A) \[{{p}^{2}}{{q}^{3}}\]
B) \[{{q}^{3}}\]
C) \[{{q}^{4}}\]
D) \[{{p}^{2}}{{r}^{2}}{{q}^{2}}\]
Correct Answer: C
Solution :
As, given \[{{q}^{2}}\] = pr \[\frac{{{p}^{2}}-{{q}^{2}}+{{r}^{2}}}{{{p}^{-2}}-{{q}^{-2}}+{{r}^{-2}}}=\]\[\frac{{{p}^{2}}-{{q}^{2}}+{{r}^{2}}}{\frac{1}{{{p}^{2}}}-\frac{1}{{{q}^{2}}}+\frac{1}{{{r}^{2}}}}=\]\[\frac{{{p}^{2}}-{{q}^{2}}+{{r}^{2}}}{\frac{{{r}^{2}}-pr+{{p}^{2}}}{{{p}^{2}}{{r}^{2}}}}\] \[=\frac{({{p}^{2}}{{r}^{2}})({{p}^{2}}-{{q}^{2}}+{{r}^{2}})}{({{r}^{2}}-pr+{{p}^{2}})}=\frac{({{p}^{2}}{{r}^{2}})({{p}^{2}}-pr+{{r}^{2}})}{({{r}^{2}}-pr+{{p}^{2}})}\] \[={{p}^{2}}{{r}^{2}}={{(pr)}^{2}}={{({{q}^{2}})}^{2}}={{q}^{4}}\]
You need to login to perform this action.
You will be redirected in
3 sec
