Solutions Iit Maths Paper I 2009
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IIT-JEE - 2009 Paper - I MATHEMATICS CODE - I
CODE - [1]
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PART - II SECTION - I Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C), and (D), for its answer, out of which ONLY ONE is correct .
21.
Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation zz 3 + zz 3 = 350 is A) 48 B) 32 C) 40 Sol : A
zz 3 + zz 3 = 350
(zz)z 2 + ( zz ) z 2 = 350 | z |2 (z 2 + z 2 ) = 350
D) 80
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(x 2 + y2 ) 2( x 2 − y 2 ) = 350 x 4 − y 4 = 175
( x + y) ( x − y ) ( x 2 + y 2 ) = 175 Vertices of the rectangle satisfying above equation are (4, 3), (4, -3), (-4, -3), (-4, 3) ∴
Area = 8 × 6 = 48
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1 If a , b , c and d are unit vectors such that ( a × b ).(c × d ) = 1 and a.c = , then 2 A) a , b , c are non-coplanar
B) b , c , d are non-coplanar
C) b , d are non-parallel
D) a , d are parallel and b , c are parallel
Sol : B
(a × b ). (c × d ) = 1
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⇒ | a ×b | | c × d | =1
sin A sin C = 1 where A = ( a , b ), C = (c , d )
⇒ sin A = 1 & sin C = 1 (or) sin A = -1 & sin C = -1
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A = C = 90° (or) A = C = -90°
1 1 a . c = ⇒ | a | | c | cos B = where B = (a, c) 2 2 B = 60°
c
b 60°
a
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∵ c ⊥ d ⇒ b, c, d are non-coplanar Option (B) is correct
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The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x 2 + 9 y 2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin O is A)
31 10
B)
29 10
C)
21 10
D)
27 10
Sol :D
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x y Equation of AB is + = 1 ⇒ x + 3 y = 3 3 1
M
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B(0, 1)
O (0, 0)
A(3, 0)
−3 6 This cuts Auxiliary circle x 2 + y 2 = 9 at M , 5 5 ∴ Area of ∆ OAM =
27 sq units 10
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15
24.
Let z = cos θ + i sin θ. Then the value of
∑
m =1
Im ( z 2 m −1 ) at θ = 2° is
1 B) 3 sin 2°
1 sin 2°
A)
1 C) 2 sin 2° Sol :D 15
∑
1 D) 4 sin 2°
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Im (cos(2m − 1)θ + i sin(2m − 1)θ)
m =1
15
= ∑ sin(2m − 1)θ m =1
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= sin θ + sin 3 θ + ........... + sin 29 θ 2θ 2 sin θ + 29 θ = 2θ 2 sin 2 sin 15
=
sin 2 15θ sin 2 30° = (∵ θ = 2°) sin θ sin 2°
=
1 4 sin 2°
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Let P(3, 2, 6) be a point in space and Q be a point on the line r = (iˆ − ˆj + 2kˆ ) + µ ( −3iˆ + ˆj + 5kˆ) Then the value of µ for which the vector PQ is parallel to the plane x - 4y +3z = 1 is A)
1 4
B) −
1 4
C)
1 8
D) −
1 8
Sol : A
OP = 3i + 2 j + 6k
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∵ PQ is parallel to the plane x - 4y + 3z = 1
⇒ [ r − (3i + 2 j + 6k)]. ( i − 4 j + 3k) = 0 ⇒ r. ( i − 4 j + 3k) = 13
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(1 − 3µ) − 4 [−1+ µ) + 3(2 + 5µ) = 13
⇒ µ=
26.
1 4
The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is A) 55 B) 66 C) 77
D) 88
Sol :C Using Five 1’s , one 2 and one 3 =
7! = 42 5!
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Using three 2’s and Four 1’s =
7! = 35 4!3!
Total required numbers = 77
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Let f be a non-negative function defined on the interval [0, 1]. If x
x
1 − ( f '(t ))2 dt = ∫ f (t ) dt ,
∫ 0
0 ≤ x ≤ 1, and f(0) = 0, then
0
1 1 1 1 A) f < and f > 2 2 3 3
1 1 1 1 B) f > and f > 2 2 3 3
1 1 1 1 C) f < and f < 2 2 3 3
1 1 1 1 D) f > and f < 2 2 3 3
Sol :C
1 − [ f '(x)]2 = f (x)
⇒
[ f (x)]
2
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+ [f '(x)]2 = 1
∵ f (0) = 0
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∴ f(x) = sin x , f '(x) = cos x
0 ≤ x ≤ 1 ⇒ sin x < x ∴ option (c) satisfies
28.
