Q1.
The
straight line y = x–2 rotates about a point where it cuts the x-axis and
becomes perpendicular to the straight line ax + by + c = 0. Then its equation
is.
(A) ax
+ by + 2a = 0
(B) ax
– by – 2a = 0
(C) bx
+ ay – 2b = 0
(D) ay
– bx + 2b = 0
Solution:
Slope of the line in the new position
is b/a, since it is perpendicular to the line ax + by + c = 0 and it cuts the
x-axis. Hence the required line passes through (2, 0) and its slope is b/a.
The required equation is y – 0 = b/a
(x-2)
or, ay = bx – 2b or, ay – bx + 2b =
0.
Hence (D) is the correct answer.
Q2.
Consider the equation
y - y1 = m(x-x1). If m and x1 are fixed and
different lines are drawn for different values of y1, then.
(A)
the
line will pass through a fixed point.
(B)
there
will be a set of parallel lines.
(C)
all
the lines intersect the line x = x1
(D)
all
the lines will be parallel to the line y=y1.
Solution:
For a fixed value of m, the given
lines form a set of parallel lines all with the slope m,
&⇒
answer (B) is correct, while (A) is incorrect. However this set will not be
parallel to the line y = y1, whose slope is zero, unless
m = 0. And (D) is therefore excluded. If, furthermore, x1 is fixed,
the above set of lines will all intersect the line x = x1. Hence
answer (C) is also correct. Hence (B) and (C) are the correct answers.
Q3.
Let
2x–3y =0 be a given line and P (sinq, 0) and Q (0, cosq)
be the two points. Then P and Q lie on the same side of the given line, if q
lies in the .
(A)
1st quadrant
(B)
2nd quadrant
(C)
3rd quadrant
(D)
4th quadrant
Solution:
P and Q lie on the same side if 2sinq
and - 3cosq have the same signs i.e. sinq
and cosq have opposite signs which is true for the 2nd and 4th
quadrant. .
Hence (B) and (D) are
the correct answers.
Q4.
The
straight lines of the family x(a + b) + y (a – b) = 2a (a and b being
parameters) are
(A)
not concurrent
(B)
Concurrent at (1, -1)
(C)
Concurrent at (1, 1)
(D)
None of these
Solution:
The given equation can be written as
a(x + y – 2) + b(x – y) = 0 or (x +
y – 2) + b/a(x – y) = 0
This is a family of lines concurrent
at x – y = 0 and x + y – 2 = 0
On solving these two equations we get
(1, 1)
Hence (C) is the correct
answer.
Q5.
Drawn
from origin are two mutually perpendicular lines forming an isosceles triangle
together with the straight line 2x+y = a. Then the area of this triangle is.
(A)
(B)
(C)
(D) None of these
Solution:
Let the two perpendiculars through
the origin intersect 2x+y = a at A and B so that the triangle OAB is isosceles.
OM
= length of perpendicular from
O
to AB, OM = .Also AM =
MB = OM
&⇒
AB = . Area of D
OAB = =
Hence
(C) is the correct answer. .
|
|
Q6.
A curve with
equation of the form y = ax4 + bx3+cx +d has zero
gradient at the point (0, 1) and also touches the x-axis at the
point ( -1, 0). Then the values of x for which the curve has
negative gradients are.
(A) x > -1
(B) x < 1
(C) x < -1
(D) -1 £
x £ 1
Solution:
y
= ax4 + bx3 +cx +d .
. . . (1).
y
touches x-axis at (-1, 0)
so,
( -1, 0) lies on it and dy/ dx =0
so,
0 = a – b – c +d . . . . (2).
From
(1) dy/ dx = 4ax3 + 3bx2 +c . . . .
(3).
Hence
&⇒
- 4a + 3b +c =0 . . . . (4)
Also
= c =0 (since
curve touches (0, 1) )
&⇒ ( 0, 1) also lies on it. Hence d
= 1.
Putting
values of c and d in (2) and (4), solving for a and b we get a = 3, b
= 4. Therefore (3) becomes = 12x3
+12x2
Now dy/dx < 0
&⇒
12 x3 + 12x2 < 0
&⇒ 12x2 (x+1) < 0
&⇒
x < -1
Hence (C) is the correct answer.
Q7.
The
area of triangle is 5. Two of its vertices are (2, 1) and (3, -2), the third
vertex is lying on y = x + 3. The co-ordinates of the third vertex can be.
(a)
(b)
(c)
(d)
Solution:
As the third vertex lies on the line
y = x+3, its co-ordinates are of the form (x, x+3). .
The area of triangle is |4x-4|
= |2x-2|
According to given
condition
&⇒ 2x-2 = ± 5
&⇒
x = -3/2, 7/2
Hence the coordinates of third vertex
can be
Hence (A) and (C) are
the correct answers.
Q8.
The orthocentre of
the triangle formed by the lines 2x2 + 3xy – 2y2 – 9x +
7y – 5 = 0, 4x + 5y – 3 = 0 lies at
(A) (3/5 , 11/5)
(B) (6/5, 11/5)
(C) (5/6, 11/5)
(D) None of these
Solution:
The pair of straight lines 2x2 + 3xy – 2y2
– 9 x + 7y– 5 = 0 are perpendicular to each other so orthocentre is point of
intersection of these lines. Hence (A) is correct.
Q9.
If
the lines x = a + m, y = -2 and y = mx are concurrent, the least value of |a|
is
(A) 0
(B) Ö2
(C) 2Ö2
(D)
None of these
Solution:
Since the lines are concurrent
&⇒
m2 + am + 2 = 0
Since m is real, a2
≥
8, |a| ≥2Ö2
Hence the least value
of |a| is 2Ö2
Hence (C) is the
correct answer.
Q10.
If the line y = x
cuts the curve x3 + y3 +3xy + 5x2 + 3y2
+ 4x + 5y – 1 = 0 at the points A, B, C then OA. OB. OC is
(A)
(B) 3+1
(C)
(D) none of these
Solution:
The Line y=passes
through the origin.
Therefore it can be written as
where r is the distance of any point
(x,y) on y = x from (0, 0)
i.e. will always
lie on the line y = x at a
distance r from (0,0)
Since it cuts the curve
x3 + y3 + 3xy +
5x2 + 3y2 + 4x + 5y – 1 = 0
we have
r3
This is a cubic in r which shows that
y = x cuts the
curve at 3 points.
\OA.
OB. OC =
Hence (A) is
the correct answer.