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

Sol:

$\begin{array}{l}Giventhat:2si{n}^2\theta =3cos\theta \\ \Rightarrow 2\left(1-co{s}^2\theta \right)=3cos\theta \\ \Rightarrow 2-2co{s}^2\theta -3cos\theta =0\\ \Rightarrow 2co{s}^2\theta +3cos\theta -2=0\\ \Rightarrow 2co{s}^2\theta +4cos\theta -cos\theta -2=0\\ \Rightarrow 2cos\theta \left(cos\theta +2\right)-1\left(cos\theta +2\right)=0\\ \Rightarrow \left(cos\theta +2\right)\left(2cos\theta -1\right)=0\\ \Rightarrow cos\theta +2=0or2cos\theta -1=0\\ \Rightarrow cos\theta \ne -2\left[-1\le cos\theta \le 1\right]\\ \therefore 2cos\theta -1=0\\ \Rightarrow cos\theta =\frac{1}{2}\\ \Rightarrow cos\theta =cos\frac{\pi }{3},cos\left(2\pi -\frac{\pi }{3}\right)\\ \Rightarrow cos\theta =cos\frac{\pi }{3},cos\frac{5\pi }{3}\\ \therefore \theta =2n\pi \pm \frac{\pi }{3}and\theta =2n\pi \pm \frac{5\pi }{3}\left(n\in Z\right)\\ Hence,thevalueof\theta are\frac{\pi }{3}and\frac{5\pi }{3}.\end{array}$

Several colleges and universities in India offer MA courses in a distance mode. Usually, admission to MA courses in distance mode does not require the candidate to qualify for an entrance exam, and most colleges offer direct admission based on merit or a first-come, first-served basis. 

Given below are some colleges that offer MA courses in various specializations in distance mode

Note: The MA admission fee structure mentioned above may vary depending on the college policies.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}\left(i\right)Giventhat:f\left(x\right)=\frac{1}{\sqrt[]{1-cosx}}\\ Weknowthat-1\le cosx\le 1\\ \Rightarrow 1\ge -cosx\ge -1\\ \Rightarrow 1+1\ge 1-cosx\ge -1+1\\ \Rightarrow 2\ge 1-cosx\ge 0\\ \Rightarrow 0\le 1-cosx\le 2\\ Forrealvalueofdomain\\ 1-cosx\ne 0\Rightarrow cosx\ne 1\\ \Rightarrow x\ne 2n\pi \forall n\in Z\\ Hence,thedomainoff=R-\left\{2n\pi ,n\in Z\right\}\\ \left(ii\right)Giventhat:f\left(x\right)=\frac{1}{\sqrt[]{x+\left|x\right|}}\\ ?x+\left|x\right|=x+x=2xifx\ge 0\\ andx+\left|x\right|=x-x=0ifx<0\\ Sofarx<0,fisnotdefined.\\ Hence,thedomainf=R^{+}.\\ \left(iii\right)Giventhat:f\left(x\right)=x\left|x\right|\\ Itisclearthatf\left(x\right)isdefinedforallx\in R.\\ Hence,thedomainf=R.\\ \left(iv\right)Giventhat:f\left(x\right)=\frac{x^3-x+3}{x^2-1}\\ Here,f\left(x\right)isonlydefinedif{x}^2\&min\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:g=\left\{\left(1,1\right),\left(2,3\right),\left(3,5\right),\left(4,7\right)\right\}\\ Sinceeveryelementofthedomaininthisrelationshasuniqueimage,sogisafunction.\\ Nowg\left(x\right)=\alpha x+\beta \\ For\left(1,1\right)g\left(1\right)=\alpha \left(1\right)+\beta =1\Rightarrow \alpha +\beta =1\dots \left(i\right)\\ For\left(2,3\right)g\left(2\right)=\alpha \left(2\right)+\beta =3\Rightarrow 2\alpha +\beta =3\dots \left(ii\right)\\ Solvingeqn.\left(i\right)and\left(ii\right)wehave\\ \alpha =2and\beta =-1\\ Hence,thevalueof\alpha =2and\beta =-1.\end{array}$

Apart from the above-mentioned courses, many colleges also offer MA degree under various streams like Design, Mass Communication, Hospitality, etc.

