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Students must have scored at least 50% aggregate marks in PCM/PCB in Class 12 with Physics and Chemistry as compulsory subjects. Reserved category candidates from Maharashtra require a minimum of 45%. A valid score in NEET 2025, MHT-CET 2025, JEE 2025, or PERA 2025 is also required.

To register for IELTS exam log on to the IDP IELTS India website. You will see Register for IELTS, click on it. It will take you to the booking page. 

BPharm programme at Faculty of Pharmacy, MIT WPU is a four-year undergraduate course. The university also offers a lateral entry (LE) option for Diploma holders, which is a three-year course. Both are divided into six to eight semesters.

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

Sol:

$\begin{array}{l}Now,takingtany=\frac{2t}{1-t^2}\\ Differentiatingbothsidesw.r.t.t,weget\\ \frac{d}{dt}\left(tany\right)=\frac{d}{dt}\left(\frac{2t}{1-t^2}\right)\\ \Rightarrow se{c}^{2}y.\frac{dy}{dt}=\frac{\left(1-t^2\right).\frac{d}{dt}\left(2t\right)-2t.\frac{d}{dt}\left(1-t^2\right)}{{\left(1-t^2\right)}^2}\\ \Rightarrow se{c}^{2}y.\frac{dy}{dt}=\frac{2.\left(1-t^2\right)-2t.\left(-2t\right)}{{\left(1-t^2\right)}^2}\\ \Rightarrow se{c}^{2}y.\frac{dy}{dt}=\frac{2-2t^2+4t^2}{{\left(1-t^2\right)}^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2+2t^2}{{\left(1-t^2\right)}^2}\times \frac{1}{se{c}^{2}y}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{1}{1+ta{n}^{2}y}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{1}{1+{\left(\frac{2t}{1-t^2}\right)}^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{1}{\frac{{\left(1-t^2\right)}^2+4t^2}{{\left(1-t^2\right)}^2}}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{{\left(1-t^2\right)}^2}{1+t^4-2t^2+4t^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{{\left(1-t^2\right)}^2}{1+t^4+2t^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{{\left(1-t^2\right)}^2}{{\left(1+t^2\right)}^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2}{1+t^2}\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{2}{1+t^2}}{\frac{2}{1+t^2}}=1\\ Hence,\frac{dy}{dx}=1\end{array}$

MBBS coarse started in 2018 with annual intake of 50, from 2020 annual intake increased to 125, so currently annual intake in MBBS coarse is 125.

This is a multiple choice answer as classified in NCERT Exemplar

Auburn University provides 17 months of two-year MBA programs in specializations such as Information Systems, Hospitality Management, Economics, Finance, Real Estate Development, Business/ Marketing Education, Business Administration, Accountancy, and more. The tuition fees of these programs range between INR 19L and INR 33L. The basic eligibility includes a bachelor's degree from a recognised institution.

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

Sol:

$\begin{array}{l}Giventhatx=3cos\theta -2co{s}^3\theta ,y=3sin\theta -2si{n}^3\theta \\ Differentiatingboththegivenparametricfunctionsw.r.t.\theta \\ \frac{dx}{d\theta }=-3sin\theta -6co{s}^2\theta .\frac{d}{d\theta }\left(cos\theta \right)\\ =-3sin\theta -6co{s}^2\theta .\left(-sin\theta \right)\\ =-3sin\theta +6co{s}^2\theta .sin\theta \\ \frac{dy}{d\theta }=3cos\theta -6si{n}^2\theta .\frac{d}{d\theta }\left(sin\theta \right)\\ =3cos\theta -6si{n}^2\theta .cos\theta \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3cos\theta -6si{n}^2\theta .cos\theta }{-3sin\theta +6co{s}^2\theta .sin\theta }\\ =\frac{cos\theta \left(3-6si{n}^2\theta \right)}{sin\theta \left(-3+6co{s}^2\theta \right)}=\frac{cos\theta \left[3-6\left(1-co{s}^2\theta \right)\right]}{sin\theta \left[-3+6co{s}^2\theta \right]}\\ =cot\theta \left(\frac{3-6+6co{s}^2\theta }{-3+6co{s}^2\theta }\right)=cot\theta \left(\frac{-3+6co{s}^2\theta }{-3+6co{s}^2\theta }\right)=cot\theta \\ Hence,\frac{dy}{dx}=cot\theta \end{array}$

