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The coefficient of x3x^3 in the expansion of 1(2+ax)5\frac1{(2+ax)^5} is 5 times the coefficient of x2x^2 in the expansion of 4−ax\sqrt{4-ax} where aa is a non-zero constant. Find the exact value of aa. marks

[2020/HCI/Prelim/I/2]

Solution

Coefficient equation

Binomial coefficients

The coefficient of x3x^3 in 1(2+ax)5\dfrac1{(2+ax)^5} is 5 times the coefficient of x2x^2 in 4−ax\sqrt{4-ax}, with a≠0a\ne0. Find the exact value of aa.

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Hint

  1. Normalise: (2+ax)−5=2−5(1+ax2)−5(2+ax)^{-5}=2^{-5}\left(1+\frac{ax}2\right)^{-5} and 4−ax=2(1−ax4)1/2\sqrt{4-ax}=2\left(1-\frac{ax}4\right)^{1/2}.

  2. Pick out one term from each.

  3. Equate and cancel.

Answer · Coefficient equation

a=4/7a=4/7.

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Where −35a³/256 meets −5a²/64

The curves touch at a = 0 (excluded) and cross at a = 4/7.

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  • Coefficient of x³ in (2+ax)⁻⁵
  • 5 times the coefficient of x² in √(4−ax)
Ideas to try
Why do the curves meet at 0 and 4/7, while only 4/7 is allowed?