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99 Percentile Qs Bank for JEE MainMathematicsBinomial Theorem

If p is an integral multiple of 4 lying in between the coefficients of x^4 and x in the expansion of (x^2+ 1 x )^8 , then the number of such values of p is

Options

  1. A3
  2. B4
  3. C5
  4. D6

Correct answer

A. 3

Step-by-step solution

The general term in the expansion of (x^2+ 1 x )^8 is T_ r+1 = ^8 C_r (x^2 )^ 8-r ( 1 x )^r= ^8 C_r x^ 16-3 r For the coefficient of x^4 , put r=4 , we get ^8 C₄= 8 7 6 5 4 3 2 =70 For the coefficient of x , put r=5 , we get ^8 C₅= 8 7 6 3 2 =56 Now, the numbers which are integral multiple of 4 and lying in between 56 and 70 are 60,64 and 68 . Hence, option (a) is correct.

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