Parabola
Short Description
Parabola for IITJEE...
Description
TOPIC: PARABOLA LEVEL – 1 1. If the normals drawn at the end points of a variable chord PQ of the parabola 2 y = 4 ax intersect at parabola, then the locus of the point of intersection of the tangent drawn at the points P and Q is (A) x +a = 0 (B) x – 2a = 0 (C) y2– 4x +6 = 0 (D) none of these 2.
If tangents at A and B on the parabola y2 = 4ax intersect at the point C, then ordinates of A, C and B are (A) always in A.P. (B) always in G.P. (C) always in H.P. (D) none of these
3.
The set of points on the axis of the parabola 2((x – 1)2 + (y – 1)2) = (x + y)2 , from which 3 distinct normals can be drawn to the parabola, is the set of points (h, k) lying on the axis of the parabola such that (A) h > 3 (B) h > 3/2 (C) k > 3/2 (D) k > 3
4.
Tangents are drawn from (–2, 0) to y = 8x, radius of circle(s) that would touch these tangents and the corresponding chord of contact, can be equal to,
2
5.
(A) 4 2 1 (B) 2 1 (C) 8 2 (D) None of these. 2 The coordinates of the point on the parabola y = x + 7x +2, which is nearest to the straight line y = 3x – 3 are (A) (–2, –8) (B) (1, 10) (C) (2, 20) (D) (–1, –4)
6.
The circle drawn with variable chord x + ay – 5 = 0 (a being a parameter) of the parabola y2 =20x as diameter will always touch the line (A) x + 5 = 0 (B) y + 5 = 0 (C) x + y + 5 = 0 (D) x – y + 5 = 0
7.
The point P on the parabola y2 = 4ax for which |PR – PQ| is maximum, where R (– a, 0), Q (0, a), is (A) (a, 2a) (B) ( a, –2a) (C) (4a, 4a) (D) (4a, –4a)
8.
If the normals at the end points of a variable chord PQ of the parabola y2 – 4y – 2x = 0 are perpendicular, then the tangents at P and Q will intersect at (A) x + y = 3 (B) 3x – 7 = 0 (C) y+3 = 0 (D) 2x + 5 = 0
9.
The number of point(s) (x, y) (where x and y both are perfect squares of integers) on the parabola y2 = px, p being a prime number, is (A) one (B) zero (C) two (D) infinite
10.
If two distinct chords of a parabola y2 = 4ax, passing through (a, 2a) are bisected on the line x + y =1, then length of the latus–rectum can be (A) 2 (B) 1 (C) 4 (D) 5
11.
A quadrilateral is inscribed in a parabola, then (A) quadrilateral may be cyclic (B) diagonals of the quadrilateral may be equal (C) all possible pairs of adjacent sides may be perpendicular (D) none of these
12.
The locus of point of intersection of any tangent to the parabola y2 = 4a (x – 2) with a line perpendicular to it and passing through the focus, is (A) The tangent to the parabola at the vertex (B) x = 20 (C) x = 0 (D) None of these
13.
The locus of the centre of the circle described on any focal chord of a parabola y2 = 4ax as diameter is (A) x2 = 2a (y – a) (B) x2 = –2a(y – a) (C) y2 = 2a (x – a) (D) y2 = –2a(x –a)
14.
Two parabolas y2 = 4a(x – 1) and x2 = 4a(y – 2) always touch each other (1, 2) being variable parameters). The their point of contact lies on a
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(A) straight line
(B) circle
(C) parabola
(D) hyperbola
15.
If the parabola y = (a – b) x2 + (b – c)x + (c –a) touches the x–axis in the intervals ( 0, 1) then the line ax + by +c = 0 (A) always passes through a fixed point (B) represents the family of parallel lines (C) data insufficient (D) none of these
16.
If one end of the diameter of a circle is (3, 4) which touches the x–axis then the locus of other end of the diameter of the circle is (A) parabola (B) hyperbola (C) ellipse (D) none of these
17.
The point (1, 2) is one extremity of focal chord of parabola y2 = 4x. The length of this focal chord is (A) 2 (B) 4 (C) 6 (D) none of these
18.
If normals at two points of a parabola y2 = 4ax intersect on the curve, then the product of ordinates is 2 2 2 2 (A) 2a (B) 4a (C) 6a (D) 8a
19.
If AFB is a focal chord of the parabola y2 = 4ax and AF = 4, FB = 5, then the latus–rectum of the parabola is equal to 80 9 (A) (B) (C) 9 (D) 80 9 80 If at x = 1, y = 2x is tangent to the parabola y = ax2 + bx + c, then respective values of a, b, c are 1 1 1 1 1 1 (A) , 1, (B) 1, , (C) , , 1 (D) None of these 2 2 2 2 2 2 The number of focal chord(s) of length 4/7 in the parabola 7y2 = 8x is (A) 1 (B) zero (C) infinite (D) None of these
20.
21.
22.
If the chord of contact of tangents from a point P(h, k) to the circle x2 + y2 = a2 touches the circle x2 + (y – a)2 = a2, then locus of P is 2 2 2 2 2 2 (A) y = a – 2ax (B) y = a + 2ax (C) x = a +2ay
2
2
(D) x = a – 2ay
2
23.
The length of the chord of the parabola x = 4y passing through the vertex and having slope cot is (A) 4 cos . cosec2 (B) 4 tan sec (C) 4 sin. sec2 (D) none of these
24.
If three normals can be drawn from (h, 2) to the parabola y2 = –4x, then (A) h < –2 (B) h > 2 (C) –2 < h < 2 (D) h is any real number
25.
The parametric coordinates of any point on the parabola y2 = x can be (A) (sin2, sin) (B) (cos2, cos) (C) (sec2, sec)
26.
Slope of tangent to x2 = 4y from (–1, –1) can be (A)
27.
1 5 2
If line y = 2x + (A) 1/ 2
28.
29.
30. 31.
(D) none of these
(B)
1 5 2
(C)
1 5 2
(D)
1 2 is tangent to y = 4ax, then a is equal to 4 (B) 1 (C) 2
1 5 2
(D) None of these
A line passing through the focus of the parabola y2 = 4ax, intersects the parabola in two distinct points. Slope of the line is (A) any real number (B) greater than 1 and less than 1 (C) less than 1 or greater than 1 (D) none of these If normals are drawn form a point P (h, k) to the parabola y2 = 4ax then the sum of the intercepts which the normals cut off from the axis of the parabola is (A) (h + a) (B) 3(h + a) (B) 2(h + a) (D) None of these 2
2
2
The shortest distance between the parabola y = 4x and the circle x + y + 6x – 12y + 20 = 0 is (A) 4 2 5 (B) 0 (C) 3 2 + 5 (D) 1 2 2 2 The equation (13x – 1) + (13y – 1) = k (5x – 12y + 1) will represent a parabola if
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(A) k = 2 32. 33.
34.
(B) k = 81
(C) k = 169
(D) k =1
The length of the common chord of the curves y2 – 4x–4 = 0 and 4x2 + 9y2 – 36 = 0 is (A) 2 3 units (B) 3 2 units (C) 4 units (D) 6 units 2 The Cartesian equation of the curve whose parametric equations are x = t + 2t + 3 and y = t + 1 is (A) y = (x – 1)2 + 2(y – 1) + 3 (B) x = (y – 1)2 + 2(y –1) + 5 2 (C) x = y + 2 (D) None of these x + y = a represents (A) a part of parabola (B) ellipse
(C) Hyperbola
(D) Line segment
35.
