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Silver Oak University offers a total of 60 seats for its BPharm course. These seats are allocated at the Silver Oak College of Pharmacy under the university's management. Interested students are required to clear the NEET or GUJCET exam for selection. 

$\begin{array}{l}Totalnumberofstudents=60\Rightarrow n\left(U\right)=60\\ NumberofstudentswhoplayCricket=25\Rightarrow n\left(C\right)=25\\ NumberofstudentswhoplayTennis=20\Rightarrow n\left(T\right)=20\\ NumberofstudentswhoplayCricketandTennisboth=10\\ \Rightarrow n\left(C\cap T\right)=10\\ \therefore n\left(C\cup T\right)=n\left(C\right)+n\left(T\right)-n\left(C\cap T\right)\\ =25+20-10=45-10=35\\ \therefore n\left(C\text{'}\cap T\text{'}\right)=n\left(U\right)-n\left(C\cup T\right)\\ =60-35=25\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

The average package offered to the MBA graduating class of 2023 stood at INR 8.5 LPA. Check out the table for more clarity:

IIT Jodhpur offers various UG, PG, and PhD courses in Engineering, Management, and Science streams. IIT Jodhpur flagship courses are MBA and BTech. Admission to these courses is based on the entrance exam. The Institute offers respective courses for durations varying from 2 years to 5 years. For more details around the respective courses, candidates must visit the official website of the Institute.

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

$\begin{array}{l}Giventhat:R=\left\{\left(x,y\right):x,y\in W,x^2+y^2=25\right\}\\ So,theorderedpairssatisfyingthegivencondition{x}^2+y^2=25are\\ \left(0,5\right),\left(3,4\right),\left(5,0\right),\left(4,3\right)\left[?x,y\in W\right]\\ Hence,thedomain=\left\{0,3,4,5\right\}andrange=\left\{0,3,4,5\right\}.\end{array}$

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

Sol:

$\begin{array}{l}L.H.S.cos\theta cos\frac{\theta }{2} -cos3\theta cos\frac{9\theta }{2}\\ =\frac{1}{2}\left[2cos\theta cos\frac{\theta }{2}\right]-\frac{1}{2}\left[2cos3\theta cos\frac{9\theta }{2}\right]\\ =\frac{1}{2}\left[cos\left(\theta +\frac{\theta }{2}\right)+cos\left(\theta -\frac{\theta }{2}\right)\right]-\frac{1}{2}\left[cos\left(3\theta +\frac{9\theta }{2}\right)+cos\left(3\theta -\frac{9\theta }{2}\right)\right]\\ =\frac{1}{2}\left[cos\frac{3\theta }{2}+cos\frac{\theta }{2}-cos\frac{15\theta }{2}-cos\left(\frac{-3\theta }{2}\right)\right]\\ =\frac{1}{2}\left[cos\frac{3\theta }{2}+cos\frac{\theta }{2}-cos\frac{15\theta }{2}-cos\frac{3\theta }{2}\right]\left[?cos\left(-\theta \right)=cos\theta \right]\\ =\frac{1}{2}\left[cos\frac{\theta }{2}-cos\frac{15\theta }{2}\right]=\frac{1}{2}\left[-2sin\left(\frac{\theta }{2}+\frac{15\theta }{2}\right).sin\left(\frac{\theta }{2}-\frac{15\theta }{2}\right)\right]\\ =-sin8\theta sin\left(-7\theta \right)=sin7\theta sin8\theta \left[?sin\left(-\theta \right)=-sin\theta \right]\\ L.H.S.=R.H.S.Henceproved.\end{array}$

BTech specialisations at Takshashila University are spread across different fields, such as Information Technology, Computer Science, and more. Listed below are some of the specialisations available at the university for students to choose from:

• Computer Science Engineering
• Artificial Intelligence and Data Science
• Information Technology
• Electronics and Computer Engineering
• Communication Systems

Yes, you can either have a BSc in Nursing or a Post Basic BSc Nursing degree to be eligible for an MSc Nursing course. However, a minimum of 1 year of work experience in a healthcare facility can be a must for some top colleges, also in a specific department, if you want to go for a specific specialization.

