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correct option (d) -15

Charle’s law states that volume of fixed mass of gas is directly proportional to absolute temperature at constant pressure.

(V₁/T₁ = V₂/T₂)_P

Answer is (c) π/2

correct option:

(A) 3,1,–2 

(i) Cu, 

(ii) U

Answer is (a) n(E) + n(S)

Answer is (b) 7

Answer is (c) π/2

Answer is (a) 4 : 3

correct option:

(B) 7 

(i) Atomic radii increases down a group in the periodic table. 

(ii) N^-3 > O^-2 > Na⁺ > Mg²⁺

(i) Four 

(ii) 18g.

Let the function be f (x) = ax + b 

f(1) = a(1) + b = a + b = 1 

⇒ f(2) = a(2) + b = 2a + b = 2 f(0) = a(0) + b = [b = -1] …..(1) 

f(-1) = a(-1) + b = -3 

∴ b – a = -3 … (2) 

(i) in (2) (-1) -a = -3 

∴ -1 + 3 = a 

∴ a = 2 

∴ f(x) = 2x – 1.

Centres of the circles are A (2,3), B(–3, – 9) 

radii are r₁ = √(4 + 9 +12) = 5 

r₂ = √(9 + 81− 26) = 8 

AB = √((2+ 3)² + (3+ 9)²) = √(25+144) = 13 = r₁ + r₂ 

∴ The circle touch externally.

The zeroes of quadratic polynominl x²+3x+2 is. (-1,-2)

Calculation:

132 = 3 × 4 × 11. 

3, 4 and 11 are factors of 132, hence if a number is divisible by 3,4 and 11, the number will be divisible by their product 132 also.

Here 3, 4 and 11 are pair-wise co-prime numbers.

If a number is not divisible by 3 or 4 or 11, then it is not divisible by 132.

⇒ 264 is divisible by 3, 4 and 11; so 264 is divisible by 132.

⇒ 396 is divisible by 3, 4 and 11; so 396 is divisible by 132.

⇒ 462 is divisible by 3 and 11 but not divisible by 4; so 462 is not divisible by 132.

⇒ 792 is divisible by 3, 4 and 11; so 792 is divisible by 132.

⇒ 968  is divisible by 4 and 11 but not divisible by 3; so 462 is not divisible by 132.

⇒ 2178  is divisible by 3 and 11 but not divisible by 4; so 462 is not divisible by 132.

⇒ 5184  is divisible by 3 and 4 but not divisible by 11; so 462 is not divisible by 132.

⇒ 6336 is divisible by 3, 4 and 11; so 6336 is divisible by 132.

Only 264, 396, 792 and 6336 are divisible by 132.

∴ 4 numbers are divisible by 132.

If ∝, β be tue uroes of Quadrate plynomal x²+7x+10 tues the value of (∝+β)2 is. 49

(a) (Termnating)

Integration of log x is x(logx - 1)

Explanation:

At first we can write log x as 1.log x

Then we will apply partial fraction to it. i.e. ∫u v dx = u∫v dx −∫u' (∫v dx) dx.
here, u = log x and v = 1
∫1.log x dx = logx∫1 dx - ∫d/dx(log x)∫(1dx) dx
xlog x - ∫1/x(x) dx 
xlog x - ∫1 dx
xlog x - x

so, x(logx - 1)

Middle term of an AP is depends on the number of elements AP have.

If AP is infinite sequence, then its mid term can't be find.

But if AP is finite then its depends on number of terms are odd & even.

(a) If number of terms (n) in AP is odd then \(\frac{n+1}{2}\)th term is mid term of AP.

(b) If number of terms (n) in AP is even then \(\frac{n}{2}\)th or \(\frac{n+2}{2}\)th terms are mid-terms of that AP.

x²(x² – 1)\(\frac{dy}{d\mathrm x}\) + x(x² + 1)y = x² – 1

\(\Rightarrow \frac{dy}{d\mathrm x}+\frac{\mathrm x(\mathrm x^2+1)}{\mathrm x^2(\mathrm x^2-1)}y\) \(=\frac{\mathrm x^2-1}{\mathrm x^2(\mathrm x^2-1)}=\frac{1}{\mathrm x^2}\)

Which is a linear differential equation.

