Q. The number of real values of a satisfying the equation a2−2asinx+1=0 is
Given Quadratic Equation: We are given the quadratic equation in terms of a:a2−2asinx+1=0This is a quadratic equation with respect to a, and it can be written in the standard form as:a2−(2sinx)a+1=0To determine the number of real values of a that satisfy this equation, we need to consider the discriminant of the quadratic equation, which is given by:D=b2−4acHere, a=1 (coefficient of a2), b=−2sinx (coefficient of a), and c=1 (constant term).
Calculate Discriminant: Calculate the discriminant using the values of a, b, and c: D=(−2sinx)2−4(1)(1) D=4sin2x−4 Since sin2x is always less than or equal to1 for all real x, the maximum value of 4sin2x is 4.
Analyzing Discriminant: Analyze the discriminant:If D > 0, there are two distinct real values of a.If D=0, there is exactly one real value of a (a repeated root).If D < 0, there are no real values of a (the roots are complex).
Range of Discriminant: Determine the range of D based on the maximum value of sin2x:Since the maximum value of 4sin2x is 4, the discriminant D can be at most 0 (when sin2x=1).Therefore, the discriminant D is always less than or equal to 0.D≤0
Concluding Real Values: Conclude the number of real values of a:Since D≤0, the equation a2−2asinx+1=0 will have at most one real value of a.When D=0 (which occurs when sin2x=1), there is exactly one real value of a.When D < 0 (which occurs when \sin^{2} x < 1), there are no real values of a.
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