Ask & Answer: India's Largest Education Community

The main criteria was you passed class 12 as a science student (PCB ,PCM ) both candidates can apply but commerce and arts students can't able to apply in this course 

When any change in matter is not responsible for changing the characteristic of the reaction,   the amount of matter we would have intensive properties. These properties, which we know as temperature or density,   remain constant even when the matter's size or quantity changes. On the other hand, we would have extensive properties in substances when they change when the amount or quantity in the system changes. 

Internal energy is the sum of all the energy in the system, as simple as that. But if we look into enthalpy, it is a state function that tells us how much internal energy is there in the system and how much work would be required to do to expand a gas against a constant pressure. 

For both work and heat, there are positive and negative signs, which mainly depend on transfer of energy. When work is done on the system, it will have a negative sign, but when work is done by the system, we will use a positive sign. This would be similar to how we view heat in chemical thermodynamics. If a system absorbs heat, it will be a positive sign. But if heat releases from a system, it will be a negative sign. 

Yes UC witnesses an incoming population of 3,000 international students each year moreover the Fall 2024 stats revealed that the university has witnessed a record of over 1,600 international UG enrollments and another interesting thing to note is that these students hail from 115 different countries.

x² – y² cosec²q = 5 $\frac{x^2}{1}-\frac{y^2}{si{n}^2\theta }=5$                        

x² cosec²q + y² = 5  $\frac{x^2}{si{n}^2\theta }+\frac{y^2}{1}=5$        

$e_H=\sqrt[]{7}{e}_e$                  

and $e_H=\sqrt[]{1+\frac{si{n}^2\theta }{1}}$  

-> $\sqrt[]{1+si{n}^2\theta }=\sqrt[]{7}\sqrt[]{1-si{n}^2\theta }$

1 + sin²q = 7 – 7 sin²q

->8sin²q = 6

-> $sin\theta =\sqrt[]{\frac{3}{4}}=\frac{\sqrt[]{3}}{2}$  

-> $\theta =\frac{\pi }{3}$  

5f(x) + 4f $\left(\frac{1}{x}\right)$  = x² – 4           .(1)

Replace x by  $\frac{1}{x}$

5f  $\left(\frac{1}{x}\right)$  + 4f(x) = $\frac{1}{x^2}$  – 4   .(2)

5 × equation (1) – 4 × equation (2)

$9f\left(x\right)=5x^{2-}\frac{4}{x^2}-4$            

$y=9f\left(x\right)\cdot x^2=\frac{5x^4-4-4x^2}{x^2}{x}^2$            

y = 5x⁴ – 4 – 4x²

y = 20x³ – 8x > 0

4x(5x² – 2) > 0

    $x\in \left(-\sqrt[]{\frac{2}{5}},0\right)\cup \left(\sqrt[]{\frac{2}{5}},\infty \right)$

Total ways to partition 5 into 4 parts are:

5 0 

4 1 0  $\frac{5!}{4!}=5$

3 2 0 $\frac{5!}{3!\cdot 2!}=10$

3 1 0 $\frac{5!}{2!2!2!}=15$

2 1 $\frac{5!}{2!\times 3!}=10$

51 Total way

(t + 1)dx = (2x + (t + 1)³)dt

$\frac{dx}{dt}-\frac{2x}{t+1}={\left(t+1\right)}^2$

I.F. $=e^{{\displaystyle \int -\frac{2}{t+1}dt}}=\frac{1}{{\left(t+1\right)}^2}$  

Solution is

$\frac{x}{{\left(t+1\right)}^2}={\displaystyle \int 1dt}$  

x = (t + c) (t + 1)²

$?$ x (0) = 2 then c = 2

x = (t + 2) (t + 1)²

 x (1) = 12

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}}$ dx

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}+\frac{8\sqrt[]{2}cosx}{\left(1+e^{-sinx}\right)\left(1+si{n}^{4}x\right)}\right)\right\}dx}$            

$=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{cosx}{1+si{n}^{4}x}dx}$            

Let sin x = t

$I=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^4}}$            

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}}dt}$       

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)dt}{{\left(t-\frac{1}{t}\right)}^2+2}}-4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(t+\frac{1}{t}\right)}^2-2}}$            

$=4\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}{\left(ta{n}^{-1}\frac{t-\frac{1}{t}}{\sqrt[]{2}}\right)}_0^1-4\sqrt[]{2}\cdot \frac{1}{2\sqrt[]{2}}{\left[log\left|\frac{t+\frac{1}{t}-\sqrt[]{2}}{t+\frac{1}{t}+\sqrt[]{2}}\right|\right]}_0^1$         

$=2\pi -2log\left|\frac{2-\sqrt[]{2}}{2+\sqrt[]{2}}\right|$        

$=2\pi +2log\left(3+2\sqrt[]{2}\right)$           

a = b = 2

${\left(\frac{C^{14}}{C^{12}}\right)}_t=\frac{{\left(\frac{C^{14}}{C^{12}}\right)}_{t=0}}{{\left(2\right)}^n}$

n = 3

t = 3 * 1580

= 4740 years

3, 7, 11, 15, 19, 23, 27, . 403 = AP₁

2, 5, 8, 11, 14, 17, 20, 23, . 401 = AP₂

so common terms A.P.

11, 23, 35, ., 395

->395 = 11 + (n – 1) 12

->395 – 11 = 12 (n – 1)

$\frac{384}{12}=n-1$    

32 = n – 1

n = 33

Sum =  $\frac{33}{2}$ [2×11+ (32)12]

=  $\frac{33}{2}$ [22 + 384]

= 6699

|A| = 3

|B| = 1

->|C| = |ABA^T| = |A|B|A⁷| = |A|²|B|

= 9

->|X| = |A|C|²|A^T|

= 3 * 9² * 3 = 9 * 9² = 729

$I={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\frac{xdx}{si{n}^4\left(2x\right)+co{s}^4\left(2x\right)}}$

           Let 2x = t then   $dx=\frac{1}{2}dt$

$I={\displaystyle \underset{}{\overset{}{\int }}\frac{\frac{t}{2}\cdot \frac{1}{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{tdt}{si{n}^{4}t+co{s}^{4}t}dt}$            

$I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-t\right)dt}{si{n}^{4}t+co{s}^{4}t}dt}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{si{n}^{4}tdt}{ta{n}^{4}t+1}}$            

Let tan t = y then

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+y^2\right)dy}{1+y^4}}$             

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2}dy}$

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+\frac{1}{y^2}\right)dy}{2+{\left(y-\frac{1}{y}\right)}^2}}$             

Let $y-\frac{1}{y}=u$  

$2I=\frac{\pi }{8}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2+u^2}}$

$=\frac{\pi }{8\sqrt[]{2}}{\left[ta{n}^{-1}\frac{4}{\sqrt[]{2}}\right]}_{-\infty }^{\infty }$                  

$I=\frac{{\pi }^2}{16\sqrt[]{2}}$