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

$\begin{array}{l}Giventhat:A=\left\{1,2,3,4,5\right\}\\ andS=\left\{\left(x,y\right):x\in A,y\in A\right\}\\ \left(i\right)x+y=5,so,theorderedpairssatisfyingthegivenconditionsare\left(1,4\right),\left(4,1\right),\left(2,3\right),\left(3,2\right).\\ \left(ii\right)x+y<5,so,theorderedpairssatisfyingthegivenconditionsare\\ \left(1,1\right),\left(1,2\right),\left(2,1\right),\left(1,3\right),\left(2,2\right),\left(3,1\right).\\ \left(iii\right)x+y>8,so,theorderedpairssatisfyingthegivenconditionsare\left(4,5\right),\left(5,4\right),\left(5,5\right).\end{array}$

No, aspirants are required to fill out the application form for SITEEE online. Aspirants cannot submit the form in-campus or send it to the institute. Aspirants must register for the exam before the deadline to avoid rejection of the applications. The exam is conducted in two phase usually in May.

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

Sol:

        $\begin{array}{l}Giventhat:tanx=\frac{b}{a}\\ \sqrt[]{\frac{a+b}{a-b}}+\sqrt[]{\frac{a-b}{a+b}}=\frac{\sqrt[]{a+b}}{\sqrt[]{a-b}}+\frac{\sqrt[]{a-b}}{\sqrt[]{a+b}}\\ =\frac{a+b+a-b}{\sqrt[]{\left(a-b\right)\left(a+b\right)}}=\frac{2a}{\sqrt[]{a^2-b^2}}=\frac{2a}{a\sqrt[]{1-\frac{b^2}{a^2}}}\\ =\frac{2}{\sqrt[]{1-ta{n}^{2}x}}\left[?tanx=\frac{b}{a}\right]\\ =\frac{2}{\sqrt[]{1-\frac{si{n}^{2}x}{co{s}^{2}x}}}=\frac{2}{\frac{\sqrt[]{co{s}^{2}x-si{n}^{2}x}}{cosx}}\\ =\frac{2cosx}{\sqrt[]{cos2x}}\left[?cos2x=co{s}^{2}x-si{n}^{2}x\right]\\ Hence,\sqrt[]{\frac{a+b}{a-b}}+\sqrt[]{\frac{a-b}{a+b}}=\frac{2cosx}{\sqrt[]{cos2x}}\end{array}$

The total tuition fees for Takshashila University BTech courses range from INR 3.4 Lacs to INR 12 lakh. This fee is provided in a range because the tuition fees differ from course to course. Moreover, in case a student fails to pay this fee at the time of admission, the university reserves the right to forfeit the admission.

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 question as classified in NCERT Exemplar

$\begin{array}{l}\left(i\right)Giventhat:\left(2a+b,a-b\right)=\left(8,3\right)\\ Comparingthedomainsandranges,weget\\ 2a+b=8\dots \left(i\right)\\ a-b=3\dots \left(ii\right)\\ Solving\left(i\right)and\left(ii\right)wegeta=\frac{11}{3}andb=\frac{2}{3}\\ \left(ii\right)Giventhat:\left(\frac{a}{4},a-2b\right)=\left(0,6+b\right)\\ Comparingthedomainsandranges,weget\\ \frac{a}{4}=0\Rightarrow a=0\\ \Rightarrow a-2b=6+b\Rightarrow a-3b=6\Rightarrow 0-3b=6\\ \therefore b=-2\\ So,a=0,b=-2.\end{array}$

As per the PTET marking scheme 3 marks is granted for each correct answer, with no negative marking. The total marks for the exam is 600, derived from 200 questions. However, the Teaching Attitude & Aptitude section has a scaled marking system, ranging from 3 to 0 marks per question. 

$\begin{array}{l}ThegivenconditionsofthequestionmayberepresentedbythefollowingVenndiagram.\\ FromthegivenVenndiagram,\\ Weclearlyconcludethat\\ S\cup T\cup C=s\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

$\begin{array}{l}Weknowthatrectangles,rhombusandsquareinaplaneisaparallelogrambuttrapeziumis\\ notparallelogram.\\ \therefore F_1=F_2\cup F_3\cup F_4\cup F_1\\ Hence,thecorrectoptionis\left(d\right).\end{array}$