IIT-JEE 2009 Mathematics Paper - I Questions and Solutions

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PART II: Mathematics Paper I

SECTION � IStraight Objective Type

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This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct

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21. Let P(3, 2, 6) be a point in space and Q be a point on the line � r (i j 2k) = − + + �(-3� � � i j 5k) + + .. Then the
value of � 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
Key (A)
Sol.: PQ= �i (-2 -3 �) + �j (� - 3) + �k (5� - 4)
PQis parallel to x - 4y + 3z = 1
1 (-2 -3�) - 4 (� - 3) + 3(5� - 4) = 0
=> � = 1/4

*22. Tangents drawn from the point P(1, 8) to the circle
x² + y² - 6x - 4y - 11 = 0
touch the circle at the points A and B. The equation of the circum circle of the triangle PAB is
(A) x² + y² + 4x - 6y + 19 = 0 (B) x² + y² - 4x - 10y + 19 = 0
(C) x² + y² - 2x + 6y - 29 = 0 (D) x² + y² - 6x - 4y + 19 = 0
Key (B)

23. Let f be a non-negative function defined on the interval [0, 1]. If

*24. 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
z�z³ + zz³= 350 is
(A) 48 (B) 32
(C) 40 (D) 80
Key (A)
z�z³ + zz³= 350
Let Z = x + iy (x, y ,Z)
(x² + y2) (x² - y2) = 175
=> x² + y2 = 25 |x² - y2 = 7
=> x = � 4, y = � 3

*25. The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse
x² + 9y2 = 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
Key (D)
Sol.:

26. If a, b, cand d are unit vectors such that ( a* b).( c*d ) =1 and (a.c)=1/2 then

27.Let z = cosθ + isinθ. Then the value of ¹⁵ ∑_m=1 (z^2m-1) at θ = 2� is
(A) 1/sin 2� (B)1/3sin 2�

(C)1/2sin 2� (D)1/4sin 2�

Key (D)

*28. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2
(A)25 (B) 66
(C) 77 (D) 88

Key (C)
Sol.:

Case I: digits used 1, 1, 1, 1, 1, 3, 2
Number of integers formed = 7!/5!= 42
Case II: digits used : 1, 1, 1, 1, 2, 2, 2
Number of integers formed = 7!/3! 4!= 35
Total number of integers formed = 77.
(C)

5!
= 42
Case II: digits used : 1, 1, 1, 1, 2, 2, 2
Number of integers formed = 7!
3! 4!
= 35
Total number of integers formed = 77.

SECTION � IIMultiple Correct Answer Type

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This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and(D), out of which ONE OR MORE is/are correct.

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29. Area of the region bounded by the curve y = ex and lines x = 0 and y = e is
(A) e - 1 (B)ln ∫¹(e + 1-y) dy s
(C) e - ₀∫ ¹e ^xdx (D)₀∫ ¹ln y dy
Key: (B, C, D)
Sol.: Area = e -₀ ∫¹ e ^xdx
=>₀ ∫^eln y dy

30. Let L = lim_{x → 0(} a-√a²-x² - x²/4)/x⁴ , a.>0. If L is finite, then
(A) a = 2 (B) a = 1
(C) L = 1/64 (D) L = 1/32
Key (A, C)
Sol.:

*31. In a triangle ABC with fixed base BC, the vertex A moves such that cosB + cosC = 4 sin² A/2
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If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively,
(A) b + c = 4a (B) b + c = 2a
(C) locus of point A is an ellipse (D) locus of point A is a pair of straight lines
Key=(B,C)
Sol.: cosB + cosC = 4 sin² A/2

*32. sin ⁴x /2+ cos ⁴x/3= 1/5 then ,

(A) tan²x = 3/2
(B) sin⁸/8 x+ cos ⁸x/27 = 1/ 125
(C) tan²x = 1/3
(D) sin⁸/8 x+ cos ⁸x/27 = 2/ 125
Key (A, B)

Sol.: sin ⁴x /2+ cos ⁴x/3= 1/5 then (sin ²x + cos ²x)²/5)
=> 9 sin⁴ x + 4cos⁴ x - 12 sin² x cos² x = 0
=> (3 sin² x - 2cos²x)2 = 0
=> sin² x = 2/5, cos² x = 3/5
=> tan²x = 2/3
and sin⁸/8 x+ cos ⁸x/27 =1/ 125

SECTION � IIILinked Comprehension Type

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This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.Paragraph for Question Nos. 33 to 35

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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.

33. The number of matrices in A is
(A) 12 (B) 6
(C) 9 (D) 3
Key (A)
Sol.: (A)
There are two cases
Case I : diagonal elements 1, 0, 0
Number of symmetric matrices
= Arrangement of main diagonal elements � Arrangement of remaining elements
3! 3!/ 2! 2!
= 9
Case II : diagonal elements 1, 1, 1
No. of symmetric matrices
3!/2! = 3
Total symmetric matrices are 9 + 3 = 12

34. The number of matrices A in A for which the system of linear equations

A	[	x	]	=	[ 1 ] 0 0
		y
		z

has a unique solution, is

(A) less than 4 (B) at least 4 but less than 7
(C) at least 7 but less than 10 (D) at least 10
Key (B)
Sol.: Consider symmetric matrix.

