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When Shares are issued at a value that is higher than the face value of the shares , the shares are said to have been Issued at Premium, i.e. Issue Price is more than Face Value. The Companies Act, 2013 requires that the amount of the Premium received should be Credited to Securities Premium Reserve Account. Securities Premium Reserve is a Capital Receipt . 

Utilisation of Securities Premium Reserve

Under Section 52 (2) of the Companies Act , 2013 , the amount of Securities Premium Reserve may be used only for the following purposes : 

– In Writing-off the Preliminary Expenses of the Company .

– For Writing-Off the Expenses, Commission or Discount Allowed on Issue of Debentures of the Company . 

– For Issuing Fully Paid Bonus Shares to the Shareholders of the Company

– For providing for the Premium Payable on Redemption of Redeemable Preference Shares or Debentures of the Company .

Increase in share premium is Source of fund. 

Indian chartered accountants institute defined it, “The statement of the changes took place in the financial position is the summary of the changes happened during a certain period where in it is include of which is sources and how these funds were used that the commercial institution has acquired the funds.”

Fund-flow statement is a statement prepared to analysise the reasons for changes in the financial positions of Company between two balancesheet. It shows the Inflow and outflow of funds.

India investment fund was established by Grindlay Bank. 

The major concern of a consumer is Satisfaction. 

Security Premium can not be applied For Paying dividend to members. 

Kandala is a free trade zone.

The term “fund” as used in fund flow analysis means Excess current assets over current Liabilities.

The two basic measures of liquidity are Current Ratio and Liquid Ratio.

Project is prepared  By promoters, By Managers and By entrepreneur.  

“Entrepreneur is a specialised group of persons who bear risks and deal with uncertainty” this definition has been given by F. H. Knight. 

Present value of money is  Npv method relation method.

The Ideal current Ratio is 2:1. 

Payback period deals with Period required for cost of investment recovery.

Project is prepared By Promoters,  By Managers and By Entrepreneurs. 

Stock turnover ratio comes under Acidity Ratio.

Of all managerial activities planning is the Beginning and end both.  

Input analysis deals with Resource Requirement.

Project life cycle is not concerned with the stabilisation stage.

A product is not concerned with Creativity.

The correct alternative is Demand Potential. 

Project is not concerned with Creative.

Risk Management a main Problem connect of Business. 

Lack of standardisation of the equipment is due to External Constraints.

Planning is Good Oriented,  Object Oriented and Mental Process. 

Project identification deals with Viable product idea. 

Planning involver What to do, When to do and How to do.

Concept:

The area between the curves y1 = f(x) and y2 = g(x) is given by:

Area enclosed = \(\rm \left|\int_{x_1}^{x_2}(y_1-y_2)dx\right|\)

Where, x1 and x2 are the intersections of curves y1 and y2 

Calculation:

Given

Curve 1: y = 2x² = f(x) (say)

Curve 2: y + 6x - 8 = 0

⇒ y = 8 - 6x = g(x) (say)

To find the intersections (or limits of the area) putting value of y from curve 1

⇒ 2x² = 8 - 6x

⇒ 2x2 + 6x - 8 = 0

⇒ 2x2 + 6x - 8 = 0

⇒ (2x - 2)( x + 4) = 0

⇒ x₁ = -4, x₂ = 1

Now the required area (A) is

A = \(\rm \left|\int_{x_1}^{x_2}[f(x)-g(x)]dx\right|\)

⇒ A = \(\rm \left|\int_{-4}^{1}[2x^2-(8-6x)]dx\right|\)

⇒ A = \(\rm \left|\int_{-4}^{1}[2x^2-8+6x]dx\right|\)

⇒ A = \(\rm \left|\left[{2x^3\over3}-8x+{6x^2\over2}\right]_{-4}^{1}\right|\)

⇒ A = \(\rm \left|{2\over3}-8+{3}-\left({2(-4)^3\over3}-8(-4)+{3(-4)^2}\right)\right|\)

⇒ A = \(\rm \left|{-13\over3}-{(-128)\over3}-32-48\right|\)

⇒ A = \(\rm \left|{(-125)\over3}\right|\)

⇒ A = \(\rm 125\over3\)

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C 

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln |x|\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to is to be determined.
• Substitution is done and integral function is then integrated.
• Finally, initial variable t, to be returned.

