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In order to be admitted to Somaiya School of Civilisation Studies, candidates must pass one of the entrance exams that the college accepts for admission to the relevant programme. In order to be eligible for the BA programme, candidates must have completed Class 12 with at least 45% aggregate. Candidates must appear for the SVUET to participate in the entrance based admission process of Somaiya School of Civilisation. 

39. L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x.

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – (cot 2x + cot x) [cot (2x + x)]

We know that,

$cot(A+B)=\frac{co\mathrm{tA}co\mathrm{tB}-1}{co\mathrm{tA}+co\mathrm{tB}}$ we can write

$L.H.S=cotxcot2x-(cot2x+cotx)\left[\frac{cot2x·cotx-1}{cot2x+cotx}\right]$

= cot x cot 2x – cot 2x cot x + 1

= R.H.S.

38. L.H.S$=\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x.}$

$=\frac{(cos4x+cos2x)+cos3x}{(sin4x+sin2x)+sin3x}.$

$=\frac{2cos\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)+cos3x}{2sin\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)sin3x}.$

$=\frac{2cos\frac{6x}{2}cos\frac{2x}{2}+cos3x}{2sin\frac{6x}{2}cos\frac{2x}{2}+sin3x}$

$=\frac{2cos3xcosx+cos3x}{2sin3xcosx+3x}$

$=\frac{cos3x}{sin3x}\times \frac{(2cosx+1)}{(2cosx+1)}$

$=cot3x=R.H.S$

The equations of the planes are

$\begin{array}{ll}2x - y + 4z = 5 \dots \left(1\right)\hfill & 5x - 2.5y + 10z = 6 \dots \left(2\right)\hfill \end{array}$

It can be seen that,

$\begin{array}{l}\frac{a_1}{a_2}=\frac{2}{5}\\ \frac{b_1}{b_2}=\frac{-1}{-2.5}=\frac{2}{5}\\ \frac{c_1}{c_2}=\frac{4}{10}=\frac{2}{5}\\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$

Therefore, the given planes are parallel.

Hence, the correct answer is B.

The equations of the planes are

$\begin{array}{lll}2x + 3y + 4z = 4\hfill & 4x + 6y + 8z = 12\hfill & \Rightarrow 2x + 3y + 4z = 6\hfill \end{array}$

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes,   $ax + by + cz = d_1 and ax + by + cz = d_2,$ is given by,

$\begin{array}{l}D=\left|\frac{d_2-d_1}{}\right|\\ \Rightarrow D=\left|\frac{6-4}{}\right|\\ D=\frac{2}{}\end{array}$

Thus, the distance between the lines is 2/√29 units.

Hence, the correct answer is D.

Students seeking admission to the MBA offered by Sambalpur University are required to present a valid score in the CPET. The fee to appear for this exam varies depending on the candidate's category. Check the below table to know the category-wise application fees:

Note: The above-mentioned application fees is sourced from various sources. However, it is subject to change.

37. $=\frac{sinx-sin3x}{si{n}^{2}x-co{s}^{2}x}$

$=\frac{2cos\left(\frac{x+3x}{2}\right)sin\frac{x-3x}{2}}{-\left(co{s}^{2}x-si{n}^{2}x\right)}$

$=\frac{2cos\frac{4x}{2}sin\left(-\frac{2x}{2}\right)}{-cos2x}\left[?cos2x=co{s}^{2}x-si{n}^{2}x\right]$

$=-\frac{2cos2xsinx}{-cos2x}[?sin(-x)=-sinx]$

= 2 sin x

= R.H.S.

With this percentile it is still hard to get a seat in private colleges also.and if u want CSE branch in private colleges then look for autonomous University that has its own exam.i suggest VIT.give their entrance exam to secure CSE branch

The final step involves confirming the admission by paying the tuition fees after receiving the official offer. The total tuition fee for the NIILM University BPharm course is INR 2,72,000. Students are advised to verify the latest fee structure with the university, as it may be subject to updates. Timely fee payment secures your seat in the programme.

The equation of a plane having intercepts a,  b,  c with x,  y, and z axes respectively is given by,

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

The distance (p) of the plane from the origin is given by,

$\begin{array}{l}p=\left|\frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{}\right|\\ \Rightarrow p=\frac{1}{}\\ \Rightarrow p^2=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\\ \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\end{array}$

36. $=\frac{sinx+sin3x}{cosx+cos3x}$

$=\frac{2\left(\frac{x+3x}{2}\right)cos\left(\frac{x-3x}{2}\right)}{2cos\left(\frac{x+3x}{2}\right)cos\left(\frac{x-3x}{2}\right)}$

$=\frac{sin\frac{4x}{2}}{cos\frac{4x}{2}}$

$=\frac{sin2x}{cos2x}$

$=tan2x=$R.H.S

Let the required line be parallel to the vector  $\overrightarrow{b}$  given by,  $\overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}$

The position vector of the point (1, 2, − 4) is  $\overrightarrow{a}=\widehat{i}+2\widehat{j}-4\widehat{k}$

The equation of the line passing through (1, 2, −4) and parallel to vector   $\overrightarrow{b}$  is

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \Rightarrow \overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-4\widehat{k}\right)+\lambda \left(b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}\right).......(1)\end{array}$

The equations of the lines are

$\begin{array}{l}\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}........(2)\\ \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}........(3)\end{array}$

Line (1) and line (2) are perpendicular to each other.

$\therefore 3b_1-16b_2+7b_3=0........(4)$

Also, line (1) and line (3) are perpendicular to each other.

$\therefore 3b_1+8b_2-5b_3=0........(5)$

From equations (4) and (5), we obtain

$\begin{array}{l}\frac{b_1}{\left(-16\right)\left(-5\right)-8\times 7}=\frac{b_2}{7\times 3-3\left(-5\right)}=\frac{b_3}{3\times 8-3\left(-16\right)}\\ \Rightarrow \frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72}\\ \Rightarrow \frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}\end{array}$

Direction ratios of    $\overrightarrow{b}$  are 2, 3, and 6.

$\therefore \overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}$

Substituting  $\overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}$  in equation (1), we obtain

$\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-4\widehat{k}\right)+\lambda \left(2\widehat{i}+3\widehat{j}+6\widehat{k}\right)$

This is the equation of the required line.

While filling out the MBA application form at Sambalpur University, candidates must upload scanned copies of various academic and personal documents. Listed below are the documents that students can keep ready beforehand:

• Class 10 and 12 marksheets
• UG transcripts
• Migration certificate
• Transfer certificate
• Income certificate
• Category proof certificate
• Identification proof
• Recent passport-size photograph

Admission to the BPharm programme at NIILM University is based on merit. Students are shortlisted primarily based on their Class 12 academic performance in the Science stream from recognised boards. There is no entrance exam for this programme. However, additional procedures such as document verification or counselling sessions may be conducted by the university if needed.

35. L.H.S: = $L.H. S:=\frac{sinx-siny}{cosx+cosy}$

$=\frac{2cos\frac{x+y}{2}sin\frac{x-y}{2}}{2cos\frac{x+y}{2}cos\frac{x-y}{2}}$

$=\frac{sin\left(\frac{x-y}{2}\right)}{cos\left(\frac{x-y}{2}\right)}$

$=tan\left(\frac{x-y}{2}\right)=R.H.S.$ = R.H.S.