If c is small in comparison with I, then $ {{\left( \dfrac{I}{I+c} \right)}^{\dfrac{1}{2}}}+{{\left( \dfrac{I}{I-c} \right)}^{\dfrac{1}{2}}}= $ .
\[\begin{align}
  & A.2+\dfrac{3c}{4I} \\
 & B.2+\dfrac{3{{c}^{2}}}{4{{I}^{2}}} \\
 & C.1+\dfrac{3{{c}^{2}}}{4{{I}^{2}}} \\
 & D.1+\dfrac{3c}{4I} \\
\end{align}\]

Hint: In this question, we need to simplify the expression $ {{\left( \dfrac{I}{I+c} \right)}^{\dfrac{1}{2}}}+{{\left( \dfrac{I}{I-c} \right)}^{\dfrac{1}{2}}} $ where c is small in comparison with I. For this, we will first use the properties of exponents which are \[{{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}\text{ and }\dfrac{1}{a}={{a}^{-1}}\]. Then we will apply the binomial expansion to evaluate our answer. Binomial expansion for $ {{\left( 1+x \right)}^{n}} $ is given by $ {{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+\cdots \cdots $ .

Complete step by step answer:
Here we are given the expression as $ {{\left( \dfrac{I}{I+c} \right)}^{\dfrac{1}{2}}}+{{\left( \dfrac{I}{I-c} \right)}^{\dfrac{1}{2}}} $ . We need to simplify this expression.
Let us first take I common from the denominator of both terms, we get $ {{\left( \dfrac{I}{I\left( 1+\dfrac{c}{I} \right)} \right)}^{\dfrac{1}{2}}}+{{\left( \dfrac{I}{I\left( 1-\dfrac{c}{I} \right)} \right)}^{\dfrac{1}{2}}} $ .
Cancelling I in the numerator and the denominator of both terms we get $ {{\left( \dfrac{I}{\left( 1+\dfrac{c}{I} \right)} \right)}^{\dfrac{1}{2}}}+{{\left( \dfrac{I}{\left( 1-\dfrac{c}{I} \right)} \right)}^{\dfrac{1}{2}}} $ .
Now using the law of exponents \[{{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}\] we get $ \dfrac{{{\left( I \right)}^{\dfrac{1}{2}}}}{{{\left( 1+\dfrac{c}{I} \right)}^{\dfrac{1}{2}}}}+\dfrac{{{\left( I \right)}^{\dfrac{1}{2}}}}{{{\left( 1-\dfrac{c}{I} \right)}^{\dfrac{1}{2}}}} $ .
As we know $ {{\left( I \right)}^{\dfrac{1}{2}}}=1 $ so we get $ \dfrac{1}{{{\left( 1+\dfrac{c}{I} \right)}^{\dfrac{1}{2}}}}+\dfrac{1}{{{\left( 1-\dfrac{c}{I} \right)}^{\dfrac{1}{2}}}} $ .
We know from the law of exponents that \[\dfrac{1}{a}={{a}^{-1}}\] using this we get $ {{\left( 1+\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}+{{\left( 1-\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}\cdots \cdots \cdots \left( 1 \right) $ .
Now let us use the binomial expansion to simplify our answer. Binomial expansion for $ {{\left( 1+x \right)}^{n}} $ is given by $ {{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+\cdots \cdots $ .
For first term $ {{\left( 1+\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}} $ we have $ n=\dfrac{-1}{2}\text{ and }x=\dfrac{c}{I} $ so we get,
\[{{\left( 1+\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1+\left( -\dfrac{1}{2} \right)\left( \dfrac{c}{I} \right)+\dfrac{-\dfrac{1}{2}\left( -\dfrac{1}{2}-1 \right)}{2!}{{\left( \dfrac{c}{I} \right)}^{2}}+\dfrac{-\dfrac{1}{2}\left( -\dfrac{1}{2}-1 \right)\left( -\dfrac{1}{2}-2 \right)}{3!}{{\left( \dfrac{c}{I} \right)}^{3}}+\cdots \cdots \]
Simplifying we get,
\[\begin{align}
