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Jaipuria School of Business, Ghaziabad provides the popular Post Graduate Diploma in Management (PGDM) course to students, which is equivalent to the contemporary MBA course. Students can receive valuable Management training as part of the PGDM curriculum. The tuition fees for Jaipuria Schol of Business, Ghaziabad PGDM is INR 9.2 lakhs. Students also have the freedom to explore the various scholarship schemes offered at Jaipuria School of Business on the basis of CAT/MAT/XAT scoring as well as the assistance-based programs.

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

Sol: 

$\begin{array}{l}Giventhatequationofthecurveisy=b.e^{-x/a}\\ andtheequationofthelineis\frac{x}{a}+\frac{y}{b}=1\\ Letthecoordinatesofthepointwherethecurveintersectsthey-axisbe\left(0,y_1\right)\\ Now,differentiatingy=b.e^{-x/a}bothsidesw.r.t.x,weget\\ \frac{dy}{dx}=b.e^{-x/a}\left(-\frac{1}{a}\right)=-\frac{b}{a}.e^{-x/a}\\ So,theslopeofthetangent,m_1=-\frac{b}{a}.e^{-x/a}\\ Differentiating\frac{x}{a}+\frac{y}{b}=1bothsidesw.r.t.x,weget\\ \therefore \frac{1}{a}+\frac{1}{b}.\frac{dy}{dx}=0\\ So,theslopeoftheline,m_2=-\frac{b}{a}\\ Ifthelinetouchesthecurve,then{m}_1=m_2\\ \Rightarrow -\frac{b}{a}.e^{-x/a}=-\frac{b}{a}\Rightarrow e^{-x/a}=1\\ \Rightarrow \frac{-x}{a}loge=log1\left(Takinglogonbothsides\right)\\ \Rightarrow \frac{-x}{a}=0\Rightarrow x=0\\ Puttingx=0inequationy=b.e^{-x/a}\\ \Rightarrow y=b.e^0=b\\ Hence,thegivenequationofthecurveintersectsat\left(0,b\right)i.e.ony-axis.\end{array}$

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

Sol: 

$\begin{array}{l}Giventhatequationofthecurveis{x}^2+y^2-2x-4y+1=0\dots \left(i\right)\\ Differentiatingbothsidesw.r.t.x,weget\\ 2x+2y.\frac{dy}{dx}-2-4.\frac{dy}{dx}=0\Rightarrow \left(2y-4\right)\frac{dy}{dx}=2-2x\\ \Rightarrow \frac{dy}{dx}=\frac{2-2x}{2y-4}\dots \left(ii\right)\\ Sincethetangenttothecurveisparalleltothey-axis.\\ \therefore Slope\frac{dy}{dx}=tan\frac{\pi }{2}=\infty =\frac{1}{0}\\ So,fromeqn.\left(ii\right)weget\\ \frac{2-2x}{2y-4}=\frac{1}{0}\Rightarrow 2y-4=0\Rightarrow y=2\\ Puttingthevalueofyineqn.\left(i\right),weget\\ \Rightarrow x^2+{\left(2\right)}^2-2x-8+1=0\\ \Rightarrow x^2-2x+4-8+1=0\Rightarrow x^2-2x-3=0\\ \Rightarrow x^2-3x-x-3=0\\ \Rightarrow x\left(x-3\right)+1\left(x-3\right)=0\Rightarrow \left(x-3\right)\left(x+1\right)=0\\ \Rightarrow \end{array}$

Yes, Nehru Gram Bharati University Kotwa offers scholarships to eligible students enrolled in the BCom programme. These financial aid options are designed to support students from economically weaker backgrounds and promote accessible, affordable education for deserving candidates.

A total 5 rounds of counselling were held for Uttarakhand medical admission 2025, round 1 to 3, stray vacancy round and round 5 for BDS seats. Each round of counselling was held online. Candidates needed to register online and declare choice of college and courses for Uttarakhand NEET counselling. Seat allotment result was released separately for each round of counselling.

The tuition fee for the full BCom programme at Nehru Gram Bharati University Kotwa ranges between INR 43,000 and 48,000. These are indicative figures and may change. Students are advised to verify the latest fee details through the official website before applying.

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

Sol:

$\begin{array}{l}Wehaveequationofthecurve3x^2-y^2=8\\ Differentiatingbothsidesw.r.t.x,weget\\ 6x-2y.\frac{dy}{dx}=0\Rightarrow -2y\frac{dy}{dx}=-6x\Rightarrow \frac{dy}{dx}=\frac{3x}{y}\\ Slopeofthetangenttothegivencurve=\frac{3x}{y}\\ \therefore Slopeofthenormaltothecurve=-\frac{1}{\frac{3x}{y}}=-\frac{y}{3x}.\\ Nowdifferentiatingbothsidesthegivenlinex+3y=4\\ \Rightarrow 1+3.\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{1}{3}\\ Sincethenormaltothecurveisparalleltothegivenlinex+3y=4.\\ \therefore -\frac{y}{3x}=-\frac{1}{3}\Rightarrow y=x\\ Puttingthevaluesofyin3x^2-y^2=8,weget\\ 3x^2-x^2=8\Rightarrow 2x^2=8\Rightarrow x^2=4\Rightarrow x=\pm 2\\ \therefore y=\pm 2\\ \therefore Thepointsonthecurveare\left(2,2\right)and\left(-2,-2\right).\\ Nowequationofthenormaltothecurveat\left(2,2\right)is\\ y-2=-\frac{1}{3}\left(x-2\right)\\ \Rightarrow 3y-6=-x+2\Rightarrow x+3y=8\\ at\left(-2,-2\right)y+2=-\frac{1}{3}\left(x+2\right)\\ \Rightarrow 3y+6=-x-2\Rightarrow x+3y=-8\\ Hence,therequiredequationsarex+3y=8andx+3y=-8orx+3y=\pm 8.\end{array}$

To secure admission to the BCom programme under NGBU courses, students must have completed Class 12 from a recognised board. Additionally, selection is based on CUET scores, making performance in this national-level entrance exam crucial for applicants.

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

Sol: 

$\begin{array}{l}Giventhattheequationofthetwocurvesare{y}^2=4x\dots \left(i\right)\\ and{x}^2+y^2-6x+1=0\dots \left(ii\right)\\ Differentiatingeq\left(i\right)w.r.t.x,weget\\ 2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}\\ Slopeofthetangentat\left(1,2\right),m_1=\frac{2}{2}=1\\ Differentiatingeq\left(ii\right)w.r.t.x,weget\\ \Rightarrow 2x+2y.\frac{dy}{dx}-6=0\Rightarrow 2y.\frac{dy}{dx}=6-2x\Rightarrow \frac{dy}{dx}=\frac{6-2x}{2y}\\ \therefore Slopeofthetangentatthesamepoint\left(1,2\right),m_2=\frac{6-2\times 1}{2\times 2}=\frac{4}{4}=1\\ Weseethat{m}_1=m_2=1atthepoint\left(1,2\right).\\ Hence,thegivencirclestoucheachotheratthesamepoint\left(1,2\right).\end{array}$