where, p \(=\frac{\mathrm x(\mathrm x^2+1)}{\mathrm x^2(\mathrm x^2-1)}\) \(=\frac{\mathrm x^2+1}{\mathrm x(\mathrm x^2-1)}\) \(=\frac{\mathrm x^2-1+2}{\mathrm x(\mathrm x^2-1)}\)

\(=\frac{1}{\mathrm x}+\frac{2}{\mathrm x(\mathrm x^2-1)}\)

\(=\frac{1}{\mathrm x}+\frac{2}{\mathrm{x(x-1)(x+1)}}\)

let \(\frac{2}{\mathrm x(\mathrm x+1)(\mathrm x+1)}\) \(=\frac{A}{\mathrm x}+\frac{B}{\mathrm x+1}+\frac{C}{\mathrm x-1}\)

Then A(x + 1)(x – 1) + Bx(x – 1) + Cx(x + 1) = 2

For x = 0, –A = 2 ⇒ A = –2

For x = 1, 2C = 2 ⇒ C = 1

For x = –1, 2B = 2 ⇒ B = 1

Hence, \(\frac{2}{\mathrm x(\mathrm x-1)(\mathrm x+1)}\) \(=\frac{-2}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\)

\(\therefore p=\frac{1}{\mathrm x}-\frac{2}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\) 

\(=-\frac{1}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\).

Now, I.F \(=e^{\int pd\mathrm x}\) \(=\int\limits_e\left(\frac{-1}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\right)d\mathrm x\)

\(=e^{-\log \mathrm x+\log(\mathrm x+1)+\log(\mathrm x-1)}\)

\(=e^{\log\left(\frac{(\mathrm x+1)(\mathrm x-1)}{\mathrm x}\right)}\)

\(=\frac{\mathrm x^2-1}{\mathrm x}\)

\(=\left(\mathrm x-\frac{1}{\mathrm x}\right)\).

Now, complete solution of given linear differential equation is

\(y.\left(\mathrm x-\frac{1}{\mathrm x}\right)\) \(=\int 2\times \left(\mathrm x-\frac{1}{\mathrm x}\right)d\mathrm x\)

\(=\int \frac{1}{\mathrm x^2}\left(\mathrm x-\frac{1}{\mathrm x}\right)d\mathrm x\)

\(=\int \left(\frac{1}{\mathrm x}-\frac{1}{\mathrm x^3}\right)d\mathrm x\)

\(=\log \mathrm x + \frac{1}{2\mathrm x^2}+C\)

Hence, solution of given differential equation is \(y\left(\mathrm x-\frac{1}{\mathrm x}\right)\) \(=\log \mathrm x + \frac{1}{2\mathrm x^2}+C\).

\(\lim\limits_{\mathrm x\to \infty}\)\(\left(1+\frac{1}{\mathrm x}\right)^{3\mathrm x}\)  [1^∞type]

= Exp\(\left\{\lim\limits_{\mathrm x\to \infty}\frac{1}{\mathrm x}\times 3\mathrm x\right\}\)   \(\Big(\because \) If \(\lim\limits_{\mathrm x \to \infty}\)\(\left(f(\mathrm x)\right)^{g(\mathrm x)}\) = [1^∞type] then \(\lim\limits_{\mathrm x \to \infty}\) \(\left(f(\mathrm x)\right)^{g(\mathrm x)}\) = Exp\(\left\{\lim\limits_{\mathrm x \to \infty}(f(\mathrm x)-1){g(\mathrm x)}\right\}\Big)\)

= Exp\(\left\{\lim\limits_{\mathrm x \to \infty}3\right\}\)

\(= e^3\)

Roots of the equation Ax² + Bx + C = 0 are α and β.

∴ Sum of roots = α + β = \(\frac{-B}{A}\)

And product of roots = αβ = \(\frac{C}{A}\)

Now,

(α + β)² = α² + β² + 2αβ

∴ α² + β² = (α + β)² - 2αβ

= \((\frac{-B}{A})^2\)- \(\frac{2C}{A}\)

= \(\frac{B^2}{A^2}\) - \(\frac{2C}{A}\)

= \(\frac{B^2-2AC}{A^2}\)

Sum of required roots \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\),

= \(\frac{\alpha}{\beta}\) + \(\frac{\beta}{\alpha}\)

= \(\frac{\alpha^2+\beta^2}{\alpha\beta}\) 

= \(\frac{\frac{B^2-2AC}{A^2}}{\frac{C}{A}}\)

= \(\frac{B^2-2AC}{AC}\)

And products of roots = \(\frac{\alpha}{\beta}\) x \(\frac{\beta}{\alpha}\) = 1.

