(i) Laspeyre's index number
\(P_{01}^{L} = \frac{\Sigma p_1q_0}{\Sigma p_0q_0}\times100\)
\(=\frac{200}{160}\times100=125\)
(ii) Paasche's Price index number
\(P_{01}^{P} = \frac{\Sigma p_1q_1}{\Sigma p_0q_1}\times100\)
\(=\frac{130}{103}\times100=126.21\)
(iii) Fisher's price index number
\(P_{01}^{F} =\sqrt {{P_{01}^{L}}\times{P_{01}^{P}}}=125.6\)

Given sample size n = 200
Sample mean \(\bar { x } \) = 0.824
Sample S.D. s = 0.042
Standard error = \(\frac { s }{ \sqrt { n } } =\frac { 0.042 }{ \sqrt { 200 } } \)
= \(\frac { 0.042 }{ 14.14 } \) = 0.00270
(a) As the level of significance is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) =  1.96
∴ 95% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (1.96) (0.00270) ≤ μ ≤ 0.824 + (1.96) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.00582
⇒ 0.818 ≤ μ ≤ 0.832
Hence, the 95% confidence limits for μ is (0.818,0.832)
(b) As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (2.58) (0.00270) ≤ μ ≤ 0.824 + (2.58) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.005825
⇒ 0.816 ≤ μ ≤ 0.832
Hence, the 99% confidence limits for μ is (0.816, 0.832)

\(E(X)=0\times \frac { 1 }{ 5 } +1(\frac { 2 }{ 5 } )+2\left( \frac { 1 }{ 5 } \right) +3\times \frac { 1 }{ 5 } \)
\(=\frac { 2 }{ 5 } +\frac { 2 }{ 5 } +\frac { 3 }{ 25 } =\frac { 7 }{ 5 } \)
\(\\ \therefore 2E(X)=\frac { 14 }{ 5 } ...(1)\)
\(E({ Y }^{ 2 })=0\times \frac { 1 }{ 5 } +{ 1 }^{ 2 }(\frac { 3 }{ 10 } )+{ 2 }^{ 2 }(\frac { 2 }{ 5 } )+{ 3 }^{ 2 }(\frac { 1 }{ 10 } )\)
\(=\frac { 3 }{ 10 } +\frac { 8 }{ 5 } +\frac { 9 }{ 10 } =\frac { 3+16+9 }{ 10 } \)
\(=\frac { 28 }{ 10 } =\frac { 14 }{ 5 } ..(2)\)
From (1) and (2), E(Y²) = 2 E(X).

Given f'(x) = a sin x + b cos x          ------(1)
ഽ f'(x) dx = a ഽ sin x dx + b ഽ cos x dx       [∵ f'(x) dx = f(x)]
⇒ f(x) = -a cos x + b sin x + c           -------(2)
Given f'(0) = 4
∴ (1) 4 = a sin 0 + b cos 0
⇒ 4 = a (0) + b (1)
[∵ sin 0 = 0 and cos 0 = 1]
⇒ 4 = b
Also, f(0) = 3
∴ (2)  3 = -a cos 0 + b sin 0 +c
⇒ 3 = -a (1) + b (0) +c
⇒ 3 = -a +c                                             --------(3)
And f\(\left( \frac { \pi }{ 2 } \right) \) = 5
∴ (2)  5 = -a cos \( \frac { \pi }{ 2 } \)+ b sin \( \frac { \pi }{ 2 } \)+ c
⇒ 5 = -a (0) + b (1) +c
[∵ sin \( \frac { \pi }{ 2 } \)=1 and cos \( \frac { \pi }{ 2 } \)= 0]
⇒ 5 = b + c
⇒ 5 = 4 + c      [∵ b = 4]
⇒ 5 - 4 = c
c = 1
Substituting c = 1 in (3) we get,
 3 = -a + 1
⇒ a = 1 - 3
⇒ a = -2
∴ From (2), f(x) = -(-2) cos x + 4 sin x + 1
⇒ f(x) = 2 cos x + 4 sin x +1