Tangents drawn from the point P(1,8) to the circle x 2 + y 2 − 6 x − 4 y − 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is A) x 2 + y 2 + 4 x − 6 y + 19 = 0
B) x 2 + y 2 − 4 x − 10 y + 19 = 0
C) x 2 + y 2 − 2 x + 6 y − 29 = 0
D) x2 + y 2 − 6 x − 4 y + 19 = 0
Sol :B P(1, 8) & centre of the given circle (3, 2) are the ends of the diameter of required circle
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Section - II Multiple Correct Answer Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for its answer, out of which ONE OR MORE is/are correct. 29.
2 In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4sin
A . 2
If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then A) b + c = 4a B) b + c = 2a
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C) locus of point A is an ellipse Sol :B, C
Cos B + Cos C = 4sin 2
D) locus of point A is a pair of straight lines
A 2
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B+C B−C 2 A ⇒ 2cos cos = 4sin 2 2 2
B−C cos 2 =2 ⇒ b+c =2 ⇒ A a sin 2 ⇒ b + c = 2a ∴ SP + S'P = 2a
⇒ Locus of A is an ellipse
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If
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sin 4 x cos 4 x 1 + = , then 2 3 5
2 A) tan x =
2 C) tan x =
2 3
1 3
B)
sin 8 x cos8 x 1 + = 8 27 125
D)
sin 8 x cos8 x 2 + = 8 27 125
Sol :A, B.
sin 4 x cos4 x 1 + = 2 3 5
⇒
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1 1 1 Tan 4 x + = sec4 x 2 3 5
⇒ 5(3T an 4 x + 2) = 6(1 + Tan 2 x)2
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⇒ 9 Tan 4 x − 12 Tan 2 x + 4 = 0
5
2 3 (3 Tan 2 x − 2) 2 = 0 ⇒ Tan x =
2 3
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Sin 8 x cos8 x 2 3 1 + = + = 8 27 625 625 125
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Let L = lim
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a − a2 − x2 −
x →0
x4
x2 4 , a > 0. If L is finite, then
A) a = 2 C) L =
B) a = 1
1 64
D) L =
1 32
Sol :A, C. 1
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x2 x2 a − a 1 − 2 − 4 is a finite value a L = Lt 4 x →0 x ⇒a =2
2
11 1 x 2 2 2 − 1 x 4 x 2 a − a 1 − . 2 + . 4 − 2 a 2! a 4 L = Lt 4 x →0 x
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1 1 = . 3 8 a 1 1 = . (∵ a = 2) 8 8
=
1 64
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Area of the region bounded by the curve y = e x and lines x = 0 and y = e is e
A) e - 1
B)
∫
ln (e +1- y)dy
1
1
e
x C) e − ∫ e dx
D)
0
∫
ln y dy
1
Sol :C, D.
PAGE
1
Area = ∫ (e − e x )dx 0
Y
y = ex
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1
X
1
= e − ∫ e x dx 0
e
= ∫ log y dy 1
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SECTION - III Comprehension Type This section contains 2 groups of questions. Each question has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D), for its answer, out of which ONLY ONE is correct. Paragraph for Question Nos. 33 to 35
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. 33.
The probability that X = 3 equals A)
25 216
Sol : A
PAGE 25 36
B)
C)
5 36
D)
125 216
D)
25 216
D)
25 36
5 5 1 25 P(X = 3) = × × = 6 6 6 216 34.
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The probability that X ≥ 3 equals A)
125 216
B)
25 36
C)
5 36
Sol :B 1 5 1 25 =1− + × = 6 6 6 36
P(X ≥ 3) = 1 − P(X < 3) 35.
The conditional probability that X ≥ 6 given X > 3 equals A)
125 216
B)
25 216
C)
5 36
Sol :D
X ≥ 6 P(X ≥ 6) P = X > 3 P(X > 3)
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1 5 52 54 1 − + 2 + 3 + ....... + 5 6 6 = 25 6 6 = 2 36 1 5 5 1− + 2 + 3 6 6 6
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Paragraph for Question Nos. 36 to 38
Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. 36.
