Question

If $ {{x}^{2}}-4x+5-\sin y=0,y\in [0,2\pi ), $ then what are the values of x and y.
(a) x=1, y=0
(b) x=1, y= $ \dfrac{\pi }{2} $
(c) x=2, y=0
(d) x=2, y= $ \dfrac{\pi }{2} $

Answer 1

Hint: As in the given equation there are two independent values x and y, and only one equation is given so we need to manipulate the equation in such a way that we find the solution at end points or extreme points.
In the given equation we have a trigonometric variable sine that needs to be manipulated to find values at extreme points i.e. $ -1\le \sin y\le 1 $

Complete step-by-step answer:
Given expression:
 $ {{x}^{2}}-4x+5-\sin y=0 $
Equation $ {{x}^{2}}-4x+5 $ can also be expressed as $ {{(x-2)}^{2}}+1 $ , hence equation becomes,
 $ {{(x-2)}^{2}}+1-\sin y=0......(1) $
Now, taking $ \sin y $ to the other side of the equation ,
\[{{(x-2)}^{2}}+1=\sin y\]
We know that $ -1\le \sin y\le 1 $ and \[{{(x-2)}^{2}}\] is a positive number, now in the above equation maximum value of the R.H.S. is 1 and on the L.H.S. we have (1+\[{{(x-2)}^{2}}\]) so \[{{(x-2)}^{2}}\] needs to be zero in order to make the both sides equal hence,
\[{{(x-2)}^{2}}=0\]
 $ \begin{align}
  & x-2=0 \\
 & x=2 \\
\end{align} $
Putting the value of $ x=2 $ in the equation 1 we get,
\[\begin{align}
  & {{(2-2)}^{2}}+1=\sin y \\
 & \sin y=1 \\
\end{align}\]
In the interval $ y\in [0,2\pi ) $ , $ \sin y=1 $ only at $ \dfrac{\pi }{2} $ hence
 $ \begin{align}
  & \sin y=1 \\
 & y=\dfrac{\pi }{2} \\
\end{align} $
Hence value of x=2 and value of y = $ \dfrac{\pi }{2} $ .
So, the correct answer is “Option D”.

Note: While solving questions which involve both trigonometric variables and algebraic variables and number of equations given are less than that of variables, these questions can be solved at extreme value cases easily.We can also solve this question easily by putting the values of x and y given in each option, but always make sure that the value you are checking lies in between the interval given in the question sometimes values satisfies the equation but they don’t lie between the given interval so make sure to check that as well.For example: x=2 and y= $ \dfrac{5\pi }{2} $ also satisfies equation but value of y is not in the interval mentioned, so it is wrong.