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$\begin{array}{l}Wearegiventhat:X=\left\{1,2,3\right\}\\ \left(i\right)\left\{4n|n\in X\right\}=\left\{4,8,12\right\}\left(ii\right)\left\{n+6|n\in X\right\}=\left\{7,8,9\right\}\\ \left(iii\right)\left\{\frac{n}{2}|n\in X\right\}=\left\{\frac{1}{2},1,\frac{3}{2}\right\}\left(iv\right)\left\{\left(n-1\right)|n\in X\right\}=\left\{0,1,2\right\}\end{array}$

The KIITEE Law 2026 slot booking process allows candidates to choose their exam date, time, and center on a first-come, first-served basis. To book a slot, visit the official website, log in using your application number and date of birth, and select your preferred options. Once confirmed, the slot cannot be changed. After successful booking, the admit card will be available for download. Slot booking is mandatory to appear for the exam; if missed, a slot is auto-assigned. It is advised to book early to secure your preferred schedule and avoid last-minute issues.

^{$\begin{array}{l}Wearegiventhat:N=\left\{1,2,3,4,5,\dots ,100\right\}\\ \left(i\right)Requiredsubsetwhoseelementsareeven\\ =\left\{2,4,6,8,\dots ,100\right\}\\ \left(ii\right)Requiredsubsetwhoseelementsareperfectsquares\\ =\left\{1,4,9,16,25,36,\dots ,100\right\}\end{array}$}

$\begin{array}{l}\left(i\right)Giventhat:A\subset UandB\subset U\\ Letx\in Aorx\in B\\ \Rightarrow x\in A\cup B\\ Hence,A\subset \left(A\cup B\right)\\ \left(ii\right)IfA\subset B\\ Thenletx\in A\cup B\\ \Rightarrow x\in Aorx\in B\\ \Rightarrow A\cup B\subset B\dots \left(i\right)\\ ButB\subset A\cup B\dots \left(ii\right)\\ Fromeq.\left(i\right)and\left(ii\right),weget\\ A\cup B=B\\ Lety\in A\Rightarrow y\in \left(A\cup B\right)\Rightarrow y\in B\\ \Rightarrow y\in B\iff A\cup B=B\\ \left(iii\right)Letx\in A\cap B\\ \Rightarrow x\in Aandx\in B\Rightarrow x\in A\\ So,A\cap B\subset A\end{array}$

$\begin{array}{l}Given,L=\left\{1,2,3,4\right\},M=\left\{3,4,5,6\right\}andN=\left\{1,3,5\right\}\\ \therefore M\cup N=\left\{1,3,4,5,6\right\}\\ L-M\cup N=\left\{2\right\}\\ Now,L-M=\left\{1,2\right\},L-N=\left\{2,4\right\}\\ \therefore \left(L-M\right)\cap \left(L-N\right)=\left\{2\right\}\\ Hence,L-M\cup N=\left(L-M\right)\cap \left(L-N\right).\end{array}$

The KIITEE Law 2026 exam schedule has not been released yet. According to multiple recent sources:

The official notification including important dates like application start, slot booking, exam date, and admit card release is expected in the first week of November 2025, via the official law KIITEE portal. Following that, the application window (Phase I) is likely to run from early November 2025 until around mid-March 2026, with the online exam to follow in March–April 2026.

$\begin{array}{l}\left(i\right)Giventhat:35\in \left\{x|xhasexactlyfourpositivefactor\right\}\\ \therefore Factorsof35are1,5,7,35\\ Hence,thestatement\left(i\right)is\text{'}True\text{'}.\\ \left(ii\right)Giventhat:128\in \left\{y|thesumofallpositivefactorsofyis2y\right\}\\ \therefore Factorsof128are1,2,4,8,16,32,64and128.\\ Sumofallfactors=1+2+4+8+16+32+64+128\\ =255\ne 2\times 128\\ Hence,thestatement\left(ii\right)is\text{'}False\text{'}.\\ \left(iii\right)Giventhat:3\in \left\{x|x^4-5x^3+2x^2-112x+6=0\right\}\\ \therefore x^4-5x^3+2x^2-112x+6=0\\ Nowforx=3,wehave\\ {\left(3\right)}^4-5{\left(3\right)}^3+2{\left(3\right)}^2-112\left(3\right)+6\\ \Rightarrow 81-135+18-336+6\Rightarrow -366\ne 0\\ Hence,thestatement\left(iii\right)is\text{'}True\text{'}.\\ \left(iv\right)Giventhat:496\notin \left\{y|thesumofallpositivefactorsofyis2y\right\}\\ \therefore Thepositivefactorsof496are1,2,4,8,16,31,62,124,248and496.\\ \therefore Thesumofallpositivefactors=1+2+4+8+16+31+62+124+248+496\\ =992=2\times 496\\ Hence,thestatement\left(iv\right)is\text{'}False\text{'}.\end{array}$