Given \(\frac { dp }{ dx } =\frac { 2{ p }^{ 3 }-{ x }^{ 3 } }{ 3x{ p }^{ 2 } } \)
The numerator and denominator are homogeneous functions of 3,
∴ Put p=vx and \(\frac { dp }{ dx } =v+x\frac { dv }{ dx } \)

= \(\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } \)
⇒ \(\frac { dv }{ dx } =\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } -v=\frac { 2{ v }^{ 3 }-1-3{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
= \(\frac { -1-{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
⇒ \(\left( \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } \right) dv=-\frac { dx }{ x } \)
Integrating, \(\int { \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } } dv=-\int { \frac { dx }{ x } } \)
⇒ log(1+v³) = -log x + log c
⇒ 1+v³ = \(\frac { c }{ x } \)
Replacing v by \(\frac { p }{ x } \) we get
\(1+\frac { { p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \Rightarrow \frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \)
⇒ \(\frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 2 } } \)= c ⇒ x³+p³ = cx²...(1)
When x = 10, p = 20
⇒ 10³ + 20³ = c(10)² ⇒ 1000 + 8000 = 100 c
⇒ 9000 = 100 c
⇒ c = 90
∴ (1) becomes,
x³+p³ = 90x²
⇒ p³ = 90x²-x³
⇒ p³ = x²(90-x) which is the required relationship.

Given that MR \(=\frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 }\)
\(\int { MR } =\int { \left( \frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 } \right) } dx\)
\(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +k\)
When x=0, R=0
\(\Rightarrow 0=\frac { { e }^{ 0 } }{ 100 } +0+0+k\)
\(k=-\frac { 1 }{ 100 } [\because { e }^{ 0 }=1]\)
∴ Revenue \(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { 1 }{ 100 } \)

Given

 Here the intervals are unequal.
∴ By Lagranges interpolation formula, we have
x₀ = 1974, x₁ = 1978, x₂ = 1982, x₃ = 1990
y₀ = 25, y₁ = 60, y₂ = 80, y₃ = 170 and x = 1986.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
\(\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times25+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times60+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times80+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times 70+\frac { (1986-1974)(1986-1982)(1986-1990) }{ (1990-1974)(1990-1978)(1990-1982) } \times 170\)

= 6.25 - 60 + 120 + 42.5
y = 108.75

We will use Newton’s backward interpolation formula to find the polynomial.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)

To find y in terms of x
\(\therefore\) x_n + nh = x, x_n = 7, h = 1 \(\Rightarrow\) n = x − 7
\({ y }_{ (x) }=29+(x-7)(7)+\frac { (x-7)(x-6) }{ 2 } (1)\)
= \(29+7x-49+\frac { 1 }{ 2 } \left( { x }^{ 2 }-13x+42 \right) \)
= \(\frac { 1 }{ 2 } \left[ { x }^{ 2 }+x+2 \right] \)

The auxiliary equation is m² + m - 6 = 0
⇒ (m + 3) (m - 2) = 0
⇒ m = -3, 2
The roots are real and different.
∴ The complementary function CF is Ae^-3x + B^ex
Particular Integral
PI₁=\(\frac { 1 }{ \phi (D) } \)f₁(x)
=\(\frac { 1 }{ ({ D }^{ 2 }+D-6) } \)e^3x
PI₁=\(\frac { { e }^{ 3x } }{ (D+3)(D-2) } =\frac { { e }^{ 3x } }{ (3+3)(3-2) } \)
= \(\frac { { e }^{ 3x } }{ 6(1) } =\frac { e^{ 3x } }{ 6 } \)
PI₂ = \(\frac { 1 }{ \phi (D0 } { f }_{ 2 }(x)=\frac { e^{ -3x } }{ (D+3)(D-2) } \)
= x\(\frac { { e }^{ -3x } }{ (-3-2) } \) [∵ when D = - 3, D + 3 = 0]
PI₂ = \(-\frac { x }{ 5 } \)e^-3x
∴ y = CF+PI₁+PI₂
∴ The general solution is
y = Ae^-3x+Be^2x+\(\frac { { e }^{ 3x } }{ 6 } -\frac { x }{ 5 } \)e^-3x

