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If \[f\left( x \right) = \left\{ {\dfrac{{{x^2} + \propto }}{{2\sqrt {{x^2} + 1 + B} }}} \right\}\] \[
  {\text{for}}\;x \geqslant 0 \\
  {\text{for}}\;x < 0 \\
\]
And $f\left( {\dfrac{1}{2}} \right) = 2$ is continuous at x =0, value of $\left( { \propto ,P} \right)$ is :
A) $\left(\dfrac{7}{4}, -\dfrac{1}{4}\right)$
B) $\left(4-\sqrt{5},2-\sqrt{5}\right)$
C) $\left(0, -1\right)$
D) $\left(\dfrac{7}{4}, \dfrac{1}{4}\right)$

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Last updated date: 13th Jun 2024
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Answer
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Hint: The function to be continuous at a particular point like say a, then we have a condition for a, where $f\left( x \right)$ function needs to have its left hand limit, Right hand limit and value equal to each other. Like \[ \Rightarrow \dfrac{{\lim }}{{x \to {a^ - }}}\;f\left( x \right) = f\left( a \right) = \;\dfrac{{\lim }}{{x \to {a^ + }}}f\left( x \right)\]

Complete step by step solution: let’s begin with the given function which is represented as
$\begin{gathered}
  f\left( x \right) = \left\{ {\dfrac{{{x^2} + \propto }}{{2\sqrt {{x^2} + 1 + B} }}} \right\}\;\;\;\;\;\;\;\; \\
    \\
\end{gathered} $ $\begin{gathered}
  x \geqslant 0 \\
  x < 0 \\
\end{gathered} $
As we know that from given data,$f\left( {{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \right) = 2,$ so for this we will use the function $f\left( x \right) = {x^2} + \propto $ because ${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$ is greater than 0
So, $f\left( {\dfrac{1}{2}} \right) = \dfrac{1}{4} + \propto = 2$
\[ \Rightarrow \; \propto \; = 2 - \dfrac{1}{4} = \dfrac{7}{4}\]
Now we need to find the value of B, and the other information given is they are continuous at x = 0 for being continuous of a function the left hand limit, Right hand limit should be equal.
$so\dfrac{{\;\lim }}{{x \to {0^ - }}}f\left( x \right) = \dfrac{{\lim }}{{x \to {0^ + }}}f\left( x \right)$
$ \Rightarrow \dfrac{{\lim }}{{x \to {0^ - }}}\;\left( {{x^2} + \propto } \right) = \dfrac{{\lim }}{{x \to {0^ + }}}2\sqrt {{x^2} + 1} + B)$
$ \Rightarrow \lim {\left( 0 \right)^2} + \propto \; = 2\sqrt {{0^2} + } 1 + B$
          $ \propto - 2 + B$
So we get,$B = \; \propto - 2\;and \propto = {\raise0.5ex\hbox{$\scriptstyle 7$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}$
$ \Rightarrow $ we get $B = \; \propto - 2 = \dfrac{7}{4} - 2 = - \dfrac{1}{4}$
Hence we get the value of \[\left( { \propto ,B} \right) = \left( {{\raise0.5ex\hbox{$\scriptstyle 7$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle {4,}$}}{\raise0.5ex\hbox{$\scriptstyle { - 1}$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}} \right)\] option A is the correct answer.

Note: we know that function to get continuous left-hand limit, Right hand limit and function value should be equal.
$\dfrac{{\lim }}{{x \to {a^ - }}}f\left( x \right) = f\left( a \right) = \dfrac{{\lim }}{{x \to {a^ + }}}f\left( x \right)$ Similarly, for differentiability of a function at point a is checked by
$\dfrac{{\lim }}{{x \to {a^ - }}}f\left( x \right) = f'\left( a \right) = \dfrac{{\lim }}{{x \to {a^ + }}}f'\left( x \right)$