If \[f(x) = \left\{ \begin{array}{l}\dfrac{{\sqrt {(1 + px)} - \sqrt {(1 - px)} }}{x}, - 1 \le x < 0\\\dfrac{{2x + 1}}{{x - 2}},0 \le x \le 1\end{array} \right.\] is continuous in the interval\[\left[ { - 1,1} \right]\], then p equal to:
A.\[ - 1\]
B.\[\dfrac{{ - 1}}{2}\]
C.\[\dfrac{1}{2}\]
D.\[1\]

Hint: Here we have to use the concept of the continuity of the function to get the value of p. firstly we will find the value of the function at \[x = 0\]. Then we will find the value of the left hand limit of the function at \[x = 0\]. Then we will equate the left hand limit and value of function at \[x = 0\] because the function is continuous at \[x = 0\]. So by this we will get the value of p.

Complete step by step solution:
Given function is\[f(x) = \left\{ \begin{array}{l}\dfrac{{\sqrt {(1 + px)} - \sqrt {(1 - px)} }}{x}, - 1 \le x < 0\\\dfrac{{2x + 1}}{{x - 2}},0 \le x \le 1\end{array} \right.\]
It is given in the question that function\[f(x)\]is a continuous function in the interval of \[\left[ { - 1,1} \right]\], which means function must be continuous at\[x = 0\]. Therefore, we get
\[f(0) = \dfrac{{2(0) + 1}}{{(0) - 2}} = \dfrac{{ - 1}}{2}\]
Now we will find the value of the left hand limit of the function\[f(x)\] at\[x = 0\] to get the value of the function in terms of p. Therefore, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 - h) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {(1 - ph)} - \sqrt {(1 + ph)} }}{{ - h}}\]
Now we have to simplify the above equation and then put the value of\[h = 0\]. We cannot put the value of h because then the denominator will become zero. So we will simplify the equation first and then put the value of h.
So, we can write the above equation as
\[ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {(1 - ph)} - \sqrt {(1 + ph)} }}{{ - h}} \times \dfrac{{\sqrt {(1 - ph)} + \sqrt {(1 + ph)} }}{{\sqrt {(1 - ph)} + \sqrt {(1 + ph)} }}\]
Now be solving this we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - ph - 1 - ph}}{{ - h\left( {\sqrt {(1 - ph)} + \sqrt {(1 + ph)} } \right)}}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2ph}}{{ - h\left( {\sqrt {(1 - ph)} + \sqrt {(1 + ph)} } \right)}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{2p}}{{\left( {\sqrt {(1 - ph)} + \sqrt {(1 + ph)} } \right)}}\]
Now we will put the limit of the function i.e. \[h = 0\]. So, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \dfrac{{2p}}{2} = p\]
Since we know that the function \[f(x)\] is continuous at \[x = 0\]. So, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0)\]
By putting their values in the above expression, we get
\[ \Rightarrow p = \dfrac{{ - 1}}{2}\]
Hence, option B is the correct option.

Note: Here we have to note that continuous function is the function whose value does not change or value remains constant and when the function is continuous at a point then the left hand limit of the function and the right hand limit of the function is equal to the value of the function at that point.
\[\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)\]
In the question we have to note when to put the value of the limit in the equation. This is because we can get infinity as the answer which is not the desired result so that is why we have to keep the note of when to put the limit in the equation.