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If $$\frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c}$$        and A = 60°, x = 15°, then the value of $${\left( {\frac{{a + c}}{b}} \right)^2} + {\left( {\frac{{a - c}}{b}} \right)^2}$$   is . . . . . . . .

Correct Answer: 4
$$\eqalign{ & \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & A = {60^ \circ },\,\,x = {15^ \circ }{\text{ given,}} \cr & \Rightarrow \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{60}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}} = \frac{a}{b} \cr & \Rightarrow \frac{{\cos {{75}^ \circ }}}{{\cos {{135}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{\sin {{15}^ \circ }}}{{ - \sin {{45}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{{{{\sin }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = 2{\sin ^2}{15^ \circ }........\left( {\text{i}} \right) \cr & \Rightarrow \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{180}^ \circ }} \right)}} = \frac{b}{c} \cr & \Rightarrow \frac{{ - \sin {{45}^ \circ }}}{{ - \cos {{15}^ \circ }}} = \frac{b}{c} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = \frac{{{{\cos }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = 2{\cos ^2}{15^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{Now, }}{\left( {\frac{a}{b} + \frac{c}{b}} \right)^2} + {\left( {\frac{a}{b} - \frac{c}{b}} \right)^2} \cr & = 2\left \cr & = 2\left \cr & = 4 \cr} $$

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?

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?

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

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

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

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

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?

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|\

The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by
$${h_1}\left( t \right) = 1,$$   $${h_2}\left( t \right) = u\left( t \right),$$   $${h_3}\left( t \right) = \frac{{u\left( t \right)}}{{t + 1}},$$    $${h_4}\left( t \right) = {e^{ - 3t}}u\left( t \right)$$
Where u(t) is the unit step function. Which of these systems is time invariant, causal, and stable?

Consider a system governed by the following equations:
$$\frac{{{\text{d}}{{\text{x}}_1}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_2}\left( {\text{t}} \right) - {{\text{x}}_1}\left( {\text{t}} \right);\,\frac{{{\text{d}}{{\text{x}}_2}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_1}\left( {\text{t}} \right) - {{\text{x}}_2}\left( {\text{t}} \right)$$
The initial conditions are such that $${{\text{x}}_1}\left( 0 \right)