Question

$If \: the \: function \\ f(x) = \begin{cases} \dfrac{ \sqrt{(1 + Px) } - \sqrt{(1 - Px)}}{ x} \: \: \: ; - 1 \leq x < 0 \\ \\ \bigg( \dfrac{2x + 1}{x - 2} \bigg) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ;0 \leq x \leq 1\end{cases}$​

is continuous [-1,1] the P=___

Answer 1

$\large\underline{\sf{Solution-}}$

Given that function

$\rm \: f(x) = \begin{cases} \dfrac{ \sqrt{(1 + Px) } - \sqrt{(1 - Px)}}{ x} \: \: \: ; - 1 \leq x < 0 \\ \\ \dfrac{2x + 1}{x - 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ;0 \leq x \leq 1\end{cases}$

is continuous on [ - 1, 1 ].

We know, A function f(x) is continuous at x = a, iff

$\color{green}\boxed{ \rm{ \:\rm \: \displaystyle\lim_{x \to a^-}\rm f(x) = \displaystyle\lim_{x \to a^ + }\rm f(x) = f(a) \: \: }}$

Since, it is given that function is continuous at x = 0.

So,

$\rm \: f(0) = \displaystyle\lim_{x \to 0^-}\rm f(x)$

$\rm \: \dfrac{2(0)+ 1}{0 - 2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{ \sqrt{(1 + Px) } - \sqrt{(1 - Px)}}{ x}$

$\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{ \sqrt{(1 + Px) } - \sqrt{(1 - Px)}}{ x} \times \frac{\sqrt{(1 + Px)} + \sqrt{(1 - Px)}}{\sqrt{(1 + Px)} + \sqrt{(1 - Px)}}$

$\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{\bigg(\sqrt{(1 + Px)}\bigg)^{2} - \bigg(\sqrt{(1 - Px)}\bigg)^{2} }{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)}$

$\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{(1 + Px) - (1 - Px)}{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)}$

$\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{1 + Px - 1 + Px}{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)}$

$\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{2Px}{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)}$

$\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{2P}{\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)}$

Now, to evaluate this limit, we use method of Substitution.

So, Substitute

$\rm \: x \: = \: 0 - h \: = \: - \: h, \: \: as \: x \: \to \: 0 \: \: as \: \: h \: \to \: 0$

So, above expression can be rewritten as

$\rm \: - \dfrac{1}{2} = \displaystyle\lim_{h \to 0}\rm \dfrac{2P}{\bigg(\sqrt{(1 - Ph)} + \sqrt{(1 + Ph)}\bigg)}$

$\rm \: - \dfrac{1}{2} = \dfrac{2P}{ \sqrt{1 - 0} + \sqrt{1 + 0} }$

$\rm \: - \dfrac{1}{2} = \dfrac{2P}{ 1 + 1}$

$\rm \: - \dfrac{1}{2} = \dfrac{2P}{2}$

$\color{green}\rm\implies \:\boxed{ \rm{ \:P \: = \: - \: \dfrac{1}{2} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

refer the given attachment