Cutoff rank for general category ranges around all India rank of 2000-2500 in 1st round, in 2nd and mop up round of counselling it can go around 3000-3200 for general category based upon number of vacant seats. For OBC category and EWS category, it can go around all India rank of 4500, for SC category around 11000 and for ST category around 20000-25000 based on number of vacant seats towards mop up round.

King George Medical College provides good placements to its graduating students. The students are given excellent placement support every year placing them in top companies. Check out the table below to know the key highlights of the King George Medical College placements 2024:

NOTE: The above data is obtained from the NIRF report 2025

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

Sol:

$\begin{array}{l}Giventhatx=e^{\theta }\left(\theta +\frac{1}{\theta }\right),y=e^{-\theta }\left(\theta -\frac{1}{\theta }\right)\\ Differentiatingboththeparametricfunctionsw.r.t.\theta \\ \frac{dx}{d\theta }=e^{\theta }\left(1-\frac{1}{{\theta }^2}\right)+\left(\theta +\frac{1}{\theta }\right).e^{\theta }\\ \frac{dx}{d\theta }=e^{\theta }\left(1-\frac{1}{{\theta }^2}+\theta +\frac{1}{\theta }\right)\Rightarrow e^{\theta }\left(\frac{{\theta }^2-1+{\theta }^3+\theta }{{\theta }^2}\right)\\ =e^{\theta }\frac{\left({\theta }^3+{\theta }^2+\theta -1\right)}{{\theta }^2}\\ y=e^{-\theta }\left(\theta -\frac{1}{\theta }\right)\\ \frac{dy}{d\theta }=e^{-\theta }\left(1+\frac{1}{{\theta }^2}\right)+\left(\theta -\frac{1}{\theta }\right).\left(-e^{-\theta }\right)\\ \frac{dy}{d\theta }=e^{-\theta }\left(1+\frac{1}{{\theta }^2}-\theta +\frac{1}{\theta }\right)\Rightarrow e^{-\theta }\left(\frac{{\theta }^2+1-{\theta }^3+\theta }{{\theta }^2}\right)\\ =e^{-\theta }\frac{\left(-{\theta }^3+{\theta }^2+\theta +1\right)}{{\theta }^2}\\ \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{e^{-\theta }\frac{\left(-{\theta }^3+{\theta }^2+\theta +1\right)}{{\theta }^2}}{e^{\theta }\frac{\left({\theta }^3+{\theta }^2+\theta -1\right)}{{\theta }^2}}\\ =e^{-2\theta }\left(\frac{-{\theta }^3+{\theta }^2+\theta +1}{{\theta }^3+{\theta }^2+\theta -1}\right)\\ Hence,\frac{dy}{dx}=e^{-2\theta }\left(\frac{-{\theta }^3+{\theta }^2+\theta +1}{{\theta }^3+{\theta }^2+\theta -1}\right)\end{array}$

Auburn University provides one-year to two-year master programs in specializations such as Polymer and Fiber Engineering; Building Construction; Industrial and Systems Engineering, Crop, Soil and Environmental Sciences; Probability and Statistics; Plant Pathology; Geology, and more.

The tuition fees of these programs range between INR 21L and INR 40L. The basic eligibility includes a bachelor's degree from a recognised institution.

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-mgL + mgL=0

As the point of application of the contact forces does not move hence work done by reaction forces will be zero.

Yes, King George Medical College has an impressive placement record. The table below provides the placement data for the past three years as per the NIRF Report 2025:

NOTE: The above data is obtained from the NIRF report 2025