }^{ 2 }) }^{ \frac { 3 }{ 2 } }=\frac { 4 }{ 3 } \times { 2 }^{ 3 }\)
\(A=\frac { 4 }{ 3 } \times 8=\frac { 32 }{ 3 } \) sq.units

The equation y² = 16x represents a parabola (Open rightward)
Required Area = 2\(\int _{ a }^{ b }{ y } dx\)
\(=2\int _{ 0 }^{ 4 }{ \sqrt { 16x } } \ dx\)
\(=8\int _{ 0 }^{ 4 }{ { x }^{ \frac { 1 }{ 2 } } } dx=8{ \left[ { \frac { { x }^{ { \frac { 3 }{ 2 } } } }{ \frac { 3 }{ 2 } } } \right] }_{ 0 }^{ 4 }=\frac { 16 }{ 3 } \left( { \left( 4 \right) }^{ \frac { 3 }{ 2 } } \right) =\frac { 128 }{ 3 } \) sq.units

\(\int { \frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } } dx =\int { \left[ \frac { 1 }{ \left( x+3 \right) } \frac { 2x }{ \left( { x }^{ 2 }+1 \right) } \right] } dx\)
\(\int { \frac { dx }{ \left( x+3 \right) } +\int { \frac { 2x }{ \left( { x }^{ 2 }+1 \right) } } } dx\)
\(=\log\left| x+3 \right| +\log\left| { x }^{ 2 }+1 \right| +c\)
\(=\log\left| \left( x+3 \right) \left( { x }^{ 2 }+1 \right) \right| +c\)
\(=\log\left| { x }^{ 3 }+{ 3x }^{ 2 }+x+3 \right| +c\)
[ By partial fractions,
\(\frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } =\frac { A }{ \left( x+3 \right) } +\frac { Bx+C }{ \left( { x }^{ 2 }+1 \right) } \Rightarrow \frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } =\frac { 1 }{ \left( x+3 \right) } +\frac { 2x }{ \left( { x }^{ 2 }+1 \right) } \)

\(\Delta =\left| \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1(2 -2) - 2(-4 -2) + 1(2 + 1)
= 1(0)-2(-6)+1(3)
= 12 + 3 = 15\(\neq \)0.
Since \(\Delta \neq 0\) Cramer's rule can be applied and thesystem is consistent with unique solution.
\({ \Delta }x=\left| \begin{matrix} 7 & 2 & 1 \\ 4 & -1 & 2 \\ -1 & 1 & -2 \end{matrix} \right| \)
= \(7\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 4 & -1 \\ -1 & 1 \end{matrix} \right| \)
= 7 (2 -2) -2 (-8 + 2) + 1 (4 - 1)
= 7 (0) - 2(-6) + 1(3)
= 12 + 3 = 15
\(\Delta y=\left| \begin{matrix} 1 & 7 & 1 \\ 2 & 4 & 2 \\ 1 & -1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| -7\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| \)
= 1 (- 8 + 2) -7(-4 -2) + 1(-2 -4)
= 1 (-6) -7 (-6) + 1 (-6)
= - 6 + 42 - 6 = 30
\(\Delta z=\left| \begin{matrix} 1 & 2 & 7 \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 4 \\ 1 & -1 \end{matrix} \right| -2\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| +7\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1 (1 - 4) - 2(- 2 - 4) + 7(2 + 1)
= 1(-3)-2(-6)+7(3)
= - 3 + 12 + 21 = 30

\(\therefore\) Solution set is {1, 2, 2}

Given non-homogeneous equations are
2x + 3y - z = 5, 3x - y + 4z = 2, x + 7y - 6z = k

Here \(\rho (A)=2\)
Since the given system is consistent, \(\rho \)(A, B) must be equal to 2.
This can happen only when
8 - k = 0 \(\Rightarrow\) k = 8

(i) Transition probability matrix T = \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \) 
After one study period, \(\left( \overset { M }{ 60\quad } \overset { E }{ 40 } \right) \)  \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \) =\(\left( \overset { M }{ 76\quad } \overset { E }{ 24 } \right) \)
So in the very next study period, there will be 76 students do maths work and 24 students do the English work.
After two study periods,
\(\left( \overset { M }{ 76\quad } \overset { E }{ 24 } \right) \) \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \) 
= (60.8+16.8 15.2+7.2)
= (77.6 22.4)
After two study periods there will be 78 (approx) students do maths work and 22 (approx) students do English work.

