Multiple Choice QuestionsIf 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
−1/2
-2
1
1
Angle between the tangents to the curve y = x2 − 5x + 6 at the points (2, 0) and (3, 0) is
π/2
π/4
π/6
π/6
The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ - θ cosθ) at any point ‘θ’ is such that
it passes through the origin
it makes angle π/2 + θ with the x-axis
it passes through (aπ/2 ,-a)
it passes through (aπ/2 ,-a)
D.
it passes through (aπ/2 ,-a)
Clearly dy/dx = an θ
⇒ slope of normal = - cot θ
Equation of normal at ‘θ’ is
y – a(sin θ - θ cos θ) = - cot θ(x – a(cos θ + θ sin θ)
⇒ y sin θ - a sin2 θ + a θ cos θ sin θ = -x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.
The line parallel to the x−axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx − 2ay − 3a = 0, where (a, b) ≠ (0, 0) is
below the x−axis at a distance of 3/2 from it
below the x−axis at a distance of 2 /3 from it
above the x−axis at a distance of 3/ 2 from it
above the x−axis at a distance of 3/ 2 from it
If the angle θ between the line 
and the plane 
is such of sin θ = 1/3 the value of λ is
5/3
-3/5
3/4
3/4
If the plane 2ax − 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres
x2 + y2 + z2 + 6x − 8y − 2z = 13 and x2 + y2 + z2 − 10x + 4y − 2z = 8, then a equals
-1
1
-2
-2
If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is
(-1, 7/3)
(-1/3, 7/3)
(1, 7/3)
(1, 7/3)
Switch
and the plane 
