\(I=\int { \cfrac { dx }{ { e }^{ x }+6+\frac { 5 }{ { e }^{ x } } } } \)
= \(\int { \cfrac { { e }^{ x }dx }{ { e }^{ 2x }+6{ e }^{ x }+5 } } \)
Put \({ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt\)
\(\Rightarrow I=\int { \cfrac { dt }{ { t }^{ 2 }+6t+5 } } \)
\(I=\int { \cfrac { dt }{ \left( t+5 \right) \left( t+1 \right) } } \)
= \(\int { \cfrac { A }{ t+5 } dt+\int { \cfrac { B }{ t+1 } dt } } \)
\(\cfrac { 1 }{ \left( t+5 \right) \left( t+1 \right) } =\cfrac { A }{ t+5 } +\cfrac { B }{ t+1 } \)
\(\Rightarrow I=A(t+1)+b(t+5)\)
Put t = - 1
\(\Rightarrow 1=B(-1+5)\)
\(\Rightarrow 1=B(4)\)
\(\Rightarrow B=\cfrac { 1 }{ 4 } \)
Put t = - 5
\(\Rightarrow 1=A(-5+1)\)
I=A(-4)
\(\Rightarrow A=-\cfrac { 1 }{ 4 } \)
From (1),\(I=\int { \cfrac { -\frac { 1 }{ 4 } dt }{ t+5 } } +\int { \cfrac { \frac { 1 }{ 4 } dt }{ t+1 } } \)
= \(-\cfrac { 1 }{ 4 } \log\left| t+5 \right| +\cfrac { 1 }{ 4 } \log|t+1|+c\)
= \(\cfrac { 1 }{ 4 } \left[ \log|t+1|-\log t+5| \right] +c\)
= \(\cfrac { 1 }{ 4 } \log\left| \cfrac { t+1 }{ t+5 } \right| +c\)

= \(\cfrac { 1 }{ 4 } \log\left| \cfrac { { e }^{ x }+1 }{ { e }^{ x }+5 } \right| +c\)
\(\left[ \because t={ e }^{ x } \right] \)

(i) Binomial distribution is symmetrical if p = q = 0.5. It is skew symmetric if p≠q. It is positively skewed if p < 0.5 and it is negatively skewed if p > 0.5.
(ii) For binomial distribution, variance is less than mean.
Variance = npq = (np)q < np < mean.

\(\int { \frac { 1 }{ \sqrt { x+2 } -\sqrt { x-2 } } } dx\)
=\(\int { \frac { \sqrt { x+2 } +\sqrt { x-2 } }{ 4 } } dx\)
\(=\frac { 1 }{ 6 } \left\{ { \left( x+2 \right) }^{ \frac { 3 }{ 2 } }+{ \left( x-2 \right) }^{ \frac { 3 }{ 2 } } \right\} +c\)

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 3 & -2 \\ 6 & -4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \end{matrix} \right) \)
AX = B

\(\therefore \)\(\rho ([A,B])=2\), \(\rho (A)=1\)
\(\rho (A)\neq \rho \left( [A,B] \right) \)
\(\therefore \) The given system is inconsistent and has no solution.

Grand average = \(\frac{67 + 61 + 63.33 + 63.37}{4}\)
= \(\frac{255}{4}=63.75\)
Seasonal index (S.I) = \(\frac{Quarterly average}{Grand average}\times100\)
Hence, S.I for I quarter = \(\frac{67}{63.75}\times100\) = 105.01
S.I for II quarter = \(\frac{61}{63.75}\times100\) = 95.68
S.I for III quarter = \(\frac{63.33}{63.75}\times100\) = 99.35
S.I for IV quarter = \(\frac{63.67}{63.75}\times100\) = 99.87

Let X be a random variable. That denote the mark obtained by a student for answering a question.
∴ X can take values 1 and -\(\frac{1}{2}\)
∴ P(X = 1) = P (answering a question correctly) 
= \(\frac{1}{4}\)
P(X = -\(\frac{1}{2}\)) = P(answering a question wrongly)
=\(1-\frac{1}{4}=\frac{3}{4}\)
∴ Probability distribution function is

