Question

If $x \in \mathbb{R}$ , the number of solutions of $\sqrt {2x + 1} - \sqrt {2x - 1} - 1$ is:
A) $1$
B) $2$
C) $4$
D) $\infty $

Answer 1

Hint:We will use the rules of indices to simplify the given expression. We will equate the given equation with zero and then simplify it for the value of the unknown. The relation between roots of an equation and the expression for calculating the cardinality of the set of the solution of the given equation.

Complete step-by-step answer:
The given expression is $\sqrt {2x + 1} - \sqrt {2x - 1} - 1$.
Let, the given expression be $f(x)$ .
Therefore, we will solve the equation $f(x) = 0$.
Consider the following:
$\sqrt {2x + 1} - \sqrt {2x - 1} - 1 = 0$
We will add $1$ on both sides of the above equation.
$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$
Squaring both sides, we get,
${\left( {\sqrt {2x + 1} - \sqrt {2x - 1} } \right)^2} = 1$
Now expand the brackets using algebraic identity $(a - b)^2= a^2+b^2-2ab$
$\left( {2x + 1} \right) - 2\sqrt {\left( {2x + 1} \right)\left( {2x - 1} \right)} + \left( {2x - 1} \right) = 1$
Simplifying the equation and arranging the terms we write:
$4x - 1 = 2\sqrt {4{x^2} - 1} $
Now again square on both sides.
${\left( {4x - 1} \right)^2} = 4\left( {4{x^2} - 1} \right)$
Expand the brackets on both the sides,
$16{x^2} + 1 - 8x = 16{x^2} - 4$
We can cancel the term $16{x^2}$ on both sides and rewrite the above equation as:
$8x - 5 = 0$
Now the above equation is a linear equation in first degree.
Therefore, it will always have a unique and only one root.

So, the correct answer is “Option A”.

Note:Here expanding the correct bracket and transferring the expressions from correct sides is important. Note that it is not necessary to solve the equation completely, instead derive it to the level where we can calculate the number of solutions for the equation. If we draw the graph of the given expression it will cut the $x$ axis at one and only one point as it has a single root.