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ACT allows you to make changes to your ACT exam appointment, if you cannot take the test on your chosen date, time and location anymore; or if you sense any discrepancy in the information you entered on your application. 

Here are the type of changes that you can make to your ACT exam registration:

1. Changing your test date
2. Changing your test centre
3. Adding/Removing Writing section
4. Updating your personal information
5. Adding/Updating colleges of your choice
6. Requesting additional score reports
7. Correcting your high school code

There are core topics that are covered in SLS PDEU BBA (H) in Marketing programme. The course curriculum is inclusive of semester-wise syllabus. Mentioned below are the topics covered in the first semester:

• Basic Mathematics
• English Communication
• Science in Everyday Life
• Introduction to Computer Science
• Leadership
• Understanding of Theatrical Arts
• Foreign Language

Hi, the table below shows the private top MHT CET accepting BTech colleges in Maharashtra along with their total tuition fees:

Nehru Arts and Science College has affordable fee structure as comapred to other institutes. The fee structure includes multiple components such as tuition fee, caution fees, application fees, hostel fee, and others. Students must ensure that they pay the fees within the stipulated timeline. NASC fee structure may vary depending on several factors, including the course and course level.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivenfunctionisf\left(x\right)=4si{n}^{3}x-6si{n}^{2}x+12sinx+100\\ f^{\text{'}}\left(x\right)=12si{n}^{2}x.cosx-12sinxcosx+12cosx\\ =12cosx\left[si{n}^{2}x-sinx+1\right]\\ =12cosx\left[si{n}^{2}x+\left(1-sinx\right)\right]\\ ?1-sinx\ge 0andsi{n}^{2}x\ge 0\\ \therefore si{n}^{2}x+1-sinx\ge 0\left(whencosx>0\right)\\ Hence,f^{\text{'}}\left(x\right)>0whencosx>0i.e.,x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\\ So,f\left(x\right)isincreasingwherex\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)f^{\text{'}}\left(x\right)<0\\ whencosx<0i.e.,x\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right)\\ Hence,f\left(x\right)isdecreasingwhenx\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right)\\ As\left(\frac{\pi }{2},\pi \right)\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right)\\ So,f\left(x\right)isdecreasingin\left(\frac{\pi }{2},\pi \right)\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

No. Waitlisted or not-selected candidates will not receive any direct communication from the WWI institution. They must contact the admissions counsellor and take an update. 

ICT Mumbai emerged to be as the No. 1 Engineering college in Maharashtra accepting MHT Score. Listed below are some of the colleges along with their NIRF rankings 2025:

Hope this helps!

This is a Objective Type Questions as classified in NCERT Exemplar

Sol. 

$\begin{array}{l}Thegivenfunctionisy=x{\left(x-3\right)}^2\\ \frac{dy}{dx}=x.2\left(x-3\right)+{\left(x-3\right)}^2.1\\ \Rightarrow \frac{dy}{dx}=2x\left(x-3\right)+{\left(x-3\right)}^2\\ Forincreasinganddecreasing\frac{dy}{dx}=0\\ \therefore 2x\left(x-3\right)+{\left(x-3\right)}^2=0\\ \Rightarrow \left(x-3\right)\left(2x+x-3\right)=0\Rightarrow \left(x-3\right)\left(3x-3\right)=0\\ \Rightarrow 3\left(x-3\right)\left(x-1\right)=0\\ \therefore x=1,x=3\\ Thepossibleintervalsare\left(-\infty ,1\right),\left(1,3\right),\left(3,\infty \right)\\ Now\frac{dy}{dx}=\left(x-3\right)\left(x-1\right)\\ \Rightarrow For\left(-\infty ,1\right)=\left(-\right)\left(-\right)=\left(+\right)increasing\\ \Rightarrow For\left(1,3\right)=\left(-\right)\left(+\right)=\left(-\right)decreasing\\ \Rightarrow For\left(3,\infty \right)=\left(+\right)\left(+\right)=\left(+\right)increasing\\ Sothefunctiondecreasesin\left(1,3\right)or1<x<3\\ Hence,\text{\&thin}\end{array}$

SLS, Pandit Deendayal Energy University registration is open. Aspirants of the BBA (Hons) programmes can visit the official website to register and apply. Students shall ensure that they apply before the application end date. Moreover, all the information provided in the application should be updated and correct. 

The WWI results for B.Des will be declared within one week of the completion of the entrance exam.