Let y2 = 4ax be a parabola and x2 + y2 + 2bx = 0 be a circle. Then condition on a and b so that parabola and circle touch each other externally is (A) ab > 0 (B) ab < 0 (C) ab < –1 (D) none of these
39.
Let y2 = 4ax be a parabola and x2 – y2 = a2 be a hyperbola. Then number of common tangents is (A) 2 for a < 0 (B) 1 for a < 0 (C) 2 for a > 0 (D) 1 for a > 0
37.
If the line y – ( 3 , 0)) 4
(A) 38.
39. 40.
32
3 x + 3 = 0 cuts the parabola y2 = x + 2 at A and B, then PA. PB is equal to (where P
4 2 3
(B)
(C)
4 3 2
2
(D)
32
3 3 3 Slope of tangent to x2 = 4y from (–1, –1) can be 1 5 1 5 1 5 1 5 (A) (B) (C) (D) 2 2 2 2 1 2 If line y = 2x + is tangent to y = 4ax, then a is equal to 4 (A) 1/ 2 (B) 1 (C) 2 (D) None of these A line passing through the focus of the parabola y2 = 4ax, intersects the parabola in two distinct points. Slope of the line is (A) any real number (B) greater than 1 and less than 1 (C) less than 1 or greater than 1 (D) none of these
41.
The equation of the line of the shortest distance between the parabola y2 = 4x and the circle x2 + y2 – 4x – 2y + 4 = 0 is (A) x + y = 3 (B) x – y = 3 (C) 2x + y = 5 (D) None of these
42.
Number of real normals drawn from the point (4, 2) to the parabola y2 = 8x is (A) 1 (B) 2 (C) 3 (D) None of these
43.
The line x – y =1 intersects the parabola y2 = 4x at A and B. Normals at A and B intersect at C. If D is the point at which line CD is normal to the parabola, then coordinates of D are (A) (4, –4) (B) (4, 4) (C) (–4, –4) (D) None of these ‘t1’ and ‘t2’ are two points on the parabola y2 = 4ax. If the focal chord joining them coincides with the normal chord, then (A) t1( t1 +t2) + 2 = 0 (B) t1 + t2 = 0 (C) t1t2 = –1 (D) none of these 2 A focal chord of the parabola y = 4ax is of length 4a. The angle subtended by it at the vertex of the parabola is –1 –1 (A) tan (–4/3) (B) /2 (C) tan (4/3) (D) none of these 2 The exhaustive set of values of k for which tangents drawn from the point (k+3, k) to the parabola y = 4x, are real, is (A) (–2, 6) (B) (–, –2) (6, ) (C) (–, –2] [6, ) (D) None of these.
44.
45.
46.
47.
If two distinct chords of a parabola y2 = 4ax, passing through (a, 2a) are bisected on the line x + y =1, then length of the latus–rectum can be (A) 2 (B) 1 (C) 4 (D) 5
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48.
The vertex V of the parabola y2 = 4ax and two points on the parabola together with the point P form a square. The coordinates of P are (A) (8a, 0) (B) (4a, 4a) (C) (8a, –4a) (D) (4a, 0)
49.
If 2x + y + = 0 is a normal to the parabola y = –8x, then equals (A) 12 (B) –12 (C) 24
2
(D) –24 2
50.
The angle between the tangents drawn from the origin to the parabola y = 4a(x – a) is (A) 90° (B) 30° (C) tan–1(1/2) (D) 45°
51.
The normal drawn at a point (at12, 2at1) of the parabola y2 = 4ax meets it again in the point (at22, 2at2). Then 2 (A) t1 = 2t2 (B) t1 = 2t2 (C) t1 = –1/t2 (D) None of these 2 The line joining the points (t1 , t1) and (t22, t2), (t1 0, t2 0) passes through the origin, if and only if. 1 1 (A) t1 + t2 = 0 (B) t1 – t2 = 0 (C) 0 (D) t1t2 = –1 t1 t 2
52.
53.
A line making an acute angle of more than 45° with the positive direction of x axis is tangent to the parabola y2 = 4x, then yintercept of the line is (A) any positive real number (B) less than 1 (C) more than 1 (D) none of these
54.
The point on the parabola y2 = 8x whose distance from the focus is 8 has x coordinate as (A) 0 (B) 2 (C) 4 (D) 6
55.
Three normals are drawn to the parabola y2 = 4ax from a given point (x1, y1). The algebraic sum of the ordinates of their feet is (A) a (B) a (C) 0 (D) 1 Two tangents, perpendicular to each other, to the parabola y2 = 4ax intersect on the line (A) x = 0 (B) x + a = 0 (C) x – 2a = 0 (D) x – a = 0
56.
57. 58.
59. 60.
The focal distance of a point on the parabola y2 = 4x and above its axis, is 5 units. Its coordinates are (A) (4, 4) (B) (– 4, 4) (C) (2, 2) (D) None of these The length of subtangent to the parabola y2 = 4x at the point whose ordinate is 2, is (A) 1 unit (B) 2 units (C) 3 units (D) 4 units If the tangents at P and Q on a parabola meet in T, then SP, ST and SQ are in (A) AP (B) GP (C) HP (D) None of these The family of parabolas having vertices at (0, a), where a is any real number, latus rectum of unit length and axis as y axis satisfies the differential equation (A)
61.
dy 1 dx 4 x
(B)
dy 1 = dx 4x
(C)
dy 2x dx
(D)
dy 4 x dx
The set of points on the axis of parabola y2 4y = 8x + 2 from where 3 distinct normals can be drawn to the parabola is 13 (A) x, 2 , x (B) {(x, 0), x > 8} (C) {(x, –3), x < 8} (D) None of these 4
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LEVEL – 1 1.
B
2.
A
3.
A, B, C, D
4.
A
5.
A
6.
A
7.
A
8.
D
9.
A
10.
A, B
11.
A, B
12.
A, B
13.
C
14.
D
15.
A
16.
A
17.
B
18.
D
19.
A
20.
A
21.
B
22.
D
23.
A
24.
A
25.
D
26.
A, B
27.
A
28.
D
29.
C
30.
A
31.
D
32.
C
33.
C
34.
A
35.
A
36.
A, C
37.
A
38.
A, B
39.
A
40.
D
41.
A
42.
A
43.
A
44.
D
45.
A
46.
B
47.
A, B, C
48.
A
49.
C
50.
A
51.
D
52.
B
53.
B
54.
D
55.
C
56.
B
57.
A
58.
B
59.
B
HINTS & SOLUTIONS HINTS & SOLUTIONS (PARABOLA) LEVEL - 1 1. B Let the points P and Q be (at12, 2at1) and (at22, 2at2) respectively. Since the normals at P and Q intersects at parabola, t1t2 = 2 Let (h, k) be the point of intersection of the tangents at P and Q. Then at1t2 = h and a( t1 + t2) = k h = 2a x – 2a = 0 2.
A A (at12, 2at1) B (at22, 2at2) Tangents at A and B will intersect at the point C, whose coordinate is given by (at1t2, a(t1+t2)) clearly ordinate of A , C and B are always in A.P.
3.
A, B, C, D The directrix of the parabola is x + y = 0 and focus is (1, 1). The length of perpendicular from (1, 1) to the directrix is 2 1 focal length is the point on the axis which is at a distance equal to focal length from the 2 focus towards the right is (3/2, 3/2). All the points to the right of this point, lying on the axis will satisfy the required condition.