\(\therefore E(x)=\sum { xp(x)=1(\frac { 1 }{ 4 } )-\frac { 1 }{ 2 } \left( \frac { 3 }{ 4 } \right) =\frac { 1 }{ 4 } -\frac { 3 }{ 8 } } \)
\(=\frac { 2-3 }{ 8 } =-\frac { 1 }{ 8 } \)
∴ Expectation of mark for answering a single question is -\(\frac{1}{8}\)
∴ Expectation of mark for answering 120 questions = 120(-\(\frac{1}{8}\)) = -15.

P(X = 0) + P(X = 1) + P(X = 2)
= 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence the given distribution of probabilities is not a probability distribution.

Min (45, 70, 53) = 45
∴ Choosing Bonds is the best decision under minimax principle.

Given f'(x) = 8x³ -2x²
∴ ∫ f'(x) dx = ∫ (8x³ - 2x²) dx
⇒ f(x) = \(\frac { { 8x }^{ 4 } }{ 4 } -\frac { { { 2x }^{ 3 } } }{ 3 } +c\)
⇒ f (x) = 2x⁴ - \(\frac { { { 2x }^{ 3 } } }{ 3 } +c\) ...(1)
Also, f(2) = 1
⇒ 1 = 2(2⁴) - \(\frac { { 2\left( { { 2 }^{ 3 } } \right) } }{ 3 } +c\)
⇒ 1 = 32 - \(\frac { { 16 } }{ 3 } +c\)
⇒ 1 - 32 + \(\frac { { 16 } }{ 3 } \) = c ⇒ -31 + \(\frac { { 16 } }{ 3 } \) =c
⇒ \(\frac { { -93+16 } }{ 3 } =c\)
⇒ c = \(\frac { { -77 } }{ 3 } \)
∴ (1) ⟶ f(x) = 2x⁴ - \(\frac { { 2x }^{ 3 } }{ 3 } -\frac { { -77 } }{ 3 } \)

The auxiliary equation is 3m² - 5m + 2 = 0
⇒ (m - 1)(3m - 2) = 0
⇒ m = 1, \(\frac { 2 }{ 3 } \)
The roots are real and different
∴ Complementary function CF is Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\)
∴ The general solution is y = Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\).

The highest derivative if of order 3 and its power is 1
∴ order is 3 and degree is 1.

Given MC = 3x² - 2x + 8
⇒ ഽMC = ഽ(3x² - 2x + 8)dx
\(\Rightarrow C=\frac { { 3x }^{ 3 } }{ 3 } -\frac { { 2x }^{ 2 } }{ 2 } +8x+k\)
⇒ C = x³ - x² + 8x + k
Since there is no fixed cost,
when x = 0, C = 0 ⇒ k = 0
∴ C = x³ - x² + 8x

Given non-homogeneous equations are x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30

Here \(\rho (A)=\rho (A,B) = 2\) 
\(\therefore\) The given system is consistent

Differentiating w.r.t 'x' we get,
x.\(\frac { dy }{ dx } \) + y(1) = 0     [Product rule]
⇒ x\(\frac { dy }{ dx } \) + y = 0 which is the required differentiated equation.

To assign the different jobs to the different machines (one job per machine) to minimize the overall cost is known as assignment problem.

The main objective in any product process is to control and maintain a satisfactory quality level of the manufactured product. This is done by Process Control. In process control the proportion of defective items in the production process is to be minimized and it is achieved through the technique of control charts.

Fisher's price index number is the geometric mean of Laspeyre's and Paasche's price index number. Hence it is weighted index number.
Fisher's price index number = \(\sqrt {\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0} }}{\times}{\frac {\sum p_{1}q_{1}}{\sum p_{0}q_{1}} \times {100}}\)

Any sample statistic which is used to estimate an unknown population parameter is called an estimator (i.e.,). an estimator is a sample statistic used to estimate a population parameter.