IIITM Gwalior JEE Main cutoff 2025 was published for admission to multiple BTech course specialisations for various All India categories. For the General AI category, admissions were closed at a rank of 11201. For the OBC AI category, the closing rank for BTech CSE was 15140. Refer to the list below to know the closing rank for all other categories. 

• SC:  69345
• ST:  933
• EWS:  12345
• PWD: 129

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l}Giventhatf\left(x\right)=2x+cosx\\ f^{\text{'}}\left(x\right)=2-sinx\\ Since{f}^{\text{'}}\left(x\right)>0\forall x\\ Sof\left(x\right)isanincreasingfunction.\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

To secure a BSc seat at Nehru Gram Bharati University Kotwa, candidates must complete Class 12 in PCM/PCB and appear for the CUET exam. Admission is granted based on CUET scores and academic performance. Interested students should follow the official admission process detailed on the university's website.

The total seat intake for the BSc programme at Nehru Gram Bharati University Kotwa ranges between 300 and 800 seats. This includes seats across various specialisations, with the highest being for BSc Chemistry. Students are advised to check the official NGBU website for the most accurate and updated intake information.

The retesting restrictions of the ACT exam suggest that a student can only take the ACT exam once in a single testing window. Taking the ACT exam two or even three times in a single testing window will not be accepted by ACT. ACT, in such a case, will provide you with the scores only from your first test attempt and not for the remaining attempts. Note that even your ACT exam registration fee paid for the 2nd and 3rd attempts will not be refunded by ACT.

ACT suggests that if you wish to take the ACT exam multiple times, then you should choose the testing dates in different testing windows.  

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: 

$\begin{array}{l}Thegivenfunctionisf\left(x\right)=2x^3+9x^2+12x-1\\ f^{\text{'}}\left(x\right)=6x^2+18x+12\\ Forincreasinganddecreasing{f}^{\text{'}}\left(x\right)=0\\ \therefore 6x^2+18x+12=0\\ \Rightarrow x^2+3x+2=0\Rightarrow x^2+2x+x+2=0\\ \Rightarrow x\left(x+2\right)+1\left(x+2\right)=0\Rightarrow \left(x+2\right)\left(x+1\right)=0\\ \Rightarrow x=-2,x=-1\\ Thepossibleintervalsare\left(-\infty ,-2\right),\left(-2,-1\right),\left(-1,\infty \right)\\ Now{f}^{\text{'}}\left(x\right)=\left(x+2\right)\left(x+1\right)\\ \Rightarrow f^{\text{'}}{\left(x\right)}_{\left(-\infty ,-2\right)}=\left(-\right)\left(-\right)=\left(+\right)increasing\\ \Rightarrow f^{\text{'}}{\left(x\right)}_{\left(-2,-1\right)}=\left(+\right)\left(-\right)=\left(-\right)decreasing\\ \Rightarrow f^{\text{'}}{\left(x\right)}_{\left(-1,-\infty \right)}=\left(+\right)\left(+\right)=\left(+\right)increasing\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol. 

$\begin{array}{l}Thegivencurveare{x}^3-3x{y}^2+2=0\left(i\right)\\ and3x^{2}y-y^3-2=0\left(ii\right)\\ Differentiatingequation\left(i\right)w.r.t,x,weget\\ 3x^2-3\left(x.2y\frac{dy}{dx}+y^2.1\right)=0\\ \Rightarrow x^2-2xy\frac{dy}{dx}-y^2=0\Rightarrow 2xy\frac{dy}{dx}=x^2-y^2\\ \therefore \frac{dy}{dx}=\frac{x^2-y^2}{2xy}\\ SoSlopeliesofthecurve{m}_1=\frac{x^2-y^2}{2xy}\\ Differentiatingequation\left(ii\right)w.r.t,x,weget\\ 3\left[x^2\frac{dy}{dx}+y.2x\right]-3y^2.\frac{dy}{dx}=0\\ x^2\frac{dy}{dx}+2xy-y^2.\frac{dy}{dx}=0\Rightarrow \left(x^2-y^2\right)\frac{dy}{dx}=-2xy\\ \therefore \frac{dy}{dx}=\frac{-2xy}{x^2-y^2}\\ SoSlopeliesofthecurve{m}_2=\frac{-2xy}{x^2-y^2}\\ Now{m}_1\times m_2=\frac{x^2-y^2}{2xy}\times \frac{-2xy}{x^2-y^2}=-1\\ Sotheanglebetweenthecurvesis\frac{\pi }{2}.\\ Hence,thecorrectoptionis\left(c\right).\end{array}$