4.
A Point ‘P’ clearly lies on the directrix of y2 = 8x. Thus slope of PA and PB are 1 and –1 respectively. Equation of PA : y =x +2, equation of PB : y = –x – 2, equation of AB : x = 2. Let the centre of the circle be ( h, 0) and radius be ‘r’ h2 h2 r 1 2
y
A
P(–2, 0)
(h, 0) O
x
B
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2
2
2
h + 4 + 4h = 2(h +4 – 4h) h – 12h + 4 = 0 12 8 2 h= = 6 4 2 | h –2| = 4( 2 –1), 2 4( 2 +1). 5.
A Any point on the parabola is (x, x2+ 7x + 2) Its distance from the line y = 3x – 3 is given by P=
3x x 2 7x 2 3 9 1
x2 4x 5 10
=
x2 4x 5 10
2
(as x +4x + 5 > 0 for all x R)
dP = 0 x = –2 dx The required point (–2, –8). 6.
A Clearly x + ay – 5 = 0 will always pass through the focus of y2 = 20x i.e. (5, 0). Thus the drawn circle will always touch the directrix of the parabola i.e. the line x + 5=0
7.
A We know any side of the triangle is more than the difference of the remaining two sides so that |PR – PQ| RQ The required point P will be the point of intersection of the line RQ with parabola which is (a, 2a) as PQ is a tangent to the parabola.
8.
D Since normals at P and Q are perpendicular, the tangents at P and Q will also be perpendicular but any two perpendicular tangents of a parabola always intersect on its directrix. The parabola is (y – 2)2 = 2(x +2). So its directrix is 2x + 5 = 0
9.
A If x is a perfect square, then px will be a perfect square only if p is a perfect square, which is not possible as p is a prime number. Hence y cannot be a perfect square. So number of such points will be only one (0, 0).
10.
A, B Any point on the line x +y =1 can be taken (t, 1 – t) equation of the chord, with this as mid–point is y(1 – t) – 2a(x +t) = (1 –t)2 –4a t, it passes through (a, 2a) 2 2 2 so t – 2t + 2a – 2a +1 = 0, this should have 2 distinct real roots so a – a< 0, 0< a< 1, so length of latus–rectum < 4.
11.
A, B As a circle can intersect a parabola in four points, so quadrilateral may be cyclic. The diagonals of the quadrilateral may be equal as the quadrilateral may be an isosceles trapezium. A rectangle cannot be inscribed in a parabola. So (C) is wrong.
12.
A, B It is a well known property of a parabola that a tangent and normal to it from focus intersect at tangent at vertex.
13.
C If A(at12 ,2 at1) , B(at22 ,2 at2) be the extremities of a focal chord for the parabola y2 = 4ax, then t1t2 = –1 . . . . . (1)
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y
A
2
y = 4ax
C S
O
x
B
We want to find locus of point C(, ) where C is the centre of the circle having AB as the diameter. a 2 = t1 t 22 ; = a( t1+ t2) 2 To eliminate t1, t2 2 2 = a2(t12 +t22 + 2t1t2) = a2 a 2
2
= 2a( – a) 2 i.e. y = 2a(x –a). 14.
D y2 =4a(x – 1), x2 = 4a(y – 2) dy dy 2y. = 4a, 2x = 4a. dx dx At the point of tangency the given curves must have the same slope 2x 4a xy = 4a2 4a 2y Locus of point of contact is hyperbola.
15.
A Solving equation of parabola with x–axis (i.e. y = 0) We get (a – b)x2 + (b – c)x +(c – a) = 0, which should have two equal values of x, as x–axis touches the parabola (b –c)2 –4 (a – b)( c– a) = 0 (b + c –2a)2 = 0 –2a + b + c = 0 ax + by +c = 0 always passes through (–2, 1).
16.
A Let other end of diameter (h, k) k4 k4 3h Hence centre is , . This circle touches x–axis means r = 2 2 2 2
= 17.
3h k 4 3 4 2 2
2
gives the equation of parabola.
B The parabola y2 = 4x. Here a =1 and focus is (1, 0). The focal chord is ASB. This is clearly latus rectum of parabola, its value = 4.
y
A(1, 2)
S(1, 0)
X
B
18.
19.
D Let the points be (at12, = 4a2t1t2 = 8a2 (since t1t2 = 2) A
2at1)
and
(at22,
2at2)
therefore
product
of
ordinates
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FA = 4, FB = 5
1 1 1 We know that a AF FB 20 80 a= 4a = . 9 9
20.
A
F (a, 0)
O B
A For x =1, y = a + b + c.
1 b y a b c ax x 1 c 2 2 Comparing with y =2x, c = a, b = 2(1 – a) Which are true for choice (A) only. Tangent at (1, a + b + c) is
21.
B Since length of latus–rectum = 8/7 Latus–rectum is the smallest focal chord. Hence focal chord of length 4/7 does not exist.
22.
D Equation of chord of contact is hx + ky = a2 ka a2 It will touch x2 + (y – a)2 = a2 , if = a k – a = h2 k 2 h2 k 2 (k – a)2 = h2 +k2 locus of (h, k) is x2 = a2 – 2ay
23.
A Let A Vertex, AP chord of x2 = 4y such that slope of AP is cot . Let P (2t, t2) t t Slope of AP = cot = t = 2 cot 2 2 Now, AP =
2
4t 2 t 4 = t 4 t 2 = 4 cot cosec = 4 cos . cosec .
24.
A Any normal to the parabola y2 = –4x of slope m is y = mx + 2m + m3. If it passes through (h, 2), m3 + (h + 2)m – 2 = 0 Let f(m) = m3 +(h + 2) m –2 Then f(m) = 3 m2 + (h + 2) If f(m) = 0 has three distinct real roots, then necessary conditions is that f(m) = 0 should have 2 distinct real roots i.e h < – 2.
25.
D No choice among (A), (B), (C) is giving all points on parabola y2 = x
26.
A, B 2 y = x/2 = m x = 2m y=m 2 So equation of tangent is y – m = m(x–2m) which passes through (–1, –1) 1 5 2 2 –1 –m = m(–1–2m) m + m – 1 = 0 m= 2
27.
A c =a/m a = 2 (1/4) = 1/2
28.
D Every focal chord other than the axis of the parabola intersects the parabola twice. C
29.
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Equation of normal: y = mx – 2am – am3 Put y = 0 We get x1 = 2a + am12 x2 = 2a + am22 x3 = 2a+ am3 2 where x1, x2, x3 are the intercepts on the axis of he parabola, The normal passes through (h,k) am3 + (2a – h) m + k = 0 m1 + m2 + m3 = 0 2a h m1 m2+ m2m3 + m3 m1 = a 2
m12 m22 m23 m1 m2 m3 2 m1m2 m2m3 m3 m1 = 2 x1 + x2 + x3
2a h a
= 6a – 2( 2a–h) = 2 ( h + a)
30.
A Normal at a point (m2, – 2m) on the parabola y2 = 4x is given by y = mx – 2m – m3. If this is normal to the circle also then it will pass through center of the circle so 6 = –3m – 2m –m3 m = – 1. Since shortest distance between parabola and circle will occur along common normal, shortest distance is 4 2 5
31.
D (13x – 1)2 + (13y – 1)2 = k (5x – 12y+ 1)2 2
2
1 1 x y k 13 13
5x 12y 1
2
2
5 12
2
k 169
5x 12y 12
This will represent a parabola if k = 1 32.