I = \(\rm ∫ {\sin (x+a)\over \sin x}\) dx

⇒ I = \(\rm ∫ {(\sin x\cos a + \cos x \sin a)\over \sin x} dx\)

⇒ I = \(\rm ∫ {(\sin x\cos a)\over\sin x} + {(\cos x \sin a)\over \sin x} dx\)

⇒ I = ∫ cos a dx + sin a ∫ \(\rm \cos x\over \sin x\) dx

Let sin x = t ⇒ cos x dx = dt 

Substituting sin x as t

⇒ I = x cos a + sin a ∫ \(\rm dt\over t\) + C

⇒ I = x cos a + sin a [ln(|t|)] + C

⇒ I = x cos a + sin a ln(|sin x|) + C

Integration by parts: Integration by parts is a method to find integrals of products. The formula for integrating by parts is given by:

⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times ∫ vdx\right)dx \) + C

where u is the function u(x) and v is the function v(x) 

ILATE rule is Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent. 

Concept:

Integral property:

• ∫ xⁿ dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm\int {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to is to be determined.
• Substitution is done and integral function is then integrated.
• Finally, initial variable t, to be returned.

Calculation:

I = \(\rm \int {6x\over(3x^2+1)^2}\)

Let x² = t

Differentiating both sides

⇒ 2x dx = dt

Now,  I =  \(\rm \int {6x\over(3x^2+1)^2}\) dx

⇒ I = \(\rm 3\int {2x\over(3x^2+1)^2}\)dx

Substituting x² by t and dt = 2x dx

⇒ I = 3 \(\rm \int {1\over(3t+1)^2}\) dt

⇒ I = \(\rm 3\left[ {(3t+1)^{-1}\over-3}\right] + c\)

⇒ I = \(\rm {-1\over(3t+1)}\) + c

Substituting t as x²

⇒ I = \(\boldsymbol{\rm -{1\over3x^2+1}+c}\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

I = \(\rm \int{3\over x^2+4x-5}\) dx

⇒ I = \(\rm 3\int{1\over x^2+4x-5}\) dx

⇒ I = \(\rm 3\int{1\over (x-1)(x+5)}\) dx

⇒ I = \(\rm 3\int{(x+5)-(x-1)\over6(x-1)(x+5)}\) dx

⇒ I = \(\rm {3\over6}\int{(x+5)\over (x-1)(x+5)}-{(x-1)\over (x-1)(x+5)}\) dx

⇒ I = \(\rm {1\over2}\int{1\over (x-1)}-{1\over (x+5)}\) dx

⇒ I = \(\rm {1\over2}\) [ln |x-1| - ln |x+5|] + C

⇒ I = \(\boldsymbol{\rm {1\over2}\ln \left|{x-1\over x+5} \right|}\) + C

Concept:

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to be determined.
• Substitution is done and integral function is then integrated.
• Finally, initial variable t, to be returned.

Calculation:

I = \(\rm \int {6\over(\sin x +\sin2x)}\) dx

Using integration by parts

⇒ I = \(\rm \int {6\over(\sin x +2\sin x \cos x)}\) dx

⇒ I = \(\rm \int {6\over\sin x(1 +2 \cos x)}\) dx

Multiply sin x both in numerator and denominator

⇒ I = \(\rm \int {6\sin x\over\sin^2 x(1 +2 \cos x)}\) dx

⇒ I = \(\rm \int {6\sin x\over(1-\cos x)(1+\cos x)(1 +2 \cos x)}\) dx

Substituting cos x = t ⇒ -sin x dx = dt

⇒ I = \(\rm \int {-6\over(1-t)(1+t)(1+2t)}\) dt

By partial fraction

⇒ I = \(\rm -6 \int {1\over6(1-t)}-{1\over2(1+t)}+{4\over3(1+2t)}\) dt

⇒ I = \(\rm -6[ {-1\over6}\ln|(1-t)|-{1\over2}\ln|(1+t)|+{4\over3\times2}\ln|(1+2t)|]\) + C

