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The current admission fee for IIIT Ranchi stands at INR 1.56 lakh. This amounts includes many types of fee including first semester tuition fee, hostel fee and institute fee along with one time payments for alumni membership fee, institutional caution fee and admission processing fee.

he line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector

$\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right|=\left(1,4,-3\right)$ Equation of line through P(1, 2, 4) and parallel to $\overrightarrow{b}\frac{x-1}{1}=\frac{y-2}{-4}=\frac{z-4}{-3}$

Let $N\equiv \left(\lambda +1,-4\lambda +2,-3\lambda +4\right)\overline{QN}=\left(\lambda ,-4\lambda +4,-3\lambda -1\right)$

$\overline{QN}$ is perpendicular to $\overrightarrow{b}\Rightarrow \left(\lambda ,-4\lambda +4,-3\lambda -1\right).\left(1,4,-3\right)=0\Rightarrow \lambda =\frac{1}{2}.$

Hence $\stackrel{}{}$$\overline{QN}\left(\frac{1}{2},2,\frac{-5}{2}\right)and\overline{\left|QN\right|}=\text{exception 25:}$

  ${\displaystyle \int \frac{\left(1-\frac{1}{\sqrt[]{3}}\right)\left(cosx-sinx\right)}{\left(1+\frac{2}{\sqrt[]{3}}sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{3}}\right)\sqrt[]{2}sin\left(\frac{\pi }{4}-x\right)}{\left(\frac{2}{\sqrt[]{3}}\right)\left(sin\frac{\pi }{3}+sin2x\right)}dx}$

$={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{\left(sin\frac{\pi }{3}+sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{sin\left(\frac{\pi }{6}+x\right)cos\left(\frac{\pi }{6}-x\right)}dx}$

$=\frac{1}{2}\left[lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{12}\right)\right|-lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{6}\right)\right|\right]+C=\frac{1}{2}lo{g}_e\left|\frac{tan\left(\frac{x}{2}+\frac{\pi }{12}\right)}{tan\left(\frac{x}{2}+\frac{\pi }{6}\right)}\right|+C$

Mean = 4 = $\mu$ = np

Variance = ${\sigma }^2=np\left(1-p\right)=\frac{4}{3}\Rightarrow 4\left(1-p\right)=\frac{4}{3}\Rightarrow p=\frac{2}{3}\Rightarrow n=6$

${=}^6{C}_0{\left(\frac{1}{3}\right)}^6{+}^6{C}_1{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^6{+}^6{C}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^4=\frac{146}{27}$

Any tangent to y² = 24x at (α, β) is βy = 12 (x + α) therefore Slope = $\frac{12}{\beta }$

and perpendicular to 2x + 2y = 5 =>12 =β and α= 6 Hence hyperbola is $\frac{x^2}{6^2}-\frac{y^2}{12^2}$ = 1 and normal is drawn at (10, 16)

therefore equation of normal $\frac{36x}{10}+\frac{144y}{16}=36+144\Rightarrow \frac{x}{50}+\frac{y}{20}=1$ This does not pass through (15, 13) out of given option.

$v=f\lambda$

$f=\frac{v}{\lambda }=\frac{3\times 10^8}{10^3\times 10^{-9}}=3\times 10^{14}{H}_z$

Channels =  $\frac{2\mathrm{\%}of3\times 10^{14}{H}_z}{8\times 10^3}$

$\frac{2\mathrm{\%}of3\times 10^{14}{H}_z}{8\times 10^3}$

$\frac{2\mathrm{\%}of3\times 10^{14}{H}_z}{8\times 10^3}$

$\frac{=\frac{2x}{100}3\times 10^{14}}{8\times 10^3}=75\times 10^7$

The IIM Trichy MBA/ PGPM selection criteria is done based on various parameters. Hence, the candidates selection is done just based on CAT score. Mentioned below are the considered parameters along with the weightage:

• CAT percentile: 50%
• PI performance: 25%
• Academic performance: 10%
• Work experience: 5%
• Female candidates: 5%
• Verbal composition: 3%
• Professional qualification: 2%

$p\iff \left(q\Rightarrow p\right)\sim \left(p\iff \left(q\iff p\right)=p\iff \sim \left(q\Rightarrow p\right)\right)$

$=\sim p\wedge \left(\sim q\vee p\right)$

$=\left(\sim p\wedge \sim q\right)$

The MA specialisations offered by Vijayanagara Sri Krishnadevaraya University are as follows:

Netaji Subhas University of Technology fee ranges between INR 17,000- INR 4.75 lakh. The fee structure at NSIT is bifurcated into various components, namely tuition, security deposit, and hostel fee. Certain fee elements are one-time payments to be made at the time of NSIT admission, whereas others are recurring as per the norms of the institution fee policy.

Joining RGUKT: Rajiv Gandhi University of Knowledge Technologies Andhra Pradesh for PhD course is generally considered a decent option. Candidates must consider factors like infrastructure, faculty experience, hostel and mess facilities, curriculum, etc. while taking admission. They can apply online or offline for admission at RGUKT: Rajiv Gandhi University of Knowledge Technologies Andhra Pradesh.