∴ Equation whose roots we are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) is :

x² - (sum of roots)x + Product of roots = 0

⇒ x² - \(\frac{(B^2-2AC)}{AC}x\) + 1 = 0

⇒ ACx² - (B²-2AC)x + AC = 0

Hence proved.

The coefficient form of the polynomial is (1,2,3)
Therefore, the index form of the polynomial is x²+2x+3

CB=OB-OC;

CD=CB/2=(OB-OC)/2=(b-c)/2;

OD=OC+CD=c+(b-c)/2=c/2+b/2;

AD=OD-OA=c/2+b/2-a;

AE=2/3*AD=2/3*(c/2+b/2-a)=c/3+b/3-2*a/3;

OE=OA+AE=a+c/3+b/3-2*a/3=a/3+b/3+c/3;

OD=c/2+b/2;

OE=a/3+b/3+c/3;

Correct answer is (A) 6

Option : (b)

f'(x) = 1 − sinx 

⇒ f'(x) > 0 ⍱x ∈ R 

⇒ no value of b exists

Option : (c)

a = b - 2 and b > 6

If b = 8, then a = 8 - 2 = 6

\(\Rightarrow(6,8)\in R\)

Correct option is: (B) (6, -7)

Option : (a)

 f(x) = \(\begin{cases} \frac{x}{-x}=-1,x<0\\ -1, x≥0 \end{cases}\) 

⇒ f(x) = −1 ⍱ x ∈ R 

⇒ f(x) is continuous ⍱ x ∈ R as it is a constant function.

Correct answer is (D) 13/12

Correct option is: (B) \(\frac{-23}{27}\)

21 Pratap Nagar
G.T. Road
Udaipur
23 Feb. 2020
The Registrar Examination Branch
BSER
Ajmer

Sub: Non-receipt of Roll Number card for the Class XII Examination.

Sir,

I, Hari Ram Bairwa S/o Sh. Ram Krishan Bairwa, resident of 21 Pratap Nagar, G.T. Road, Udaipur, haven’t received the Roll Number card for the Class XII examination. I have enclosed a photocopy of the board examination form and of the fee receipt for your kind consideration.
Please, send it to me immediately.

Thanking you

Yours faithfully

Hari Ram Bairwa

Answer: c. Demand

A task of developing a technical blueprint and specifications for a solution that fulfills the business requirements is undertaken in the system design phase of the system development process.

'Bhimrao Ramji Ambedkar' was born on 14 April 1891 in Mhow town of Madhya Pradesh, India. He was the son of Ramji Maloji Sakpal and Bhimabai. His father served in the Indian Army at the Mhow cantonment. He became one of the first Dalit (untouchables) to obtain a college education in India, eventually earning a law degree and doctorates for his study and research in law, economics and political science.

Bhimrao Ramji Ambedkar is popularly known as 'Babasaheb'. He was an Indian jurist, political leader, philosopher, anthropologist, historian, orator, economist, teacher, editor, prolific writer, revolutionary and a revivalist for Buddhism in India. He became the 1st Law Minister of India. He became the Chairman of the Constitution Drafting Committee. For his contributions, he was awarded with 'Bharat Ratna'.   Ambedkar died on 6 December 1956 at his home in Delhi.  

Explanation:

Turnaround document:

A document designed to initiate action and then be returned to use in processing the completion of the transaction is known as a turnaround document i.e., A turnaround document is one which is output from one computerized system and will be the input to another. Examples are optically readable documents, stubs, or punched cards included with customer billings with the request that they be returned with payment.  The turnaround document assists in positive identification and reduces errors because the document with the required feedback information is already prepared, often in machine-readable form.

