Ask & Answer: India's Largest Education Community

Sagar Institute of Research and Technology is known for its engineering curriculum and approved by AICTE. The flagship programmes offered at Sagar Institute of Research and Technology Bhopal are BTech and MBA. The college offers these programmes to students at the UG and PG levels. Further, the college has also been accredited by NAAC with an A grade.

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$  

Coefficient of x^-1 ⇒ r = 10 ⇒ m = ¹⁵C₁₀ $\cdot 2^5$  

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$               

now mn² = ¹⁵C_r   $\cdot 2^r$

⇒ r = 5

For admission to SIRT Bhopal, candidates must register on the official website first and also enrol for the DTE Counseling from the official website. The application process is conducted online; however, all other steps are carried forward offline.

Here are the steps to register for SIRT Bhopal admissions.

Step 1: Visit the official website of the Sagar Institute of Research and Technology, Bhopal

Step 2: Scroll to the end of the page and look for ‘Register Now. ’

Step 3: Click on ‘Register Now’ and register for the candidature for the preferred courses.

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (2, 2)

Given a > b

Area common to x2 + y2 $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$

is $\pi a^2-\pi ab=30\pi ..............\left(i\right)$

Similarly $\pi ab-\pi b^2=18\pi .................\left(ii\right)$

Equation (i) and equation (ii) $\frac{a}{b}=\frac{5}{3}$

Equation (i) + equation (ii) $a^2-b^2=48$

a² = 75, b² = 27

International students are required to demonstrate their English language proficiency by tests such as IELTS or TOEFL or PTE. 

Given below are the minimum scores required in these ELP tests: 

sin x = 1 – sin² x

sin x = $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$

draw y = sin x

y = $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

First common term to both AP’s is 9

t₇₈ of $\left(3,6,9,......\right)=78\times 3=234$

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$

n^th common term $\le 234$

9 + (n – 1) 12 $\le$ 234

n < $\frac{237}{12}\Rightarrow n=19$

Now sum of 19 terms with a = 9, d = 12

$=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$

 $\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$

= $e^{2x}\cdot {\left(x-1\right)}^2$

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

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

y (2) = $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$

y (3) = $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$

Yes, BTech programme at Sagar Institute of Research and Technology is good as it is known for its engineering curriculum and is approved by AICTE. The college is accredited by the NBA for the year 2022-2025 for Computer Science Engineering, Electronics and Communications, and Computer Science and Information Technology. Further, the college has also been accredited by NAAC with an A grade. 

In order to obtain a seat in IHM Goa, one must apply, both online and offline and satisfy the eligibility requirements laid out by the course. In BSc, you will have to pass NCHMCT JEE and go to its counselling. In Diploma and certificate courses admission is on merit basis.

Data contradiction.

$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$

International students at this university go through intense competition, as the University of Hamburg acceptance rate stands at 36% (unofficial sources), meaning that out of every 100 applicants, only 36 were rolled out an offer letter.

The University of Hamburg is committed to maintaining a high-quality student body that is best qualified and most likely to succeed in its meticulous programs.

Apart from this, the university follows a holistic process to choose candidates. Applications are considered individually. A strong record of academics is required to get into the university in the first place.

For taking admission in the MSc course at Ramakrishna Mission Vivekananda Centenary College, candidates must fulfil the eligibility criteria required for the course. This is- male candidates who have passed or appeared in the BSc (Honours) in the corresponding subject during the year 2023 or 2024 or 2025. On the basis of merit of this graduation, they are selected for admission. 

Candidates must have prepared documents such as a Class 10 and Class 12 marksheet and certificate, recent passport-size photograph, Aadhar card, and caste certificate (in case of reserved categories). In the online form it is mandatory to upload these documents after scanning them. Acquiring clarity of the documents and proper format is the key to successful application

Yes, you can pursue a B.Ed from IGNOU. They offer a very respected course through distance learning which makes it accessible for students from all over India.
Below mentioned is a quick overview that the interested candidates can check:

• Eligibility: Hold a Bachelor's degree in any discipline with at least 50 percent marks.
• Entrance Exam: You need to take the IGNOU B.Ed Entrance Exam that is conducted once a year.
• Duration: The programme is 2 years long but you have a maximum of 5 years to complete it.
• Specializations: Choose from various specializations such as English, Hindi, Mathematics, Social Sciences etc.

The NCHMCT JEE 2025 application fee for General, OBC-NCL, and Gen-EWS candidates is INR 1,000, while it is INR 450 for SC, ST, PwD, and transgender candidates. The fee must be paid online after uploading the required documents. The application process is not complete until the payment is successfully submitted. Candidates should keep a receipt for reference.