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Binomial Theorem question

2005 · Shift 0 · Q100
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Binomial Theorem question

2005 · Shift 0 · Q100

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If xxx is so small that x3{x^3}x3 and higher powers of xxx may be neglected, then (1+x)32−(1+12x)3(1−x)12{{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}(1−x)21​(1+x)23​−(1+21​x)3​ may be approximated as
  1. A
    1−38x21 - {3 \over 8}{x^2}1−83​x2
  2. B
    3x+38x23x + {3 \over 8}{x^2}3x+83​x2
  3. C
    −38x2- {3 \over 8}{x^2}−83​x2
  4. D
    x2−38x2{x \over 2} - {3 \over 8}{x^2}2x​−83​x2
View written solutionFree

Correct answer: C

  1. We need to approximate
(1+x)3/2−(1+x2)3(1−x)1/2\frac{(1+x)^{3/2}-(1+\tfrac{x}{2})^3}{(1-x)^{1/2}}(1−x)1/2(1+x)3/2−(1+2x​)3​

neglecting x3x^3x3 and higher powers.

So we expand each factor up to x2x^2x2.

  1. Expand (1+x)3/2(1+x)^{3/2}(1+x)3/2 using binomial theorem:
(1+x)3/2=1+32x+32(12)2!x2+⋯=1+32x+38x2(1+x)^{3/2}=1+\frac{3}{2}x+\frac{\frac32\left(\frac12\right)}{2!}x^2+\cdots =1+\frac{3}{2}x+\frac{3}{8}x^2(1+x)3/2=1+23​x+2!23​(21​)​x2+⋯=1+23​x+83​x2
  1. Expand (1+x2)3\left(1+\frac{x}{2}\right)^3(1+2x​)3:
(1+x2)3=1+3⋅x2+3(x2)2+(x2)3\left(1+\frac{x}{2}\right)^3 =1+3\cdot \frac{x}{2}+3\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3(1+2x​)3=1+3⋅2x​+3(2x​)2+(2x​)3

Ignoring x3x^3x3 term,

(1+x2)3≈1+32x+34x2\left(1+\frac{x}{2}\right)^3\approx 1+\frac{3}{2}x+\frac{3}{4}x^2(1+2x​)3≈1+23​x+43​x2
  1. Subtract the two expansions:
(1+x)3/2−(1+x2)3(1+x)^{3/2}-\left(1+\frac{x}{2}\right)^3(1+x)3/2−(1+2x​)3 =(1+32x+38x2)−(1+32x+34x2)=\left(1+\frac{3}{2}x+\frac{3}{8}x^2\right)-\left(1+\frac{3}{2}x+\frac{3}{4}x^2\right)=(1+23​x+83​x2)−(1+23​x+43​x2) =38x2−68x2=−38x2=\frac{3}{8}x^2-\frac{6}{8}x^2=-\frac{3}{8}x^2=83​x2−86​x2=−83​x2

So the numerator is already of order x2x^2x2.

  1. Expand denominator:
(1−x)1/2=1−x2−x28+⋯(1-x)^{1/2}=1-\frac{x}{2}-\frac{x^2}{8}+\cdots(1−x)1/2=1−2x​−8x2​+⋯

Hence,

1(1−x)1/2=(1−x)−1/2=1+x2+38x2+⋯\frac{1}{(1-x)^{1/2}}=(1-x)^{-1/2}=1+\frac{x}{2}+\frac{3}{8}x^2+\cdots(1−x)1/21​=(1−x)−1/2=1+2x​+83​x2+⋯
  1. Now divide:
−38x2(1−x)1/2;=  −38x2 (1−x)−1/2\frac{-\frac{3}{8}x^2}{(1-x)^{1/2}} ;=\;-\frac{3}{8}x^2\,(1-x)^{-1/2}(1−x)1/2−83​x2​;=−83​x2(1−x)−1/2 ≈−38x2(1+x2+38x2)\approx -\frac{3}{8}x^2\left(1+\frac{x}{2}+\frac{3}{8}x^2\right)≈−83​x2(1+2x​+83​x2)

Since we neglect x3x^3x3 and higher powers, only the leading term survives:

≈−38x2\approx -\frac{3}{8}x^2≈−83​x2
  1. Therefore the approximation is
−38x2-\frac{3}{8}x^2−83​x2

which matches option C\boxed{\text{C}}C​.

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