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The position of reflection of the point (4, 1) about the line y = x -1 is

1) (1, 2) 2) (3, 4) 3) (-1, 0) 4) (2, 3) Solution: Let Q be the image of P. Join PQ and it cuts the line mirror y = x-1 at point R. Let R... View Article

The equation of bisectors of the angles between the lines |x| = |y| are

1) y = ± x and x = 0 2) x = 1/2 and y = 1/2 3) y = 0 and x = 0 4) None of these Solution: Given |x| = |y| => |y| = ± x => y = x or y... View Article

If one of the lines of my² + (1 – m²)xy – mx² = 0 is a bisector of the angle between the lines xy = 0, then m is/are

1) - 1/2 2) - 2 3) ± 1 4) 2 Solution: Given xy = 0 => x = 0 or y = 0 ie y axis or x axis. So the coordinate axis is bisected by the line... View Article

The vertices of a triangle are A (3, 7), B (3, 4) and C (5, 4). The equation of the bisector of ∠ABC is

1) y = x + 1 2) y = x - 1 3) y = 3x - 5 4) y = x Solution: Given A (3, 7), B (3, 4), and C (5, 4) are the vertices of the triangle ABC.... View Article

If α and α², are the roots of x² + x + 1 = 0, then the equation, whose roots are α³¹ and α⁶², is

1) x2 - x + 1 = 0 2) x2 + x - 1 = 0 3) x2 + x + 1 = 0 4) x60 + x30 + 1 = 0 Solution: (3) x2 + x + 1 = 0 Given α and α2 are the roots of the... View Article

If α and β are the roots of the equation x² + px + q = 0 and the sum (α + β) – x – (α² + β²) / 2 (x²) + (α³ + β³) / 3 (x³) – ….. exists, then it is =

1) log(x2 + px + q) 2) log(x2 - px + q) 3) log(1 + px + qx2) 4) log(1 - px + qx2) 5) log (x2 qx + q) Solution: (4) log(1 - px + qx2) The... View Article

If sec α and cosec α are the roots of the equation x² + px + q = 0

1) p2 = p + 2q 2) q2 = p + 2q 3) p2 = q(q +2) 4) q2 = p (p + 2) 5) p2 = q (q - 2) Solution: (3) p2 = q (q +2) Since sec α + cosec α = p and... View Article

If α and β are the roots of x² + 5x + 4 = 0, then the equation, whose roots are (α+ 2)/3 and (β+2)/3, is

1) 9x2 + 3x + 2 = 0 2) 9x2 - 3x - 2 = 0 3) 9x2 + 3x - 2 = 0 4) 9x2 - 3x + 2 = 0 Solution: (4) 9x2 - 3x + 2 = 0 α + β = - 5, αβ = 4 The roots... View Article

If one root of the quadratic equation ax² + bx + c = 0 is equal to nth power of the other root, then the value of (aⁿc)^1/(n+1) + (acⁿ)^1/(n+1) is equal to

1) b 2) - b 3) 1/ (bxn+1) 4) - (1/bxn+1) Solution: (1) b The given equation is ax2 + bx + c = 0. Let α and αn be the roots of the equation,... View Article

If α and β are the roots of the equation x² – px + r = 0 and α/2, 2β are the roots of the equation x² – qx + r = 0. Then, the value of r is

1) 2/9 (p - q) (2q - p) 2) 2/9 (q- p) (2p - q) 3) 2/9 (q - 2p) (2q - p) 4) 2/9 (2p - q) (2q - p) Solution: (4) 2/9 (2p - q) (2q - p) α + β =... View Article

If α and β are the roots of the equation x² + ax + b = 0, then (1 / α²) + (1 / β²) =

1) [a2 - 2b] / b2 2) [b2 - 2a] / b2 3) [a2 + 2b] / b2 4) [b2 + 2a] / b2 Solution: (1) [a2 - 2b] / b2 (α + β) = α, α β = b (1 / α2) + (1 /... View Article

If α and β are the roots of the equation ax² + bx + c = 0, αβ = 3, and a, b, c are in AP, then α + β is equal to

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

If α, β and γ are the roots of the equation x³ – 6x² + 11 x + 6 = 0, then ∑α²β + ∑αβ² is equal to

1) 80 2) 168 3) 90 4) - 84 Solution: (2) 168 Since ∑ɑ = 6 ∑αβ = 11 and ɑβɣ = - 6 ∑α2β + ∑αβ2 = 2 (α2β + β2ɣ + ɣ2α + βɣ2 + ɣα2) = 2 [αβ (α +... View Article

If α and β are the roots of the quadratic equation ax² + bx + c = 0, observe the lists given below

1) A = 5 B = 2 C = 4 D = 6 2) A = 5 B = 2 C = 1 D =4 3) A = 5 B = 4 C = 2 D = 6 4) A = 5 B = 2 C = 4 D = 1 Solution: (4) A = 5 B = 2 C = 4... View Article

The cubic equation, whose roots are thrice to each of the roots of x³ + 2x² – 4x + 1 = 0, is

1) x3 - 6x2 + 36x + 27 = 0 2) x3 + 6x2 + 36x + 27 = 0 3) x3 + 6x2 + 36x + 27 = 0 4) x3 + 6x2 - 36x + 27 = 0 Solution: (4) x3 + 6x2 - 36x + 27... View Article

If α, β and γ are the roots of the equation x³ – 8 x + 8 = 0, then ∑α² and ∑ 1/αβ are respectively

1) 0 and - 16 2) 16 and 18 3) - 16 and 0 4) 16 and 0 Solution: (4) 16 and 0 α + β + γ = 0 = ∑ ɑβ = - 8 = ɑβɣ = - 8 ∑ ɑ2 = (α + β + γ)2 - 2 ∑... View Article

If α and β are the roots of the equation ax² + bx + c = 0, then α / [ɑβ + b] + β / [aɑ + b] =

1) 2/a 2) 2/b 3) 2/c 4) - (2/a) Solution: (4) - (2 / a) It is given that α and β are the roots of the equation ax2 + bx + c = 0. aɑ2 + bɑ +... View Article

If α and β are the roots of x² – 2x cos ∅ + 1 = 0, then the equation, whose roots are αⁿ + βⁿ

1) x2 - 2 xcos n∅ - 1 = 0 2) x2 - 2 xcos n∅ + 1 = 0 3) x2 - 2 xsin n∅ + 1 = 0 4) x2 + 2 xsin n∅ - 1 = 0 Solution: (2) x2 - 2 xcos n∅ + 1 = 0... View Article

If two persons A and B solve the equation x² + ax + b = 0, while solving, A commits a mistake the coefficient of x was taken as 15 in place of -9 and finds the roots as – 7 and – 2 Then, the equation is

1) x2 + 9x + 14 = 0 2) x2 - 9x + 14 = 0 3) x2 + 9x - 14 = 0 4) x2 - 9x - 14 = 0 5) None of these Solution: (2) x2 - 9x + 14 = 0 The... View Article

If α and β are the roots of the equations x² – (1 + n²) x+ 1 / 2 (1 + n² + n⁴) = 0, then α² + β² is equal to

1) n2 2) - n2 3) n4 4) - n4 Solution: (1) n2 (α + β) = 1 + n2; (αβ) = [1 + n2 + n4] / 2 (α2 + β2) = (α + β)2 - 2αβ = (1 + n2)2 - 2 (1 / 2)... View Article

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