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SLS, Pandit Deendayal Energy University offers entrance-based admission to its BBA (Hons) programme. Candidates seeking admission have to first check eligibility and then submit their applications. The application form can be submitted in an online mode. Students then have to appear for the entrance conducted by the university followed by a personal interview round. Additionally, the students are also selected based on merit in SAT India. Candidates who make it to the merit list can get their documents verified and pay the admission fee amount.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: 

$\begin{array}{l}Thegivencurveisx=t^2+3t-8andy=2t^2-2t-5\\ Differentiatingbothequationsw.r.t,t\\ \frac{dx}{dt}=2t+3and\frac{dy}{dt}=4t-2\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4t-2}{2t+3}\\ Now\left(2,-1\right)liesonthecurve\\ \therefore 2=t^2+3t-8\Rightarrow t^2+3t-10=0\\ \Rightarrow t^2+5t-2t-10=0\Rightarrow t\left(t+5\right)-2\left(t+5\right)=0\\ \Rightarrow \left(t+5\right)\left(t-2\right)=0\\ \therefore t=2,t=-5and-1=2t^2-2t-5\\ \Rightarrow 2t^2-2t-4=0\\ \Rightarrow t^2-t-2=0\Rightarrow t^2-2t+t-2=0\\ \Rightarrow t\left(t-2\right)+1\left(t-2\right)=0\Rightarrow \left(t+1\right)\left(t-2\right)=0\\ \Rightarrow t=-1andt=2\\ Sot=2iscommonvalue\\ \therefore Slope={\frac{dy}{dx}}_{x=2}=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

Students can start taking the ACT test as soon as they reach grade 6. Most students prefer to take the ACT exam for the first time during spring of their junior year and or the fall their senior year. This allows them to retake the exam if required, and make improvements in the areas of weaknesses. 

There is no specific information that has been released regarding the Surana College Post Graduate Departments MBA application deadline. However, the application for the MBA programme for the academic year 2025-27 is currently ongoing. Candidates with acceptable scores from an entrance examination are eligible to apply. To avoid missing any deadlines, candidates must stay updated with the upcoming events.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Equationofthecurveisy=e^{2x}\\ Slopeofthetangent\frac{dy}{dx}=2e^{2x}\Rightarrow {\frac{dy}{dx}}_{\left(0,1\right)}=2e^0=2\\ \therefore Equationoftangenttothecurveat\left(0,1\right)is\\ y-1=2\left(x-0\right)\\ \Rightarrow y-1=2x\Rightarrow y-2x=1\\ Sincethetangentmeetx-axis,wherey=0\\ \therefore 0-2x=1\Rightarrow x=-\frac{1}{2}\\ So,thePointis\left(-\frac{1}{2},0\right)\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

Is Nehru Arts and Science College is generally considered a Tier 2. The institute is approved by UGC and is ISO certified and  accredited with an A Grade by NAAC. The institute offers undergrdaute courses across various streams such as Microbiology, Food Science & Nutrition, and others. NASC offer good quality education, infrastructure, and placement opportunities. 

This is a Objective Type Questions as classified in NCERT Exemplar

Sol. 

$\begin{array}{l}Giventhaty=x^3-12x+18\\ Differentiatingbothsidesw.r.t.x,wehave\\ \Rightarrow \frac{dy}{dx}=3x^2-12\\ Sincethetangentsareparalleltox-axis,then\frac{dy}{dx}=0\\ \therefore 3x^2-12=0\Rightarrow x=\pm 2\\ \therefore y_{x=2}={\left(2\right)}^3-12\left(2\right)+18=8-24+18=2\\ y_{x=-2}={\left(-2\right)}^3-12\left(-2\right)+18=-8+24+18=34\\ \therefore Pointsare\left(2,2\right)and\left(-2,34\right)\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

For the admission cycle of 2024-25, the number of student applications received by the University of Central Florida crossed 55,100, whereas the number of applications accepted was around 24,600. This means that the UCF acceptance rate 2025 is 44.7%, making the admission criteria of the university moderately selective. 

