Quadratic equations-2 (XAT-2008)

The number of real solutions of y in equation y²-2ycosx+1=0 are:

(A)0     (B)1     (C)2     (D)3     (E) None of these

Solution follows here:

Solution:

Funda:

Roots of the quadratic equation ax²+bx+c=0 are -b±√(b²-4ac) / 2a  

If b²-4ac ≥ 0, roots are real,

If b²-4ac < 0, roots are imaginary.

Determinant = b²-4ac = (2cosx)²-4 = 4(cos²x-1) = -4sin²x

For getting real solutions, Determinant ≥ 0 => -4sin²x ≥ 0

This is possible only for sin²x = 0 => cos²x = 1-sin²x = 1 => cosx = ±1

=> the given equation becomes  y²-2y(±1)+1=0

=> y²-2y+1=0 (or) y²+2y+1=0

=> (y-1)² = 0 (or) (y+1)² = 0

=> y = 1 (or) -1

Two real solutions   

Answer (C)