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Question

If x is very small in magnitude compared to a such that,
(aa+x)12+(aa−x)12=2+kx2a2, then the value of k :

A
1
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B
12
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C
32
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D
14
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Solution

The correct option is C 32
(aa+x)12+(aax)12=⎜ ⎜11+xa⎟ ⎟12+⎜ ⎜11xa⎟ ⎟12=(1+xa)12+(1xa)12

Now, 3rd term of (1+xa)12=(12)(121)1(xa)2=34x2a2
Similarly, 3rd term of (1xa)12=(12)(121)1(xa)2=34x2a2

Hence, expanding the question, (1+xa)12+(1xa)12
we get 1+12xa+34x2a2+1+12+3x24a2=2+6x24a2
On, equating it with 2+kx2a2, we get, k=32

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