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

Let I = ∫ e^x(1+x) log(xe^x)dx
Put t = x e^x
⇒ dt = (x.e^x + e^x(1))dx
= e^x(x+1)dx
∴ I = ∫ log t.dt
Let u = log t; dv = dt
\(du=\frac { 1 }{ t } dt;v=t\)
∴Using integration by parts we get,
I = ∫udv = vu - ∫vdu
= t log t - ∫dt = t logt - t + c
= t log t - t + c
= xe^x log(xe^x) - (xe^x) + c  [∵ t = xe^x]
= xe^x (log(xe^x)-1)+c

Mean of the random variableX = E(X) = \({ \sum_x { x{ P }_{ x }(x) } } \)
= (1 × 0.15) + (2 × 0.10) + (3 × 0.10) + (4 × 0.01) + (5 × 0.08) + (6 × 0.01) +(7 × 0.05) + (8 × 0.02) + (9 × 0.28) + (10 × 0.20)
E(X) = 6.18
E(X²) = \(\sum _{ x }^{ }{ { x }^{ 2 } } { P }_{ X }(x)\)
= (1² × 0.15) + (2² × 0.10) + (3² × 0.10) + (4² × 0.01) + (5² × 0.08) + (6² × 0.01) + (7² × 0.05) + (8² × 0.02) + (9² × 0.28) + (10² × 0.20).
= 50.38
Variance of the Random Variagble X = V(X) = E(X²)-[E(X)]²
=50.38-(6.56)²
= 12.19
Therefore, the mean and variance of the given discrete distribution are 6.18 and 12.19 respectively.

Given \( f(x)\begin{cases} { Ae }^{ -x/5 },\quad \text{for}\quad x\ge 0 \\ 0 \quad ,\quad \text{otherwise }\end{cases}\)
a) since f(x) is a p.d.f.,
\(\int _{ 0 }^{ \infty }{ { Ae }^{ \frac { -x }{ 5 } }dx=1\Rightarrow A.\frac { { \left[ { e }^{ \frac { -x }{ 5 } } \right] }_{ 0 }^{ \infty } }{ \frac { - }{ 5 } } =1 } \)
\(\Rightarrow -5A[{ e }^{ -\infty }-{ e }^{ 0 }]=1\)
\(\Rightarrow -5A[0-1]=1\quad [\because { e }^{ -\infty }=0\quad and\quad { e }^{ 0 }=1]\)
\(\Rightarrow 5A=1\Rightarrow A=\frac { 1 }{ 5 } \)
\(\therefore A=\frac { 1 }{ 5 } \)
b) i) Probability that the person will talk over the phone more than 10 minutes is P(X >10)
∴ P(X>10) =\(\\ \int _{ 10 }^{ \infty }{ \frac { 1 }{ 5 } { e }^{ \frac { -x }{ 5 } }dx } [\because A=\frac { 1 }{ 5 } ]\)
\(={ \frac { 1 }{ 5 } \left[ \frac { { e }^{ \frac { -x }{ 5 } } }{ \frac { -1 }{ 5 } } \right] }_{ 10 }^{ \infty }\)
\(=-\left[ { e }^{ -\infty }-{ e }^{ \frac { -10 }{ 5 } } \right] =-\left[ 0-{ e }^{ -2 } \right] \)
\(={ e }^{ -2 }=\frac { 1 }{ { e }^{ 2 } } \left[ \because { e }^{ -\infty }=0 \right] \)
ii) Probability that the person will take over the phone less that 5 minutes is P(X < 5)
\(\therefore P(X<5)=\int _{ 0 }^{ 5 }{ { e }^{ \frac { -x }{ 5 } }dx } [\because A=\frac { 1 }{ 5 } ]\)
\(=\frac { 1 }{ 5 } { \left[ \frac { { e }^{ \frac { -x }{ 4 } } }{ \frac { -1 }{ 5 } } \right] }_{ 0 }^{ 5 }=-{ e }^{ \frac { -5 }{ 5 } }{ -e }^{ 0 }\)
\(=-\left( { e }^{ -1 }-1 \right) \left[ \because { e }^{ 0 }=1 \right] \)
\(=1-{ e }^{ -1 }-\frac { 1 }{ e } =\frac { e-1 }{ e } \)
\(\therefore P(X<5)=\frac { e-1 }{ e } \)
iii) The probability that the person will take over the phone between 5 and 10 minutes is P(5 < X < 10)
\(=-\left[ { e }^{ \frac { -10 }{ 5 } }-{ e }^{ \frac { -5 }{ 5 } } \right] \)
\(=-\left[ { e }^{ -2 }-{ e }^{ -1 } \right] \)
\(={ e }^{ -1 }-{ e }^{ -2 }=\frac { 1 }{ e } -\frac { 1 }{ { e }^{ 2 } } \)