The number of matrices in A is a) 12 B) 6
C) 9
D) 3
Sol :A 37.
The number of matrices A in A for which the system of linear equations
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x 1 A y = 0 z 0 has a unique solution, is A) less than 4
B) at least 4 but less than 7
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C) at least 7 but less than 10 Sol :B 38.
The number of matrices A is A for which the system of linear equations
x 1 A y = 0 z 0 is inconsistent, is A) 0
B) more than 2
C) 2
D) 1
Sol :B Sol : 36 to 38
1 1 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 0 1 0 0 1 0 1 0 0 1 0 0 0 0 1 0 0 1 1 0 1 0 1 1 0 1 0 1 0 0 0 1 0
PAGE
0 1 0 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 1
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SECTION-IV Matrix-Match Type This section contains 2 questions. Each question contains statements given in two columns which have to be matched. Statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example. If the correct matches are A-p, s and t; B-q and r; C-p and q; and D - s and t; then the correct darkening of bubbles will look like the following.
39.
Match the conics in Column I with the statements/expressions in Column II. Column I A) Circle
B) Parabola
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Column II p) The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x2 + y2 = 4 q) Points z in the complex plane satisfying z + 2 − z − 2 = ±3 r) Points of the conic have parametric 1− t2 2t ,y= representation x = 3 2 1+ t2 1+ t s) The eccentricity of the conic lies in the interval 1 ≤ x < ∞ t) points z in the complex plane satisfying Re( z + 1) 2 =| z |2 +1
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C) Ellipse
D) Hyperbola
Sol : A → p ; B → t, s ; C → r ; D → q, s
P → (1) 2 = 4(h 2 + k 2 ) 1 1 ⇒ circle is x 2 + y2 = 4 4 q → ± 3 < Distance between (-2, 0) & (2, 0) h2 + k2 =
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i.e ±3 < 4 ∴ locus of z is Hyperbola
r → let t = tan θ x = cos 2θ, y = sin 2θ 3 2
2
x y + =1 3 1
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∴ locus is ellipse
s →1≤ x < ∞ ⇒ If e = 1, The conic is parabola ⇒ If e > 1, The conic is Hyperbola
t → z = x + iy (z + 1)2 = [(x + 1) + iy]
2
= (x + 1)2 − y2 + 2i(x + 1)y | z |2 = x 2 + y2 Re(z + 1) 2 = | z |2 +1
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(x + 1) 2 − y 2 = x 2 + y 2 + 1
x 2 + 2x + 1 − y2 = x 2 + y2 + 1 2y 2 = 2x + 1 ∴ locus is parabola
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Mathc the statements/expression in Column I with the open intervals in Column II. Column I Column II A) Interval contained in the domain of
π π p) − , 2 2
definition of non-zero solutions of the differential equation ( x − 3)2 y '+ y = 0 π B) Interval containing the value of the integral q) 0, 2
PAGE
5
∫
( x − 1)( x − 2)( x − 3)( x − 4)( x − 5) dx
1
C) Interval in which at least one of the points
r)
π 5π , 8 4
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of local maximum of cos 2 x + sin x lies D) Interval in which tan −1 (sin x + cos x)
π s) 0, 8
is increasing t)
( −π, π )
Sol : A → p,q,s ; B → p,t ; C → p, q, r, t ; D → s
(x − 3)2
∫
−
dy = −y dx
1 1 dy = ∫ dx y (x − 3) 2
1 = log ky x −3 Domain x ≠ 3
⇒ A → p,q,s
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5
B)
∫
(x − 1)(x − 2)(x − 3)(x − 4)(x − 5) = 0
1
Since f (a + b - x) = - f(x)
⇒ B → p, t C) f (x ) = cos 2 x + sin x
f '(x) = −2 cos x sin x + cos x f ''(x) = −(2 cos 2x + sin x)
f(x) is maximum if x =
⇒ C → p, q, r, t
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π 5π , 6 6
1 D) 1 + (sin x + cos x)2 (cos x − sin x) > 0
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cos x > sin x
⇒ x ∈ [0, π / 4) ⇒D→s
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CODE - [1]
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PART - II SECTION - I Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C), and (D), for its answer, out of which ONLY ONE is correct .