Let point P : (h, k)

Therefore according to question,  ${\left(h-1\right)}^2+{\left(k-2\right)}^2+{\left(h+2\right)}^2+{\left(k-1\right)}^2=14$

$\therefore$ locus of P (h, k) is $x^2+y^2+x-3y-2=0$

Now intersection with x – axis are $x^2+x-2=0\Rightarrow x=-2,1$

Now intersection with y – axis are $y^2-3y-2=0\Rightarrow y=\frac{3\pm \sqrt[]{17}}{2}$

Therefore are of the quadrilateral ABCD is = $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}\times 3\times \sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$

 $Let\frac{si{n}^{-1}x}{\alpha }=\frac{co{s}^{-1}x}{\beta }=k\Rightarrow si{n}^{-1}x+co{s}^{-1}x=k\left(\alpha +\beta \right)\Rightarrow \alpha +\beta =\frac{\pi }{2k}$

Now $\frac{2\pi \alpha }{\alpha +\beta }=\frac{2\pi \alpha }{\frac{\pi }{2k}}\Rightarrow 4k\alpha =4si{n}^{-1}x.Heresin\left(\frac{2\pi \alpha }{\alpha +\beta }\right)=sin\left(4si{n}^{-1}x\right)$

$\Rightarrow sin4\theta =4x\sqrt[]{1-x^2}\left(1-2x^2\right)$

No, admission to the PhD course at RGUKT: Rajiv Gandhi University of Knowledge Technologies Andhra Pradesh is not merit-based. Candidates are granted admission on the basis of the marks scored by them in the entrance exams accepted by the university. These are- GATE, UGC NET, CSIR NET. 

 $\frac{dy}{dx}+2ytanx=sinx,\therefore I.F.e^{{\displaystyle \int 2tanxdx}}=se{c}^{2}x$

= cos x – 2 cos² x= $-2{\left(cosx-\frac{1}{4}\right)}^2+\frac{1}{8}\therefore y_{max}=\frac{1}{8}$

$If{\widehat{n}}_1$ is a vector normal to the plane determined by $\widehat{i}and\widehat{i}+\widehat{j}then{\widehat{n}}_1=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right|=\widehat{k}$

$If{\widehat{n}}_2$ is a vector normal to the pane determined by $\widehat{i}-\widehat{j},$ and $\widehat{i}+\widehat{k}$ then ${\widehat{n}}_2$ = $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}+\widehat{k}$

Vector $\widehat{a}$ is parallel to ${\widehat{n}}_1\times {\widehat{n}}_2$ i.e. $\widehat{a}$ is parallel to $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}$

Given $\overrightarrow{b}=\widehat{i}-2\widehat{j}+2\widehat{k}$

consine of acute angle between $\widehat{a}and\widehat{b}$ = $\left|\frac{\widehat{a}.\widehat{b}}{\left|\widehat{a}\right|.\left|\widehat{b}\right|}\right|=\frac{1}{\sqrt[]{2}}$

obtuse angle between $\widehat{a}and\widehat{b}=\frac{3\pi }{4}$

${\beta }_1=\frac{{\lambda }_1{D}_1}{d_1}$

${\beta }_2=\frac{{\lambda }_2{D}_2}{d_2}$

$\frac{{\beta }_1}{{\beta }_2}=\frac{{\lambda }_1}{d_1}\times \frac{d_2}{{\lambda }_2}$

$\frac{0.5}{{\beta }_2}=\frac{5000\times d_1}{d_1\times 600}$

${\beta }_2=\frac{0.5\times 6}{10}$

2 = 0.3mm

Yes, aspirants of MBA/ PGPM at IIM Trichy have to appear for CAT and secure a valid score. There is no other provision for admission in the PG programme. Moreover, candidates shortlisted based on CAT score are evaluated based on other factors as well, which are as follows:

• Performance in personal interview round
• Educational background
• Work experience (if any)
• Gender diversity

 $a=\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{2n}{n^2+k^2}=\underset{n\to \infty }{lim}\frac{1}{n}}{\displaystyle \sum _{k=1}^n\frac{2}{1+{\left(\frac{k}{n}\right)}^2}}$

$\therefore a={\displaystyle {\int }_0^1\frac{2}{1+x^2}dx=2ta{n}^{-1}x}]{}_0^1=\frac{\pi }{2}$

$f\text{'}\left(\frac{a}{2}\right)=\sqrt[]{2}f\left(\frac{a}{2}\right)$

Given hyperbola : $\frac{k{x}^2}{6}-\frac{y^2}{6}=1$ so eccentricity e = $\sqrt[]{1+k}$ and directrices $x=\pm \frac{a}{e}$

$\Rightarrow x=\pm \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}\Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}=1$

k = 2 therefore equation of hyperbola is $\frac{x^2}{3}-\frac{y^2}{6}=1$

hence it passes through the point $\left(\sqrt[]{5},-2\right)$