C The common chord is y –axis by drawing the graph of each curves, so length of common chord is equal to 4 units
33.
C x = t2 + 2t + 3 = (t + 1)2 + 2 = y2 + 2
34.
A x + y = a x+ y – a = –2 x y (x + y – a)2 = 4xy x2–2xy + y2 – 2ax – 2ay + a2 = 0 which represents a parabola
35.
A The given parabola and circle will touch each other externally if they are on the opposite sides of y – axis, which will happen only if a and b are of same sign i.e. ab > 0
36.
A, C Let a common tangent be y = mx + c a Then C = a m2 1 m 4 2 or, m – m – 1 = 0 1 5 2 Thus there are always two common tangents irrespective of value of a
or, m =
37.
A y–
3 x + 3 = 0 can be rewritten as
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y0 x 3 (1) r 1 3 2 2 Solving (1) with the parabola y2 = x + 2 3r 2 r 3 20 4 2
4
PA.PB = r1.r2 =
3 2
3
38.
A, B y = x/2 = m x = 2m 2 y=m 2 So equation of tangent is y – m = m(x–2m) which passes through (–1, –1) 2 –1 –m = m(–1–2m) 2 m +m–1=0 1 5 m= 2
39.
A c =a/m a = 2 (1/4) = 1/2
40.
D Every focal chord other than the axis of the parabola intersects the parabola twice.
41.
A Line of shortest distance is normal for both parabola and circle Centre of circle is (2, 1) Equation of normal to circle is y – 1 = m(x – 2) y = mx + (1 – 2m) ….. (1) Equation of normal for a parabola is y = mx – 2am – am3 ….(2) Comparing (1) and (2), am3 = –1 m3 = –1 m = –1 (a = 1) Equation is y– 1 = –x +2 x + y = 3. A Equation of normal to parabola is y = mx – 4m – 2m3 (since a = 2) it passes through (4, 2) 3 2m = – 2 m = –1 Only one normal is positive.
42.
43.
Since y = x –1 solving this equation of parabola (x –1)2 = 4x x2 – 6x + 1 = 0 x=3 8 y=2 8 Suppose point D is (x3, y3) then y1 +y2 + y3 = 0 2+ 8 2 8 + y3 = 0 y3 = –4 then x3 = 4 The point is (4, –4).
44.
D For a focal chord t1 t2 = –1 and for the normal t1( t1 + t2) + 2 = 0 2 2 t1 + t1 t2 + 2 = 0 t1 = –1 t1 is imaginary.
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45.
A (a, 2a) 2a (0, 0) 2a (a, –2a)
m1 = 2 and m2 = –2 m m2 4 4 tan = 1 1 m1m2 1 4 3
46.
4 = tan–1 . 3 B The point (k+3, k) for chord of contact yk = 2( x + k+ 3) yk – 2x = 2k + 6 2
2x 2k 6 Solving with the equation of parabola 4x 0 k x2 + (2k + 6 – k2) x + (k + 3)2 = 0 (2k + 6 – k2)2 – 4(k + 3)2 > 0 k4 – 4k3 – 12k2 > 0 k2 (k–6) (k +2) > 0 k (–, –2) (6, )
47.
A, B, C (a, 2a) (a, 0)
O
Let the point on the line where it is bisected be (, ) +=1 Equation of chord T = S, y – 2a(x + ) = 2 = 4a 2 2 + 2a(x +) = = 4a 2 2 + 2a – 2a = 0 (since = 1 – ) Disc 0 8a(1 – a) 0 a1 4a 4 (A), (B) and (C) are the correct answers. 48.
A 2
A(at1 , 2at1) 2a (0, 0)
P 2a 2
B(at1 , –2at1)
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Since angle between the sides = 90 2 2 1 t1 t1 t1 = 2 Hence the coordinates of A and B are (4a, 4a) and (4a, –4a). Coordinates of P will be (8a, 0). 49.
C y = mx + c is a normal to y2 = 4ax if c = –2am – am3 = 24
50.
A Joint equation of the pair of tangents from (0, 0) is (y2 – 4ax + 4a2)4a2 = (x – 2a)2.4a2 2 2 y –x =0 y = x = 90°.
51.
D The equation of the normal at (at12, 2at1) is y – 2at1 = – t1(x – at12). It passes through (at22, 2at2). 2 = – t1(t1 + t2)
52.
B t1 0 1 . t12 0 t1
Slope of OT1 =
t2 0 1 t 22 0 t 2 Since line T1T2 passes through origin. t1 = t2.
Slope of OT2 =
53. 54. 55. 56.
B D C B
57.
A Focal distance of (t2, 2t) = (t2 + 1) = 5 t = 2 So point is (4, 4).
58.
B y y Length of subtangent = = = 2 units dy / dx x1,y1 2x / y 1,2
59.
B Let parabola be y2 = 4ax 2 2 Let P = (at1 , 2at1), Q = (at2 , 2at2) Then, T = (at1t2, a(t1 + t2))
2
SP = a2 t12 1 4a2 t12 = a t12 1 Similarly SQ = ST =
a
2
2 a(t2
t1t 2 1
2
+ 1) a2 t1 t 2
2
=
a2 t12 1 t 22 1
SP, ST, SQ are in G.P.
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LEVEL - II 1. Two perpendicular tangents to y2 = 4ax always intersection on the line (A) x + 4a = 0 (B) x + 2a = 0 (C) x = a 2.
The line y = mx + c, touches the parabola x2 = 4ay if a (A) c = –am (B) c = (C) c = –am2 m
(D) x + a = 0
(D) c =
a m2
2
3.
If the normals at points t1 and t2 of the parabola y = 4ax meets it again at the point t3, then 2 (A) t2 = –t1 – (B) t1t2 = –1 (C) t1t2 = 2 (D) none of these t1
4.
The numcer of distinct normals that can be drawn from the point (–2, 1) to the parabola y2 + 6x – 2y + 13 = 0 (A) 1 (B) 2 (C) 3 (D) none of these
5.
The angle between the tangents drawn from the origin to the parabola y2 = 4b(x – b) is 1 (A) (B) (C) (D) tan1 3 4 2 2
6.
The pole of the line 2x + y + 8 = 0 w.r.t the parabola y2 – 4x is (A) (4, 1) (B) (8, 1) (C) (4k, k); k I
(D) (1, 4)
7.
The equation of the diameter of the parabola x2 = 4ay, if the slope of the series of parallel chords is m, is 2a 2a (A) y = (B) x = (C) x = 2am (D) y = 2am m m
8.
Equation x2 – 2x – 2y + 5 = 0 represents (A) a circle with centre (1, 1) 5 (C) a parabola with directrix y = 2
9.
10.
(B) a parabola with vertex (1, 2) (D) a parabola with directrix y =
The point on the curve y2 = 4x which is nearest to the point (2, 1) is (A) (1, –2) (B) (–2, 1) (C) (1, 22)
1 3
(D) (1, 2)
2
The vertex of the parabola x + 8x + 12y + 4 = 0 is (A) (–4, 1) (B) (4, –1) (C) (–4, –1)
(D) (4, 1)
11.
Consider the parabola y2 = 8x; if the normal at a point P on the parabola meets it again at a point Q, then the least distance of Q from the tangent at the vertex of the parabola is (A) 16 (B) 8 (C) 4 (D) none of these
12.
