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$A\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)$

$A\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}-1\\ 0\\ 1\end{array}\right)$

$A\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)$ $\Rightarrow A\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

$AB=C\Rightarrow A=C{B}^{-1}$

$\left|A-2l\right|=\left(-4\right)\left(-1\right)+3\left(-1\right)+1\left(-1\right)$

= 4 – 3 – 1 = 0

$\mathcal{l}\left(-1,0,1\right)+m\left(-1,1,0\right)=\left(-4,3,1\right)\Rightarrow -\mathcal{l}-m=-4$

$m=3,\mathcal{l}=1$

Any point on line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$       

Seneca Polytechnic is one of the top universities for international students in the Canada. Seneca Polytechnic placement rate stands at is 79.1%, and the graduates are employed within 6 months of graduation. As per unofficial sources, the average graduate salary of the Seneca Polytechnic is CAD 53,092 (INR 33.90 L), each year. Seneca graduates are working for top companies such as TD, CIBC, RBC, Scotiabank, and more. Seneca graduates are highly skilled in Communication, Leadership, Microsoft Excel, and others.

The SITEEE 2026 application form is available from December 12, 2025 to April 15, 2026. Candidates can fill the SITEEE registration form on the official website, set-test.org.

Some of the top recruiters at the Seneca Polytechnic are given below:

• Seneca Polytechnic
• TD
• CIBC
• RBC
• Scotiabank

The total tuition fee for the MBBS course at KUHS ranges from INR 1.1 Lacs to INR 5.4 lakh. Additionally, for admission to MBBS course at NTRUHS, the tuition fee for MBBS ranges from INR 54,700 to INR 56.3 lakh. The tuition fee mentioned here is for thr university and its affiliated colleges. The fee amount mentioned is sourced from the official website/sanctioning body and is subject to change. Therefore, it is indicative.

Candidates can apply for the SITEEE exam 2026 in online mode only. The authority will release the SITEEE application form on the official website, set-test.org.

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$   

  $=3\pi -\frac{3\sqrt[]{3}}{2}$           

$\therefore$ Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$

The exam conducting authority does not have any age limit to appear for the SITEEE exam. Candidates who fulfil the minimum scores in their qualifying exams are eligible to apply for the SITEEE BTech exam.

$f\left(x\right)=\left[x-1\right]cos\left(\frac{2x-}{2}\right)\pi$

$Ifx=k,k\in I$

then f(x) = 0 as $cos\left(\frac{2k-1}{2}\right)\pi =0,\forall k\in I$  

LHL = $\underset{h\to 0}{Lt}\left[k-h-1\right]cos\left(\frac{2k-2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-2\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$  

$RHL=\underset{h\to 0}{Lt}\left[k+h-1\right]cos\left(\frac{2k+2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-1\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$           

  $\therefore f\left(x\right)$  is continuous $\forall x\in R$  

The JEE Advanced exam is held in computer based test mode. The exam is not held in offline mode. Candidates have to visit the allotted exam centre to take the exam. The allotted exam centre will be mentioned on the admit card of JEE Advanced issued to the caandidates a week before the exam.

No, IIT Hyderabad is yet to reveal the BTech placements for the year 2025. The BTech placement data for IIT Hyderabad in 2024 shows that the IIT Hyderabad BTech the highest and average package stood at INR 90 and 22.5 respectively. Candidates are advised to visit the official website of IIT Hyderabad to remain updated on any news on IIT Hyderabad BTech placements 2025.

Yes. As per the SITEEE eligibility criteria mentioned by the University, applicants who passed D.Voc. Stream in the same or allied sector can apply for the BTech exam. SIU will offer suitable bridge courses for subjects like Mathematics, Physics, Engineering drawing, etc., for those who come from diverse educational backgrounds.

$f\left(x+y\right)=2f\left(x\right)f\left(y\right),x,y\in N$

$f\left(1\right)=2$

${\displaystyle \sum _{k=1}^{10}f\left(\alpha +k\right)={\displaystyle \sum _{k=1}^{10}2f\left(\alpha \right)f\left(k\right)}}$

$=2f\left(\alpha \right){\displaystyle \sum _{k=1}^{10}f\left(k\right)=2.2^{2\alpha -1}.\frac{2}{3}\left(4^{10}-1\right)}$

$=\frac{2^{2\alpha +1}}{3}.\left(4^{10}-1\right)$

$\Rightarrow 2\alpha +1=9$

= 4

The JEE Advanced exam pattern is dynamic. It changes every year. The only known aspect of the JEE Advanced exam pattern are:

1. The number of papers: There are two papers - paper 1 and paper 2.
2. The exam mode: The JEE Advanced exam is held in computer based test mode.
3. Duration: The exam is held for 6 hours. Each paper is of 3 hours.
4. Subjects: Each question papers has questions from three sections - Maths, Physics and Chemistry

JEE Advanced pattern will change every year in terms of making scheme and the kinds of questions asked (MCQs, match the following, paragraph type, numerical answer type.)

  $f\left(x\right)=2x-1g:R-\left\{1\right\}\to R$       

  $g\left(x\right)=\frac{x-1/2}{x-1}$         

  $f\left\{g\left(x\right)\right\}=2\left(\frac{x-1/2}{x-1}\right)-1=\frac{2x-1-x+1}{x-1}=\frac{x}{x-1},x\ne 1$          

$y=\frac{x}{x-1}\Rightarrow x=\frac{y}{y-1}\therefore y\ne 1$           

One – one but not onto

For mastering fundamentals and concepts, following NCERT book for VITEEE preparation is great, as syllabus is completely based on NCERT. If basics are strong, you will be able to solve any type of questions easily.