Bissoy.com
Doctors Medicines MCQs Q&A

Related Questions

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 :
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaaigdacaaIXaGaaiilaiaabckacaqG0bGaaeiAaiaa % bwgacaqGUbGaaeiOaiaabAgacaqGPbGaaeOBaiaabsgacaqGGcGaae % iDaiaabIgacaqGLbGaaeiOaiaabAhacaqGHbGaaeiBaiaabwhacaqG % LbGaaeiOaiaab+gacaqGMbGaaeiOamaabmaapaqaa8qacaqG4bGaey % OeI0IaaGymaiaaiodaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGa % aGynaaaakiaabckacqGHRaWkcaqGGcWaaSaaa8aabaWdbiaaigdaa8 % aabaWdbmaabmaapaqaa8qacaWG4bGaeyOeI0IaaGymaiaaigdaaiaa % wIcacaGLPaaapaWaaWbaaSqabeaapeGaaGynaaaaaaGccaGGGcGaai % Olaaaa!6E72! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = 11,{\rm{\;then\;find\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)If x + 1/(x - 13) = 11, then what will be the value of (x – 13)5 + 1/(x – 11)5?
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?
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
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 critical reasoning question is based on a short argument, a set of statements, or a plan of action. For each question, select the best answer of the choices given and explain why the chosen answer is the right fit. Question: Every Indian citizen has certain fundamental rights which include right to equality, right to freedom, right against exploitation, right to freedom of religion, cultural and educational rights, and right to freedom of religion. Which of the following would be infringement on one of these rights?