⇒ I = \(\rm \ln|(1-t)|+3\ln|(1+t)|-4\ln|(1+2t)|\) + C

⇒ I = \(\boldsymbol{\rm \ln|1-\cos x|+3\ln|1+\cos x|-4\ln|1+2\cos x|}\) + C

 Integration by parts: Integration by parts is a method to find integrals of products. The formula for integrating by parts is given by:

⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times \int vdx\right)dx \) + C

where u is the function u(x) and v is the function v(x) 

ILATE rule is Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Calculation:

I = \(\rm \int {2\sin x\over \cos^2x + \cos x -6 }\) dx

⇒ I = \(\rm -2\int {(-\sin x)\over \cos^2x + \cos x -6 }\) dx

By substitution let cos x = t ⇒ (- sin x) dx = dt

⇒ I = \(\rm -2\int {1\over t^2 + t -6}\) dt

⇒ I = \(\rm -2\int {1\over (t-2)(t+3)}\) dt

⇒ I =  \(\rm -2\int {1\over5}\left[{1\over (t-2)}-{1\over(t+3)}\right]\) dt

⇒ I = \(\rm {2\over5}\int\left[{1\over (t+3)}-{1\over(t-2)}\right] dt\)

⇒ I = \(\rm {2\over5}\left[\ln|t+3|-\ln|t-2|\right]\) + C

⇒ I = \(\rm {2\over5}\left[\ln|\cos x+3|-\ln|\cos x-2|\right]\) + C

∵ -1 ≤ cos x ≤ 1 

⇒ I = \(\boldsymbol{\rm {2\over5}\left[\ln(\cos x+3)-\ln(2-\cos x)\right]}\) + C

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Calculation:

I = \(\rm \int{2x+5\over x^2+5x+6}\) dx

⇒ I = \(\rm \int {(x+3)+(x+2)\over x^2+5x+6}\) dx

⇒ I = \(\rm \int {(x+3)+(x+2)\over (x+3)(x+2)}\) dx

⇒ I = \(\rm \int {(x+3)\over (x+3)(x+2)} + {(x+2)\over (x+3)(x+2)}\) dx

⇒ I =  \(\rm \int {1\over (x+2)} + {1\over (x+3)}\) dx

⇒ I = \(\rm \left[\ln|x+2|+\ln|x+3|\right]\) + C

Putting the limits [0, 2]

⇒ I = \(\rm \left[\ln|2+2|+\ln|2+3| - (\ln|0+2|+\ln|0+3|)\right]\)

⇒ I = \(\rm \left[\ln20- \ln6\right]\)

⇒ I = ln \(10\over3\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C
• ∫\(\rm {1\over1+x^2}\) dx = tan^-1 + C

Calculation:

y = \(\rm ∫{x\over1+x^4}\) dx

⇒ y = \(\rm {1\over2}∫{2x\over1+x^4}\) dx

By substitution let x² = t ⇒ 2x dx = dt

⇒ y = \(\rm {1\over2}∫{1\over1+t^2}\) dt

⇒ y = \(\rm {1\over2}\tan ^{-1}t\) + C

⇒ y = \(\rm {1\over2}\tan ^{-1}x^2\) + C

Given y(0) = 2

⇒ \(\rm {1\over2}\tan ^{-1}0\) + C = 2

⇒ C = 2

Now y = \(\rm {1\over2}\tan ^{-1}x^2\) + 2 

y(2) = \(\rm {1\over2}\tan ^{-1}2^2\) + 2

⇒ y(2) = \(\boldsymbol{\rm {1\over2}\tan ^{-1}4}\) + 2

Concept: 

