Question

Let [x] be the greatest integer function. Then the equation sin x= [1+sinx] + [1-cosx] has
A.One solution in \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\]
B.One solution in \[\left[ {\dfrac{\pi }{2},\pi } \right]\]
C.One solution in R
D.No solution in R

Answer 1

Hint: In this question for the given equation we will substitute the values of function x between the range \[\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]\], one by one and then we will check the solution of the function for which the LHS of the given equation is equal to the RHS of the equation and the points at which the LHS and RHS of the equation will be equal will be the solution of the given equation.

Complete step-by-step answer:
Given the equation
\[\sin x = \left[ {1 + \sin x} \right] + \left[ {1 - \cos x} \right] - - (i)\]
This equation can be written as
\[\sin x = 2 + \sin x - \cos x - - (ii)\]
Now we substitute the values of x in the equation (ii) and will check the solution at which the equation (ii) has RHS and LHS equal
Let \[x = - \dfrac{\pi }{2}\]
So we substitute the value of x in equation (ii), so by substituting
\[
  \sin x = 2 + \sin x - \cos x \\
\Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = 2 + \sin \left( { - \dfrac{\pi }{2}} \right) + \cos \left( { - \dfrac{\pi }{2}} \right) \\
   - 1 = 2 + - 1 + 0 \\
   - 1 \ne 1 \;
 \]
Hence we can say LHS is not equal to RHS
Now again let \[x = 0\]
So we again substitute the value of x in equation (ii), so by substituting we get
\[
  \sin x = 2 + \sin x - \cos x \\
\Rightarrow \sin \left( 0 \right) = 2 + \sin \left( 0 \right) + \cos \left( 0 \right) \\
  0 = 2 + 0 + 1 \\
  0 \ne 3 \;
 \]
Hence we can say LHS is not equal to RHS
Now again let \[x = \dfrac{\pi }{2}\]
So we again substitute the value of x in equation (ii), so by substituting we get
\[
  \sin x = 2 + \sin x - \cos x \\
\Rightarrow \sin \left( {\dfrac{\pi }{2}} \right) = 2 + \sin \left( {\dfrac{\pi }{2}} \right) + \cos \left( {\dfrac{\pi }{2}} \right) \\
  1 = 2 + 1 + 0 \\
  1 \ne 3 \;
 \]
Hence we can say LHS is not equal to RHS
So we can conclude that the RHS and LHS are not equal for any value of x, hence we can say the function has no solution for any real number.
So, the correct answer is “Option D”.

Note: Students can also find the solution of the given function by plotting the graph of the given equation and check the number of points which satisfies the equation. The value of the sine function lies within the range \[\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]\].