Ask & Answer: India's Largest Education Community

${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$           

IBSAT 2025 result was announced on December 30. Candidates who appeared for exam can check their results online when released. IBSAT 2025 exam was held on December 27 and December 28 in remote proctored mode. To download the IBSAT scorecrad, candidates must enter their login details on official website- general.ibsindia.org.

log x = t
(t² - 1/t)¹²
T? = ¹²C? (t²)¹²? (-1/t)? = ¹²C? t²? ³? (-1)?
For constant term r = 8 & coefficient ¹²C?
= (12×11×10×9) / 24 = 45×11 = 495

Yes, students in their last year of graduation, are eligible to apply for NIMCET 2026.

The ACJ entrance exam will have two sections: Language Skills and General Knowledge. Both sections will be of 100 marks each. Candidates will be given 0 marks for any incorrect answer. Therefore,  there is no negative marking in ACJ entrance exam.

Joining Chandola Homoeopathic Medical College and Hospital for BHMS course is generally considered a decent option. Candidates must consider factors such as infrastructure, faculty experience, fees, curriculum, etc. while taking admission. They can apply online or offline for admission at Chandola Homoeopathic Medical College and Hospital.

TANCET answer key is the documents released after the exam for test takers to check what were the answers to the questions that appeared in the exam. Candidates find all the answers for the questions that were available in the TANCET exam.

Since in JEE Advanced, the application form correction window does not open, if you have made any mistake aftyer the form is submitted you cannot change it. In such case please contact the coordinating IIT as mentioned on the official brochure.

However, if you have not submitted the form, the you can make correction in the form as many times. Just ensure that all the details in the form has been rectified before the final submission.

DNA and RNA both are the types of nucleic acids.

• DNA (Deoxyribonucleic acid) store the genetic  information.

• RNA (Ribonucleic acid) helps in protein synthesis. It carry and translates the genetic codes.

tan? ¹ (x) = t
t² - 4t + 3 > 0
t ∈ (-∞, 1) U (3, ∞)
tan? ¹ (x) ∈ (-π/2, 1)
x ∈ (-∞, tan1)
the largest integral x = 1

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$           

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$            

$PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$           

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$          

Yes, Chandola Homoeopathic Medical College and Hospital offers the BHMS course for a duration of 5.5 years. Admissions to this course are entrance-based and scores of the UPMT exam is accepted. Candidates seeking admission in the course must appear for the exam followed by the counselling process.

Candidates can download the SITEEE 2026 admit card using their SITEEE ID/Registered email ID and password. 

$V_1=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{R}=30$

$V_2=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{2R}=15$

$V_3=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{3R}+\frac{k{Q}_1}{3R}=10$

Eq. (1) $\&$  (2) $\to \frac{k{Q}_1}{R}=30$  

Eq. (2) $\&$  (3) $\to \frac{k{Q}_2}{6R}+\frac{k{Q}_1}{6R}=5\Rightarrow \frac{k{Q}_2}{R}=-\frac{k{Q}_1}{R}+30=0$

By (3) $\to \frac{k{Q}_3}{R}=30-\frac{k{Q}_2}{R}-\frac{k{Q}_1}{R}=0$

Potential of innermost shell (earthed) $=\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_3}{3R}=0$

$\Rightarrow \mathrm{}\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}=0\mathrm{}\therefore \mathrm{}{Q}_1^{\mathrm{\text{'}}}=0$

Yes, you can. But mostly top MA colleges in India will require you to provide entrance exam scores. For example, if you want to get admission to a DU college for an MA course, you'll need to clear the CUET-PG exam. There are some private colleges that offer admission on the merit system though. 

There is no form correction facility provided in JEE Advanced exam. Details once fille in the form and the application fee is paid, the form correction window does not open. Hence the candidates have to make sure to fill in all the details correctly before paying the application fee and submitting the form.

yes, candidates can take IBSAT exam more than once. The IBSAT exam is conducted annually. If candidate is not able to appear for the current session, he/she may apply for the next session examinations. ICFAI has not specified the number of attempts for IBSAT exam.

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 × 4² = 48

No, to get admission into the BBA course at the Krupanidhi Group of Institutions, there is no requirement of the entrance examination. As per the information gathered from the official sources, the institution grants merit-based admission to the BBA course. Thus, applicants can apply for the course as per their scores in Class 12. Further, the institution releases the merit list of shortlisted candidates according to their scores.

  $\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}.....\left(i\right)$

  $\overrightarrow{a}.\overrightarrow{c}=7\overrightarrow{b}.\overrightarrow{c}=0$         

$\overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}\Rightarrow \left|\overrightarrow{a}\right|=\sqrt[]{3}$           

$\overrightarrow{b}=2\widehat{i}+\widehat{k}\Rightarrow \left|\overrightarrow{b}\right|=\sqrt[]{5}\overrightarrow{a}.\overrightarrow{b}=-2+1=-1$           

From   $\left(i\right)\overrightarrow{a}.\overrightarrow{c}=\alpha {\left|\overrightarrow{a}\right|}^2-\beta$

$3\alpha -\beta =7..........\left(ii\right)$           

$\overline{b}.\overline{c}=\alpha \overline{b}.\overline{a}+\beta {\left|\overrightarrow{b}\right|}^2\Rightarrow -\alpha +5\beta =0.......\left(iii\right)$           

Solving $\alpha =\frac{5}{2}and\beta =\frac{1}{2}$

$\Rightarrow 2{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=75$