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${\left(2x^3+\frac{3}{x^k}\right)}^{12}$

gen term = 

${=}^{12}{C}_r{2}^{12-r}.3^r.x^{36-3r-rk}$  

For constant term

36 – 3r – rk = 0

$\Rightarrow k=\frac{36-3r}{r}$  

for r = 1, 2, 4   

Possible values of k = 3, 1

$\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}$

$\begin{array}{l}\overrightarrow{b}=3\widehat{i}+3\widehat{j}+\widehat{k}\\ \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}\end{array}$

$Coplnanar\Rightarrow \left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right|=0$

$\begin{array}{l}\Rightarrow -8c_1+7c_2+12c_3=0........(i)\\ \overrightarrow{a}.\overrightarrow{c}=5\Rightarrow 2c_1+c_2+3c_3=5........(ii)\\ \overrightarrow{b}.\overrightarrow{c}=0\Rightarrow 3c_1+3c_2+c_3=0........(iii)\end{array}$

Solving (i), (ii), (iii)

$C_1=\frac{10}{122},c_2=\frac{-85}{122},c_3=\frac{225}{122}$

$\therefore 122\left(c_1+c_2+c_3\right)=150$

$e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$

$\left(e^{2x}+\frac{1}{e^{2x}}\right)+4\left(e^x+\frac{1}{e^x}\right)-58=0$

$\Rightarrow {\left(e^x+\frac{1}{e^x}+2\right)}^2=64$

$e^x=\frac{6\pm \sqrt[]{32}}{2}=3\pm 2\sqrt[]{2}$

A = {1, 2, 3, ….50}

R₁ = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R₂ = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

$\therefore R_1-R_2$  contains only 9 elements.

Candidates appearing for their final year exam can apply for FPM at Jaipuria Institute of Management, Jaipur. If selected, such aspirants are provisionally admitted provided they complete all the requirements in obtaining their master's degree before September 30 of the year of admission. The admission of such aspirants remains provisional until they produce the marksheet proving their eligibility criteria. The deadline to submit the final year marksheet is December 31.

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Jaipuria Institute of Management conducts JFPMAT or Jaipuria Fellow Programme in Management Aptitude Test to shortlist candidates for FPM programme. The institute tests the knowledge of the candidates in the domain of Business and Management, General Awareness, Research Methodology, General and  Teaching Aptitude. Once the test results are out, selected candidates have to further pass the personal interview round for final admission.

Jaipuria Institute of Management, Jaipur offers admission based on entrance exam scores of the aspirants. Those who meet the eligibility criteria have to first apply for the preferred programme. Aspirants must have a valid score in CAT, MAT, XAT or CMAT for PGDM programme. Admission to FPM is based on score of the candidates in the aptitude test conducted by the institute. Further, aspirants also have to pass a personal interview and other selection rounds for admission.