1) positive 2) negative 3) real 4) imaginary Solution: (3) real (x - a) ( x - b) + (x - b) (x - c) + (x - c) (x - a) = 0 3x2 - 2 (a + b + c)... View Article
1) log(2 + √5) 2) - log(2 + √5) 3) log( - 2 + √5) 4) None of the above Solution: (4) None of the above ecos x - e-cos x = 4 Put ecosx = t t... View Article
1) 0, 1 2) - 1, 1 3) 0, - 1 4) - 1, 2 Solution: (3) 0, - 1 It is given that 1 - p is a root of x2 + px + 1 - p = 0. => (1 - p)2 + p (1 -... View Article
1) real and equal 2) unequal and rational 3) unequal and irrational 4) nothing can be said - Solution: (4) nothing can be said (p - q)x2 +... View Article
1) no real root 2) one real root 3) two real roots 4) four real roots Solution: (4) four real roots The given equation is x2 - 3|x| + 2 = 0... View Article
1) equal 2) imaginary 3) real 4) None of these Solution: (3) real Given a + b + c = 0, 4ax2 + 3bx + 2c = 0 D = 9b2 - 4 (4a) (2c) = 9 (a +... View Article
1) ± a 2) ± (1/2) a 3) ± (3/2)a 4) ± 2a Solution: (2) ± (1 / 2) a The equation given is (a + 1)x2 - (a + 2) x + (a + 3) = 0. The roots are... View Article
1) real and distinct 2) imaginary 3) equal 4) rational and equal Solution: (2) imaginary It is given that f (x) = x2 + ax + b has imaginary... View Article
1) equal 2) imaginary 3) real and distinct 4) rational and equal Solution: (3) real and distinct (x - a) (x - a - 1) + (x - a - 1) (x - a -... View Article
1) |(r / p) - 7| ≥ 4 √3 2)|(p / r) - 7| < 4 √3 3) all p and r 4) no p and r Solution: (1) |(r / p) - 7| ≥ 4 √3 2q = p + r px2 + qx + r =... View Article