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IIT Palakkad MSc cutoff 2025 has been released for admission to different specialisations. The cutoff was released as opening and closing ranks for different categories of students belonging to the AI quota. During the IIT Palakkad JAM cutoff 2025, the first round closing ranks varied between 478 and 609 for General AI category students. Of all the MSc specialisations, the most competitive was the MSc in Mathematics. For the rest of the other MSc specialisations, the IIT JAM first and last round closing ranks are tabulated below for the General AI quota:

B.Sc Perfusion Technology: 3-year course, fees INR 50,000– INR 1L/year. Need 12th pass with PCB. Offered in medical colleges, the entrance exam may be applied for.

There are multiple components included in the total fee structure. SBBS University BCom tuition fee amount is INR 86,650. Selected candidates have to pay the first semester of prescribed admission fees for seat confirmation. The university reserves the right to revise the fee structure in an academic year. It must be noted that the given amount is as per the official website/ sanctioning body. It is still subject to changes and hence, is indicative.

Smt. Kishoritai Bhoyar College of Pharmacy offers various courses at the UG and PG level. The selection criteria are merit-based or entrance based as per course requirement. The application process at Smt. KBCP is conducted online. To secure a seat at Smt. KBCP., students can check the following steps presented below:

1. Visit the official website and complete the online application process. 
2. Candidates can also apply online for various courses.
3. Ensure eligibility and selection criteria for all programme.
4. The selection criteria is merit or entrance-based.
5. Upon selected, pay the specified fee to confirm and secure your seat.

PTT courses provide you with the skills and information required to be a successful primary school teacher. Learning about child development, classroom management, and teaching approaches are all part of this.

If you enjoy dealing with children and assisting them in their learning, a PTT course may be a suitable choice for you. Qualified primary school teachers are in high demand.

Teaching is a difficult yet rewarding career. Primary school instructors have a significant impact on the lives of their children. They assist youngsters in developing the academic and social skills required for success in school and in life.

Self-driven placement: alumni secure jobs through personal networking or off-campus applications.

Entry-level roles: such as tutoring, translating, research, NGOs, or journalism—often around ?8–10k/month initially.

Further education: many pursue postgraduate degrees (e.g., M.A., law, civil services) rather than immediate job placements.

You can use Shiksha Free IELTS Classes, IDP official website, and British Council practice tests. Also, on YouTube, many free videos are there with IELTS tips, tricks, and sample answers.

If you study about 2 to 3 hours daily, it's enough. Give time to all 4 parts: Listening, Reading, Writing, Speaking. You can change subjects every day so it don't feel boring.

You need the Cambridge books for practice tests. You need a timer or watch to stop yourself in 60 minutes. Also watch some YouTube videos to understand how to solve tricky questions. Consistancy is key.

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \left(1+sin\theta \right)\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1+cos\theta \hfill \end{array}\right|\\ C_1\to C_1-C_2,C_2\to C_2-C_3\\ =\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -sin\theta \hfill & \hfill sin\theta \hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -cos\theta \hfill & \hfill 1+cos\theta \hfill \end{array}\right|\\ Expandingalong{C}_3\\ =1\left|\begin{array}{cc}\hfill -sin\theta \hfill & \hfill sin\theta \hfill \\ \hfill 0\hfill & \hfill -cos\theta \hfill \end{array}\right|=sin\theta cos\theta -0=sin\theta cos\theta \\ =\frac{1}{2}.2sin\theta cos\theta =\frac{1}{2}sin2\theta \\ =\frac{1}{2}\times 1=\frac{1}{2}\left[Maximumvalueofsin2\theta =1\right]\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\underset{x\to 4^{-}}{lim}f\left(x\right)=\frac{x-4}{\left|x-4\right|}+a=\underset{h\to 0}{lim}\frac{4-h-4}{\left|4-h-4\right|}+a\\ =\underset{h\to 0}{lim}\frac{-h}{h}+a=-1+a\\ \underset{x\to 4}{lim}f\left(x\right)=a+b\\ \underset{x\to 4^{+}}{lim}f\left(x\right)=\frac{x-4}{\left|x-4\right|}+b=\underset{h\to 0}{lim}\frac{4+h-4}{\left|4+h-4\right|}+b\\ =\underset{h\to 0}{lim}\frac{h}{h}+b=1+b\\ Asthefunctioniscontinuousatx=4.\\ \therefore \underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{x\to 4}{lim}f\left(x\right)=\underset{x\to 4^{+}}{lim}f\left(x\right)\\ -1+a=a+b=1+b\\ \therefore -1+a=a+b\Rightarrow b=-1\\ 1+b=a+b\Rightarrow a=1\\ Hence,thevalueofa=1andb=-1.\end{array}$

Yes, many students can prepare for IELTS at home without any coaching. You just need books or online material, a simple plan, and practice every day. If you try hard and follow your plan, you can get good IELTS score by yourself.

