Question

If \(\alpha,\beta,5\) are the roots of the equation \[ x^3-ax+a= \frac{\sin^2x+\cos^4x} {\cos^2x+\sin^4x}, \] then \(a(\alpha+\beta)=\)

• A. \(-150\)
• B. \(-155\)
• C. \(75\)
• D. \(105\)

Hint:

If \[ \sin^2x+\cos^2x=1, \] then \[ \boxed{ \sin^2x+\cos^4x = \cos^2x+\sin^4x, } \] so the given trigonometric expression simplifies directly to \[ \boxed{1.} \] Then apply Vieta's formulas to the resulting polynomial.

Answer 1

The Correct Option is B

Step 1: Simplify the right-hand side. Let \[ s=\sin^2x,\qquad c=\cos^2x. \] Since \[ s+c=1, \] we have \[ \sin^2x+\cos^4x =s+c^2 =(1-c)+c^2 =1-c+c^2, \] and \[ \cos^2x+\sin^4x =c+s^2 =c+(1-c)^2 =1-c+c^2. \] Hence, \[ \frac{\sin^2x+\cos^4x} {\cos^2x+\sin^4x} =1. \] Therefore, the equation becomes \[ x^3-ax+a=1, \] or \[ \boxed{x^3-ax+(a-1)=0.} \]

Step 2: Use Vieta's formula. The roots are \[ \alpha,\beta,5. \] Since the coefficient of \(x^2\) is zero, \[ \alpha+\beta+5=0. \] Thus, \[ \boxed{\alpha+\beta=-5.} \] Also, \[ \alpha\beta\cdot5=-(a-1). \] Using the sum of pairwise products, \[ \alpha\beta+5(\alpha+\beta)=-a. \] Substituting \[ \alpha+\beta=-5, \] we get \[ \alpha\beta-25=-a. \] Hence, \[ \alpha\beta=25-a. \] Now, \[ 5(25-a)=1-a. \] Therefore, \[ 125-5a=1-a, \] \[ 124=4a, \] \[ \boxed{a=31.} \]

Step 3: Find \(a(\alpha+\beta)\). Since \[ \alpha+\beta=-5, \] we obtain \[ a(\alpha+\beta) = 31(-5) = \boxed{-155.} \] Hence, \[ \boxed{(B)} \] is the correct answer.