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The placement rate recorded during Siliguri Institute of Technology placements stood at 84.92%. The table below states the same:

Let f(x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$                   Non zero finite

So, d = e = f = 0

f(x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$          Non zero finite

f’(x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$  

f’(1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 .(i)

f’(-1) = 0

-6 + 5a – 4b + 3 = 0 .(ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$  

The placements at Siliguri Institute of Technology are good. The following table presents the placement statistics for Siliguri Institute of Technology:

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$  

K =    $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$   

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e² – 1)

$e=\sqrt[]{4}=2$      

According to the KUK Engineering NIRF report, 129 BTech students were placed out of the total 302 graduating students during KUK BTech placements 2024. The total number of students who graduated and were placed during Kurukshetra University placements between 2023 and 2024 is presented below:

According to the KUK Engineering NIRF report,  the median package offered to the BTech batch 2024 was INR 4.31 LPA. The median packages offered to BTech students during KUK placement between 2023and 2024 are presented below:

n = 1 mole

$W_{AB}=nRTln\frac{v_2}{v_1}=1\times RTln\frac{2v_1}{v_1}=RTln2$

$W_{BC}=0$

$W_{CA}=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{\frac{P_1}{4}\times 2V_1-P_1\times V_1}{\gamma }$

$\Rightarrow W_{CA}=\frac{-RT}{2\left(\gamma -1\right)}$

$\therefore W_{total}=RT\left[\mathcal{l}n2-\frac{1}{2\left(\gamma -1\right)}\right]$

$f\left(x\right)=\frac{5^x}{5^x+5}$

$f\left(2-x\right)=\frac{5^{2-x}}{5^{2-x}+5}=\frac{25/5^x}{\frac{25}{5^x}+5}=\frac{25}{25+5^{x+1}}=\frac{5}{5+5^x}$

$f\left(x\right)+f\left(2-x\right)=1$

$S={\displaystyle \sum _{k=1}^{39}f\left(\frac{k}{20}\right).....\left(i\right)}$

$S={\displaystyle \sum _{k=1}^{39}f}\left(2-\frac{k}{20}\right)..........\left(ii\right)$

(i) + (ii) 2S = ${\displaystyle \sum _{k=1}^{39}\left(f\left(\frac{k}{20}\right)+f\left(2-\frac{k}{20}\right)\right)}$

$S=\frac{39}{2}$

CPGET 2025 final answer key is expected to be released in August. Osmania University will release the answer key for different courses on its official website at cpget.tsche.ac.in. The CPGET provisional answer key is already released and is available on the OU website. Candidates can use the CPGET answer key to calculate their scores. 

ACJ Entrance Exam 2026 total score is 200 marks. There are two sections in the ACJ online test: English Language and General Knowledge. Each section of the question paper will be of 100 marks. The general knowledge section will be divided into two parts: current affairs and general knowledge.

Kurukshetra University offers good placements to its BTech students annually. Check out the key highlights of Kurukshetra University placements for the BTech Class of 2024 from the table below:

Note: The above-mentioned data has been taken from the KUK Engineering NIRF report 2025.

The Ranchi University placement packges are out on the official website. Check out the table for more information:

Let a_n be the side of square A_n

$a_n=\sqrt[]{2}{a}_{n+1}$       

a₁ = 12

a_n = 12 $\times {\left(\frac{1}{\sqrt[]{2}}\right)}^{n-1}$  

${\left(a_n\right)}^2<1$

$\frac{144}{2^{\left(n-1\right)}}<1$

$\Rightarrow 2^{\left(n-1\right)}>144$   

$\Rightarrow n-1\ge 8$

$\Rightarrow n\ge 9$        

$1+{\alpha }^2=1\Rightarrow \alpha =0$

$\&{\alpha }^2+{\beta }^2=1\Rightarrow {\beta }^2=\pm 1$

$\&\alpha -\alpha \beta =0\Rightarrow \alpha \left(1-\beta \right)=0\Rightarrow \alpha =0or\beta =1$

= 0 & = 1 or = 0 & = 1

4 + 4 = 1

Let x = A sin $\omega t$ $$ & v = $A\omega cos\omega t$ $$

$\therefore v=\omega \sqrt[]{A^2-x^2}$

$v^2={\omega }^2{x}^2-{\omega }^2{A}^2$

$\frac{v^2}{{\omega }^2{A}^2}+\frac{x^2}{A^2}=1$

$A=\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill y\hfill & \hfill z\hfill & \hfill x\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill y\hfill \end{array}\right],\left|A\right|=3xyz-\left(x^3+y^3+z^3\right)=-\left(x+y+z\right)\left[{\left(x+y+z\right)}^2-3\left(xy+yz+zx\right)\right]$

A² = l

A. A’ = l    (as A = A’)

$\therefore x^2+y^2+z^2=1andxy+yz+zx=0$         

$x^3+y^3+z^3=3\times 2+1\times \left(1-0\right)=7$             

V = 1.24 × 10⁶ volt

${\lambda }_{min}=?$

${\lambda }_{min}=\frac{hc}{eV}=\frac{1242nm}{e\left(1.24\times 10^6\right)}=\frac{1000\times 10^{-9}}{10^6}\Rightarrow {\lambda }_{min}=10^{-3}nm$

BAOU offers seats to candidates based on their merit. The university does not accept CUET scores and aspirants can get admission based on their score in the qualifying exam. Candidates meeting the eligibility criteria can apply for the preferred course and pay the programme fee to confirm their seat in the university.

$\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j},\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k}$      

-> $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}$

$\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{a}$

Now, $0=\overrightarrow{b}.\overrightarrow{c}+\lambda \overrightarrow{a}.\overrightarrow{b}as\overrightarrow{r}.\overrightarrow{b}=0$  

$\lambda =\frac{-\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}=2$        

$\therefore \overrightarrow{r}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{c}+2a^2=12$           

Slope = 1

$\frac{dy}{dx}=x=1$

$P\left(1,\frac{1}{2}\right)$

Equation of tangent at P : y - $\frac{1}{2}$ = 1 (x – 1)

$\Rightarrow y=x-\frac{1}{2}$

dist= $\frac{\frac{1}{2}}{\sqrt[]{2}}=\frac{1}{2\sqrt[]{2}}$