Question 19 of 20advanced🔧 ApplyLong Answer5 marks

Expand using Binomial Theorem (1+x2−2x)4\left(1 + \dfrac{x}{2} - \dfrac{2}{x}\right)^4, x≠0x \neq 0.

Correct Answer

To expand (1+x2−2x)4\left(1 + \dfrac{x}{2} - \dfrac{2}{x}\right)^4, we first rewrite it as a binomial: [(1+x2)+(−2x)]4\left[\left(1 + \dfrac{x}{2}\right) + \left(-\dfrac{2}{x}\right)\right]^4.

Step 1: Apply the Binomial Theorem with a=(1+x2)a = \left(1 + \dfrac{x}{2}\right) and b=(−2x)b = \left(-\dfrac{2}{x}\right):

(1+x2−2x)4=4C0(1+x2)4+4C1(1+x2)3(−2x)+4C2(1+x2)2(−2x)2+4C3(1+x2)(−2x)3+4C4(−2x)4\left(1 + \dfrac{x}{2} - \dfrac{2}{x}\right)^4 = {}^4C_0\left(1 + \dfrac{x}{2}\right)^4 + {}^4C_1\left(1 + \dfrac{x}{2}\right)^3\left(-\dfrac{2}{x}\right) + {}^4C_2\left(1 + \dfrac{x}{2}\right)^2\left(-\dfrac{2}{x}\right)^2 + {}^4C_3\left(1 + \dfrac{x}{2}\right)\left(-\dfrac{2}{x}\right)^3 + {}^4C_4\left(-\dfrac{2}{x}\right)^4

Step 2: Expand each (1+x2)n\left(1 + \dfrac{x}{2}\right)^n term using binomial theorem:

(1+x2)4=1+2x+3x22+x32+x416\left(1 + \dfrac{x}{2}\right)^4 = 1 + 2x + \dfrac{3x^2}{2} + \dfrac{x^3}{2} + \dfrac{x^4}{16} (1+x2)3=1+3x2+3x24+x38\left(1 + \dfrac{x}{2}\right)^3 = 1 + \dfrac{3x}{2} + \dfrac{3x^2}{4} + \dfrac{x^3}{8} (1+x2)2=1+x+x24\left(1 + \dfrac{x}{2}\right)^2 = 1 + x + \dfrac{x^2}{4} (1+x2)=1+x2\left(1 + \dfrac{x}{2}\right) = 1 + \dfrac{x}{2}

Step 3: Substitute and simplify each term:

=1(1+2x+3x22+x32+x416)+4(1+3x2+3x24+x38)(−2x)+6(1+x+x24)(4x2)+4(1+x2)(−8x3)+1(16x4)= 1\left(1 + 2x + \dfrac{3x^2}{2} + \dfrac{x^3}{2} + \dfrac{x^4}{16}\right) + 4\left(1 + \dfrac{3x}{2} + \dfrac{3x^2}{4} + \dfrac{x^3}{8}\right)\left(-\dfrac{2}{x}\right) + 6\left(1 + x + \dfrac{x^2}{4}\right)\left(\dfrac{4}{x^2}\right) + 4\left(1 + \dfrac{x}{2}\right)\left(-\dfrac{8}{x^3}\right) + 1\left(\dfrac{16}{x^4}\right) =(1+2x+3x22+x32+x416)+(−8x−12−6x−x2)+(24x2+24x+6)+(−32x3−16x2)+16x4= \left(1 + 2x + \dfrac{3x^2}{2} + \dfrac{x^3}{2} + \dfrac{x^4}{16}\right) + \left(-\dfrac{8}{x} - 12 - 6x - x^2\right) + \left(\dfrac{24}{x^2} + \dfrac{24}{x} + 6\right) + \left(-\dfrac{32}{x^3} - \dfrac{16}{x^2}\right) + \dfrac{16}{x^4}

Step 4: Combine like terms:

=x416+x32+x22−4x−5+16x+8x2−32x3+16x4= \dfrac{x^4}{16} + \dfrac{x^3}{2} + \dfrac{x^2}{2} - 4x - 5 + \dfrac{16}{x} + \dfrac{8}{x^2} - \dfrac{32}{x^3} + \dfrac{16}{x^4}
Exercise: Miscellaneous Exercise on Chapter 7 | Q: 5 | (Page 8)
For More Understanding

Explanation

This question tests the application of Binomial Theorem to a trinomial expression. The key insight is to group terms to convert it into a binomial form.

Students must carefully expand each component and combine like terms systematically. The context provides the formula and Example 1 demonstrates a similar expansion with fractional powers.

Solution Steps

  1. Step 1: Rewrite (1+x2−2x)4\left(1 + \dfrac{x}{2} - \dfrac{2}{x}\right)^4 as [(1+x2)+(−2x)]4\left[\left(1 + \dfrac{x}{2}\right) + \left(-\dfrac{2}{x}\right)\right]^4

  2. Step 2: Apply binomial expansion with coefficients 4C0=1{}^4C_0=1, 4C1=4{}^4C_1=4, 4C2=6{}^4C_2=6, 4C3=4{}^4C_3=4, 4C4=1{}^4C_4=1

  3. Step 3: Expand each (1+x2)n\left(1 + \dfrac{x}{2}\right)^n term separately

  4. Step 4: Multiply and simplify each of the 5 terms

  5. Step 5: Combine like terms to get the final expansion

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