  & {{\left( 1+\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1-\dfrac{c}{2I}+\dfrac{-\dfrac{1}{2}\left( -\dfrac{3}{2} \right)}{2}{{\left( \dfrac{c}{I} \right)}^{2}}+\dfrac{-\dfrac{1}{2}\left( -\dfrac{3}{2} \right)\left( -\dfrac{5}{2} \right)}{6}{{\left( \dfrac{c}{I} \right)}^{3}}+\cdots \cdots \\
 & \Rightarrow {{\left( 1+\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1-\dfrac{c}{2I}+\dfrac{3}{8}\dfrac{{{c}^{2}}}{{{I}^{2}}}-\dfrac{15}{48}\dfrac{{{c}^{3}}}{{{I}^{3}}}+\cdots \cdots \cdots \cdots \left( 2 \right) \\
\end{align}\]
For second term $ {{\left( 1-\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}} $ we have $ n=\dfrac{-1}{2}\text{ and }x=\dfrac{-c}{I} $ so we get,
\[{{\left( 1-\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1-\left( -\dfrac{1}{2} \right)\left( \dfrac{-c}{I} \right)+\dfrac{-\dfrac{1}{2}\left( -\dfrac{1}{2}-1 \right)}{2!}{{\left( \dfrac{-c}{I} \right)}^{2}}+\dfrac{-\dfrac{1}{2}\left( -\dfrac{1}{2}-1 \right)\left( -\dfrac{1}{2}-2 \right)}{3!}{{\left( \dfrac{-c}{I} \right)}^{3}}+\cdots \cdots \]
Simplifying we get,
\[\begin{align}
  & {{\left( 1-\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1+\dfrac{c}{2I}+\dfrac{\left( -\dfrac{1}{2} \right)\left( -\dfrac{3}{2} \right)}{2}{{\left( \dfrac{c}{I} \right)}^{2}}-\dfrac{\left( -\dfrac{1}{2} \right)\left( -\dfrac{3}{2} \right)\left( -\dfrac{5}{2} \right)}{6}{{\left( \dfrac{c}{I} \right)}^{3}}+\cdots \cdots \\
 & \Rightarrow {{\left( 1-\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1+\dfrac{c}{2I}+\dfrac{3}{8}\dfrac{{{c}^{2}}}{{{I}^{2}}}+\dfrac{15}{48}\dfrac{{{c}^{3}}}{{{I}^{3}}}+\cdots \cdots \cdots \cdots \left( 3 \right) \\
\end{align}\]
Putting the values from (2) and (3) in (1) we get,
\[{{\left( 1+\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}+{{\left( 1-\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1-\dfrac{c}{2I}+\dfrac{3}{8}\dfrac{{{c}^{2}}}{{{I}^{2}}}-\dfrac{15}{48}\dfrac{{{c}^{3}}}{{{I}^{3}}}+\cdots 1+\dfrac{c}{2I}+\dfrac{3}{8}\dfrac{{{c}^{2}}}{{{I}^{2}}}+\dfrac{15}{48}\dfrac{{{c}^{3}}}{{{I}^{3}}}+\cdots \]
Cancelling the terms with opposite signs we get,
\[\begin{align}
  & {{\left( 1+\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}+{{\left( 1-\dfrac{c}{I} \right)}^{-\dfrac{1}{2}}}=1+\dfrac{3}{8}\dfrac{{{c}^{2}}}{{{I}^{2}}}+1+\dfrac{3}{8}\dfrac{{{c}^{2}}}{{{I}^{2}}}+\cdots \\
 & \Rightarrow 2+\dfrac{2\times 3}{8}\dfrac{{{c}^{2}}}{{{I}^{2}}}+\cdots \\
 & \Rightarrow 2+\dfrac{3}{4}\dfrac{{{c}^{2}}}{{{I}^{2}}}+\cdots \\
\end{align}\]
Now as we are given that c is small in comparison with I, so $ \dfrac{c}{I} $ will be small term and $ \dfrac{{{c}^{3}}}{{{I}^{3}}},\dfrac{{{c}^{4}}}{{{I}^{3}}} $ will become very small such that they can be neglected. So ignoring higher powers of $ \dfrac{c}{I} $ we have our answer as,
 $ {{\left( \dfrac{I}{I+c} \right)}^{\dfrac{1}{2}}}+{{\left( \dfrac{I}{I-c} \right)}^{\dfrac{1}{2}}}=2+\dfrac{3}{4}\dfrac{{{c}^{2}}}{{{I}^{2}}} $ .
Hence option B is the correct answer.

Note:
 Students should take care of the signs while solving the sum. Students can make mistakes in signs while applying binomial expansion. Students should note that $ \dfrac{a}{b} $ is less than 1 if a is smaller than b, so rising power of numbers less than 1 decreases them more and more.