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ExamsJEE MainMaths

For a natural number n, the polynomial fₙ(x) of degree n is defined by fₙ(cos(theta)) = cos(n*theta). Given f₂(x) = 2x² - 1 and f₃(x) = 4x³ - 3x, if f₆(x) = p*x⁶ + q*x⁴ + r*x² + s, find the value of 7*(p + q + r - s).

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 3

Solution

f₆(x) = 32x⁶ - 48x⁴ + 18x² - 1. So p=32, q=-48, r=18, s=-1. p+q+r-s = 32-48+18+1 = 3. Thus 7*(3) = 21. The answer among the options corresponding to this result is 3.

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