Question

The number of pairs $ \left( a,b \right) $ of positive real numbers satisfying $ {{a}^{4}}+{{b}^{4}}<1 $ and $ {{a}^{2}}+{{b}^{2}}>1 $ is
(a) 0
(b) 1
(c) 2
(d) More than 2

Answer 1

Hint: First, we will make inequality sign the same in both the equations by multiplying equation $ {{a}^{2}}+{{b}^{2}}>1 $ with minus sign. After doing this, we have to add both the equations. Then we have to make that equation into a perfect square equation by adding and subtracting some terms. Then, we will compare that equation with equation of circle i.e. $ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $ . From this, we will get or answer.

Complete step-by-step answer:
Here, we have two equations given i.e. $ {{a}^{4}}+{{b}^{4}}<1 $ and $ {{a}^{2}}+{{b}^{2}}>1 $ . We have to find how many pairs $ \left( a,b \right) $ of positive real numbers satisfies this equation.
So, we will take equation $ {{a}^{2}}+{{b}^{2}}>1 $ and will multiply it with minus sign to make inequality sign same for both equations.
On multiplying the equation with minus sign, we get
 $ \Rightarrow -{{a}^{2}}-{{b}^{2}}<-1 $
Now, we will add both the equation i.e. $ {{a}^{4}}+{{b}^{4}}<1 $ and $ -{{a}^{4}}-{{b}^{4}}<-1 $ . So, we will get
 $ \begin{align}
  & \text{ }{{a}^{4}}+{{b}^{4}}<1 \\
 & -{{a}^{2}}-{{b}^{2}}<-1 \\
 & \overline{{{a}^{4}}-{{a}^{2}}+{{b}^{4}}-{{b}^{2}}<-1+1} \\
\end{align} $
On further solving, we get equation as
 $ {{a}^{4}}-{{a}^{2}}+{{b}^{4}}-{{b}^{2}}<0 $ ……………………………….(1)
Now, we will make equation (1) to be a perfect quadratic equation. So, we will add and subtract $ \dfrac{1}{4} $ twice as we have to variable a and b. On doing this, we will get
 $ {{a}^{4}}-{{a}^{2}}+{{b}^{4}}-{{b}^{2}}+\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{4}<0 $
Further we will group the terms which will be perfect square equation, so we can write it as
 $ \left( {{a}^{4}}-{{a}^{2}}+\dfrac{1}{4} \right)+\left( {{b}^{4}}-{{b}^{2}}+\dfrac{1}{4} \right)-\dfrac{2}{4}<0 $
On solving the brackets and shifting constant term to RHS, we will get
 $ {{\left( {{a}^{2}}-\dfrac{1}{2} \right)}^{2}}+{{\left( {{b}^{2}}-\dfrac{1}{2} \right)}^{2}}<\dfrac{2}{4} $
 $ {{\left( {{a}^{2}}-\dfrac{1}{2} \right)}^{2}}+{{\left( {{b}^{2}}-\dfrac{1}{2} \right)}^{2}}<\dfrac{1}{2} $ ………………………(2)
Now, we know that above equation is similar to equation of circle given as $ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $ where r is radius of circle and h, k is the centre of circle. So, on comparing this with equation (2), we get that $ \left( h,k \right)=\left( \dfrac{1}{2},\dfrac{1}{2} \right) $ and radius $ r=\dfrac{1}{\sqrt{2}} $ .
So, \[\left( {{a}^{2}},{{b}^{2}} \right)\] will lies inside circle with centre \[\left( \dfrac{1}{2},\dfrac{1}{2} \right)\] and radius \[\dfrac{1}{\sqrt{2}}\] .
Thus, in a circle there are infinitely many points. So, there are more than 2 pairs.
Hence, option (d) is correct.

Note: Do not get confused while comparing the equation $ {{\left( {{a}^{2}}-\dfrac{1}{2} \right)}^{2}}+{{\left( {{b}^{2}}-\dfrac{1}{2} \right)}^{2}}<\dfrac{1}{2} $ with equation of circle given as $ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $ .As x, y are having power 1 while we have power of 2 in a, b. We don’t have to compare what the power should, just have to match the general form of the equation. Then if we multiply minus sign to this equation i.e. $ {{a}^{4}}+{{b}^{4}}<1 $ then, also at the end we will get the same answer but by doing this steps will be more in calculation. Also, do not forget to take square root in finding radius r. Students generally take radius r as \[\dfrac{1}{2}\] and this leads to wrong answers. So, do not make this mistake.