In an MA programme with specialization, there would be core and elective subjects related to the specialization. The MA degree course syllabus is designed by experts to enhance the knowledge and valuable skills of the candidates. The important subjects included in the syllabus and subject lists of different MA courses are tabulated below:

Disclaimer: The information provided in the above table is taken from external sources and may vary.

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

Sol:

$\begin{array}{l}Giventhat:cot\theta +tan\theta =2cosec\theta \\ \Rightarrow \frac{cos\theta }{sin\theta }+\frac{sin\theta }{cos\theta }=\frac{2}{sin\theta }\\ \Rightarrow \frac{co{s}^2\theta +si{n}^2\theta }{sin\theta cos\theta }=\frac{2}{sin\theta }\\ \Rightarrow \frac{1}{sin\theta cos\theta }=\frac{2}{sin\theta }\\ \Rightarrow 2sin\theta cos\theta =sin\theta \\ \Rightarrow 2sin\theta cos\theta -sin\theta =0\\ \Rightarrow sin\theta \left(2cos\theta -1\right)=0\\ \Rightarrow sin\theta \ne 0or2cos\theta -1=0orcos\theta =\frac{1}{2}\\ \Rightarrow cos\theta =cos\frac{\pi }{3}\\ \therefore \theta =2n\pi \pm \frac{\pi }{3}\\ Hence,thegeneralvalueof\theta are2n\pi \pm \frac{\pi }{3}.\end{array}$

Master of Arts (MA) courses duration is two years and it is available in different specializations including Political Science, English, Hindi, Economics, History etc. Students usually take up the teaching line after attaining this postgraduate degree. Students can also pursue a doctoral degree after completing their MA course from a certified university of India. 

The following table shows the list of some of the main MA Subjects which can be pursued after graduation degree.

Disclaimer: The information provided in the above table is taken from external sources and may vary.

Candidates seeking admission to the certificate programme must have completed Class 12. The college offers certificate programmes to students in both short- and long-term modes. Gamics Institute, Indore admissions are merit-based. Furthermore, the selection criteria for all courses are based on merit. The courses are available in various streams such as Animation, Design, Mass Communication, and Media.

Takshashila University charges different total fees for different BTech courses. This happens because the tuition fees charged differ for most of the BTech specialisations. Apart from the tuition fees, students are also required to pay other fee components, such as hostel fees. Check out the below table to know the complete fee structure:

Note: The fee mentioned above is as per the official sources. However, it is indicative and subject to change.

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

Sol:

$\begin{array}{l}Giventhat:tan\theta =-1andcos\theta =\frac{1}{\sqrt[]{2}}\\ tan\theta =-1\\ \Rightarrow tan\theta =tan\left(\frac{-\pi }{4}\right)\\ \Rightarrow tan\theta =tan\left(2\pi -\frac{\pi }{4}\right)\Rightarrow tan\theta =tan\frac{7\pi }{4}\\ \therefore \theta =\frac{7\pi }{4}\\ Now,cos\theta =\frac{1}{\sqrt[]{2}}\Rightarrow cos\theta =cos\frac{\pi }{4}\\ \Rightarrow cos\theta =cos\left(2\pi -\frac{\pi }{4}\right)\\ \Rightarrow cos\theta =cos\frac{7\pi }{4}\\ \therefore \theta =\frac{7\pi }{4}\left[tan\theta andcos\theta arepositivein4thquadrant\right]\\ Hence,themostgeneralvalueof\theta are2n\pi +\frac{7\pi }{2}.\end{array}$

$\begin{array}{l}Here,A=\left\{0,1,2\right\},Bisasethavingallrealnumbersfrom0to2.SoA\ne B.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$