When it comes to international enrollments, the number of applications received vs. number of applications accepted were 1,800 and 440 respectively, making the UCF acceptance rate for international students a mere 23.3%.

Yes, Yenepoya University, Bangalore Campus admissions 2025 are open for various courses. Interested students can visit the official website to apply for the respective courses. The university admits students based on the entrance exam called the Y-entrance exam for UG and PG courses. Further, the university offers courses in multiple streams such as Business and Management, Science, Arts, Design and Creativity, etc. 

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhaty\left(1+x^2\right)=2-x\dots \left(i\right)\\ Ifitcutsx-axis,theny-coordinateis0.\\ \therefore 0\left(1+x^2\right)=2-x\Rightarrow x=2\\ Putx=2inequation\left(i\right)\\ y\left(1+4\right)=2-2\Rightarrow y\left(5\right)=0\Rightarrow y=0\\ Pointofcontact=\left(2,0\right)\\ Differentiatingeq.\left(i\right)w.r.t.x,wehave\\ y\times 2x+\left(1+x^2\right)\frac{dy}{dx}=-1\\ \Rightarrow 2xy+\left(1+x^2\right)\frac{dy}{dx}=-1\Rightarrow \left(1+x^2\right)\frac{dy}{dx}=-1-2xy\\ \therefore \frac{dy}{dx}=\frac{-1-2xy}{\left(1+x^2\right)}\Rightarrow {\frac{dy}{dx}}_{\left(2,0\right)}=\frac{-1}{\left(1+4\right)}=\frac{-1}{5}\\ Equationoftangentisy-0=\frac{-1}{5}\left(x-2\right)\\ \Rightarrow 5y=-x+2\Rightarrow x+5y=2\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

Yes, the NIOS 10th board exam is equivalent to the CBSE, ICSE and other state boards. Students can join regular school for 11th after successful completion of NIOS 10th board exam. 

Students may apply online to the Surana College Post Graduate Departments through the official website. For admission to the MBA programme, they need to first register and fill out the application form. The steps below are included in the MBA admission process of the Surana College Post Graduate Departments:

Step 1: Visit the official website of Surana College.

Step 2: Fill out the application form and upload the required documents.

Step 3: Pay the application fee and submit the form.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol. 

$\begin{array}{l}Giventhaty=x^4-10\\ \frac{dy}{dx}=4x^3\\ \Delta x=2.00-1.99=0.01\\ \therefore \Delta y=\frac{dy}{dx}.\Delta x=4x^3.\Delta x\\ =4\times {\left(2\right)}^3\times 0.01=32\times 0.01=0.32\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

Candidates applying for BBA (Hons) courses of SLS, Pandit Deendayal Energy University have to pay the application fees. An application form get successfully submitted on fee payment only. The fee payment can be done via credit card/ debit card or through net banking. Mentioned below is the application fee amount for the UG courses:

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: 

$\begin{array}{l}Equationofthegivencurvesareay+x^2=7\dots \left(i\right)\\ and{x}^3=y\dots \left(ii\right)\\ Differentiatingeq.\left(i\right)w.r.t.x,wehave\\ a.\frac{dy}{dx}+2x=0\\ \Rightarrow \frac{dy}{dx}=-\frac{2x}{a}\\ \therefore m_1=-\frac{2x}{a}\left(m_1=\frac{dy}{dx}\right)\\ Now,differentiatingeq.\left(ii\right)w.r.t.t,wehave\\ 3x^2=\frac{dy}{dx}\Rightarrow m_2=3x^2\left(m_2=\frac{dy}{dx}\right)\\ Thetwocurvesaresaidtobeorthogonaliftheanglebetweenthetangentsatthepointof\\ intersectionis90^0.\\ \therefore m_1\times m_2=-1\\ \Rightarrow -\frac{2x}{a}\times 3x^2=-1\Rightarrow -\frac{6x^3}{a}=-1\Rightarrow 6x^3=a\\ \left(1,1\right)isthepointofintersectionoftwocurves.\\ \therefore 6{\left(1\right)}^3=a\Rightarrow a=6\\ Hence,thecorrectoptionis\left(d\right).\end{array}$