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}True.\\ Giventhat\Delta =\left|\begin{array}{ccc}\hfill a\hfill & \hfill p\hfill & \hfill x\hfill \\ \hfill b\hfill & \hfill q\hfill & \hfill y\hfill \\ \hfill c\hfill & \hfill r\hfill & \hfill z\hfill \end{array}\right|=16\\ L.H.S.{\Delta }_1=\left|\begin{array}{ccc}\hfill p+x\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill q+y\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill r+z\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill 2p+2x+2a\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill 2q+2y+2b\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill 2r+2z+2c\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\\ =2\left|\begin{array}{ccc}\hfill p+x+a\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill q+y+b\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill r+z+c\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\left[Taking2commonfrom{C}_1\right]\\ C_1\to C_1-C_2\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill q\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill r\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\\ C_3\to C_3-C_1\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill a+x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill b+y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill c+z\hfill & \hfill c\hfill \end{array}\right|\\ Splittingup{C}_2\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill b\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill c\hfill & \hfill c\hfill \end{array}\right|+2\left|\begin{array}{ccc}\hfill p\hfill & \hfill x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill z\hfill & \hfill c\hfill \end{array}\right|=2\left(0\right)+2\left|\begin{array}{ccc}\hfill p\hfill & \hfill x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill z\hfill & \hfill c\hfill \end{array}\right|\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill z\hfill & \hfill c\hfill \end{array}\right|\Rightarrow 2\left|\begin{array}{ccc}\hfill a\hfill & \hfill p\hfill & \hfill x\hfill \\ \hfill b\hfill & \hfill q\hfill & \hfill y\hfill \\ \hfill c\hfill & \hfill r\hfill & \hfill z\hfill \end{array}\right|\left(C_1\leftrightarrow C_{3}and{C}_2\leftrightarrow C_3\right)\\ =2\times 16=32\end{array}$

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill x+a\hfill & \hfill p+u\hfill & \hfill l+f\hfill \\ \hfill y+b\hfill & \hfill q+v\hfill & \hfill m+g\hfill \\ \hfill z+c\hfill & \hfill r+w\hfill & \hfill n+h\hfill \end{array}\right|\\ Splittingup{C}_1\\ =\left|\begin{array}{ccc}\hfill x\hfill & \hfill p+u\hfill & \hfill l+f\hfill \\ \hfill y\hfill & \hfill q+v\hfill & \hfill m+g\hfill \\ \hfill z\hfill & \hfill r+w\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill a\hfill & \hfill p+u\hfill & \hfill l+f\hfill \\ \hfill b\hfill & \hfill q+v\hfill & \hfill m+g\hfill \\ \hfill c\hfill & \hfill r+w\hfill & \hfill n+h\hfill \end{array}\right|\\ Splittingup{C}_{2}inboth\text{\hspace{0.17em}determinants}\\ =\left|\begin{array}{ccc}\hfill x\hfill & \hfill p\hfill & \hfill l+f\hfill \\ \hfill y\hfill & \hfill q\hfill & \hfill m+g\hfill \\ \hfill z\hfill & \hfill r\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill x\hfill & \hfill u\hfill & \hfill l+f\hfill \\ \hfill y\hfill & \hfill v\hfill & \hfill m+g\hfill \\ \hfill z\hfill & \hfill w\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill a\hfill & \hfill p\hfill & \hfill l+f\hfill \\ \hfill b\hfill & \hfill q\hfill & \hfill m+g\hfill \\ \hfill c\hfill & \hfill r\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill a\hfill & \hfill u\hfill & \hfill l+f\hfill \\ \hfill b\hfill & \hfill v\hfill & \hfill m+g\hfill \\ \hfill c\hfill & \hfill w\hfill & \hfill n+h\hfill \end{array}\right|\\ Similarlybysplitting{C}_{3}ineach\text{\hspace{0.17em}determinants},wewillget\mathrm{8determinants}.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{x}{\left|x\right|+2x^2}=\underset{h\to 0}{lim}\frac{0-h}{\left|0-h\right|+2{\left(0-h\right)}^2}\\ =\underset{h\to 0}{lim}\frac{-h}{h+2h^2}=\underset{h\to 0}{lim}\frac{-h}{h\left(1+2h\right)}\\ =\underset{h\to 0}{lim}\frac{-1}{1+2h}=\frac{-1}{1+2\left(0\right)}=-1\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{x}{\left|x\right|+2x^2}=\underset{h\to 0}{lim}\frac{0+h}{\left|0+h\right|+2{\left(0+h\right)}^2}\\ =\underset{h\to 0}{lim}\frac{h}{h+2h^2}=\underset{h\to 0}{lim}\frac{h}{h\left(1+2h\right)}=\frac{1}{1+0}=1\\ As\underset{x\to 0^{-}}{lim}f\left(x\right)\ne \underset{x\to 0^{+}}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=0regardlessthechoiceofk.\end{array}$

Yes, candidates belonging to the General AI quota with a cutoff rank of 500 can get an admission to MSc courses at IIT Tirupati. However, please note that the cutoff ranks vary for different rounds, specialisations, and categories. Therefore, admission will be granted to only those MSc specialisations whose Round 1 closing rank is above 500. Tabulated below are the IIT JAM first round closing ranks for the General AI category students: 

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill sinA+cosB\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill sinB+cosB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill sinC+cosB\hfill \end{array}\right|\\ Splittingup{C}_3\\ =\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill sinA\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill sinB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill sinC\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill cosB\hfill \end{array}\right|\\ =0+\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill cosB\hfill \end{array}\right|\left[?C_{1}and{C}_{3}areidentical\right]\\ =cosAcosB\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill sinB\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill sinC\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|\left[TakingcosAandcosBcommonfrom{C}_{2}and{C}_{3}respectively\right]\\ =cosAcosB\left(0\right)\left[?C_{2}and{C}_{3}areidentical\right]\\ =0\end{array}$