If c is small in comparison with I, then (II+c)12+(II-C)12 =.

Here we are given the expression as (II+c)12+(II-C)12 . We need to simplify this expression.Let us first take I common from the denominator of both terms, we get (II+c1)12+(II-c1)12.Canceling I in the numerator and the denominator of both terms we get (II+c1)12+(II-c1)12.Now using the law of exponents (ab)m= ambm we get (I)12(I+cI)12+(I)12(I-cI)12.As we know (I)12=1 so we get 1(I+cI)12+1(I-cI)12. .We know from the law of exponents that 1a=a-1 using this we get I+cI-12+(I-cI)-12⋯⋯⋯(1) .Now let us use the binomial expansion to simplify our answer. Binomial expansion for (1+x)n is given by (1+x)n=1+nx+n(n-1)2!x2+nn-1(n-2)3!x3⋯⋯.For first term I+cI-12 we have n=-12 and x=cI so we get,I+cI-12=1+-12cI+-12(-12-1)2!cI2+-12-12-1(-12-2)3!cI3+…Simplifying we get,I+cI-12=1−c2I+-12(-32)2!cI2+-12-32(-52)3!cI3+⋯⋯⇒I+cI-12=1−c2I+38c2I2−1548c3I3+⋯⋯⋯⋯(2)For second term I-cI-12 we have n=-12 and x=-cI so we get,I-cI-12=1-(-12)( -cI)+ -12(-12-1)2!-cI2+-12-12-1(-12-2)3!-cI3+…Simplifying we get, I-cI-12=1+c2I+-12-322!cI2- -12-32-523!cI3+… ⇒I-cI-12=1+c2I+38c2I2+1548c3I3+⋯⋯⋯⋯(3)Putting the values from (2) and (3) in (1) we get,(II+c)12+(II-C)12 =1- c2I++38c2I2 - 1548c3I3+⋯⋯⋯⋯1+c2I+38c2I2+1548c3I3+⋯⋯⋯⋯Canceling the terms with opposite signs we get,(II+c)12+(II-C)12=1+38c2I2+1+38c2I2 +…..⇒2+2×38c2I2+⋯⇒2+34c2I2+⋯Now as we are given that c is small in comparison with I, so cI will be small term and c3I3, c4I3 will become very small such that they can be neglected. So ignoring higher powers of cI we have our answer as, (II+c)12+(II-C)12=2+34c2I2Hence option 2 is the correct answer.

If c is small in comparison with I, then (II+c)12+(II-C)12 =.

Here we are given the expression as (II+c)12+(II-C)12 . We need to simplify this expression.Let us first take I common from the denominator of both terms, we get (II+c1)12+(II-c1)12.Canceling I in the numerator and the denominator of both terms we get (II+c1)12+(II-c1)12.Now using the law of exponents (ab)m= ambm we get (I)12(I+cI)12+(I)12(I-cI)12.As we know (I)12=1 so we get 1(I+cI)12+1(I-cI)12. .We know from the law of exponents that 1a=a-1 using this we get I+cI-12+(I-cI)-12⋯⋯⋯(1) .Now let us use the binomial expansion to simplify our answer. Binomial expansion for (1+x)n is given by (1+x)n=1+nx+n(n-1)2!x2+nn-1(n-2)3!x3⋯⋯.For first term I+cI-12 we have n=-12 and x=cI so we get,I+cI-12=1+-12cI+-12(-12-1)2!cI2+-12-12-1(-12-2)3!cI3+…Simplifying we get,I+cI-12=1−c2I+-12(-32)2!cI2+-12-32(-52)3!cI3+⋯⋯⇒I+cI-12=1−c2I+38c2I2−1548c3I3+⋯⋯⋯⋯(2)For second term I-cI-12 we have n=-12 and x=-cI so we get,I-cI-12=1-(-12)( -cI)+ -12(-12-1)2!-cI2+-12-12-1(-12-2)3!-cI3+…Simplifying we get, I-cI-12=1+c2I+-12-322!cI2- -12-32-523!cI3+… ⇒I-cI-12=1+c2I+38c2I2+1548c3I3+⋯⋯⋯⋯(3)Putting the values from (2) and (3) in (1) we get,(II+c)12+(II-C)12 =1- c2I++38c2I2 - 1548c3I3+⋯⋯⋯⋯1+c2I+38c2I2+1548c3I3+⋯⋯⋯⋯Canceling the terms with opposite signs we get,(II+c)12+(II-C)12=1+38c2I2+1+38c2I2 +…..⇒2+2×38c2I2+⋯⇒2+34c2I2+⋯Now as we are given that c is small in comparison with I, so cI will be small term and c3I3, c4I3 will become very small such that they can be neglected. So ignoring higher powers of cI we have our answer as, (II+c)12+(II-C)12=2+34c2I2Hence option 2 is the correct answer.

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If c is small in comparison with I, then ${(\frac{I}{I + c})}^{\frac{1}{2}} + {(\frac{I}{I - C})}^{\frac{1}{2}}$ =.

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\frac{3 c}{4 I}

\frac{3 c^{2}}{4 I^{2}}

\frac{3 c^{2}}{4 I^{2}}

\frac{3 c}{4 I}

answer is B.

