1.If a, b, c are unequal and different from
1 such that the points\[\left(\frac{a^{3}}{a-1},\frac{a^{2}-3}{a-1}\right),\left(\frac{b^{3}}{b-1},\frac{b^{2}-3}{b-1}\right)\] and \[\left(\frac{c^{3}}{c-1},\frac{c^{2}-3}{c-1}\right)\]
are collinear, then
a) bc + ca + ab + abc = 0
b) a + b + c = abc
c) bc + ca + ab = abc
d) bc + ca + ab – abc = 3 (a + b + c)
Explanation: Suppose the given points lie on the line
2. For a > b > c > 0, the distance between
(1, 1) and the point of intersection of the lines ax + by
+ c = 0 and bx + ay + c = 0 is less than \[2\sqrt{2}.\] Then
a) a + b – c > 0
b) a – b + c < 0
c) a – b + c > 0
d) a + b – c < 0
Explanation: The lines ax + by + c = 0 and bx + ay + c
3. The line L has intercepts a and b on
the coordinate axes. The coordinate axes are rotated through a fixed angles, keeping the origin
fixed .if p and q are the intercepts of the line L on the new axes, then
\[\frac{1}{a^{2}}-\frac{1}{p^{2}}+\frac{1}{b^{2}}-\frac{1}{q^{2}}\] is equal to
a) -1
b) 0
c) 1
d) none of these
Explanation: Equation of the line L in the two coordinate
4. If P is a point (x, y) on the line. y = – 3x
such that P and the point (3, 4) are on the opposite sides of
the line 3x – 4y = 8, then
a) x > 8/15, y < – 8/5
b) x > 8/5, y < – 8/15
c) x = 8/15, y = – 8/5
d) none of these
Explanation: Let k = 3x – 4y – 8
5. The area enclosed by \[2|x| + 3|y| \leq 6\] is
a) 3 sq units
b) 4 sq units
c) 12 sq units
d) 24 sq units
Explanation: The given inequality is equivalent to the following system of inequalities
6. Let O be the origin, A (1, 0) and B (0, 1)
and P (x, y) are points such that xy > 0 and x + y < 1, then
a) P lies either inside the triangle OAB or in the
third quadrant
b) P can not lie inside the triangle OAB
c) P lies inside the triangle OAB
d) P lies in the first quadrant only
Explanation: Since xy > 0, P either lies in the first quadrant or in the third quadrant. The inequality x + y < 1 represents all points below the line x + y = 1. So that xy > 0 and x + y < 1 imply that either P lies inside the triangle OAB or in the third quadrant.
7. If a line joining two points A (2, 0) and
B (3, 1) is rotated about A in anticlockwise direction
through an angle 15º, then equation of the line in the new
position is
a) \[\sqrt{3}x+y=2\sqrt{3}\]
b) \[\sqrt{3}x-y=2\sqrt{3}\]
c) \[x+\sqrt{3}y=2\sqrt{3}\]
d) \[x-\sqrt{3}y=2\sqrt{3}\]
Explanation:
8. An equation of a line through the point
(1, 2) whose distance from the point (3, 1) has the greatest
value is
a) y = 2x
b) y = x + 1
c) x + 2y = 5
d) y = 3x – 1
Explanation: Let the equation of the line through (1, 2) be
9. Let \[0<\alpha<\pi/2\] be a fixed angle .
If \[P=\left(\cos\theta,\sin\theta\right)\] and \[Q=\left(\cos\left(\alpha-\theta\right),\sin\left(\alpha-\theta\right)\right)\]
then
Q is obtained from P by
a) clockwise rotation around the origin through an
angle \[\alpha\]
b) anticlockwise rotation around the origin through an
angle \[\alpha\]
c) reflection in the line through origin with slope \[\tan\alpha\]
d) reflection in the line through origin with slope \[\tan\left(\alpha/2\right)\]
Explanation: OP makes an angle \[\theta\] with the positive direction of x-axis and OQ makes an angle ( \[\alpha\] – \[\theta\] ) with the positive direction of x-axis
So that \[\angle\] POQ = \[\alpha\] and thus Q is obtained from P by clockwise rotation through an angle \[\alpha\] around the origin
10. On the portion of the straight line x +
y = 2 which is intercepted between the axes, a square is
constructed, away from the origin, with this portion as
one of its side. If p denotes the perpendicular distance of
a side of this square from the origin, then the maximum
value of p is
a) \[2\sqrt{3}\]
b) \[3\sqrt{2}\]
c) \[2/\sqrt{3}\]
d) \[3/\sqrt{2}\]
Explanation: Clearly p = perpendicular distance from (0, 0) to AD + side of the square