21.
Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation zz 3 + zz 3 = 350 is A) 48 B) 32 C) 40 Sol : A
zz 3 + zz 3 = 350
(zz)z 2 + ( zz ) z 2 = 350 | z |2 (z 2 + z 2 ) = 350
D) 80
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(x 2 + y2 ) 2( x 2 − y 2 ) = 350 x 4 − y 4 = 175
( x + y) ( x − y ) ( x 2 + y 2 ) = 175 Vertices of the rectangle satisfying above equation are (4, 3), (4, -3), (-4, -3), (-4, 3) ∴
Area = 8 × 6 = 48
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1 If a , b , c and d are unit vectors such that ( a × b ).(c × d ) = 1 and a.c = , then 2 A) a , b , c are non-coplanar
B) b , c , d are non-coplanar
C) b , d are non-parallel
D) a , d are parallel and b , c are parallel
Sol : B
(a × b ). (c × d ) = 1
PAGE
⇒ | a ×b | | c × d | =1
sin A sin C = 1 where A = ( a , b ), C = (c , d )
⇒ sin A = 1 & sin C = 1 (or) sin A = -1 & sin C = -1
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A = C = 90° (or) A = C = -90°
1 1 a . c = ⇒ | a | | c | cos B = where B = (a, c) 2 2 B = 60°
c
b 60°
a
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∵ c ⊥ d ⇒ b, c, d are non-coplanar Option (B) is correct
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The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x 2 + 9 y 2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin O is A)
31 10
B)
29 10
C)
21 10
D)
27 10
Sol :D
PAGE
x y Equation of AB is + = 1 ⇒ x + 3 y = 3 3 1
M
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B(0, 1)
O (0, 0)
A(3, 0)
−3 6 This cuts Auxiliary circle x 2 + y 2 = 9 at M , 5 5 ∴ Area of ∆ OAM =
27 sq units 10
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15
24.
Let z = cos θ + i sin θ. Then the value of
∑
m =1
Im ( z 2 m −1 ) at θ = 2° is
1 B) 3 sin 2°
1 sin 2°
A)
1 C) 2 sin 2° Sol :D 15
∑
1 D) 4 sin 2°
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Im (cos(2m − 1)θ + i sin(2m − 1)θ)
m =1
15
= ∑ sin(2m − 1)θ m =1
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= sin θ + sin 3 θ + ........... + sin 29 θ 2θ 2 sin θ + 29 θ = 2θ 2 sin 2 sin 15
=
sin 2 15θ sin 2 30° = (∵ θ = 2°) sin θ sin 2°
=
1 4 sin 2°
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Let P(3, 2, 6) be a point in space and Q be a point on the line r = (iˆ − ˆj + 2kˆ ) + µ ( −3iˆ + ˆj + 5kˆ) Then the value of µ for which the vector PQ is parallel to the plane x - 4y +3z = 1 is A)
1 4
B) −
1 4
C)
1 8
D) −
1 8
Sol : A
OP = 3i + 2 j + 6k
PAGE
∵ PQ is parallel to the plane x - 4y + 3z = 1
⇒ [ r − (3i + 2 j + 6k)]. ( i − 4 j + 3k) = 0 ⇒ r. ( i − 4 j + 3k) = 13
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(1 − 3µ) − 4 [−1+ µ) + 3(2 + 5µ) = 13
⇒ µ=
26.
1 4
The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is A) 55 B) 66 C) 77
D) 88
Sol :C Using Five 1’s , one 2 and one 3 =
7! = 42 5!
PAGE
Using three 2’s and Four 1’s =
7! = 35 4!3!
Total required numbers = 77
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Let f be a non-negative function defined on the interval [0, 1]. If x
x
1 − ( f '(t ))2 dt = ∫ f (t ) dt ,
∫ 0
0 ≤ x ≤ 1, and f(0) = 0, then
0
1 1 1 1 A) f < and f > 2 2 3 3
1 1 1 1 B) f > and f > 2 2 3 3
1 1 1 1 C) f < and f < 2 2 3 3
1 1 1 1 D) f > and f < 2 2 3 3
Sol :C
1 − [ f '(x)]2 = f (x)
⇒
[ f (x)]
2
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+ [f '(x)]2 = 1
∵ f (0) = 0
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∴ f(x) = sin x , f '(x) = cos x
0 ≤ x ≤ 1 ⇒ sin x < x ∴ option (c) satisfies
28.