[	a	h	g	]
	h	b	f
	g	f	c

|A| = abc + 2fgh � af2 � bg2 � ch2
Case I : when a = b = c = 1 then out f, g, h two are '0' and remaining '1'

=> |A| = 0
=> there are three such matrices.
Case II : when either of a, b or c = 1 and other two are zero
then abc = 0, fgh = 0

=> |A| = - af2 or -bg2 or - ch2
when a = 1, and f = 0
=> |A| = 0
=> there are three such matrices
=>total number of matrices such that |A| = 0
= 3 + 3 = 6
=>Total number of matrices such that |A| ≠ 0 is 12 - 6 = 6

35. The number of matrices A in A for which the system of linear equations

A	[	x	]	=	[ 1 ] 0 0
		y
		z

is inconsistent, is
(A) 0 (B) more than 2
(C) 2 (D) 1
Key (B)
Sol.:

A	[	x	]	=	[ 1 ] 0 0
		y
		z

ax + hy + gz = 1
hx + by + fz = 0
gx + fy + cz = 0
when a = b = c = 1
then system will be inconsistent when h = 1 or g = 1.
When a = b = 0, c = 1
Then system will be inconsistent when h = 0
Hence more than 2 matrices.

Paragraph for Question Nos. 36 to 38
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required.

36. The probability that X = 3 equals
(A) 25/216 (B) 25/36
(C) 5/36 (D) 125/216
Key (A)
Sol.: P(X = 3) =5 /6.5 /6.1 /6 = 25/216

37. The probability that X ≥ 3 equals
(A) 125/216 (B) 25/216
(C) 5/36 (D) 25/36
Key (B)
Sol.: P(X ≥ 3) = 5 /6.5/6 = 25/36

38. The conditional probability that X ≥ 6 given X > 3 equals
(A) 125/216 (B) 25/216
(C) 5/36 (D) 25/36
Key (D)
Sol.:P(X≥6/X>3) = (5/6)³/(5/6)³ = 25/36

SECTION � IV

Matrix Match Type

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This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled 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:

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39. Match the statements/expressions in Column I with the open intervals in Column II.

Column I Column II.

(A) Interval contained in the domain of definition of non-zero solutions of the differential equation (x - 3)2 y′ + y = 0	( p) (- Π/2,Π/2,)
(B) Interval containing the value of the interval 1∫ 5(x 1) (x 2) (x 3) (x 4) (x 5)	(q) (0,Π/2,)
(C) Interval in which at least one of the points of local maximum	(r) (Π/8,5Π/4,)
(D) Interval in which tan-1 (sinx + cosx) is increasing	(s) (0,Π/8) (t) (-Π,Π)

Key (A-p, q, s), (B-p, t), (C-p, q, r, t) (D-s)

Sol.: (A) dy /dx + dx/ (x- 3)² = 0
ln | y | - 1 /x-3 = ln c
y= ce^1/x-3 , domain R � {3}
p,q,s

(B) Let I =¹ ∫⁵(x+2)(x+1)x(x-2)(x-1)dx ∫
I = ¹ ∫⁵(6x-1)(6-x-2)(6- x-3)(6- x- 4)(6 -x -5)dx
I = -¹ ∫⁵(x+ 2)(x+ 1)x(x- 2)(x- 1)dx

=> I = - I => I = 0
p, s, t

(C) f(x) = cos²x + sin x
f(x) = cos x(1 � 2 sinx)
points of max. are Π/6,5Π/6

p, q, r, t

(D)s

*40. Match the conics in Column I with the statements/expressions in Column II.

Column I Column II.

(A) Circle	(p) The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x2 + y2 = 4
(B) Parabola	(q) Points z in the complex plane satisfying |z + 2| |z - 2| = � 3
(C) Ellipse	(r) Points of the conic have parametric representation x =2
(D) Hyperbola	(s) The eccentricity of the conic lies int he interval 1 ≤ x < ∞ (t) Points z in the complex plane satisfying Re (z + 1)2 = |z|2 + 1

Key (A-p), (B-s, t) (C-r) (D-q, s)

Sol.: (p) hx + ky = 1 touches x² + y2 = 4

=>h²+k² = 1/4
Clarly Circle

(q) ||z + 2| − |z � 2|| = 3
2ae = 4, 2a = 3
=>4/3>1
= > i.e. (hyperbola)

(r)Ellipse

(s) Clearly parabola or hyperbola

(t) Let z = x + iy
Re(z + 1)2 = (x + 1)2 � y2 = x² + y2 + 1
y2 = x (parabola)