More than 280 companies visited the campus and 380 offers were made during UG placements 2024. The overall placement rate recorded during IIT Jodhpur BTech placements 2024 was 92%. IIT Jodhpur MBA average package received during placements 2024 stood at INR 11.77 LPA. Whereas, the highest package offered during MBA placements 2024 was INR 24.81 LPA.

For fee details related to the BCA course offered at GB Shah College for Technology and Management, students must visit the official website and ask for the same. Till then, keep applying on the official website for relevant courses. Follow the application process. 

$\begin{array}{l}Given,Y=\left\{x|xisapositivefactorofthenumber{2}^{p-1}\left(2^p-1\right),where{2}^{p-1}isaprimenumber\right\}.\\ Since,thefactorsof{2}^{p-1}are1,2,2^2,2^3,\dots ,2^{p-1}andfactorsof{2}^p-1are1and{2}^{p-1}\\ \therefore Y=\left\{1,2,2^2,2^3,\dots ,2^{p-1},2^p-1\right\}\\ =2\left(2^p-1\right),2^2\left(2^p-1\right),\dots ,2^{p-1}\left(2^p-1\right)\end{array}$

The KIITEE Law 2026 exam date has not been announced yet. The official notification is expected to be released in the first week of November 2025, and the exam itself is anticipated to be scheduled in March–April 2026  Keep an eye on the official KIIT Law websites—kiit.ac.in or law.kiit.ac.in—starting late October 2025.

$\begin{array}{l}\left(i\right)Wehave,D=\left\{t|t^3=t,t\in R\right\}\\ \therefore t^3=t\\ \Rightarrow t^3-t=0\Rightarrow t\left(t^2-1\right)=0\\ \Rightarrow t\left(t-1\right)\left(t+1\right)=0\Rightarrow t=0,1,-1\\ \therefore D=\left\{-1,0,1\right\}\\ \left(ii\right)Wehave,E=\left\{w|\frac{w-2}{w+3}=3,w\in R\right\}\\ \therefore \frac{w-2}{w+3}=3\\ \Rightarrow w-2=3w+9\Rightarrow w-3w=9+2\\ \Rightarrow -2w=11\Rightarrow w=\frac{-11}{2}\\ \therefore E=\left\{\frac{-11}{2}\right\}\\ \left(iii\right)Wehave,F=\left\{x|x^4-5x^2+6=0,x\in R\right\}\\ \therefore x^4-5x^2+6=0\\ \Rightarrow x^4-3x^2-2x^2+6=0\Rightarrow x^2\left(x^2-3\right)-2\left(x^2-3\right)=0\\ \Rightarrow \left(x^2-3\right)\left(x^2-2\right)=0\Rightarrow x=\pm \sqrt[]{3},\pm \sqrt[]{2}\\ \therefore F=\left\{-\sqrt[]{3},-\sqrt[]{2},\sqrt[]{2},\sqrt[]{3}\right\}\end{array}$

$\begin{array}{l}\left(i\right)Wehave,A=\left\{x:x\in R,2x+11=15\right\}\\ \therefore 2x+11=15\\ \Rightarrow 2x=15-11\\ \Rightarrow 2x=4\Rightarrow x=2\\ \therefore A=\left\{2\right\}\\ \left(ii\right)Wehave,B=\left\{x|x^2=x,x\in R\right\}\\ \therefore x^2=x\\ \Rightarrow x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow x=0,1\\ \therefore B=\left\{0,1\right\}\\ \left(iii\right)Wehave,C=\left\{x|xisapositivefactorofprimenumberp.\right\}\\ Since,positivefactorsofaprimenumberare1andthenumberitself.\\ \therefore C=\left\{1,p\right\}\end{array}$