Question

If the square root of \[\left( {\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}}} \right)\] is \[ \pm \left( {\dfrac{x}{y} + \dfrac{y}{x} - \dfrac{i}{m}} \right)\], then the value of m is
a. 2
b. 3
c. 4
d. 5

Answer 1

Hint: Here the question is related to the square root concept. The term which is present in the square root id in the form of algebraic expression. So by using the concepts of squares and square root first we simplify. For the further simplification we use arithmetic operations and hence we determine the value of m.

Complete step-by-step answer:
The square and square root are inverse to each other. The square of a term is written as a number which has power 2. The square root is denoted as \[\sqrt {} \].
Now consider the given question and it is written as
\[ \Rightarrow \sqrt {\left( {\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}}} \right)} = \pm \left( {\dfrac{x}{y} + \dfrac{y}{x} - \dfrac{i}{m}} \right)\]
On applying the square on both sides, we have
\[ \Rightarrow {\left( {\sqrt {\left( {\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}}} \right)} } \right)^2} = {\left( { \pm \left( {\dfrac{x}{y} + \dfrac{y}{x} - \dfrac{i}{m}} \right)} \right)^2}\]
Since the square and square root are inverse to each other it will get cancels and then we have
\[ \Rightarrow \left( {\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}}} \right) = {\left( {\dfrac{x}{y} + \dfrac{y}{x} - \dfrac{i}{m}} \right)^2}\]
We simplify the RHS by using the formula \[{(a + b - c)^2} = {a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ac\], so now we have
\[ \Rightarrow \left( {\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}}} \right) = \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{{{i^2}}}{{{m^2}}} + 2.\dfrac{x}{y}.\dfrac{y}{x} - 2\dfrac{y}{x}.\dfrac{i}{m} - 2.\dfrac{x}{y}.\dfrac{i}{m}\]
On further simplifying we have
\[ \Rightarrow \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}} = \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{{{i^2}}}{{{m^2}}} + 2 - 2\dfrac{y}{x}.\dfrac{i}{m} - 2.\dfrac{x}{y}.\dfrac{i}{m}\]
As we know that \[{i^2} = - 1\], on substituting this value and on multiplying the some terms in above inequality we have
\[ \Rightarrow \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} + \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}} = \dfrac{{{x^2}}}{{{y^2}}} + \dfrac{{{y^2}}}{{{x^2}}} - \dfrac{1}{{{m^2}}} + 2 - 2\dfrac{{iy}}{{xm}} - 2\dfrac{{ix}}{{ym}}\]
Cancelling the terms which are present in both LHS and RHS so we have
\[ \Rightarrow \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}} = - \dfrac{1}{{{m^2}}} + 2 - 2\dfrac{{iy}}{{xm}} - 2\dfrac{{ix}}{{ym}}\]
For the last two terms we can multiply and divide by \[i\] so we have
\[ \Rightarrow \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}} = - \dfrac{1}{{{m^2}}} + 2 - 2\dfrac{{{i^2}y}}{{ixm}} - 2\dfrac{{{i^2}x}}{{iym}}\]
As we know that \[{i^2} = - 1\], on substituting this value in above inequality we have
\[ \Rightarrow \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}} = - \dfrac{1}{{{m^2}}} + 2 + 2\dfrac{y}{{ixm}} + 2\dfrac{x}{{iym}}\]
Taking the common terms in the last two terms and we have
\[ \Rightarrow \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \dfrac{{31}}{{16}} = - \dfrac{1}{{{m^2}}} + 2 + \dfrac{2}{{im}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right)\]
On comparing the terms we have
\[ \Rightarrow \dfrac{1}{{2i}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) = \dfrac{2}{{im}}\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right)\] and \[\dfrac{{31}}{{16}} = 2 - \dfrac{1}{{{m^2}}}\]
On simplifying we have
\[ \Rightarrow \dfrac{1}{2} = \dfrac{2}{m}\] and \[\dfrac{1}{{{m^2}}} = 2 - \dfrac{{31}}{{16}}\]
On further simplification we have
\[ \Rightarrow \dfrac{1}{2} = \dfrac{2}{m}\] and \[\dfrac{1}{{{m^2}}} = \dfrac{1}{{16}}\]
On further simplifying the value of m will be 4.
Hence option c is the correct one.
So, the correct answer is “Option C”.

Note: Sometimes to solve the problem we have to compare the terms what and which and all terms will be equated. Because of simplifying we may not get the answer in an easier way. we can substitute the value of m and we can verify the problem.