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VSAT, which stands Vignan's Scholastic Aptitude Test, it is a national-level entrance exam which is comnducted to shortlist candidates for various UG/PG courses at its university.

The VSAT registrations generally lasts over a month and candidates meeting the minimum eligibility criteria can apply till the last date. In many cases, the exam body also extends the registrations, so if candidates miss the deadline, they can apply till then also. However, it is not advised to wait till the last moment and apply as early as the application form is out.

To keep track of the VSAT events candidates can visit the official website, or they can also bookmark this page as we will be updating all the relevand and updated information here as and when it is released.

No, a B.Sc. Radiology graduate is not considered a doctor. After completing B.Sc. Radiology, you become a radiology technologist or radiographer, whose main role is to operate imaging equipment like X-ray, MRI, CT scan, ultrasound machines, etc., and assist doctors in diagnosis.

To be called a “doctor” in this field, you need to pursue an MBBS and then specialize in Radiology (MD/DNB in Radiology).

So in short, B.Sc. Radiology makes you a healthcare professional, but not a doct

or.

Yes, if candidates have got admission through JEE/EAMCET, they can still appear for VSAT counselling, provided they have qualified in the VSAT exam and hold a good rank. However, before finalising the college, they must compare the college based on course, fee and facilities.

The VSAT counselling is a process thorugh which candidates are shortlisted for the courses. During the counselling process, candidates are alloted courses based on their rank of merit.

The VSAT will be declared after the exam is conducted. It is expected that the result will bs declared after the final answer key is out.

The difficulty level of the VSAT syllabus is easy to moderate. It is based on the 10+2 level, however the questions are easy to solve and if students have prepared well, they can qualify in the exam.

$\frac{dy}{dx}=\frac{1}{1+sin2x}$

$dy=\frac{se{c}^{2}xdx}{{\left(1+tanx\right)}^2}$

$\Rightarrow y=-\frac{1}{1+tanx}+c$

when   $x=\frac{\pi }{4},y=\frac{1}{2}$ gives c = 1

so $x+\frac{\pi }{4}=\frac{5\pi }{6}or\frac{13\pi }{6}\Rightarrow x=\frac{7\pi }{12}or\frac{23\pi }{12}$

sum of all solutions =

$\pi +\frac{7\pi }{12}+\frac{23\pi }{12}=\frac{42\pi }{12}$

Hence k = 42

Each element of ordered pair (i, j) is either present in A or in B.

              So, A + B = Sum of all elements of all ordered pairs {i, j} for $1\le i\le 10$ and $1\le j\le 10$  

              = 20 (1 + 2 + 3 + … + 10) = 1100

$l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[{\left(\frac{\pi }{2}-x\right)}^3-\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Using ${\displaystyle {\int }_a^{b}f\left(x\right)dx={\displaystyle {\int }_a^{b}f\left(a+b-x\right)dx}}$

we get $l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[-{\left(\frac{\pi }{2}-x\right)}^3+\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Adding these two equations, we get

$\Rightarrow l=\frac{12}{\pi }{\left[-ta{n}^{-1}\left(cosx\right)\right]}_0^{\pi }=\frac{12}{\pi }.\frac{\pi }{2}=6$

Sum of all elements of $A\cap B=2$   [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

$=2\left[\frac{100\times 101}{2}-3\left(\frac{33\times 34}{2}\right)-5\left(\frac{20\times 21}{2}\right)+15\left(\frac{6\times 7}{2}\right)\right]$

= 10100 – 3366 – 2100 + 630

              = 5264

$\left(sin10°.sin50°.sin70°\right).\left(sin10°.sin20°.sin40°\right)$

$=\left(\frac{1}{4}sin30°\right).\left[\frac{1}{2}sin10°\left(cos20°-cos60°\right)\right]$

$=\frac{1}{32}\left[sin30°-sin10°-sin10°\right]$

$\frac{1}{64}-\frac{1}{16}sin10°$

Clearly $\alpha =\frac{1}{64}$  

              Hence 16 + a^-1 = 80

$\left(4+x^2\right)dy-2x\left(x^2+3y+4\right)dx=0$

$\Rightarrow \frac{dy}{dx}=\left(\frac{6x}{x^2+4}\right)y+2x$

$e^{-3\mathcal{l}n\left(x^2+4\right)}=\frac{1}{{\left(x^2+4\right)}^3}$

so $\frac{y}{{\left(x^2+4\right)}^3}={\displaystyle \int \frac{2x}{{\left(x^2+4\right)}^3}dx+c}$

$\Rightarrow y=-\frac{1}{2}\left(x^2+4\right)+c{\left(x^2+4\right)}^3$

When x = 0, y = 0 gives $c=\frac{1}{32}$

So, for x = 2, y = 12

$\therefore {\displaystyle \underset{-6}{\overset{0}{\int }}f\left(x\right)dx=2\times \frac{1}{2}\left(2+5\right)\times 3=21}$

Boys (10)            Girls (5)

  (3)                        (3)

B₁ & B₂ should not be selected together

Total number of ways    

= (56 + 56) × 10 = 1120

$x^4-3x^3-2x^2+3x+1=0$

$\Rightarrow x=\pm 1$

and let a, b are roots of x² – 3x – 1 = 0

$\therefore \alpha +\beta =3\alpha \beta =-1$

$\therefore 1^3+{\left(-1\right)}^3+{\alpha }^3+{\beta }^3$

= 27 + 3 (3) = 36