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Yes, the IELTS test fee is same in both ways. Only the payment mode is different. Online you pay with card or UPI, and offline you pay with cash, DD or bank slip.

The overall average package offered at IFIM School of Management placement in 2024 stood at INR 8.95 LPA. Check out the table for reference:

Online IELTS registration is faster and easy. You can book from home in 10–15 mins. But if you don't have internet or need help, then offline method is also good.

The IFIM School of Management has been providing successful placements. Check out the table below to know the overall key highlights of the IFIM School of Management placement in 2024:

It is better to apply at least 2 or 3 weeks before your IELTS test. If you wait too late, your preferred date or centre may not be available for offline or even online IELTS registration.

The IFIM School of Management placement report is out on the official website for the year 2024. Check out the tabulated data given below to know more about the overall IFIM School of Management placement in 2024:

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

Sol:

$\begin{array}{l}Giventhatx=asin2t\left(1+cos2t\right),y=bcos2t\left(1-cos2t\right)\\ Differentiatingboththeparametricfunctionsw.r.t.t,weget\\ \Rightarrow \frac{dx}{dt}=a\left[sin2t.\frac{d}{dt}\left(1+cos2t\right)+\left(1+cos2t\right).\frac{d}{dt}\left(sin2t\right)\right]\\ \Rightarrow =a\left[sin2t.\left(-sin2t\right).2+\left(1+cos2t\right)\left(cos2t\right).2\right]\\ \Rightarrow =a\left[-2si{n}^{2}2t+2cos2t+2co{s}^{2}2t\right]\\ \Rightarrow =a\left[2\left(co{s}^{2}2t-si{n}^{2}2t\right)+2cos2t\right]\\ \Rightarrow =a\left[2cos4t+2cos2t\right]\left[?cos2x=co{s}^{2}x-si{n}^{2}x\right]\\ \Rightarrow =2a\left[cos4t+cos2t\right]\\ Now,y=bcos2t\left(1-cos2t\right)\\ \Rightarrow \frac{dy}{dt}=b\left[cos2t.\frac{d}{dt}\left(1-cos2t\right)+\left(1-cos2t\right).\frac{d}{dt}\left(cos2t\right)\right]\\ \Rightarrow =b\left[cos2t.\left(sin2t\right).2+\left(1-cos2t\right)\left(-sin2t\right).2\right]\\ \Rightarrow =b\left[2sin2tcos2t-2sin2t+2sin2tcos2t\right]\\ \Rightarrow =b\left[sin4t-2sin2t+sin4t\right]\left[?sin2x=2sinxcosx\right]\\ \Rightarrow =b\left[2sin4t-2sin2t\right]=2b\left(sin4t-sin2t\right)\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2b\left[sin4t-sin2t\right]}{2a\left[cos4t+cos2t\right]}=\frac{b}{a}\left[\frac{sin4t-sin2t}{cos4t+cos2t}\right]\\ Putt=\frac{\pi }{4}\\ \therefore {\left(\frac{dy}{dx}\right)}_{att=\frac{\pi }{4}}=\frac{b}{a}\left[\frac{sin4\left(\frac{\pi }{4}\right)-sin2\left(\frac{\pi }{4}\right)}{cos4\left(\frac{\pi }{4}\right)+cos2\left(\frac{\pi }{4}\right)}\right]=\frac{b}{a}\left[\frac{sin\pi -sin\frac{\pi }{2}}{cos\pi +cos\frac{\pi }{2}}\right]\\ \text{\&thi}\end{array}$

More than 200 companies took part in the IFIM School of Management placements 2024. Check out the table for reference:

The IELTS fee is INR 17,000 for paper-based or computer-based test. This is same for both online and offline IELTS registration. 

Yes, someone else can submit your IELTS documents for you. But make sure all the forms are filled, and your passport copy and payment proof are correct.

To get into IHM Hajipur for B.Sc. in Hospitality and Hotel Administration, candidate must obtain a valid rank as per the cutoff list released. As pet the IHM Hajipur cutoff 2025, the round 1 cutoff for the General AI category candidates was 7739. Therefore, the lowest rank to get into IHM Hajipur is 7739; and student must obtain rank of or below 7739 to get into the college. 