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Detailed Solution

Here we are given the expression as

{(\frac{I}{I + c})}^{\frac{1}{2}} + {(\frac{I}{I - C})}^{\frac{1}{2}}

. We need to simplify this expression.
Let us first take I common from the denominator of both terms, we get

{(\frac{I}{I + \frac{c}{1}})}^{\frac{1}{2}} + {(\frac{I}{I - \frac{c}{1}})}^{\frac{1}{2}}

.
Canceling I in the numerator and the denominator of both terms we get

{(\frac{I}{I + \frac{c}{1}})}^{\frac{1}{2}} + {(\frac{I}{I - \frac{c}{1}})}^{\frac{1}{2}}

.
Now using the law of exponents

({\frac{a}{b})}^{m}

\frac{a^{m}}{b^{m}}

we get

\frac{{(I)}^{\frac{1}{2}}}{{(I + \frac{c}{I})}^{\frac{1}{2}}} + \frac{{(I)}^{\frac{1}{2}}}{{(I - \frac{c}{I})}^{\frac{1}{2}}}

.
As we know

{(I)}^{\frac{1}{2}} = 1

so we get

\frac{1}{{(I + \frac{c}{I})}^{\frac{1}{2}}} + \frac{1}{{(I - \frac{c}{I})}^{\frac{1}{2}}}

. .
We know from the law of exponents that

\frac{1}{a} = a^{- 1}

using this we get

{(I + \frac{c}{I})}^{- \frac{1}{2}} + {(I - \frac{c}{I})}^{- \frac{1}{2}}

⋯⋯⋯(1) .
Now let us use the binomial expansion to simplify our answer. Binomial expansion for

{(1 + x)}^{n}

is given by

{(1 + x)}^{n} = 1 + nx + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3}

⋯⋯.
For first term

{(I + \frac{c}{I})}^{- \frac{1}{2}}

we have n=

- \frac{1}{2}

and x=

\frac{c}{I}

so we get,

{(I + \frac{c}{I})}^{- \frac{1}{2}} = 1 + (- \frac{1}{2}) (\frac{c}{I}) + \frac{- \frac{1}{2} (- \frac{1}{2} - 1)}{2!} {(\frac{c}{I})}^{2} + \frac{- \frac{1}{2} (- \frac{1}{2} - 1) (- \frac{1}{2} - 2)}{3!} {(\frac{c}{I})}^{3} + \dots

Simplifying we get,

{(I + \frac{c}{I})}^{- \frac{1}{2}}

=1−

\frac{c}{2 I}

\frac{- \frac{1}{2} (- \frac{3}{2})}{2!} {(\frac{c}{I})}^{2}

\frac{- \frac{1}{2} (- \frac{3}{2}) (- \frac{5}{2})}{3!} {(\frac{c}{I})}^{3}

+⋯⋯
⇒

{(I + \frac{c}{I})}^{- \frac{1}{2}}

=1−

\frac{c}{2 I}

\frac{3}{8} \frac{c^{2}}{I^{2}}

−

\frac{15}{48} \frac{c^{3}}{I^{3}}

+⋯⋯⋯⋯(2)

For second term

{(I - \frac{c}{I})}^{- \frac{1}{2}}

we have n=

- \frac{1}{2}

and x=

\frac{- c}{I}

so we get,

{(I - \frac{c}{I})}^{- \frac{1}{2}} = 1 - (- \frac{1}{2}

)(

\frac{- c}{I}) + \frac{- \frac{1}{2} (- \frac{1}{2} - 1)}{2!} {(\frac{- c}{I})}^{2} + \frac{- \frac{1}{2} (- \frac{1}{2} - 1) (- \frac{1}{2} - 2)}{3!} {(\frac{- c}{I})}^{3} + \dots

Simplifying we get,

{(I - \frac{c}{I})}^{- \frac{1}{2}}

=1+

\frac{c}{2 I}

\frac{(- \frac{1}{2}) (- \frac{3}{2})}{2!} {(\frac{c}{I})}^{2} -

\frac{(- \frac{1}{2}) (- \frac{3}{2}) (- \frac{5}{2})}{3!} {(\frac{c}{I})}^{3} + \dots

⇒

{(I - \frac{c}{I})}^{- \frac{1}{2}}

=1+

\frac{c}{2 I}

\frac{3}{8} \frac{c^{2}}{I^{2}}

\frac{15}{48} \frac{c^{3}}{I^{3}}

+⋯⋯⋯⋯(3)
Putting the values from (2) and (3) in (1) we get,

{(\frac{I}{I + c})}^{\frac{1}{2}} + {(\frac{I}{I - C})}^{\frac{1}{2}}

=1-

\frac{c}{2 I} +

\frac{3}{8} \frac{c^{2}}{I^{2}}

\frac{15}{48} \frac{c^{3}}{I^{3}}

+⋯⋯⋯⋯1+

\frac{c}{2 I}

\frac{3}{8} \frac{c^{2}}{I^{2}}

\frac{15}{48} \frac{c^{3}}{I^{3}}

+⋯⋯⋯⋯
Canceling the terms with opposite signs we get,

{(\frac{I}{I + c})}^{\frac{1}{2}} + {(\frac{I}{I - C})}^{\frac{1}{2}} = 1 + \frac{3}{8} \frac{c^{2}}{I^{2}} + 1 + \frac{3}{8} \frac{c^{2}}{I^{2}}

+…..
⇒2+

\frac{2 \times 3}{8} \frac{c^{2}}{I^{2}}

+⋯
⇒2+

\frac{3}{4} \frac{c^{2}}{I^{2}}

+⋯
Now as we are given that c is small in comparison with I, so

\frac{c}{I}

will be small term and

\frac{c^{3}}{I^{3}}

\frac{c^{4}}{I^{3}}

will become very small such that they can be neglected. So ignoring higher powers of

\frac{c}{I}

we have our answer as,

{(\frac{I}{I + c})}^{\frac{1}{2}} + {(\frac{I}{I - C})}^{\frac{1}{2}} = 2 + \frac{3}{4} \frac{c^{2}}{I^{2}}

Hence option 2 is the correct answer.

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If c is small in comparison with I, then ${(\frac{I}{I + c})}^{\frac{1}{2}} + {(\frac{I}{I - C})}^{\frac{1}{2}}$ =.