We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \),since f(x) is a density function
\(\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k{ \left[ \frac { { e }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }=1\)
⇒ k = 2
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x{ e }^{ -2x } } dx\)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx } \)
\(=2\left\{ { \left[ \frac { { xe }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }\int _{ 0 }^{ \infty }{ \frac { { e }^{ -2x } }{ -2 } dx } \right\} \)
\((\because \int { udv=uv-\int { udv } } )\)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -2x }dx } \) \(=\frac { 1 }{ 2 } \)

Let p be the additional product produced for additional of x labour,
\(\frac{dp}{dx}\) = 300 - 5x^2/3
\(p=\int _{ 0 }^{ 64 }{ { \left( 300-5{ x }^{ \frac { 2 }{ 3 } } \right) } } dx\)
\(={ \left[ 300x-3{ x }^{ \frac { 5 }{ 3 } } \right] }_{ 0 }^{ 64 }\)
= 300 \(\times\) 64 -3(64)^5/3
= 16128
∴ The number of additional units produced 16128
Total number of units produced by 264 workers
= 50,000 + 16,128 = 66128 units

\(\int { \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } dx } =\int { \left[ \frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \right] dx } \)
\(=20\int { \frac { dx }{ x-3 } -13\int { \frac { dx }{ x-2 } } } \)
\(20log\left| x-3 \right| -13 \log\left| x-2 \right| +c\)
[ By partial fractions,
\(\frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { A }{ x-3 } +\frac { B }{ x-2 } \Rightarrow \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \)]

The equations are
2x + 3y = 7
3x + 5y = 9
Here \(\triangle =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =1\)
\(\neq 0\)
\(\therefore \) we can apply Cramer’s Rule
Now \({ \triangle }_{ x }=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =8\)  \({ \triangle }_{ y }=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =-3\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }_{ X } }{ \triangle } =\frac { 8 }{ 1 } =8\)   \(y=\frac { { \triangle }_{ y } }{ \triangle } =\frac { -3 }{ 1 } =-3\)
\(\therefore \) Solution is x = 8, y = −3

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

Laspeyre’s price index number
\({ P }_{ 01 }^{ L }=\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 1110 }{ 952 } \times 100=116.60\)
On an average, there is an increase of 16.60 % in the price of the commodities when the year 2000 compared with the year 2010.
Paasche’s price index number
\({ P }_{ 01 }^{ P }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 1 } } } \times 100=\frac { 1460 }{ 1236 } \times 100=118.12\)
On an average, there is an increase of 18.12 % in the price of the commodities when the year 2000 compared with the year 2010.
Fisher’s price index number
\({ P }_{ 01 }^{ F }=\sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \times 100=\sqrt { \frac { 1110\times 1460 }{ 952\times 1236 } } \times 100=117.36\)
On an average, there is an increase of 17.36 % in the price of the commodities when the year 2000 compared with the year 2010.

Given 3e^x tan y dx + (1 + e^x)sec²y dy = 0
3e^x tan y dx = −(1 + e^x)sec² y dy
\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx=-\frac { { sec }^{ 2 }y }{ tany } dy\)
Integrating, we get 3ഽ\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx\) = -ഽ\(\frac { { sec }^{ 2 }y }{ tany } dy\) + c
3log(1+ e^x ) = −log tan y + log c   \(\left[ \therefore \int { \frac { f'(x) }{ f(x) } dx=logf(x) } \right] \)
log(1+e^x)³ + log tan y = log c
log [(1 + e^x)³ tan y ] = log c
(1+e^x)³ tan y = c (1)
Given y(0) = \(\frac { \pi }{ 4 } \) (i.e) y = \(\frac { \pi }{ 4 } \) at x =0
(1) ⇒ (1 +e⁰ )³ tan \(\frac { \pi }{ 4 } \) = c
2³ (1) = c
⇒ c = 8
Hence the required solution is (1  +e^x)³ tan y = 8

Sample size n = 50
Sample mean \(\bar { x } =9.3\) minutes
Sample S.D s = 1.6 minutes
Population mean  μ = 8.9 minutes
Null hypothesis H₀:  μ = 8.9
Alternative hypothesis H₁:  μ = 8.9 (two tail)
Level of significance  μ = 0.05 
Test statistic  \(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(\\ Z=\frac { 9.3-8.9 }{ \frac { 1.6 }{ \sqrt { 50 } } } =\frac { 0.4 }{ 0.2263 } =1.7676\)
Calculated value Z = 1.7676
Critical value at 5% level of significance is \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Inference: Since the calculated value is less than table value i.e., \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) at 5% level of significance, the null hypothesis is accepted.
Therefore we conclude that an ambulance service claims on the average 8.9 minutes to reach its destination in emergency calls.

P = 1/5/100 = 1/500 = 0.002 n = 10  λ = np = 0.02
\(p(x)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -0.02 }{ (0.02) }^{ x } }{ x! } \)
(i) Number of packets containing no defective = N p(o) = 1,00,000 × e^–0.02
= 98020
(ii) Number of packets containing one defective = N p(1) = 1,00,000 × 0.9802 × 0.02
= 1960
(iii) Number of packets containing 2 defectives = N p(2) = 20

Let ‘x’ be the cost of a Business Mathematics book
Let ‘y’ be the cost of a Accountancy book.
Let ‘z’ be the cost of a Commerce book.
\(\therefore \) 3x + 2y + z = 840
2x + y + z = 570
x + y + 2z = 630
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} \right| =-2\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 840 & 2 & 1 \\ 570 & 1 & 1 \\ 630 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right| =-240 \)
\({ \triangle }_{ y }=\left| \begin{matrix} 3 & 840 & 1 \\ 2 & 570 & 1 \\ 1 & 630 & 2 \end{matrix} \right| =-300 \)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 840 \\ 2 & 1 & 570 \\ 1 & 1 & 630 \end{matrix} \right| =-360\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -240 }{ -2 } =120 \)
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { 300 }{ -2 } =150 \)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { 360 }{ -2 } =180\) 
\(\therefore \) The cost of a Business Mathematics book is Rs. 120,
the cost of a Accountancy book is Rs. 150 and
the cost of a Commerce book is Rs. 180.