Tangents drawn from the point A(–4, 8) to the parabola y = 16x meet the parabola at P and Q. Then locus of the centre of the circle described on PQ as diameter is (A) x2 = –2a(y – a) (B) y2 = –2a(x – a) (C) x2 = 2a(y – a) (D) y2 = 2a(x – a)
13.
The total number of chords that can be drawn from the point (a, a) to the circle x + y = 2a such that 2 they are bisected by the parabola y = 4ax is (A) 1 (B) 4 (C) 2 (D) 0
14.
Number of the normals drawn from the point (6, –8) to the parabola y2 – 12y – 4x + 4 = 0 is (A) 3 (B) 1 (C) 2 (D) 0
15.
A ray of light moving parallel to the x-axis gets reflected from a parabolic mirror whose equation is (y – 2)2 = 4(x + 1). After reflection, the ray must pass through the point (A) (2, 0) (B) (–1, 2) (C) (0, –2) (D) none of these
2
2
2
2
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2
16.
Let P (, ) be any point on parabola y = 4x (0 2). M is the foot of perpendicular from the focus S to the tangent at P, then the maximum value of area of PMS = (A) 1 (B) 2 (C) (D) 3 6
17.
If P is the pole of chord AB of a parabola and the length of perpendiculars from A, P, B on any tangent to the curve be p1, p2 and p3, then (A) p3 is the G.M. between p1 and p2 (B) p3 is the A.M. between p1 and p2 (C) p1 is the G.M. between p2 and p3 (D) none of these
18.
If tangents at A and B on the parabola y2 = 4ax intersect at point C, then ordinates of A, C and B are in (A) A.P (B) G.P (C) H.P (D) none of these
19.
If the normals at three points A, B, C of a parabola meet in a point, then the centroid of ABC lies on (A) a line parallel to y-axis (B) a line perpendicular to y-axis different from axis of the parabola (C) axis of the parabola (D) the line y = x
20.
If PQ be a focal chord of a parabola, then the tangent at P and normal at Q are (A) perpendicular (B) make an angle of 45 (C) parallel (D) make an angle of 60 1.
D
2.
C
3.
C
4.
A
5.
C
6.
A
7.
C
8.
B, C
9.
D
10.
A
11.
A
12.
D
13.
C
14.
B
15.
D
16.
A
17.
D
18.
A
19.
C
20.
C
LEVEL – III 1. Two perpendicular tangents to y2 = 4ax always intersection on the line (A) x + 4a = 0 (B) x + 2a = 0 (C) x = a 2.
(D) x + a = 0
2
The line y = mx + c, touches the parabola x = 4ay if a (A) c = –am (B) c = (C) c = –am2 m
(D) c =
a m2
3.
If the normals at points t1 and t2 of the parabola y2 = 4ax meets it again at the point t3, then 2 (A) t2 = –t1 – (B) t1t2 = –1 (C) t1t2 = 2 (D) none of these t1
4.
The numcer of distinct normals that can be drawn from the point (–2, 1) to the parabola 2 y + 6x – 2y + 13 = 0 (A) 1 (B) 2 (C) 3 (D) none of these
5.
The angle between the tangents drawn from the origin to the parabola y2 = 4b(x – b) is 1 (A) (B) (C) (D) tan1 3 4 2 2
6.
The pole of the line 2x + y + 8 = 0 w.r.t the parabola y2 – 4x is (A) (4, 1) (B) (8, 1) (C) (4k, k); k I
(D) (1, 4)
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7.
The equation of the diameter of the parabola x2 = 4ay, if the slope of the series of parallel chords is m, is 2a 2a (A) y = (B) x = (C) x = 2am (D) y = 2am m m
8.
Equation x – 2x – 2y + 5 = 0 represents (A) a circle with centre (1, 1) 5 (C) a parabola with directrix y = 2
9.
10.
2
(B) a parabola with vertex (1, 2) (D) a parabola with directrix y =
The point on the curve y2 = 4x which is nearest to the point (2, 1) is (A) (1, –2) (B) (–2, 1) (C) (1, 22)
(D) (1, 2)
The vertex of the parabola x2 + 8x + 12y + 4 = 0 is (A) (–4, 1) (B) (4, –1) (C) (–4, –1)
(D) (4, 1)
1 3
11.
Consider the parabola y2 = 8x; if the normal at a point P on the parabola meets it again at a point Q, then the least distance of Q from the tangent at the vertex of the parabola is (A) 16 (B) 8 (C) 4 (D) none of these
12.
Tangents drawn from the point A(–4, 8) to the parabola y2 = 16x meet the parabola at P and Q. Then locus of the centre of the circle described on PQ as diameter is (A) x2 = –2a(y – a) (B) y2 = –2a(x – a) (C) x2 = 2a(y – a) (D) y2 = 2a(x – a)
13.
The total number of chords that can be drawn from the point (a, a) to the circle x2 + y2 = 2a2 such that they are bisected by the parabola y2 = 4ax is (A) 1 (B) 4 (C) 2 (D) 0
14.
Number of the normals drawn from the point (6, –8) to the parabola y2 – 12y – 4x + 4 = 0 is (A) 3 (B) 1 (C) 2 (D) 0
15.
A ray of light moving parallel to the x-axis gets reflected from a parabolic mirror whose equation is (y – 2)2 = 4(x + 1). After reflection, the ray must pass through the point (A) (2, 0) (B) (–1, 2) (C) (0, –2) (D) none of these
16.
Let P (, ) be any point on parabola y2 = 4x (0 2). M is the foot of perpendicular from the focus S to the tangent at P, then the maximum value of area of PMS = (A) 1 (B) 2 (C) (D) 3 6
17.
If P is the pole of chord AB of a parabola and the length of perpendiculars from A, P, B on any tangent to the curve be p1, p2 and p3, then (A) p3 is the G.M. between p1 and p2 (B) p3 is the A.M. between p1 and p2 (C) p1 is the G.M. between p2 and p3 (D) none of these
18.
If tangents at A and B on the parabola y2 = 4ax intersect at point C, then ordinates of A, C and B are in (A) A.P (B) G.P (C) H.P (D) none of these
19.
If the normals at three points A, B, C of a parabola meet in a point, then the centroid of ABC lies on (A) a line parallel to y-axis (B) a line perpendicular to y-axis different from axis of the parabola (C) axis of the parabola (D) the line y = x
20.
If PQ be a focal chord of a parabola, then the tangent at P and normal at Q are (A) perpendicular (B) make an angle of 45 (C) parallel (D) make an angle of 60
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ANSWERS 1.
D
2.
C
3.
C
4.
A
5.
C
6.
A
7.
C
8.
B, C
9.
D
10.
A
11.
A
12.
D
13.
C
14.
B
15.
D
16.
A
17.
D
18.
A
19.
C
20.
C
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COMPLEX NO. ROTATION L-I 1. 2.
3.
If z1, z2, z3 are non-zero complex numbers such that
2 1 1 then prove that z1, z1 z 2 z3
z2, z3
lie on a circle passing through the origin. If a, b, c and u, v, w are the complex numbers representing the vertices of two triangle such that c = (1 – r) a + rb, w = (1 – r) u + rv where r is non-real complex then prove that triangles are similar. If z1, z2 are two complex numbers representing consecutive vertices of a regular hexagon then find the complex number z3 representing the vertex adjacent to z2.
4.
If the two triangles whose vertices are z1, z2, z3 and w1, w2, w3 are similar then prove w1(z2-z3) + w2(z3-z1) + w3(z1 – z2) = 0.
5.
