Question

If x is small so that \[{x^2}\] and higher powers can be neglected, then the approximate value for \[\dfrac{{{{\left( {1 - 2x} \right)}^{ - 1}}{{\left( {1 - 3x} \right)}^{ - 2}}}}{{{{\left( {1 - 4x} \right)}^{ - 3}}}}\] is ?
(A) \[1 - 2x\]
(B) \[1 - 3x\]
(C) \[1 - 4x\]
(D) \[1 - 5x\]

Answer 1

First of all, convert all negative power into positive then expand the equation using formula \[(a + b) = ({a^2} + 2ab + {b^2})\]. Then by neglecting the \[{x^2}\] and the higher powers we get the answer.

Complete step-by-step answer:

Let’s assume,
 \[f(x) = \dfrac{{{{\left( {1 - 2x} \right)}^{ - 1}}{{\left( {1 - 3x} \right)}^{ - 2}}}}{{{{\left( {1 - 4x} \right)}^{ - 3}}}}\]
         \[ = \dfrac{{{{\left( {1 - 4x} \right)}^3}}}{{{{\left( {1 - 2x} \right)}^1}{{\left( {1 - 3x} \right)}^2}}}\]
In the above equation convert all the negative power into positive.
         \[ = \dfrac{{{{\left( {1 - 4x} \right)}^2}(1 - 4x)}}{{{{\left( {1 - 2x} \right)}^1}{{\left( {1 - 3x} \right)}^2}}}\]
Now expand the equation \[{\left( {1 - 4x} \right)^2}\]and \[{\left( {1 - 3x} \right)^2}\]by using formula \[(a + b) = ({a^2} + 2ab + {b^2})\]. So, we get,
          \[ = \dfrac{{(1 - 8x + 16{x^2})(1 - 4x)}}{{(1 - 2x)(1 + 6x + 9{x^2})}}\]
Then neglected the \[{x^2}\]and the higher powers terms
          \[ = \dfrac{{(1 - 8x)(1 - 4x)}}{{(1 - 2x)(1 - 6x)}}\]
Now multiply the equation \[(1 - 2x)\]and \[(1 - 6x)\]
         \[ = \dfrac{{(1 - 8x)(1 - 4x)}}{{(1 - 8x + 12{x^2})}}\]
Again, neglecting the \[{x^2}\]and the higher powers terms
         \[ = \dfrac{{(1 - 8x)(1 - 4x)}}{{(1 - 8x)}}\]
         \[ = (1 - 4x)\]
Hence Option C is the correct answer.

Note: At the time of expanding the square equation or neglecting the terms of power of x take attention that on numerator and denominator same equation can happen then try that way so you get the answer. For example in this question if we directly expand the equation \[{\left( {1 - 4x} \right)^3}\] then we didn’t get the answer but take two part \[{\left( {1 - 4x} \right)^2}\] and \[\left( {1 - 4x} \right)\] so at the end of equation on numerator and denominator we get the same equation \[\left( {1 - 8x} \right)\] and we get the answer of this question. So by seeing the option we solve this question so we get the answer to one of the options.