Let the amount invested in the rate of 2%, 3% and 6% be Rs. x, Rs. y and Rs. z respectively
By the given data,
x+ y+z = 8500
\(\cfrac { 2x }{ 100 } +\cfrac { 3y }{ 100 } +\cfrac { 6z }{ 100 } =380\)
\(\Rightarrow \cfrac { 2x+3y+6z }{ 100 } =380\)
\(\because Interest=\cfrac { PNR }{ 100 } =\cfrac { x\times 1\times 2 }{ 100 } =\cfrac { 2x }{ 100 } \)
\(\Rightarrow 2x+3y+6z=38000\)
Also,z = x+y
x+y-z  =0
\(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{matrix} \right| \)
\(1\left| \begin{matrix} 3 & 6 \\ 1 & -1 \end{matrix} \right| -1\left| \begin{matrix} 2 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= 1(-3 - 6) - 1(-2 -6) + 1 (2 - 3)
= 1(-9) - 1 (-8) + 1(-1)
= -9 + 8 - 1 = 2 \(\neq \) 0 
Since \(\Delta \neq 0\),Cramer's rule can be applied and the system is consistent with unique solution
\({ \Delta x }=\left| \begin{matrix} 8500 & 1 & 1 \\ 38000 & 3 & 6 \\ 0 & 1 & -1 \end{matrix} \right| \)
= \(8500\left| \begin{matrix} 3 & 6 \\ 1 & -1 \end{matrix} \right| -1\left| \begin{matrix} 38000 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 38000 & 3 \\ 0 & 1 \end{matrix} \right| \)
= 8500 (- 3 - 6) - 1(-38000 -0) + 1(38000 - 0)
= 8500(-9) - 1(-38000) + 1(38000)
= - 76500 + 38000 + 38000
= -500
\(\Delta y=\left| \begin{matrix} 1 & 8500 & 1 \\ 2 & 38000 & 6 \\ 1 & 0 & -1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 38000 & 6 \\ 0 & -1 \end{matrix} \right| -8500\left| \begin{matrix} 2 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 38000 \\ 1 & 0 \end{matrix} \right| \)
= 1 (-38000 - 0) - 8500 (-2 -6) + 1(0 - 38000)
= - 38000 - 8500 (-8) - 38000
= - 38000 + 68000 - 38000
= - 8000
\(\Delta z=\left| \begin{matrix} 1 & 1 & 8500 \\ 2 & 3 & 38000 \\ 1 & 1 & 0 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 38000 & 6 \\ 0 & -1 \end{matrix} \right| -1\left| \begin{matrix} 2 & 38000 \\ 1 & 0 \end{matrix} \right| +8500\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= 1 (0 - 38000) - 1(0 -38000) +85000 (2 - 3)
= - 38000 + 38000 + 8500 (-1)
= - 8500


Hence, the amount invested in the three accounts are Rs. 250, Rs. 4000 and Rs. 4250 respectively.

Here \(\triangle =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =13\neq 0\)
\(\therefore \) We can apply Cramer’s Rule and the system is consistent and it has unique solution.
\({ \triangle }_{ x }=\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =-13\)
\( { \triangle }_{ y }=\left| \begin{matrix} 1 & 4 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & -1 \end{matrix} \right| =39\)
\( { \triangle }_{ z }=\left| \begin{matrix} 1 & 1 & 4 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{matrix} \right| =26\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -13 }{ 13 } =-1\)
\( y=\frac { { \triangle }y }{ { \triangle } } =\frac { 39 }{ 13 } =3\)
\( z=\frac { { \triangle }z }{ { \triangle } } =\frac { 26 }{ 13 } =2\)
\(\therefore \) The solution is (x, y, z) = (−1, 3, 2)

Let the amount of investment in each bond be
Rs. x, Rs. y, Rs. z respectively.
Given x + y + z = 5000    ..(1)
Also \(\cfrac { 6x }{ 100 } +\cfrac { 7y }{ 100 } +\cfrac { 8z }{ 100 } =358\)
∴ Interes \(= \cfrac { PNR }{ 100 } =\cfrac { x\times 1\times 6 }{ 100 } =\cfrac { 6x }{ 100 } \)
\(\Rightarrow \cfrac { 6x+7y+8z }{ 100 } =358\)
\(\Rightarrow 6x+7y++8z=35800\)
Given that \(\cfrac { 6x }{ 100 } +\cfrac { 7y }{ 100 } =70+\cfrac { 8z }{ 100 } \)
\(\Rightarrow 6x+7y=7008z\)
\(\Rightarrow 6x+7y-8z=7000\)
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right) \)