If the complex numbers z1, z2, z3 are represented by the vertices of an equilateral triangle whose circumcentre represents the complex number z0 then prove that 3z02 = z12 + z22 + z32.
6.
If two consecutive vertices of a regular hexagon be z and z , find the other vertices if centre is z = 0
7.
Prove that z z1 z z 2
8.
Complex numbers z1, z2 and z3 are the vertices A, B and C respectively of an isosceles right angled triangle with right angle at C. Show that (z1 – z2)2 = 2(z1 – z3) (z3 – z2). and z12 + z22 + 2z32 = 2 (z1 + z2) z3
9.
If z1, z2, z3 be the vertices of an equilateral triangle occurring in the anticlockwise sense then prove that 1 1 1 (i) z12 + z22 + z32 = z1 z2 + z2 z3 + z3 z1 (ii) + 0 z1 z2 z 2 z3 z3 z1 (iii) z1 + w z2 + w2 z3 = 0 where w is a nonreal cube root of unity.
10.
If z0 represents the orthocentre of a triangle whose circumcentre is represented by
2
Prove that z0 11.
2
= k will represent a circle if z1 z2
z1(z2 z3 )
z1
2
2
that
the
2 k.
z0.
(z 2 z3 ) .
Let z1 = 10 + 6i and z2 = 4 + 6i. If z be a complex number such that arg
z z1 z z2 4
then
prove that | z 7 9i | 3 2 . 12. 13. 14. 15.
If |z – 3i| <
5 then prove that the complex number z also satisfies the inequality
| i(z 1) 1| 2 5 . If the vertices of a square are z1, z2, z3, z4 then prove that (i) z3 = –iz1 + (1 + i)z2 (ii) z4 = iz2 + (1 – i)z1. z6 5 z2 and 1. If so, find z. If there any z such that z 4i 3 z4
Find minimum |z| satisfying z
1 2. 2
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16.
Find z satisfying |z – 5i| 1 such that amp z is minimum.
L-2
Subjective:
1.
Complex numbers z1, z2, z3 are the vertices A, B, C respectively of an isosceles right angled triangle with right angle at C. Show that (z1 – z2)2 = 2 (z1 – z2)(z3 – z2)
2.
Let z1 and z2 be roots of the equation z + pz + q = 0, where the coefficients p and q may be complex numbers. Let A and B represents z1 and z2 in the complex plane. If AOB = 0 and OA = OB, when O 4qcos2 2 is the origin, prove that p = 2
3.
If A, B, C represent the complex numbers z1, z2, z3 respectively on the complex plane and the angles at 1 B and C are each equal , then prove that 2 2 (z2 – z3) = 4(z3 – z1) (z1 – z2) sin3 . 2
4.
If is the nth root of unity and z1 and z2 are any two complex numbers. Prove that
2
n1
z
1
k 0
k z2
2
2
n z1 z2
2
(n N)
5.
Two different non parallel lines meet the circle |z| = r in the points a, b and c, d respectively. Prove that a1 b 1 c 1 d1 these lines meet in the point z given by z = a1b1 c 1d1
6.
A, B, C are the points representing the complex numbers z1, z2, z3 respectively on the complex plane and the circumcentre of the triangle ABC lies at the origin. If the altitude of the triangle through the vertex z z A meets the circumcircle again at P, then prove that P represents the complex numbers 2 3 . z1
7.
It is known that z
8.
Simplify the expressions of the sums 2 3 n (a) cot 2 cot 2 cot 2 cot 2 2n 1 2n 1 2n 1 2n 1 2 3 n (b) cosec 2 cosec 2 cosec 2 cosec 2 2n 1 2n 1 2n 1 2n 1
1 = a, where z is a complex number. What are the greatest and least values of the z modulus |z| of the complex number z ?
Objective: 9.
10.
If (3 + i) z z – (2 + i) z z + 14i = 0, then zz is equal to (A) 10 (B) 8 (C) –9
(D) –10
The centre of a square ABCD is at z = 0. A is z1. Then the centroid of triangle ABC is z (A) z1(cos i sin ) (B) 1 (cos i sin ) 3
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(C) z1 cos isin 2 2
(D)
z1 cos isin 3 2 2
11.
If z satisfies |z – 1| < |z + 3|, then = 2z + 3 – i satisfies (A) | – 5 – i| < | + 3 + i| (B) | – 5| < | + 3| (C) lm (i) > 1 (D) |arg ( – 1)| < 2
12.
If z = (A) iz
13.
(C) z
(D) none of these
If , and are the roots of x3 – 3x2 + 3x + 7 = 0 ( is cube roots of unity), then (A)
14.
3 i 101 101 103 , then (i + z ) equals 2 (B) z
3
(B) 2
(C) 22
1 1 1 is 1 1 1
(D) 32
Let a and b be two fixed non zero complex numbers and z is a variable complex number. If the lines az az 1 0 and bz bz 1 0 are mutually perpendicular, then (A) ab ab 0
(B) ab ab 0
(C) ab ab 0
(D) ab ab 0
15.
The set of points in an Argand diagram which satisfy both |z| 4 and arg z is (A) a circle and a line (B) a radius of a circle (C) a sector of a circle (D) an infinite part line
16.
If f(x) = g(x3) + xh (x3) is divisible by x2 + x + 1, then (A) g(x) is divisible by (x – 1) but not h(x) (B) h(x) is divisible by (x – 1) but not g(x) (C) both g(x) and h(x) are divisible by (x – 1) (D) none of these
17.
If the point represented by complex numbers z1 = a + ib, z2 = c + id and z1 – z2 are collinear, then (A) ad + bc = 0 (B) ad – bc = 0 (C) ab + cd = 0 (D) ab – cd = 0
18.
Let C denote the set of complex numbers and R the set of real numbers. Let the function f: C R be defined by f(z) = |z|. Then (A) f is injective but not surjective (B) f is surjective but not injective (C) f is neither injective nor surjective (D) f is both injective and surjective
19.
Let A, B and C represent the complex numbers z1, z2, z3 respectively on the complex plane. If the circumcentre of the triangle ABC lies at the origin. Then the nine point centre is represented by the complex number z z2 z z 2 z3 z z 2 z3 z z z3 (A) 1 z3 (B) 1 (C) 1 (D) 1 2 2 2 2 2
20.
Let and be two distinct complex numbers such that || = ||. If real part of is positive and imaginary part of is negative, then the complex number may be (A) zero
(B) real and negative
(C) real and positive
(D) purely imaginary
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21.
The complex number z satisfies the condition z ordinates to the point z is (A) 25 (B) 30
25 = 24. The maximum distance from the origin of coz
(C) 32
(D) none of these
22.
The points A, B and C represent the complex numbers z1, z2, (1 – i) z1 + iz2 respectively on the complex plane. The triangle ABC is (A) isosceles but not right-angled (B) right-angled but not isosceles (C) isosceles and right-angled (D) none of these
23.
The system of equations (A) no solution
z 1 i 2 has z 3 (B) one solution
(C) two solutions
(D) none of these
24.
Dividing f(z) by z – i, we obtain the remainder 1 – i and dividing it by z + i we get the remainder 1 + i. Then the remainder upon the division of f(z) by z2 + 1 in (A) i + z (B) 1 + z (C) 1 – z (D) none of these
25.
The centre of the circle represented by |z + 1| = 2|z – 1| on the complex plane is 5 1 (A) 0 (B) (C) (D) none of these 3 3
26.
