Ask & Answer: India's Largest Education Community

ISBM offers a variety of courses at the UG, PG, and Diploma levels. The fee structure for the programmes have a variety of components such as the hostel fee, admision deposit, and the tuition fee. Check out the table below to learn more about the fee for the various programmes at ISBM University: 

RRB Group D mock tests are offered in multiple languages too including Hindi, English, Tamil, Telugu, Bengali, etc., just like the actual exam. The question paper of RRB Group D is also framed in the regional languages.

There is no limit in the number of attempts in the IBPS PO exam. But important is to confirm the standards given in the notification. Candidate can give the exam as many times but the age limit is from 20 to 30 years. That is in between he or she can appear in the exam as many times they want.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Weknowthatanysquarematrixcanbeexpressedasthesumofsymmetricandskewsymmetricmatrixi.e.\\ A=\frac{1}{2}\left[A+A^{\text{'}}\right]+\frac{1}{2}\left[A-A^{\text{'}}\right]\\ SoP=\frac{1}{2}\left[\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right)+\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 2\hfill \end{array}\right)\right]\\ =\frac{1}{2}\left[\begin{array}{ccc}\hfill 2+2\hfill & \hfill 3+1\hfill & \hfill 1+4\hfill \\ \hfill 1+3\hfill & \hfill -1-1\hfill & \hfill 2+1\hfill \\ \hfill 4+1\hfill & \hfill 1+2\hfill & \hfill 2+2\hfill \end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right]\\ =\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 2\hfill & \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 2\hfill \end{array}\right]\\ P^{\text{'}}=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 2\hfill & \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 2\hfill \end{array}\right]=P\\ As{P}^{\text{'}}=P\therefore Pisasymmetricmatrix.\\ Now,Q=\frac{1}{2}\left[A-A^{\text{'}}\right]\\ =\frac{1}{2}\left[\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right)-\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 2\hfill \end{array}\right)\right]\\ =\frac{1}{2}\left[\begin{array}{ccc}\hfill 2-2\hfill & \hfill 3-1\hfill & \hfill 1-4\hfill \\ \hfill 1-3\hfill & \hfill -1+1\hfill & \hfill 2-1\hfill \\ \hfill 4-1\hfill & \hfill 1-2\hfill & \hfill 2-2\hfill \end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\\ =\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \end{array}\right]=-\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \end{array}\right]=-Q\\ AsQ=-Q\therefore Qisaskewsymmetricmatrix.\\ SoA=P+Q\\ A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 2\hfill & \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 2\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \end{array}\right]\\ =[\begin{array}{ccc}\hfill 2+0\hfill & \hfill 2+1\hfill & \hfill \frac{5}{2}-\frac{3}{2}\hfill \\ \hfill 2-1\hfill & \hfill -1+0\hfill & \hfill \frac{3}{2}+\frac{1}{2}\hfill \\ \hfill \frac{5}{2}+\frac{3}{2}\hfill & \hfill \frac{3}{2}\&min\hfill \end{array}\end{array}$

RRB Group D exam carries 100 questions. The duration of the test is 90 minutes. Candidates must know the syllabus. They should also understand the exam pattern.  Questions are of objective type.

Jyothy Institute of Commerce and Management delivers the MBA programme to the interested students.  It is important to meet the required eligibility criteria to develop deeper knowledge about the subject. The table below discusses the eligibility criteria:

The IBPS PO 2025 exam is conducted online that is Computer-Based Test. Prelims, Mains and Descriptive exam is conducted in a CBT mode.

Check out the table for the average packages offered sector-wise during the Amrita Business School placements:

Yes, the RRB Group D mock test is free.

In Germany, there is no course called BTech. Instead, students apply for:

• Bachelor of Engineering (B.Eng.), or

• Bachelor of Science (B.Sc.)

These are same as BTech and cover all the engineering subjects like Mechanical, Electrical, Civil, and Computer Science.

Answer: (a) A catalyst favours the formation of products

No, JEE Mains is not directly accepted in Germany. But, if you are an Indian student and you have:

• Finished Class 12 with required subjects

• Cleared JEE Advanced, or

• Done 1 year of college in India, or

• Completed a Studienkolleg (foundation course) in Germany

Then you can apply for engineering courses like B.Eng. or B.Sc. in Germany. You also need to show your English or German language skills, depending on your course.

Yes, Germany is very good for engineering studies. It has many top-ranked engineering universities, and most of them are tuition-free. You also get good job options after graduation.

Many German colleges work with big companies like BMW, Siemens, Bosch, so students get practical training and chances for internships. Also, the degrees from Germany are accepted worldwide.

Answer: (d) decreases

Answer: (a) increases

Yes, engineering colleges in Germany are mostly free if you apply to public universities. These universities don't charge tuition fees, even if you are an international student. But students have to pay a semester fee, which is between EUR 150 to EUR 300 every semester.

If you apply to a private university, then you have to pay tuition fees, which can be between EUR 9,000 – EUR 13,000 per year.

Some the top recruiters for Amrita School of Business placement are listed below:

(i) f(x)=sin x cos x

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{sin(x+h)cos(x+h)?sinxcosx}{h}$

$=\underset{h?0}{lim}\frac{1}{2h}*[2sin(x+h)cos(x+h)?2sinxcosx]$

$=\underset{h?0}{lim}\frac{1}{2h}[sin2(x+h)?sin2x]$

$=\underset{h?0}{lim}\frac{1}{2h}\left[2cos\frac{2(x+h)+2x}{2}sin\frac{2(x+h)?2x}{2}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[cos\left(2x+h\right)sinh\right]$

$=\underset{h?0}{lim}cos(2x+h)*\underset{h?0}{lim}\frac{sinh}{h}$

$=cos(2x+0)$

$=cos2x$

$\mathrm{(ii)\; f}(x)=secx$

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{1}{h}[sec(x+h)?secx]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{?2sin\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{(?1sin\left(\frac{2x+h}{2}\right)}{cos(x+h)cosx}*\underset{h?0}{lim}(?1)\frac{sinh/2}{h/2}$

$=\frac{sinx}{cosx?cosx}*1$

$=tanx?secx.$

(iii) Given f(x)=5 sec x+4 cosx.

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$\begin{array}{l}\\ =\underset{h?0}{lim}\frac{5sec(x+h)+4cos(x+h)?[5secx+4cosx]}{h}.\end{array}$

$=\underset{h?0}{lim}\frac{5}{h}[sec(x+h)?secx]+\underset{h?0}{lim}\frac{4}{h}[cos(x+h)?cosx]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h?x}{2}\right)\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)(cosx)}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{2x+h}{2}\right)sin\frac{h}{2}\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]?4\underset{h?0}{lim}sin\left(\frac{2x+h}{2}\right)\underset{h?0}{lim}\frac{sinh/2}{h/2}$

$=\frac{sin\left(\frac{2x+0}{2}\right)}{cos(x+0)cosx}*1?4sin\left(\frac{2x}{2}\right)$

$=5\frac{sinx}{cosx}?\frac{1}{cosx}?4sinx$

$=5tanx?secx?4sinx$

$\mathrm{(iv)\; Given\; f}(x)=cosecx$

$f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{1}{h}[cosec(x+h)?cosecx]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{1}{sin(x+h)}?\frac{1}{sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{sinx?sin(x+h)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)]}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{2x+h}{2}\right)sin(?h/2)]}{sin(x+h)sinx}\right]$