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Calculation:

I = ∫ 27x² - 8x + 3 dx

⇒ I = \(\rm 27 \left[{x^3\over3}\right] - 8 \left[x^2\over2\right]+ 3x\)

⇒ I = 9x³ - 4x² + 3x

Putting the limits

⇒ I = [9x3 - 4x2 + 3x]\(\rm _0^3\)

⇒ I = 9(3)³ - 4(3)² + 3(3)

⇒ I = 243 - 36 + 9

⇒ I = 216

A knight in King Arthur's court

The division of labour is the separation of tasks in any economic system or organisation so that participants may specialise (specialisation). Individuals, organizations, and nations are endowed with or acquire specialised capabilities and either form combinations or trade to take advantage of the capabilities of others in addition to their own. Specialised capabilities may include equipment or natural resources as well as skills and training and combinations of such assets acting together are often important. For example, an individual may specialise by acquiring tools and the skills to use them effectively just as an organization may specialize by acquiring specialised equipment and hiring or training skilled operators. The division of labour is the motive for trade and the source of economic interdependence.

Historically, an increasing division of labour is associated with the growth of total output and trade, the rise of capitalism, and the increasing complexity of industrialised processes. The concept and implementation of division of labour has been observed in ancient Sumerian (Mesopotamian) culture, where assignment of jobs in some cities coincided with an increase in trade and economic interdependence. Division of labour generally also increases both producer and individual worker productivity.

After the Neolithic Revolution, pastoralism and agriculture led to more reliable and abundant food supplies, which increased the population and led to specialisation of labour, including new classes of artisans, warriors, and the development of elites. This specialistion was furthered by the process of industrialisation, and Industrial Revolution-era factories. Accordingly, many classical economists as well as some mechanical engineers such as Charles Babbage were proponents of division of labour. Also, having workers perform single or limited tasks eliminated the long training period required to train craftsmen, who were replaced with lesser paid but more productive unskilled workers.

It was a cold, foggy morning, and Julie was in no mood to take her dog, Snoopy, out for a walk in the bitter cold. Snoopy himself seemed reluctant to leave his bed near the fireplace. Julie got up and sat with Snoopy near the window of their house, soaking in the warm sunlight. Suddenly, she heard some commotion outside her front door, and hurriedly went to see what it was. As she opened the door, a young, scruffy looking man started running, with the milk bottles he had stolen from her front porch. Snoopy took after the man, ignoring the bitter cold, while Julie ran after them, still in her night robe and slippers. Snoopy tripped two newspaper boys and overturned a garbage can as he pursued the fleeing thief. As Julie turned a corner, she saw the thief slip on the hard ice, milk bottles flying, as Snoopy leaped and caught one of them in his mouth. The thief got away, but Snoopy came back happily, his tail wagging, to return the bottle to an exhausted Julie, expecting his breakfast.

The poet, William Blake, revealed his anger to his friend and the anger ended. But when the poet concealed his anger from his enemy, the anger grew. It grew like a tree that bore a bright apple.

Mary Kom was not accustomed to American food. The greatest disadvantage was her loss of appetite. She could not eat food however hard she tried. She started losing weight. She was just 46 kg before the finals. This probably cost her the dream of winning the gold in the finals.

She was heartbroken because she lost the gold. The coaches were kind and appreciated her for winning silver to medal hungry India. She was the only person in the team to get a medal. The biggest thing she took away from World Women’s boxing championship in Pennsylvania, USA was the conviction that she could take on any boxer in the world.

Prerana: Who are you talking about?

Sakshi: Well, a it’s about, Nandini. 

Nandini is not only clever but honest and reliable. No wonder then she is the star of our college,

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