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

Sol:

$\begin{array}{l}Weknowthaty=f\left(x\right)willbecontinuousatx=aif\\ \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)\\ Givenf\left(x\right)=x^3+2x^2-1\\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\left(1+h\right)}^3+2{\left(1+h\right)}^2-1=1+2-1=2\\ \underset{x\to 1}{lim}f\left(x\right)={\left(1\right)}^3+2{\left(1\right)}^2-1=1+2-1=2\\ \underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\left(1+h\right)}^3+2{\left(1+h\right)}^2-1=1+2-1=2\\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=2.\\ Hence,f\left(x\right)iscontinuousatx=1.\end{array}$

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

Sol:

You must have a valid passport. While booking, you have to upload passport photo page. On test day, you must carry same passport with you.

Following are the main subjects covered in Diploma in AHTM courses:

• Introduction to Aviation Industry

• Airport Operations and Ground Handling

• Airline Customer Service

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• Safety and Emergency Procedures

• Airfare & Ticketing (GDS – Galileo/Amadeus Basics)

Candidates opting for BTech in Mechanical Engineering at DYPCOE must meet the eligibility criteria to seek admission. The Institute offers BTech in Mechanical Engineering for the duration of 4 years with an annual intake of 120 students. Furthermore, the Institute offers admission to BTech in Mechanical Engineering on the basis of Class 12 aggregate of 45% along with valid scores in accepted entrance exams such as MHT CET, JEE Main and others. For final admission, candidates must meet the cutoff criteria and seek final admission.

Yes, if you book paper-based IELTS, you can pick your Speaking test slot 5 to 7 days before the exam. If you don't pick, IDP will allott it for you. For computer-based test, speaking is on same day or close to your main test.

A total of 2792 job offers during the are provided at RK University placement. Check out the table for reference:

The RK University placement report is out on the official website for the year 2025. Check out the tabulated data given below to know more about RK University placement in 2025:

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

Sol:

$\begin{array}{l}Giventhatx=sintandy=sinpt\\ Differentiatingbothsidesw.r.t.t\\ \Rightarrow \frac{dx}{dt}=costand\frac{dy}{dt}=cospt.p=p.cospt\\ \Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{p.cospt}{cost}\\ \Rightarrow \frac{dy}{dx}=\frac{p.cospt}{cost}\\ Againdifferentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=p.\frac{d}{dx}\left(\frac{cospt}{cost}\right)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\frac{d}{dx}\left(cospt\right)-cospt.\frac{d}{dx}\left(cost\right)}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\left(-sinpt\right).p\frac{dt}{dx}-cospt.\left(-sint\right).\frac{dt}{dx}}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{dt}{dx}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{1}{cost}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\\ Nowwehavetoprovethat\\ \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+ p^{2}y=0\\ L.H.S.=\left(1-x^2\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-x\left(\frac{p.cospt}{cost}\right)+ p^{2}y\\ \Rightarrow =\left(1-si{n}^{2}t\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =co{s}^{2}t\left[\frac{-p^{2}cost.sinpt+pcospt.sint}{co{s}^{3}t}\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint}{cost}-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint-p.sintcospt+p^2.sinptcost}{cost}\\ \Rightarrow =\frac{0}{cost}=0=R.H.S.\\ Hence,proved.\end{array}$

The highest package offered at RK University placements 2025 stood at INR 9 LPA. Refer to the table to know the same:

ISMS Pune admissions are based on entrance exam. Candidates can register online for admission to their preferred course. Students can check the table  below to know course-wise selection criteria:

Top recruiters for BA students at Faculty of Liberal Arts, MIT WPU include Infosys, TCS, IBM, Accenture, Amazon, and Capgemini. These companies offer roles in various sectors such as consulting, human resources, research, and content development, showcasing the versatility of liberal arts education and the university's effective placement support system.

A total of 19,838 vacancies are announced for the Bihar Police Constable 2025 exam along with the notification. Scroll down for more detail.

The RK University provides good placement opportunities to its graduating students. Check out the table below to know the key highlights of theRK University placement in 2025:

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

Sol:

$\begin{array}{l}\left(i\right)Giventhat{x}^m.y^n={\left(x+y\right)}^{m+n}\\ Takinglogonbothsides,weget,\\ \Rightarrow log{x}^m.y^n=log{\left(x+y\right)}^{m+n}\left[?logxy=logx+logy\right]\\ \Rightarrow log{x}^m+log{y}^n=\left(m+n\right)log\left(x+y\right)\\ \Rightarrow mlogx+nlogy=\left(m+n\right)log\left(x+y\right)\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow m.\frac{d}{dx}logx+n.\frac{d}{dx}logy=\left(m+n\right)\frac{d}{dx}log\left(x+y\right)\\ \Rightarrow m.\frac{1}{x}+n.\frac{1}{y}.\frac{dy}{dx}=\left(m+n\right)\frac{1}{x+y}\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}.\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}.\frac{dy}{dx}\\ \Rightarrow \frac{n}{y}.\frac{dy}{dx}-\frac{m+n}{x+y}.\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{n}{y}-\frac{m+n}{x+y}\right).\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{nx+ny-my-ny}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{mx+nx-mx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \left(\frac{nx-my}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{nx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \frac{dy}{dx}=\frac{nx-my}{x\left(x+y\right)}\times \frac{y\left(x+y\right)}{nx-my}=\frac{y}{x}\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}Hence,proved.\end{array}$