We have F(x) = f(x) ≥ 0, where F(x) is the distribution function and f(x) is the probability density function.
Here F(x) = 0 for x ≤ 1  f(x) = 0 for x ≤ 1 
Again F(x) = 1 for x > 3 
f(x) = d/dx (1) = 0 for x > 3 
In 1 < x ≤ 3, F(x) = k(x – 1)4 
f(x) = d/dx (k(x – 1)4) = 4k(x – 1)3 
\(\therefore f(x)=4k{ (x-1) }^{ 3 }for\quad 1\le x\le 3\)
i) Since f(x) is a probability density function,
\(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ { 4k(x-1) }^{ 3 }dx=1 } \)

\(\Rightarrow k[{ (3-1) }^{ 4 }-{ (0) }^{ 4 }]=1\)
\(\Rightarrow k({ 2 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
ii) \(\therefore\)p.d.f
\(f(x)=\frac { 4\times 1 }{ 16 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)
\(f(x)=\frac { 1 }{ 4 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)

Given marginal revenue function.
\(MR=\frac { DR }{ dx } =20{ e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) \)
\(dR={ 20e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) dx\)
\(R=\int { { 20e }^{ \frac { -x }{ 10 } } } \left( 1-\frac { x }{ 10 } \right) dx\)
We know that \(\int { { e }^{ ax }[af(x)+f'(x)]dx={ e }^{ ax }f(x) } +c\)
Here \(\\ a=\frac { -1 }{ 10 } ,f(x)=x,\quad f'(x)=1\)
\(R=20\int { { e }^{ \frac { -x }{ 10 } } } \left[ -\frac { 1 }{ 10 } x+1 \right] dx=20{ e }^{ \frac { -x }{ 10 } }x+k\)
\(\Rightarrow R=20{ e }^{ \frac { -x }{ 10 } }x+k\quad ...(1)\)
When x = 0, R = 0
0 = 0 + k ⇒ k = 0
(1) becomes 
\(R=20x{ e }^{ \frac { -x }{ 10 } }\)
Demand function P
\(=\frac { R }{ x } =\frac { 20x{ e }^{ \frac { -x }{ 10 } } }{ x } \)
\(\Rightarrow P=20{ e }^{ \frac { -x }{ 10 } }\)

(a) We know that
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\)
\(\int _{ 0 }^{ 10 }{ Axdx } +\int _{ 10 }^{ 20 }{ A(20-x)dx=1 } \)
\(A\left\{ { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }+{ \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 } \right\} =1\)
A[(50-0)+(400-200)-(200-50)] = 1
\(A=\frac{1}{100}\)
(b) (i) The probability that the number of pounds of bread that will be sold tomorrow is more than 10 pounds is given by
\(P(10\le X\le 20)=\int _{ 10 }^{ 20 }{ \frac { 1 }{ 100 } (20-x) } dx\)
\(=\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 }\)
\(=\frac { 1 }{ 100 } [(400-200)-(200-50)]\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is less than 10 pounds, is given by
\(P(0\le X\le 20)=\int _{ 0 }^{ 10 }{ \frac { 1 }{ 100 } } xdx\)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }\)
\(=\frac { 1 }{ 100 } (50-0)\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is between 5 and 15 pounds is
\(P(5\le X \le15)=\int _{ 5 }^{ 10 }{ \frac { 1 }{ 100 } xdx } +\int _{ 10 }^{ 15 }{ \frac { 1 }{ 100 } (20-x)dx } \)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 5 }^{ 10 }+\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 15 }\)
= 0.75