Tangents drawn from the point P(1,8) to the circle x 2 + y 2 − 6 x − 4 y − 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is A) x 2 + y 2 + 4 x − 6 y + 19 = 0
B) x 2 + y 2 − 4 x − 10 y + 19 = 0
C) x 2 + y 2 − 2 x + 6 y − 29 = 0
D) x2 + y 2 − 6 x − 4 y + 19 = 0
Sol :B P(1, 8) & centre of the given circle (3, 2) are the ends of the diameter of required circle
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Section - II Multiple Correct Answer Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for its answer, out of which ONE OR MORE is/are correct. 29.
2 In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4sin
A . 2
If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then A) b + c = 4a B) b + c = 2a
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C) locus of point A is an ellipse Sol :B, C
Cos B + Cos C = 4sin 2
D) locus of point A is a pair of straight lines
A 2
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B+C B−C 2 A ⇒ 2cos cos = 4sin 2 2 2
B−C cos 2 =2 ⇒ b+c =2 ⇒ A a sin 2 ⇒ b + c = 2a ∴ SP + S'P = 2a
⇒ Locus of A is an ellipse
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P AGE 30.
If
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sin 4 x cos 4 x 1 + = , then 2 3 5
2 A) tan x =
2 C) tan x =
2 3
1 3
B)
sin 8 x cos8 x 1 + = 8 27 125
D)
sin 8 x cos8 x 2 + = 8 27 125
Sol :A, B.
sin 4 x cos4 x 1 + = 2 3 5
⇒
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1 1 1 Tan 4 x + = sec4 x 2 3 5
⇒ 5(3T an 4 x + 2) = 6(1 + Tan 2 x)2
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⇒ 9 Tan 4 x − 12 Tan 2 x + 4 = 0
5
2 3 (3 Tan 2 x − 2) 2 = 0 ⇒ Tan x =
2 3
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Sin 8 x cos8 x 2 3 1 + = + = 8 27 625 625 125
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P AGE 31.
Let L = lim
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a − a2 − x2 −
x →0
x4
x2 4 , a > 0. If L is finite, then
A) a = 2 C) L =
B) a = 1
1 64
D) L =
1 32
Sol :A, C. 1
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x2 x2 a − a 1 − 2 − 4 is a finite value a L = Lt 4 x →0 x ⇒a =2
2
11 1 x 2 2 2 − 1 x 4 x 2 a − a 1 − . 2 + . 4 − 2 a 2! a 4 L = Lt 4 x →0 x
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1 1 = . 3 8 a 1 1 = . (∵ a = 2) 8 8
=
1 64
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P AGE 32.
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Area of the region bounded by the curve y = e x and lines x = 0 and y = e is e
A) e - 1
B)
∫
ln (e +1- y)dy
1
1
e
x C) e − ∫ e dx
D)
0
∫
ln y dy
1
Sol :C, D.
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1
Area = ∫ (e − e x )dx 0
Y
y = ex
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1
X
1
= e − ∫ e x dx 0
e
= ∫ log y dy 1
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P AGE
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SECTION - III Comprehension Type This section contains 2 groups of questions. Each question has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D), for its answer, out of which ONLY ONE is correct. Paragraph for Question Nos. 33 to 35
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. 33.
The probability that X = 3 equals A)
25 216
Sol : A
PAGE 25 36
B)
C)
5 36
D)
125 216
D)
25 216
D)
25 36
5 5 1 25 P(X = 3) = × × = 6 6 6 216 34.
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The probability that X ≥ 3 equals A)
125 216
B)
25 36
C)
5 36
Sol :B 1 5 1 25 =1− + × = 6 6 6 36
P(X ≥ 3) = 1 − P(X < 3) 35.
The conditional probability that X ≥ 6 given X > 3 equals A)
125 216
B)
25 216
C)
5 36
Sol :D
X ≥ 6 P(X ≥ 6) P = X > 3 P(X > 3)
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1 5 52 54 1 − + 2 + 3 + ....... + 5 6 6 = 25 6 6 = 2 36 1 5 5 1− + 2 + 3 6 6 6
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P AGE
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Paragraph for Question Nos. 36 to 38
Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. 36.
