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(a) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(b) The set of quantum numbers is possible.
(c) The set of quantum numbers is not possible because, for n = 1, l cannot be equal to 1. It can have 0 value.
(d) The set of quantum numbers is possible.
(e) The set of quantum numbers is not possible because, for n = 3, l cannot be 3.
(f) The set of quantum numbers is possible

Sasurie College of Engineering provides good placements to its graduating BE students with packages reaching up to INR 4.7 LPA. Check out the table below to know the key highlights of the Sasurie College of Engineering placements for BE/BTech between 2023- 2025:

Kindly go through the solution

Yes, with 85%, admissions are possible in LSR. However, students must qualify the entrance exams and look for quota options if possible as the competition for this college is quite high. Look for options like admissions under Management quota, sports Quota, and others. For other details, please visit the official website of the college. 

Yes, the Digital SAT exam fee costs around INR 11k and above for Indian applicants. The fee can change as per College Board discretion. Since it is an expensive exam, retaking it should be considered. 

Candidates interested to take admission to IILET for 5-year integrated LLB courses can register for the exam through the website. Candidates will have to enter the following details during registration- 

• Candidate's Name 
• Mobile Number 
• Select the course - BA LLB (Hons.)/ BBA LLB (Hons.)/ BBA LLB (Hons.) in Global and Transnational Studies
• Enter a valid Email ID 
• Current City 
• Reference Institute 

The Sasurie College of Engineering has had a decent placement record. Check out the tabulated data given below to know more about Sasurie College of Engineering placements for BE/BTech during 2023- 2025:

The Indore Institute of Law (IIL) conducts an entrance exam known as IILET for admissions to its five-year integrated LLB courses. IILET is conducted in a paper-based test mode for a duration of two hours. The exam has 150 MCQs carrying 1 mark each. 

Let I $={\displaystyle ?f^{?}(ax+b)·[f]{(ax+b)}^n}dx.$

$=\frac{1}{a}{\displaystyle ?[f]{(ax+b)}^n*\left[a{f}^{?}(ax+b)\right]}dx.$

Putting f (ax + b) = t.

$f^{?}(ax+b)\frac{d}{dx}(ax+b)=\frac{dt}{dx}$

af'  (ax + b) dx = dt

$?=\frac{1}{a}{\displaystyle ?t^{x}dt}$

$=\frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]$

$=\frac{1}{a(x+1)}[f]{(ax+b)}^{n+1}+\; C$

Let I$={\displaystyle \int e^{3logx}{\left(x^4+1\right)}^{-1}dx}$

$={\displaystyle \int \frac{e^{log{x}^3}}{\left(x^4+1\right)}dx}$

$={\displaystyle \int \frac{x^3}{x^4+1}}dx=\frac{1}{4}{\displaystyle \int \frac{4x^3}{x^4+1}}dx$

$=\frac{1}{4}log\left|x^4+1\right|+\; C$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

Specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or 1 Kelvin). That is why it is an intensive property which does not depend on mass.

No, Indian students or international students cannot pay SAT fee though EMI or through any part-payment option. Applicants have to pay full SAT fee in one go when registering for SAT exam on College Board's website.

Let I = $\int co{s}^{3}x\cdot e^{logsinx}dx.$

$={\displaystyle \int co{s}^{3}xsinxdx}\left\{?e^{logx}=x\right\}$

Putting cos x = t =>-sin x dx = dt =>sin x dx = -dt.

$I\; ={\displaystyle \int t^3(-dt)}=-{\displaystyle \int t^{3}dt}=-\frac{t^4}{4}+C=-\frac{co{s}^{4}x}{4}+C.$

Indian students planning to give SAT exam to study in India and come from a low family income can apply for Indian Scholars Program Scholarship and get a fee waiver. However, those appearing for SAT to study abroad, cannot take SAT exam for free. 

Let I = $\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}dx$

Putting x²= y then,

1 - A (y + 4) + 1B (y + 1)

Comparing the co-efficient,

A + B = 0 _____ (1)

4A + B = 1 ______ (2)

Equation (2) - (1),

4A + B - A - B = 1 - 0

3A = 1 =>A = $\frac{1}{3}.$

and B = - A = -$\frac{1}{3}.$

$\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{3}\frac{1}{(y+1)}-\frac{1}{3}\frac{1}{(y+4)}$

$=\frac{1}{3}\frac{1}{\left(x^2+1\right)}-\frac{1}{3}\frac{1}{\left(x^2+4\right)}$

$\therefore I=\frac{1}{3}\int \frac{dx}{x^2+1}-\frac{1}{3}\int \frac{dx}{x^2+2^2}$

$=\frac{1}{3}ta{n}^{-1}x-\frac{1}{6}ta{n}^{-1}\frac{x}{2}+\; C$

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i)

Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached.

Let I$=\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}dx$

Putting e^x = t=>e^xdx = dt.

$=\int \frac{dt}{(1+t)(2+t)}=\int \frac{(t+2)-(t+1)}{(t+1)(t+2)}dt$

$=\int \frac{dt}{t+1}-\int \frac{dt}{t+2}$

$=log|t+1|-log|t+2|+C$

$=log\left|\frac{t+1}{t+2}\right|+C.$

$=log\left[\frac{e^x+1}{e^x+2}\right]+C.$