The value of the expression 2(1 + )(1 + 2) + 3(2 + 1)(22 + 1) + 4(3 + 1)(32 + 1) + ........ + (n – 1)(n + 1)(n2 + 1) is ( is the cube roots of unity) (A)
27.
28.
n2 (n 1)2 4
2
n(n 1) (B) n 2
2
n(n 1) (C) n 2
(D) none of these
z 1 is a purely imaginary number, then z lies on a zi (A) straight line (B) circle 1 (C) circle with radius = (D) circle passing through the origin 2 8 8 2 3 4 5 If = cos isin 11 , then Re( + + + + ) is 11
If
(A)
1 2
(B)
1 2
a
a
(C) 0
(D) none of these
L-2 ANSWERS 2
4
; least value: a
2
4 a
7.
greatest value:
9.
A
10.
D
11.
B, C, D
12.
D
13.
D
14.
D
15.
C
16.
C
17.
B
18.
B
19.
C
20.
D
21.
A
22.
C
23.
A
24.
C
25.
B
26.
B
27.
B, C, D
28.
B
2
2
FIITJEE (Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004
DEFINITE INTEGRALS
1.
If a, b, c R satisfy (A) (3, 4)
7a 3b 2 c 0 , then atleast one root of the equation ax + bx + c = 0 lies between 3 2 3 3 (B) (0, 2) (C) ,2 (D) 1, 2 2
2
2.
cot 1 x dx , (where [] represents the greatest integer function) is equal to
/ 2
(A) – cot 1 + cot 2
(B) + cot 1 – cot 2
(C) + cot 1 + cot 2
(D) none of these
1
3.
If f (0) = 2, f (x) = f (x) , (x) = x + f(x) then
f(x)(x)dx
is
0
(A) e2
(B) 2e2 x
4.
If 5x x 2 6
5 (B) x (–, 2) (3, ) (C) x ,3 2
7
The sum
(D) none of these
2
2 2 tan x 6 dx tan x 18x 75 dx is equal to
3
6
(A) 0 6.
(D) none of these
dt x sin2 tdt , then 2 0 0
(A) x (2, 3) 5.
(C) 2e
(B) 1
(C) 2
(D) none of these
Let a, b, c be non zero numbers such that 3
0
5
x2 x 1 ax 2 bx c dx 1 x 2 x ax 2 bx c dx . Then quadratic equation 0
ax2 + bx + c = 0 has (A) no root in (0, 3) (C) a double root in (0, 3)
(B) at least one root in (3, 5) (D) two imaginary roots
1
7.
x
2
x 5 dx where {} denotes the fractional part of x, is equal to
1
(A) 1 + 3 5
(B)
1 1 3 5 6
(C) 3 5 1
(D)
1 1 3 5 6
x
8.
Let f (x) be a function defined by f (x) = x x 2 3x 2 dx,1 x 4 , Then the range of f (x) is 1
1 63 (A) , 4 4 2
9.
1 (B) ,2 4
1 63 (C) , 4 2
(D) none of these
(C) 0
(D) none of these
5 x2 dx is equal to 2 4
ln 6x x 1
(A) ln 2
FIITJEE
(B) – ln 2
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
10.
Let T> 0 be a fixed real number. Suppose f is a continuous function such that for all x R T
3 3T
f (x + T) = f (x). If I f(x)dx , then the value of
3 (A) I 2
(C) 3I
0
11.
(B) 2I
f(2x)dx is
3
(D) 6I
Let f (x) be a function satisfying f (x) = f (x) with f (0) = 1 and g be the function satisfying f (x) = g(x) = x2. The 1
value of the integral
f(x)g(x)dx
is
0
1 (A) (e 7) 4
1 (B) (e 2) 4 /3
12.
The value of the integral
dx
1 tan
5
/ 6
(A) 1
(B)
/2
13.
If I
dx 1 sin3 x
0
is
12
(B) I 100
The value of
(D) none of these
(C)
6
(D) none of these
, then
(A) 0 < I < 1 14.
x
1 (C) (e 3) 2
2 2
(C)I 2
(D) I > 2
x dx (where {x} is the fractional part of x) is
0
(A) 50
(B) 1
n
15.
If I1
f | cos x | dx
and I2
0
(A) nI1 = 5I2 16.
17.
(C) 100
(D) none of these
5
f(| cos x |)dx
then (n N)
0
(B) I1 + I2 = n + 5
(C)
I1 I2 n 5
(D) none of these
3 1 2 The value of I [x] x x dx , where [] denotes the greatest integer function, is equal to 3 3 0 (A) 10 (B) 11 (C) 12 (D) none of these
If f(x) (A)
1 x2
x
4t
2
2f '(t) dt then f (4) is equal to
4
32 9
(B) 32
(C)
32 3
(D) none of these
18.
x f(sin x)dx is equal to 0
(A)
f(sin x)dx 2 0
FIITJEE
/2
(B) f(sin x)dx 0
/2
(C) 2
f(sin x)dx
(D) none of these
0
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
sin2 x
19.
The value of
cos2 x
sin
1
0
(A)
20.
21.
t dt
cos1
t dt
is
0
4
(B) 0
(C)
|x| , x0 3[x] 5 If f (x) = then x 2, x 0 11 7 (A) (B) 2 2
2
(D) none of these
2
f(x)dx is equal to ([] denote the greatest integer function)
3 / 2
(C) –6
(D)
17 2
1 x0 A function is defined as f(x) , then the value of 1, x 0 1
[x] is x f(x) n 1 x f(x) n 2 n n
x f(x) n 1
1
n
(A)
1 3
(B)
/ 2
22.
0
dx 1 e 2 cos x 4
(A) If
3
(B)
(D) None of these
4
6
(D) none of these
(C)
/2
sin
4
x cos4 x dx k
0
sin
4
x cos4 x dx , then k is equal to
0
(A) 10 24.
(C) 1
is equal to
11 / 2
23.
1 4
(B) 11
(C) 22
(D) 24
Let (a, b) and (, ) be two point on the curve y = f(x). if the slope of the tangent to the curve at (x, y) be (x)
then
(x)dx
is equal to
a
(A) – a
(B) – b
(C) + – a – b
(D) none of these
(C) 2 2
(D) none of these
25.
| sin 4x | | cos 4x | dx is equal to 0
(A) 2
(B) 4 b
26.
The value of the integral (A) b – a
|x| dx(a b) is equal to x a
(B) a – b 4
27.
The value of
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(D) none of these
(C) 0
(D) none of these
1 x x2 1 x x 2 dx is
4
(A) >0
(C) b + a
(B) < 0
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
n 1
28.
The value of
f(r 1 x)dx
is equal to
r 1 0
1
1
(A) f(x)dx
n
(B) n f(x)dx
0
(C) (n 1) f(x)dx
0
n
(D)
0
f(x)dx 0
9
29.
The value of x dx , where [] is the greatest integer function, is 0
(A) 13
(B) 9
(C) 8
(D) none of these
30.
The value of
asin
3
x b tan x 3 c 2 dx is
(A) dependent on a, b, c (C) dependent on b and c 1
31.
The value of the integral
(B) dependent on a and b (D) dependent on c only x3
1 x
8
dx is equal to
0
(A)
32.
16
(B)
, then 2
(B) ln 2 10
The value of
8
(D) none of these
2
f(x)dx
is equal to
0
(C) 2 ln 2
(D) none of these
20 ln3
(D) none of these
x
3 dx is equal to 3[x ] 10
(A) 20 35.
(C)
Let f (x) = min (tanx, cotx), 0 x (A) ln 2
33.
