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If (x² + px + 1) is a factor of (ax³ + bx + c), then

1) a2 + c2 = - ab 2) a2 - c2 = - ab 3) a2 - c2 = ab 4) none of these Solution: (3) a2 - c2 = ab Since (x2 + px + 1) is a factor of (ax3 +... View Article

If the roots of the equation ax² + bx + c = 0 are of the form [k + 1] / k and [k + 2] / [k + 1] then (a + b + c)² =

1) b2 - 4ac 2) b2 - 2 ac 3) 2b2 - ac 4) ∑a2 Solution: (1) b2 - 4ac [k + 1] / k + [k + 2] / [k + 1] = - b / a, [k + 2] / k = c / a 1 + (1 /... View Article

If the arithmetic mean of the roots of a quadratic equation is 8 and the geometric mean is 5, then the equation is

1) x2 - 16 x - 25 = 0 2) x2 + 16 x - 25 = 0 3) x2 - 16 x + 25 = 0 4) x2 - 8 x - 5 = 0 Solution: (3) x2 - 16 x + 25 = 0 [α + β] / 2 = 8, √αβ... View Article

If the roots of ax² + bx + c = 0 are sin α and cos β for some α, then which one of the following is correct?

1) a2 + b2 = 2ac 2) b2 - c2 = 2ab 3) b2 - a2 = 2ac 4) b2 + c2 = 2ab Solution: (3) b2 - a2 = 2ac sin α + cos α = - b / a and sin α cos α = c... View Article

If α and β are the roots of the equation ax² + bx + c = 0 and S_n = αⁿ + βⁿ and a S_n+1 + b S_n + c S_n-1 is equal to

1) 0 2) abc 3) a + b + c 4) None of these Solution: (1) 0 α + β = - b / a, αβ = c / a Sn = αn + βn Sn+1 = αn+1 + βn+1 αn+1 + βn+1 + αnβ +... View Article

Let f (x) = 2x² + 5x + 1. If we write f (x ) as f (x) = [a (x + 1) (x – 2) + b (x – 2) (x – 1) + c(x – 1)(x + 1)] for real numbers a,b and c, then

1) there are infinite numbers of choices for a, b and c 2) only one choice for a but infinite numbers of choices for b and c 3) exactly one... View Article

If α, β are the roots of ax² + bx + c = 0 (a ≠ 0) and α + h, β + h are the roots of px² + qx + r = 0 (p ≠ 0), then the ratio of the squares of their discriminants is

1) a2 : p2 2) a : p2 3) a2 : p 4) a : 2p Solution: (1) a2 : p2 Let α, β are the roots of the equation. ax2 + bx + c = 0 and α + h, β + h are... View Article

If a, b and c are positive numbers in a GP, then the roots of the quadratic equation (log_e a) x² – 2 (log_e b) x + (log_e c) = 0 are

1) - 1 and [loge c] / [loge a] 2) 1 and - [loge c] / [loge a] 3) 1 and loga c 4) - 1 and logc a Solution: (3) 1 and loga c If a + b + c =... View Article

If the roots of the equation x² + 2bx + c = 0 are α and β, then b² – c is equal to

1) (α - β)2 / 4 2) (α + β)2 - αβ 3) (α + β)2 + αβ 4) [(α - β)2 / 2] + αβ 5) [(α + β)2 / 2] + αβ Solution: (1) (α - β)2 / 4 (α + β) = 2b,... View Article

If α and β are the roots of the equation x² + 3x – 4 = 0, then 1/α + 1/β is equal to

1) -3/4 2) 3/4 3) -4/3 4) 4/3 5) 3/2 Solution: (2) 3 / 4 α + β = - 3, αβ = - 4 1 / α + 1 / β = [α + β] / [αβ] = - 3 / - 4 = 3 / 4... View Article

If the roots of x² – ax + b = 0 are two consecutive odd integers, then a² – 4b is equal to

1) 3 2) 4 3) 5 4) 6 5) 7 Solution: (2) 4 Let α, β are the roots of the equation. x2 - ax + b = 0 α + β = a, αβ = b Consider a2 - 4b = (αβ)2... View Article

If α and β are the roots of x² – ax + b² = 0, then α² + β² is equal to

1) a2 + 2b2 2) a2 - 2b2 3) a2 - 2b 4) a2 + 2b 5) a2 - b2 Solution: (2) a2 - 2b2 Given α, β are the roots of the equation α + β = a, αβ = b... View Article

The value of a, so that the sum of squares of the root of the equations x² – (a – 2) x – a + 1 = 0 assume the least value, is

1) 2 2) 0 3) 3 4) 1 Solution: (4) 1 x2 - (a - 2) x - a + 1 = 0 Let ɑ, β be the roots of the given equation ɑ + β = ɑ - 2, ɑβ = 1 - ɑ ɑ2 + β2... View Article

If a ϵ R and the equation -3(x)- [x])² + 2(x – [x]) + a² = 0 (where, [x]denotes the greatest integer ≤ x] has no integral solution, then all possible values lie in the interval )

1) (-1, 0) ∪ (0,1) 2) (1,2) 3) (-2, -1) 4) (- ∞,- 2) ∪ (2, ∞) Solution: (1) (-1, 0) ∪ (0, 1) -3(x)- [x])2 + 2(x - [x]) + a2 = 0 Put x - [x]... View Article

If x, y, z are in GP and a^x = b^y = c^z, then

1) loga c = logb a 2) logb a = logc b 3) logc b = loga c 4) None of these Solution: (2) logb a = logc b ax = by = cz Let ax = by = cz = k... View Article

Equation of the bisector of the acute angle between lines 3x – 4y + 7 = 0 and 12x + 5y – 2 = 0 is

1) 21x - 77y + 100 = 0 2) 99x - 27y + 81 = 0 3) 99x + 27y + 30 = 0 4) 21x - 77y - 100 = 0 Solution: Given 3x - 4y + 7 = 0…(i) 12x + 5y - 2 =... View Article

If two pairs of lines x² – 2mxy – y² = 0 and x²– 2nxy – y² = 0 are such that one of them represents the bisector of the angles between the other, then

1) mn = 1 2) m + n = mn 3) mn = - 1 4) m - n = mn Solution: Given x2 - 2mxy - y2 = 0 …(i) x2 - 2nxy - y2 = 0 …(ii) Equation of bisector of... View Article

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