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Official notification of UCEED 2026 was released by October 01, 2025, by IIT Bombay on the official website. The dates are out and was conducted on January 18, 2026.

The course curriculum of BVoc course at Rustomjee Academy for Global Careers, Thane includes various topics depending upon the specialisation:-

• Basic Electrical Systems
• Automotive Systems
• Thermodynamics and Heat Transfer
• Mechanics of Machines, etc.

$\underset{x\to 0}{lim}\frac{\alpha x\left(1+x+\frac{x^2}{2}+....\right)-\beta \left(x-\frac{x^2}{2}+\frac{x^3}{3}+....\right)+\gamma x^2\left(1-x\right)}{x^3}=10$

For limit to exist a - b = 0 . (i), $\alpha +\frac{\beta }{2}+\gamma =0.........(ii)$

and $\frac{\alpha }{2}-\frac{\beta }{2}-\gamma =10$ . (iii)

Solving (i), (ii) and (iii) we get α = 6, β = 6 and $\lambda =-9\Rightarrow \alpha +\beta +\lambda =3$

  $C{H}_3-C{H}_2-C{H}_2-C-N{H}_2\to \pi -\mathcal{l}pconjugation$  

              $C{H}_3-C{H}_2-CH=CH-C{H}_2-N{H}_2\to$ No conjugation

          $C{H}_3-C{H}_2-O-CH=C{H}_2\to \pi -\mathcal{l}p$     conjugation

$a_{n+1}=\frac{a_{n+2}-a_n}{2}letp={\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}$

$\Rightarrow 64\frac{a_{n+2}}{8^{n+2}}=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$

$\Rightarrow 64{\displaystyle \sum _{n=1}^{\infty }\frac{a_{n+2}}{8^{n+2}}=}{\displaystyle \sum _{n=1}^{\infty }\frac{16a_{n+1}}{8^{n+1}}+}{\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}\Rightarrow 64\left(p-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(p-\frac{a_1}{8}\right)+p$

$\Rightarrow 64\left(p-\frac{1}{8}-\frac{1}{8^2}\right)=16\left(p-\frac{1}{8}\right)+p\Rightarrow 64p-8-1=16p-2+p$

Joining BVoc course at Rustomjee Academy for Global Careers, Thane is generally considered a decent option. Candidates must consider factors like infrastructure, faculty experience, curriculum, etc. while taking admission. They can apply online or offline for admission in BVoc course at Rustomjee Academy for Global Careers, Thane.

The following specialisations are offered under BVoc course at Rustomjee Academy for Global Careers, Thane:-

• Automobile
• Hotel and Hospitality Management
• Electrical Engineering
• Construction Technology

MBA admission at MSRUAS is entrance-based. Candidates are selected based on their performance in the KMAT Entrance Exam. Only those with a valid KMAT scorecard are eligible for the counseling round, where final admission is confirmed after document verification and payment of the required fees within the given timeline.  

The final stage of MBA admission at MSRUAS involves counseling, where shortlisted candidates undergo document verification. After successful verification, they must confirm their admission by paying the tuition fee (ranging from INR 2 Lacs to INR 9 Lakh), along with applicable registration, security, and other charges within the specified deadline.

While specific hours aren't listed, it is recommended to call the helpline during regular office hours (10:00 AM to 5:00 PM on working days).

Official address of WBJEEB for PUBDET 2025 related queries is:

Address: Controller of Examinations, West Bengal Joint Entrance Examinations board, AQ-13/1, Sector –V, Salt Lake City, Kolkata-700091

Phone No.: 1800-3450-050

Candidates who have completed Class 12 with a 50% aggregate with Physics and Mathematics/English as compulsory subjects can apply for NFSC BTech programme. Besides, aspirants must also qualify for the entrance exam to be accepted for admission. The institute admits students based on merit + JEE Main Paper one + the weightage ratio for these will be decided by NFSC's competent authority. Post merit list, the institute will verify documents and conduct a medical examination. Shortlisted aspirants need to come for document verification and pay the National Fire Service College fee to confirm their seats at the institute.

  $P\left(\overline{A}\cap B\right)+P\left(A\cap \overline{B}\right)=1-k,$

$P\left(A\cap B\cap C\right)=k^2$             

$P\left(A\right)+P\left(B\right)-2P\left(A\cap B\right)=1-k$ .(i)

$P\left(B\right)+P\left(C\right)-2P\left(B\cap C\right)=1-k$ .(ii)

$P\left(A\right)+P\left(C\right)-2P\left(A\cap C\right)=1-2k$ .(iii)

Adding (i), (ii) and (iii) we get $P\left(A\cup B\cup C\right)=\frac{-4k+3}{2}+k^2$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{2k^2-4k+2+1}{2}=\frac{2{\left(k-1\right)}^2+1}{2}\Rightarrow P\left(A\cup B\cup C\right)>\frac{1}{2}$

1, 2 and 3 talks about a black hole and its distinctive features. However, 4^th sentence talks about a star and its life.

National Fire Service College offers a four-year BTech programme at the UG level. Candidates who have completed Class 12 with a 50% aggregate with Physics and Mathematics/English as compulsory subjects can apply for NFSC BTech programme. The institute admits students based on merit + JEE Main Paper one + the weightage ratio for these will be decided by NFSC's competent authority.