4
(B)
40 ln3
(C)
If f (x) and g(x) are real valued functions such that f(x) > 0 x R and g(x) is differentiable every where and g( x)
h(x) =
f(t) dt , then
0
(A) h(x) is increasing function when g(x) is decreasing function (B) h(x) is increasing function when g(x) is increasing function (C) h(x) is decreasing when g(x) is decreasing (D) Nothing can be said in general about the behaviour of h(x) / 4
36.
If In
n
tan
d , then for any positive integer the value of n(In–1 + In+1) is
0
(A) 1
(B) 2
(C) /4
(D)
(C) 4
(D) 4 2
2
37.
The value of
cos x sin x dx
is equal to
0
(A) 2 2
FIITJEE
(B) 2
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
KEY
1.
B
2.
C
3.
B
4.
A
5.
A
6.
B
7.
B
8.
A
9.
C
10.
C
11.
D
12.
B
13.
C
14.
D
15.
C
16.
C
17.
A
18.
A, B
A
22.
B
25.
B
26.
D
19. 23. 27.
A B C
20. 24. 28.
A B D
29.
A
30.
D
31.
A
32.
A,C
34.
B
35.
B,C
36.
A
37.
D
21.
SOLUTIONS 1.
B 2
2
Since ax 2 bx c dx 1
2.
ax 3 bx 2 7a 3b cx c 0 , roots lie between (1, 2). 3 2 3 2 1
C cot 2
Here
cot1
cot 1 x dx
/ 2
2
cot 1 x dx
cot 2
cot 2
cot 1 x dx = 2
cot1
cot1
1dx
/ 2
1dx 0
cot 2
= 2 cot 2 cot1 cot 2 cot1 cot 2 . 2 3.
B Here
f '(x) 1 f(x) ex c f(x) c
f (0) = 2 = e c = ln 2 1
1
x
f (x) = 2e so that I 2e x (x 2e x )dx (2xex 4e2x )dx 0
0
= 2 e e 1 2 e 1 2 2e 2 2e . 2
4.
2
2
A
5x x 2 6
It is given that
x 1 cos 2t x dt 2 2 0
5x x 2 6 0 x 2 5x 6 0 5.
(x –2) (x –3) < 0 x (2, 3)
A 7
2
Let tan(x 2 6)dx 3
tan (x 9)
2
6 dx
= I1 I2 where
6
2
I2
tan(x 9)
2
6 dx
Put x + 9 = – t
6 7
=
7
2 2 tan t 6 ( dt) tan(x 6)dx I1
3
3
Hence I = 0.
FIITJEE
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
6.
B 5
The given equation implies 1 x 2 x ax 2 bx c dx 0 . 3
This is possible only if f (x) cross the x-axis at least once because algebraic sum of the area from 3 to 5 is zero. Function ax2 + bx + c has atleast one zero in (3, 5). 7.
B 1
1
The given integral is same as
x 2 x dx
1
1
1
x3 x 2 x dx =2 3 0 1 0
2 = ( 1)dx 3 1
8.
1
0dx
0
1dx
5 1 2
1 1 3 5 6
1
5 1 2
0
2 = x 2 x dx 3 1
5 1 2
1
1
x 2 x dx x 2 x dx
1
x 2 x dx
0
x 2 x dx
5 1 2
A Differentiating the given function, we get f (x) = x(x –1) (x –2) = 0. 2
Clearly the function ha its minimum value at x = 2, that is fmn x(x 2 3x 2)dx 1
4
When x = 1, f (1) = 0
when x = 4
f (4) =
xx
2
3x 2 dx
1
1 4
63 4
1 63 Hence the range of f (x) is , . 4 4 9.
C 2
2
Let I ln(5 (x 3)2 )dx ln(5 (x 3)2 )dx 1
1
2
2
= ln(5 x 2 )dx ln(5 (1 2 x 3)2 )dx 0 1
10.
1
C 3 3T
Let L
f(2x)dx .
3
Put 2x = t so that
L
1 2
6 6T
6
a T a T f(t)dt f(x)dx n f(x)dx a 0
T
= 11.
6 f(t)dt 3I 2 0
D Here f (x) = f (x)
FIITJEE
f '(x)
f(x) dx dx
or ln|f (x)| + c = x c = 0 (given f (0) = 1)
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
x
2
x
f (x) = e , g(x) = x – e . 1
1
1
Now I f(x)g(x)dx e x x 2 e x dx = 0
12.
0
e x x
2
e2 x dx e
0
e2 3 2 2
B b
b
Using the property f(x)dx f(a b x)dx , the given integral a
3
a
3 dx I 5 1 tan x
3 dx dx so that 1 cot 5 x 5 1 tan x 6 6 3 6
6
3
2I dx I 6
13.
1 2 3 6 12
C
Since x 0, 1 1 sin3 x 2 2 2
1 1 1 2 1 sin3 x
14.
2 2
I
0
2
2
dx 0
2
dx 1 sin3 x
dx 0
2
D 100
Given integral
0
100
=
=
=
x x dx
(by the def. of {x})
i2
10
100
2 3/2 2 x dx 3 x
xdx
i1 (i1)
0
15.
1
0
10
i
2 t[t]dt i1 i1
2 3 10 1 2000 2 0.t dt 1.t dt 2.t dt ...... 9.t dt 3 0 1 2 9
2000 t2 3
2
2 3
2 4
1
2
3
2t 3t
10
..... 9t 2
9
(where t2 = x)
155 615 2000 3 3
C
We have I1 n f(| cos x |)dx; I2 5 f(| cos x |)dx 0
= 16.
0
I1 I2 . n 5
C 1/ 3
Let I
0
2/3
0.dx
1/ 3
1
1.dx
2/3
4/3
2.dx
1
5/3
3.dx
4/3
6/3
4.dx
5/3
7/3
5.dx
6/3
8/3
6.dx
7/3
9/3
7 / dx
8.dx
8/3
1 = (1 2 3 4 5 6 7 8) 12 . 3
FIITJEE
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
17.
A We have f(x)
f '(x) f '(4) 18.
1 x2
x
4t
2
2f '(t) dt
4
1 2 4x 2 2f '(x) 3 x2 x
x
4t
2
2f(t) dt
4
1 1 32 64 2f '(4) 4 f '(4) . 16 8 9
A, B
Let I x f(sin x)dx ( x)f sin( x)dx f(sin x)dx I 0
0
0
I 19.
f(sin x)dx . Again I f(sin x)dx = 2 2 0 20 2
/2
/2
f(sin x)dx f(sin x)dx 0
0
A sin2 x
Let f (x) =
cos2 x
sin
0
1
t dt
cos1
t dt .
0
Here f (x) = 0 f (x) = c (a constant) 1/ 2
But f (/4) =
sin
1
t cos 1 t dt
0
20.
1/ 2
2 dt 4 0
A We know that 3[x] 5
|x| 3[x] 5 , if x >0, x
2
3 f(x)dx = 1 1 2(1) 1( 5) ( 2) 2 3 / 2
=
1 11 252 2 2
y 2 1 2
1
3/2 1
2 x
-2 -5 21.
A
22. 23. 24. 25.
B B B B
FIITJEE
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
26. 27. 28. 29.
D C D A
30. 31. 32. 34. 35. 36. 37.
D A A,C B B,C A D
FIITJEE
(Hyderabad Classes) Limited., 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad–500 063. Phone: 040-66777000-03 Fax: 040-66777004
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