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

\(y=\left| x+3 \right| =\begin{cases} x+3\quad if\quad x\ge -3\quad \\ -(x+3)\quad if\quad x<-3 \end{cases}\)
Required area = \(\int _{ b }^{ a }{ y } dx=\int _{ -6 }^{ 0 }{ y } dx\)
= \(\int _{ -6 }^{ -3 }{y}\ dx+\int _{ -3 }^{ 0 }{ y} dx\)
= \(\int _{ -6 }^{ -3 }{ -(x+3) } dx+\int _{ -3 }^{ 0 }{ (x+3) } dx\)
= \(-{ \left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -6 }^{ -3 }{ +\left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -3 }^{ 0 }\) 
\(=-\left[ 0-\frac { 9 }{ 2 } \right] +\left[ \frac { 9 }{ 2 } -0 \right] \)
= 9 sq. units

 

\(\int { { \left( \log x \right) }^{ 2 } } dx= \int { udv } \)
= uv − \(\int { } \)vdu
= x (log x)² − 2\(\int { } \) logxdx...(*)
\(=x(\log x)^{ 2 }-2\int { udv } \)
\(=x(\log x)^{ 2 }-2[uv-\int { udv } ]\)
\(=x(\log x)^{ 2 }-2[x \log x-\int { dx] } \)
\(=x(\log x)^{ 2 }-2x \log x+x+c\)
\(=x[(\log{ ) }^{ 2 }-\log{ x }^{ 2 }+2]+c\)

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

Transition probability matrix
(A B) T = (A B)

Where A represents the percent of people those who bought soap A and B represents the percent of people those who bought soap B.
By the given data
A = 15% = ·15
and B = 85% = ·85
Percentage after one year is
\(\left( \cdot 15\quad \cdot 85 \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) \)
= ((.15)(·65) + (·85)(-45) ·15(-35)+ ·85(-55))
= (-0975 + ·3825 ·0525 + -4675)
= (-48 ·52)
Hence, market share after one year is 48% and 52% At equilibrium,
\(\left( A\quad B \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) =(A\quad B)\)
(-65A + A5B ·35A +·55B) = (A B)
Equating the corresponding entries on both sides we get
\(\Rightarrow \cdot 65A+\cdot 45B=A\)
\(\Rightarrow \cdot 65A+\cdot 45(1-A)=A\)
[Since A+B = 1   B = 1-A]
\(\Rightarrow \cdot 65A+\cdot 45-\cdot 45A=A\)
\(\Rightarrow \cdot 45=A-\cdot 65A+\cdot 45A\)
\(\Rightarrow \cdot 45=A\left( \cdot 35+45 \right) \)
\(\Rightarrow \cdot 45=A(\cdot 35+45)\)
\(\Rightarrow \cdot 45=A(-8)\)
\(\Rightarrow A=\cfrac { \cdot 45 }{ \cdot 8 } =\cdot 5625=56.25\)
\(\therefore B=1-A=1-\cdot 5625=\cdot 4375\)
= 43.75%
\(\therefore\) Equilibrium is reached when A = 56.25% and B = 43.75%

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.

Given non-homogeneous equations are
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 11 & 4 & 9 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ 4 \\ 6 \end{matrix} \right) \)
AX = B

Clearly the last equivalent matrix is in echelon form and it has three non-zero rows
\(\therefore \rho (A)=3\quad \rho \left( \left[ A,B \right] \right) =3\)
\(\rho (A)=\rho \left( \left[ A,B \right] \right) =3\)
\(\therefore\) The given system is consistent and has unique solution.
To find the solution, let us rewrite the above echelon form into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ 1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow\) x + y + z = 3
y + 2z = 1
\((3)\Rightarrow 2x=0\Rightarrow z=\cfrac { 0 }{ 2 } =0\)
\((2)\Rightarrow y+2(0)=1\Rightarrow y+0=1\Rightarrow y=1-0=1\)
\(\left( 1 \right) \Rightarrow x+1+0=3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=3-1\)
\(\Rightarrow x=2\)
\(\therefore\) Solution set [2, 1, 0]