\(\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \sin x } dx={ \left[ -\cos x \right] }_{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)
\(=-\left( \cos\frac { \pi }{ 3 } -\cos\frac { \pi }{ 6 } \right) \)
\(=\frac { 1 }{ 2 } (\sqrt { 3 } -1)\)

Mathematical expectation E(X) is an average of the values, that the random variable takes on, where each value is weighted by the probability that the random variable is equal to that value. Values that are most probable receive more weight. Each value x is multiplied by the approximate probability that X equals the valuex.

A random variable is a real valued function defined on a sample space S and taking values in (-∞, ∞) or whose possible values are numerical outcomes of a random experiment.

2y + x = 8

Given 2y + x = 8
2y = 8-x
y = \(\frac{1}{2}\) (8-x)
Given limits are x = 2 and x = 4
Area of the shaded region between the given limits
\(A=\int _{ a }^{ b }{ y\quad dx } =\int _{ 2 }^{ 4 }{ \frac { 1 }{ 2 } (8-x)dx } \)
\(\frac { 1 }{ 2 } \int _{ 2 }^{ 4 }{ (8-x)dx } =\frac { 1 }{ 2 } { \left[ 8x-\frac { { x }^{ 2 } }{ x } \right] }_{ 2 }^{ 4 }\)
\(=\frac { 1 }{ 2 } \left[ \left( 8(4)-\frac { { 4 }^{ 2 } }{ 2 } \right) \left( 8(2)-\frac { { 2 }^{ 2 } }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } [(32-8)-(16-21)]\)
\(=\frac { 1 }{ 2 } [24-14]\)

A = 5 sq. units.

∫(2sin x − 5cos x )dx = 2∫sin x dx − 5∫cos x dx
= −2cos x −5sin x + c

Given

Here the intervals are unequal
∴ By Lagranges interpolation formula, we have
x₀ = 0, x₁ = 1, x₂ = 3, x₃ = 4
y₀ = -12, y₁ = 0, y₂ = 6, y₃ = 12 and x = x.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (x-1)(x-3)(x-4) }{ (0-1)(0-3)(0-4) } (-12)+\frac { (x-0)(x-3)(x-4) }{ (1-0)(1-3)(1-4) } (0)+\frac { (x-0)(x-1)(x-4) }{ (3-0)(3-1)(3-4) } (6)+\frac { (x-1)(x-3)(x-4) }{ (4-0)(4-1)(4-3) } (12)\)
= \(\frac { (x-1)(x-3)(x-4) }{ (-1)(-3)(-4) } (-12)+0+\frac { x(x-1)(x-4) }{ (3)(2)(-1) } (6)+\frac { x(x-1)(x-3) }{ (4)(3)(1) } (12)\)
= +[(x - 1)(x - 3)(x - 4)] - x (x - 1)(x -  4) + x(x - 1)(x - 3)
= +[(x³-4x+3)(x-4)] -x(x²-5x+4) + x(x²-4x + 3)
= - (x³ - 8x²+ 19x - 12) - 4x² + 3x
= (x - 4)(x² - 4x + 3) - x (x² - 5x + 4) + x(x² - 4x + 3)

= x³ - 7x² + 19x - 12.

Sample size n = 100
Sample mean \(\bar { X } \)= 72
Population mean μ = 76
Population standard deviation σ = 8
Null Hypotheses H₀:
μ = 76(i.e., There is no Significant difference between the state scores and the national scores)
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 72-76 }{ \frac { 8 }{ \sqrt { 100 } } } =\frac { -4 }{ \frac { 8 }{ 10 } } =\frac { -4 }{ 8 } =-5\)
\(\Rightarrow |Z|=5\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(Z>{ Z }_{ \frac { \alpha }{ 2 } }i.e.,\ 5>1.96\)
Inference : Since \(Z>{ Z }_{ \frac { \alpha }{ 2 } }\)  at 5% level of significance, the null hypothesis H₀ is rejected. 
Hence, we conclude that there is significant difference between the state scores and the national scores.