The tuition fee for the BPharm course at Faculty of Pharmacy, MIT WPU is approximately INR 9.4 Lakh. This does not include additional costs such as security deposit, hostel charges, and mess fees. The total fee structure may vary and is subject to change.

IFIM School of Management provides good placements to its graduating students. Check out the table below to know the overall key highlights of the IFIM School of Management placement in 2024:

If you go to IDP office to book IELTS slot in offline mode, soon after the payment of IELTS fees, they will give you a confirmation letter right away. If you send by courier, you will get SMS or email confirmation once IDP checks your documents.

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

Sol:

$\begin{array}{l}Giventhatx=e^{cos2t},y=e^{sin2t}\\ \Rightarrow cos2t=logxandsin2t=logy\\ Differentiatingboththeparametricfunctionsw.r.t.t,weget\\ \Rightarrow \frac{dx}{dt}=e^{cos2t}.\frac{d}{dt}\left(cos2t\right)\\ \Rightarrow =e^{cos2t}\left(-sin2t\right).\frac{d}{dt}\left(2t\right)\\ \Rightarrow =e^{cos2t}\left(-sin2t\right).2=-2e^{cos2t}sin2t\\ Now,y=e^{sin2t}\\ \Rightarrow \frac{dy}{dt}=e^{sin2t}.\frac{d}{dt}\left(sin2t\right)\\ \Rightarrow =e^{sin2t}\left(cos2t\right).\frac{d}{dt}\left(2t\right)\\ \Rightarrow =e^{sin2t}\left(cos2t\right).2\\ \Rightarrow =2e^{sin2t}cos2t\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2e^{sin2t}cos2t}{-2e^{cos2t}sin2t}=\frac{e^{sin2t}cos2t}{-e^{cos2t}sin2t}\\ =\frac{ycos2t}{-xsin2t}=\frac{ylogx}{-xlogy}\left[\begin{array}{l}?cos2t=logx\\ sin2t=logy\end{array}\right]\\ Hence,\frac{dy}{dx}=\frac{ylogx}{-xlogy}\end{array}$

IFIM School of Management is known for its excellent placement record. Check out the tabulated data given below to know more about the overall IFIM School of Management placements in 2024:

For lateral entry at MIT WPU for BPharm, students must have a minimum 50% aggregate in a DPharm course from an institution approved by the Pharmacy Council of India under Section 12 of the Pharmacy Act. Reserved category and persons with disabilities need 45%.

Yes, you can do IELTS offline application. You can visit any IDP office, go to an IDP partner centre, or send your documents by courier to book your IELTS test.

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

Sol:

$\begin{array}{l}Giventhatx=\frac{1+logt}{t^2},y=\frac{3+2logt}{t}\\ Differentiatingboththeparametricfunctionsw.r.t.t,weget\\ \Rightarrow \frac{dx}{dt}=\frac{t^2.\frac{d}{dt}\left(1+logt\right)-\left(1+logt\right).\frac{d}{dt}\left(t^2\right)}{t^4}\\ \Rightarrow =\frac{t^2.\left(\frac{1}{t}\right)-\left(1+logt\right).2t}{t^4}\\ \Rightarrow =\frac{t-\left(1+logt\right).2t}{t^4}\\ \Rightarrow =\frac{t\left[1-2-2logt\right]}{t^4}=\frac{\left[-1-2logt\right]}{t^3}\\ \Rightarrow \frac{dx}{dt}=\frac{-\left(1+2logt\right)}{t^3}\\ Now,y=\frac{3+2logt}{t}\\ \Rightarrow \frac{dy}{dt}=\frac{t.\frac{d}{dt}\left(3+2logt\right)-\left(3+2logt\right).\frac{d}{dt}\left(t\right)}{t^2}\\ \Rightarrow =\frac{t.\left(\frac{2}{t}\right)-\left(3+2logt\right).1}{t^2}\\ \Rightarrow =\frac{2-3-2logt}{t^2}\\ \Rightarrow =\frac{-\left(1+2logt\right)}{t^2}\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-\left(1+2logt\right)}{t^2}}{\frac{-\left(1+2logt\right)}{t^3}}=\frac{t^3}{t^2}=t\\ Hence,\frac{dy}{dx}=t\end{array}$