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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 VSAT events visit official website, or bookmark this page - shisha.com. We will be updating all relevant and updated information here.

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

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.

VSAT is easy to moderate level. It is based on 10+2 level, however questions are easy to solve and if students have prepared well, they can qualify easily.

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

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

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

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