The number of matrices in A is a) 12 B) 6
C) 9
D) 3
Sol :A 37.
The number of matrices A in A for which the system of linear equations
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x 1 A y = 0 z 0 has a unique solution, is A) less than 4
B) at least 4 but less than 7
D) at least 10 www.aieeepage.com
C) at least 7 but less than 10 Sol :B 38.
The number of matrices A is A for which the system of linear equations
x 1 A y = 0 z 0 is inconsistent, is A) 0
B) more than 2
C) 2
D) 1
Sol :B Sol : 36 to 38
1 1 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 0 1 0 0 1 0 1 0 0 1 0 0 0 0 1 0 0 1 1 0 1 0 1 1 0 1 0 1 0 0 0 1 0
PAGE
0 1 0 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 1
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P AGE
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SECTION-IV Matrix-Match Type This section contains 2 questions. Each question contains statements given in two columns which have to be matched. Statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example. If the correct matches are A-p, s and t; B-q and r; C-p and q; and D - s and t; then the correct darkening of bubbles will look like the following.
39.
Match the conics in Column I with the statements/expressions in Column II. Column I A) Circle
B) Parabola
PAGE
Column II p) The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x2 + y2 = 4 q) Points z in the complex plane satisfying z + 2 − z − 2 = ±3 r) Points of the conic have parametric 1− t2 2t ,y= representation x = 3 2 1+ t2 1+ t s) The eccentricity of the conic lies in the interval 1 ≤ x < ∞ t) points z in the complex plane satisfying Re( z + 1) 2 =| z |2 +1
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C) Ellipse
D) Hyperbola
Sol : A → p ; B → t, s ; C → r ; D → q, s
P → (1) 2 = 4(h 2 + k 2 ) 1 1 ⇒ circle is x 2 + y2 = 4 4 q → ± 3 < Distance between (-2, 0) & (2, 0) h2 + k2 =
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i.e ±3 < 4 ∴ locus of z is Hyperbola
r → let t = tan θ x = cos 2θ, y = sin 2θ 3 2
2
x y + =1 3 1
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P AGE
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∴ locus is ellipse
s →1≤ x < ∞ ⇒ If e = 1, The conic is parabola ⇒ If e > 1, The conic is Hyperbola
t → z = x + iy (z + 1)2 = [(x + 1) + iy]
2
= (x + 1)2 − y2 + 2i(x + 1)y | z |2 = x 2 + y2 Re(z + 1) 2 = | z |2 +1
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(x + 1) 2 − y 2 = x 2 + y 2 + 1
x 2 + 2x + 1 − y2 = x 2 + y2 + 1 2y 2 = 2x + 1 ∴ locus is parabola
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P AGE 40.
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Mathc the statements/expression in Column I with the open intervals in Column II. Column I Column II A) Interval contained in the domain of
π π p) − , 2 2
definition of non-zero solutions of the differential equation ( x − 3)2 y '+ y = 0 π B) Interval containing the value of the integral q) 0, 2
PAGE
5
∫
( x − 1)( x − 2)( x − 3)( x − 4)( x − 5) dx
1
C) Interval in which at least one of the points
r)
π 5π , 8 4
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of local maximum of cos 2 x + sin x lies D) Interval in which tan −1 (sin x + cos x)
π s) 0, 8
is increasing t)
( −π, π )
Sol : A → p,q,s ; B → p,t ; C → p, q, r, t ; D → s
(x − 3)2
∫
−
dy = −y dx
1 1 dy = ∫ dx y (x − 3) 2
1 = log ky x −3 Domain x ≠ 3
⇒ A → p,q,s
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5
B)
∫
(x − 1)(x − 2)(x − 3)(x − 4)(x − 5) = 0
1
Since f (a + b - x) = - f(x)
⇒ B → p, t C) f (x ) = cos 2 x + sin x
f '(x) = −2 cos x sin x + cos x f ''(x) = −(2 cos 2x + sin x)
f(x) is maximum if x =
⇒ C → p, q, r, t
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π 5π , 6 6
1 D) 1 + (sin x + cos x)2 (cos x − sin x) > 0
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cos x > sin x
⇒ x ∈ [0, π / 4) ⇒D→s
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