The last equivalent matrix is in echelon form and ρ(A) = ρ([A,B]) = 3 = Number of unknowns
Thus, the given system is consistent with unique solution. To find the solution, let us rewrite the above echelon form into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & -16 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 5800 \\ -28800 \end{matrix} \right) \)
\(\Rightarrow x+y+z=5000\)   ...(1)
\(\Rightarrow y+2z=5800\)      ..(2)
\(\Rightarrow -16z=-28800\)    ...(3)
\((3)\Rightarrow -16z=-28800\)
\(\Rightarrow z=\cfrac { 28800 }{ -16 } =1800\)
Substituting z = 1800 in (2) we get,
y + 2(1800) = 5800
\(\Rightarrow\) y + 3600 = 5800
\(\Rightarrow\)  y=5800-3600
\(\Rightarrow\) y = 2200
Substituting y = 2200 and z = 1800 in (1) we get
\(\Rightarrow\) x + 2200 + 1800 = 5000
\(\Rightarrow\)x + 4000 = 5000
\(\Rightarrow\)x = 5000 - 4000
\(\Rightarrow\)x = 1000
Hence, the amount of investment in each bond is Rs. 1000, Rs. 2200 and Rs. 1800 respectively

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \)
AX = B

Case (i) For no solution:
The system possesses no solution only when \(\rho (A)\neq ([A,B])\) which is possible only when a−3 = 0 and b −10 \(\neq \) 0.
Hence for a = 3, b \(\neq \) 10, the system possesses no solution.
Case (ii) For a unique solution:
The system possesses a unique solution only when \(\rho (A)= ([A,B])\)=number of unknowns.
i.e when \(\rho (A)=\rho ([A,B])\) = 3
Which is possible only when a−3 \(\neq \) 0 and b may be any real number as we can observe .
Hence for a \(\neq \) and b \(\in \) R, the system possesses a unique solution.
Case (iii) For an infinite number of solutions:
The system possesses an infinite number of solutions only when
\(\rho (A)=\rho ([A,B])\) i,e when \(\rho (A)=\rho ([A,B])\) = 2 < 3 ( number of unknowns) which is possible only when a − 3 = 0, b−10 = 0
Hence for a = 3, b = 10, the system possesses infinite number of solutions.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & -4 & 7 \\ 3 & 8 & -2 \\ 7 & -8 & 26 \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 14 \\ 13 \\ 5 \end{matrix} \right) \)
AX = B

The last equivalent matrix is in the echelon form. [A, B] has 3 non-zero rows and [A] has 2 non-zero rows.
\(\therefore \rho ([A,B])=3\),\(\rho (A)=2,\)
\(\rho (A)\neq ([A,B])\)
The system is inconsistent and has no solution.

\(\int _{ -1 }^{ 1 }{ x\sqrt { x+1 } } dx=\int _{ 0 }^{ 2 }{ (t-1)\sqrt { t } dt } \)
\(=\int _{ 0 }^{ 2 }{ \left( { t }^{ \frac { 3 }{ 2 } }-{ t }^{ \frac { 1 }{ 2 } } \right) } dt\)
\(={ \left[ \frac { { 2t }^{ \frac { 5 }{ 2 } } }{ 5 } -\frac { { 2t }^{ \frac { 3 }{ 2 } } }{ 3 } \right] }_{ 0 }^{ 2 }\)
\(=\frac { 8\sqrt { 2 } }{ 5 } -\frac { 4\sqrt { 2 } }{ 3 } \)
\(=\frac { 4\sqrt { 2 } }{ 15 } \)

\(\int { \frac { \left( 3x+2 \right) dx }{ \left( x-2 \right) \left( x-3 \right) } } \)
\(\frac { 3x+2 }{ \left( x-2 \right) \left( x-3 \right) } =\frac { A }{ x-2 } +\frac { B }{ x-3 } \)
⇒ 3x+2 = A (x-3)+B(x-2)
Put x = 3
9+2 = B(1) ⇒ B = 11
Put x = 2
8 = A(-1) ⇒ A = -8
\(\frac { 3x+2 }{ \left( x-2 \right) \left( x-3 \right) } =\frac { -8 }{ x-2 } +\frac { 11 }{ x-3 } \)
\(=\int { \left( \frac { -8 }{ x-2 } +\frac { 11 }{ x-3 } \right) } dx\)
\(=-8\log { \left| x-2 \right| } +11\log { \left| x-3 \right| } +c\)
\(=-11\log { \left| x-3 \right| } -8\log { \left| x-2 \right| } +c\)