Multiple choice

$(a^{4}-1)x^{2}-(a^{2}+1)(sin^{-1}sin^{3} 2)x+(cos^{-1}cos2)(a^{2}-1) =0$. .Find the set of values of a so that above equation have roots of opposite in sign.

  1. [-2 , 3]

  2. [-1 , 1]

  3. [-2 ,1]
  4. $\phi $
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D Correct answer
Explanation

For roots to be opposite in sign, the product of roots (c/a) must be negative. The equation is (a^4-1)x^2 - (a^2+1)(sin^-1(sin^3(2)))x + (cos^-1(cos 2))(a^2-1) = 0. Since sin^-1(sin 2) = pi-2 and cos^-1(cos 2) = 2, the constant term is 2(a^2-1). For the product to be negative, 2(a^2-1)/(a^4-1) < 0. Simplifying, 2/(a^2+1) < 0, which is impossible for any real a.

AI explanation

For the roots to be opposite in sign, the product of the roots must be negative, so (c divided by a) must be less than zero, which gives ((cos^(-1)(cos 2))((a^2 - 1))) divided by (a^4 - 1) is less than 0. Since 2 is in the second quadrant, cos^(-1)(cos 2) equals 2, which is positive. The expression simplifies to 2 divided by (a^2 + 1), which is always strictly positive for all real values of a. Therefore, the condition is never satisfied and the set of values is the empty set.

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