NIMCET 2014 Question Paper with solutions

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Question : 30 of 120

Marks: +1, -0

If x is so small that

x^2

and higher powers of x can be neglected, then

{(9+2x)^{1\/2}(3+4x)}/(1-x)^{1\/5}

is approximately equal to

9 + ${74}/{15}$ x
9 + ${74}/{5}$ x
3 + ${74}/{15}$ x
3 + ${74}/{5}$ x

Solution:

We have,

{(9+2x)^{1\/2}(3+4x)}/(1-x)^{1\/5}

=

{9(1+2/9x)^{1\/2}(1+4/3x)}/(1-x)^{1\/5}

=

{9[1+2/9.1/2x+...][1+4/3x]}/(1-x)^{1\/5}

=

{9[1+x/9+4/3x]}/(1-x)^{1\/5}

=

{9[1+{13x}/9]}/(1-x)^{1\/5}

= 9

(1-x)^{-13/9-1/5}

= 9

(1-x)^{-74/45}

= 9

(1+{74}/{45}x)

[neglecting higher powers of x]
= 9 +

{74x}/5

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