• Some transaction records are distributed to other departments in the organization to provide background information for recipients in the event that they need to respond to inquiries or need them for other references. With online systems, a reference copy of the transaction can be stored in a computer file and may be retrieved via a terminal by anyone who is authorized and has need of the information.
• Transaction documents may also be used for managerial information or control scanning, as when a purchasing manager scans all purchase orders to spot unusual occurrences. In general, however, managerial information purposes are better met by reports or analyses which summarize transactions. 
• When transactions are processed, a listing of data about each transaction is usually prepared. The listing includes control totals for the number of transactions processed, the total rupees amount of transactions, etc The listing represents a batch of transactions or, for online processing, processing during a period of time It provides a means of processing reference and error control

This is a business involved in the selling of services and products for profit, usually professional services. Although a firm can provide its products and services in more than one location, it is unified under the same owners hence must have the same Employer Identification Number. 

Examples of firms include accounting firms, consulting firms, law firms and graphic design firms. The IRS does not control the operations of a firm.

This is a business involved in any income-generating activity involving the sale of goods and services and includes all business trades and structures. Companies can either be corporations, sole proprietorships and limited liability companies, with each having different tax benefits and liabilities. Also, they must be registered under the Companies Act.

Types of companies include private limited company, a public limited company or a one-person company.

When a promissory note or bill of exchange has been dishonoured by non-acceptance or non-payment, the holder may cause such dishonour to be noted by a notary public upon the instrument or upon a paper attached thereto or partly upon each (Section 99). The holder may also within a reasonable time of the dishonour of the note or bill, get the instrument protested by notary public (Section 100).

B. Brand Mark

Correct answer is C. Logo

c. Manpower Plan

Effects of endorsement: 

The legal effect of negotiation by endorsement and delivery is, 

(i) To transfer property in the instrument from the endorser to the endorsee. 

(ii) To vest in the latter the right of further negotiation, and 

(iii) A right to sue on the instrument in his own name against all the other parties (Section 50).

Term loan: that is repaid in regular payments over a set period of time .

Types of term loans based on repayment is as follows: 

1. Short Term Loan: Repayable in 1 to 3 years. It is generally used for short term credit like cash credit, personal loan etc. 

2. Medium term Loan : Repayable in a period of more than 3 years but within 5 years. These are issued to finance furniture, fixtures, vehicles etc. 

3. Long Term Loan: It is repayable in a period of more than 5 years example is Home Loan.

Report on the Scene of Examination Hall
(by Navin, Class XII A)

Jodhpur, March 10
It was the first day of our Senior Secondary Examination. Students entered the examination hall. Some students had an anxious look, while some others were busy reading notices. Only well-prepared students were looking calm. At last, the bell rang. We all took our seats. Invigilators distributed answer books and question papers. In a moment, a hush fell on all. The students were busy solving questions. Hours passed and the warning bell rang. There were fifteen minutes left. A few fingers moved faster. Some diligent students were revising their answers. When the final bell rang, students handed over their answer books to the invigilators and came out of the hall.

जोधपुर, 10 मार्च
यह दिन हमारे Senior Secondary Examination का प्रथम दिन था। विद्यार्थियों ने परीक्षा भवन में प्रवेश किया। कुछ विद्यार्थी चिंतित मुद्रा में थे जबकि कुछ दूसरे विद्यार्थी नोटिस पढ़ने में व्यस्त थे। केवल पूर्ण तैयारी किये हुए विद्यार्थी ही धैर्यपूर्ण मुद्रा में थे। अंत में घंटी बजी। हम लोगों ने अपना-अपना स्थान ग्रहण किया। इनविजिलेटरों ने उत्तर पुस्तिका और प्रश्न-पत्र वितरित किये। कुछ ही समय में चारों ओर शांति छा गयी। विद्यार्थी प्रश्नों को हल करने में व्यस्त हो गए। घंटे बीत गये और चेतावनी की घंटी बजी। अब केवल पन्द्रह मिनट बचे थे। कुछ अंगुलियाँ तेज गति से चली। कुछ परिश्रमी विद्यार्थी अपने-अपने उत्तरों को दुहरा रहे थे। जब अंतिम घंटी बजी, विद्यार्थियों ने अपनी-अपनी उत्तर पुस्तिकाओं को इनविजिलेटर को सौंप दी और परीक्षा भवन से बाहर चले गये।

Commercial Application