Let $${\text{f}}\left( {{\text{x, y}}} \right) = \frac{{{\text{a}}{{\text{x}}^2} + {\text{b}}{{\text{y}}^2}}}{{{\text{xy}}}},$$ where a and b are constants. If $$\frac{{\partial {\text{f}}}}{{\partial {\text{x}}}} = \frac{{\partial {\text{f}}}}{{\partial {\text{y}}}}$$ at x = 1 and y = 2, then the relation between a and b is

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$ + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$ + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$ + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$ is?

A

0

B

5

C

1

D

4

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$ + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$ + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$ + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$ is?

A

0

B

1

C

4

D

1 + abcd

Let the function
\[{\text{f}}\left( \theta \right) = \left| {\begin{array}{*{20}{c}} {\sin \theta }&{\cos \theta }&{\tan \theta } \\ {\sin \left( {\frac{\pi }{6}} \right)}&{\cos \left( {\frac{\pi }{6}} \right)}&{\tan \left( {\frac{\pi }{6}} \right)} \\ {\sin \left( {\frac{\pi }{3}} \right)}&{\cos \left( {\frac{\pi }{3}} \right)}&{\tan \left( {\frac{\pi }{3}} \right)} \end{array}} \right|\

A

where \\

B

and \ denote the derivative of f with respect to \. Which of the following statements is/are TRUE? I. There exists \ such that \ II. There exists \ such that\

C

A. l only

D

ll only

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$ where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$ is the angular momentum. The Hamiltonian does not commute with

A

$$\overrightarrow {\bf{L}} + \overrightarrow {\bf{S}} $$

B

$$\overrightarrow {{{\bf{S}}^2}} $$

C

$${L_z}$$

D

$$\overrightarrow {{{\bf{L}}^2}} $$

The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$ = ?

A

0

B

3

C

$$\frac{1}{3}$$

D

2

$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$ $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$ $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$ is simplified to :

A

0

B

1

C

2

D

3

The quark content of $$\sum {^ + } ,\,{K^ - },\,{\pi ^ - }$$ and p is indicated: $$\left| {\sum {^ + } } \right\rangle = \left| {uus} \right\rangle ;\,\left| {{K^ + }} \right\rangle = \left| {s\overline u } \right\rangle ;\,\left| \pi \right\rangle = \left| d \right\rangle ;\,\left| p \right\rangle = \left| {uud} \right\rangle $$
In the process, $${\pi ^ - } + p \to {K^ - } + \sum {^ + } ,$$ considering strong interactions only, which of the following statements is true?

A

The process is allowed because ΔS = 0

B

The process is allowed because $$\Delta {I_3} = 0$$

C

The process is not allowed because ΔS ≠ 1 and $$\Delta {I_3} \ne 0$$

D

The process is not allowed because the Baryon number is violated

Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$ with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$ and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$ is

A

-2

B

-1

C

0

D

1

A system in a normalized state $$\left| \psi \right\rangle = {c_1}\left| {{\alpha _1}} \right\rangle + {c_2}\left| {{\alpha _2}} \right\rangle $$ with $$\left| {{\alpha _1}} \right\rangle $$ and $$\left| {{\alpha _2}} \right\rangle $$ representing two different eigen states of the system requires that the constants c₁ and c₂ must satisfy the condition

A

$$\left| {{c_1}} \right| \cdot \left| {{c_2}} \right| = 1$$

B

$$\left| {{c_1}} \right| + \left| {{c_2}} \right| = 1$$

C

$${\left( {\left| {{c_1}} \right| + \left| {{c_2}} \right|} \right)^2} = 1$$

D

$${\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} = 1$$

Consider the line integral $$I = \int_{\text{c}} {\left( {{{\text{x}}^2} + {\text{i}}{{\text{y}}^2}} \right){\text{dz,}}} $$ where z = x + iy. The line c is shown in the figure below

The value of $$I$$ is

A

$$\frac{1}{2}{\text{i}}$$

B

$$\frac{2}{3}{\text{i}}$$

C

$$\frac{3}{4}{\text{i}}$$

D

$$\frac{4}{5}{\text{i}}$$