Class 12th

Yes, after passing class 12 you can go to BITS Pilani Goa by passing in BITSAT. Your marks in Physics, Chemistry and Maths/Biology should be 75 per cent or more. Your admission in the programme will depend on the score in the entrance exam. As well, be sure to comply with both academic and exam eligibility.

These are the difference between the alpha, beta and gamma decay:

• Alpha decay: It reduces mass number by 4 and atomic number by 2. It is the emission of a helium nucleus (2 protons, 2 neutrons).
• Beta decay: It changes atomic number by ±1 but there is no change in the mass number. It involves a neutron converting into a proton (or vice versa).
• Gamma decay: Here, there is no change in the atomic or mass number, just the emission of high-energy photons.

Within a nucleus, how tightly nucleons are bound is the binding energy per nucleon. If the binding energy per nucleon is higher then it means more stable nuclei. Iron-56 is the most stable as it has one of the highest values. Both the fusion of light nuclei and the fission of heavy nuclei release energy by moving toward higher binding energy per nucleon.

The difference between the mass of a nucleus and the sum of the masses of its individual protons and neutrons is known as the mass defect. Mass defect is the key concept in understanding the nuclear stability and energy release in nuclear reactions. The binding energy is converted from the "missing" mass. The binding energy holds the nucleus together. 

Consider, $I={\displaystyle {\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)}dx$

$\begin{array}{l}\Rightarrow I={\displaystyle {\int }_0^{1}ta{n}^{-1}\left(\frac{x-\left(1-x\right)}{1+x\left(1-x\right)}\right)}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}x-ta{n}^{-1}\left(1-x\right)\right]}dx-----(1)\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(1-1+x\right)\right]}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}x\right]}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(x\right)\right]}dx-----(2)\end{array}$

Adding (1) and (2), we get

$\begin{array}{l}\Rightarrow 2I={\displaystyle {\int }_0^1\left[ta{n}^{-1}x-ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}x\right]}dx\\ \Rightarrow 2I=0\\ \Rightarrow I=0\end{array}$

Thus, the correct option is B.

Giver, f(a + bx) = f(x). _________ (1)

Let I $={\displaystyle {\int }_a^{b}x}+(x)={\displaystyle {\int }_a^b(}a+b-x)f(a+b-x)dx\_\_\_\_\_\left(2\right)$

$\{?{\displaystyle {\int }_a^{b}f}(x)dx={\displaystyle {\int }_a^{b}f}(a+b-x)dx.$

$={\displaystyle {\int }_a^b(}a+b-x)f(x)dx.$ ___________ (3) {because Equation (1)}

$+={\displaystyle {\int }_a^b[(a+b-x)f(x)+xf(x)]}dx$

$\Rightarrow \mathrm{2I}={\displaystyle {\int }_a^b(}a+b-x+x)\stackrel{}{f}(x)dx$

$\Rightarrow \mathrm{2I}={\displaystyle {\int }_a^b(}a+b)f(x)dx$

$\Rightarrow =\frac{a+b}{2}{\displaystyle {\int }_a^{b}f}(x)dx.$

therefore, Option D is correct.

Let $=\int \frac{cos2xdx}{{(sinx+cosx)}^2}$

$=\int \frac{co{s}^{2}x-si{n}^{2}x}{{(sinx+cosx)}^2}dx\{?cos2x=co{s}^{2}x-si{n}^{2}x$

$=\int \frac{(cosx+sinx)(cosx-sinx)}{{(sinx+cosx)}^2}dx\{?a^2-b^2=(x+b)(x-b)$

$=\int \frac{(cosx-sinx)}{sinx+cosx.}dx$

$=log|sinx+cosx|+c\{{\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=log|f(x)|+c}$

So, option B is correct.

Let $=\int \frac{dx}{e^x+e^{-x}}$

$=\int \frac{dx}{e^x+\frac{1}{e^x.}}=\int \frac{e^{x}dx}{e^x·e^x+1}$

$=\int \frac{e^{x}dx}{e^{2x}+1}$

Putting e^x = t

e^xdx = dt.

$\therefore =\int \frac{dt}{t^2+1.}$

= tan^{- 1} t + c

= tan^{- 1} (e^x) + c

therefore, Option A is correct.

Let $I={\displaystyle {\int }_0^1{e}^{2-3x}dx}$

We know that,

${\displaystyle {\int }_a^{b}f\left(x\right)}dx=\left(b-a\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(a\right)+f\left(a+h\right)+...+f\left(a\left(n-1\right)h\right)\right]$

Where, $h=\frac{b-a}{n}$

Here, $a=0,b=1$ and $f\left(x\right)e^{2-3x}$

$\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$

$\therefore {\displaystyle {\int }_0^1{e}^{2-3x}dx}=\left(1-0\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(0\right)+f\left(0+h\right)+...+f\left(0+\left(n-1\right)h\right)\right]$

$\begin{array}{l}=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2+e^{2-3x}+...+e^{2-3\left(n-1\right)h}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{1+e^{-3h}+e^{-6h}+e^{-9h}+...+e^{-3\left(n-1\right)h}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-{\left(e^{-3h}\right)}^n}{1-\left(e^{-3h}\right)}\right\}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-e^{\frac{-3}{n}n}}{1-e^{\frac{-3}{n}}}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\end{array}$

$\begin{array}{l}=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\left(-\frac{1}{3}\right)\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]=\frac{e^2\left(e^{-3}-1\right)}{3}\underset{n\to \infty }{lim}\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]\\ =\frac{-e^2\left(e^{-3}-1\right)}{3}(1)\\ =\frac{-e^{-1}+e^2}{3}\\ =\frac{1}{3}\left(e^2-\frac{1}{e}\right)\end{array}$

Kindly go through the solution