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Which of the following contain the right usage of a relative pronoun ?

Which of the following contain the right usage of a relative pronoun ? Correct Answer Tehrin who loves the boy is going to marry Tabinda.

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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 :
You install and run a third-party 32-bit application named Application on your Windows 2000 Server computer. After several days, the application stops responding. You open Task Manager and find that the CPU usage is at 100 percent. The normal range of CPU usage on the server is from 20 percent to 30 percent You end the application. However, you see that the CPU on the server is still at 100 percent. Task Manager shows no other applications running. You then examine the Processes page in Task Manager and confirm that the Application.exe process is no longer running. You want to return the CPU usage to its normal range. What should you do?
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?
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaaigdacaaIXaGaaiilaiaabckacaqG0bGaaeiAaiaa % bwgacaqGUbGaaeiOaiaabAgacaqGPbGaaeOBaiaabsgacaqGGcGaae % iDaiaabIgacaqGLbGaaeiOaiaabAhacaqGHbGaaeiBaiaabwhacaqG % LbGaaeiOaiaab+gacaqGMbGaaeiOamaabmaapaqaa8qacaqG4bGaey % OeI0IaaGymaiaaiodaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGa % aGynaaaakiaabckacqGHRaWkcaqGGcWaaSaaa8aabaWdbiaaigdaa8 % aabaWdbmaabmaapaqaa8qacaWG4bGaeyOeI0IaaGymaiaaigdaaiaa % wIcacaGLPaaapaWaaWbaaSqabeaapeGaaGynaaaaaaGccaGGGcGaai % Olaaaa!6E72! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = 11,{\rm{\;then\;find\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)If x + 1/(x - 13) = 11, then what will be the value of (x – 13)5 + 1/(x – 11)5?
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|\