JEE Mathematics CENGAGE LEARNING

Appendix

B

Mock Test 1

PaPer 1

Section a: Only One Option Correct Type

1. For qŒÊ_ËÁ0 pˆ_¯˜

2

, , the value of definite integral

ln (1 tan tan ) 0 +

Ú

q q x dx is equal to (a) q ln (sec q) (b) q ln(cosec q) (c) q ln 2

2 (d) 2q ln sec q

2. The focal length of a mirror is given by 1 1 2

v u- = .f

If errors made in measuring u and v are a, then the relative error in f is (a) 2 a (b) a 1 1ÊËÁu v+ ˆ¯˜ (c) a 1 1 u v -Ê ËÁ ˆ¯˜ (d) None of these

3. If the function f (x) = 2 tan x + (2a +1) loge | sec x | +

(a –2) x is increasing on R, then

(a) a Œ (1/2, •) (b) a Œ (–1/2, 1/2) (c) a = 1/2 (d) a Œ R

4. 3x + 4y – 7 = 0 and 3x – 4y –7 = 0 are equations of asymptotes of a hyperbola H. From a point P(3, 4), pair of tangents are drawn to hyperbola H in such a way that both tangents touch the same branch of hyperbola H. Then its eccentricity is

(a) 4 3 (b) 7 3 (c) 5 3 (d) 5 4

5. A parabola touches two given straight lines originat-ing from a given point. The locus of the mid point of the portion of any tangent, which is intercepted between the given straight lines, is a/an

(a) parabola (b) ellipse (c) straight line (d) hyperbola

6. If x2 x 4 2 -p +sin - = | x2 _{– 2 | + | sin x | +} p 4 , then (a) x Œ( ,0 2) (b) x Œ -( 2 2, ) (c) x ŒR (d) x Œ -( 2 0, )

7. If A and B are different matrices satisfying A3 _{= B}3 and A2_{B = B}2_{A, then}

(a) det (A2 _{+ B}2_{) must be zero.} (b) det (A – B) must be zero.

(c) det (A2 _{+ B}2_{) as well as det (A – B) must be zero} (d) At least one of det (A2 _{+ B}2_{) or det (A – B) must}

be zero.

8. A fair dice is thrown 3 times. The probability that the product of the three outcomes is a prime number is

(a) 1 24 (b) 1 36 (c) 1 32 (d) 1 8

9. The number of real solution of equation 16 sin–1 _{x tan}–1 _{x cosec}–1 _{x = p}3 _is/are

(a) 0 (b) 1 (c) 2 (d) infinite 10. Let a, b, c are non-zero constant numbers. Then

Lim r Æ •

cos cos cos

sin sin a r b r c r b r c r equals (a) a b c bc 2 2 2 2 + -(b) c a b bc 2 2 2 2 + -(c) b c a bc 2 2 2 2 + -(d) Independent of a, b and c

Section B: One or More Options Correct

Type

11. Given 2 functions f and g which are integrable on every interval and satisfy

(i) f is odd, g is even (ii) g(x) = f (x + 5), then (a) f (x – 5) = g(x) (b) f (x – 5) = – g(x) (c) f t dt( ) g( t dt) 0 5 0 5 5

Ú

=

Ú

-(d) f t dt( ) g( t dt) 0 5 0 5 5

Ú

= -

Ú

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12. A continuous function f (x) on R Æ R satisfies the relation f (x) + f (2x + y) + 5xy = f (3x – y) + 2x2 _{+ 1} for " x, y Œ R, then which of the following hold(s) good?

(a) f is many-one (b) f has no minima (c) f is neither odd nor even (d) f is bounded

13. The equation x2 _{– 4x + a sin a = 0 has real roots} (a) for all values of a

(b) for all values of a provided –p/4 < a < p/4 (c) for all values of a ≥ 4 provided p £ a £ 2p (d) for all a provided | a | £ 4

14. Which of the following statement(s) is/are not correct? (a) If the roots of a quadratic equation are imagi-nary, then these are conjugates of each other. (b) If a continuous function is strictly monotonic,

then it is differentiable.

(c) If f (x) is periodic, then | f (x) | is also periodic. (d) sin x/(2x – 2np – p) and cos x tan x/(2x – 2np

Рp), where n ΠZ, are identical functions. 15. If the curve y = ax1/2 _{+ bx passes through the point}

(1, 2) and lies above the x-axis for 0 £ x £ 9 and the

area enclosed by the curve, the x-axis and the line x = 4 is 8 sq. units, then

(a) a2 _{+ b}2 _{= 6 (b) a/b = 1} (c) a – b = 4 (d) ab = –3

Section C: Integer Value Correct Type

16. The value of Lim

n k n n k Æ • = +

Â

1 2 2 1 ( ) is 17. If the slope of the curve y = ax

b x- at the point (1, 1)

is 2, then the value of a + b is

18. Given a = 3i + j + 2k, b = i – 2j – 4k are the posi-tion vectors of point A and B. Then the distance of point –i + j + k from the plane passing through B and perpendicular to AB is

19. If z = x + iy (x, y Œ R, x π –1/2), then the number of values of z satisfying | z |n _{= z}2 _{| z |}n–2 _{+ z | z |}n–2 _{+ 1,}

(n ΠN, n > 1) is

20. If x, y, z are distinct positive numbers such that

x

y y z z x

+ = + = +1 1 1 , then the value of xyz is

PaPer 2

Section a: One or More Options Correct

Type

1. For natural numbers m and n, if (1 – y)m _{(1 + y)}n

= 1 + a1 y + a2 y2 + L, and a1 = a2 = 10, then (a) m < n (b) m > n

(c) m + n = 80 (d) m – n = 20

2. Let P(x) = x2 _{+ bx + c, where b and c are integer. If} P(x) is a factor of both x4 _{+ 6x}2 _{+ 25 and 3x}4 _{+ 4x}2 ₊ 28x + 5, then

(a) P(x) = 0 has imaginary roots (b) P(x) = 0 has roots of opposite sign (c) P(1) = 4

(d) P(1) = 6

3. Let tan x – tan2 _{x > 0 and | 2 sin x | < 1. Then the} int-ersection of which of the following two sets satisfies both inequalities?

(a) x > np + p/6, n ΠZ (b) x > np Рp/6, n ΠZ (c) x < np Рp/4, n ΠZ (d) x < np + p/4, n ΠZ

4. A bag initially contains one red and two blue balls. An experiment consisting of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour. If three such trials are made, then

(a) probability that atleast one blue ball is drawn is 0.9.

(b) probability that exactly one blue ball is drawn is 0.2.

(c) probability that all the drawn balls are red given that all the drawn balls are of same colour is 0.2. (d) probability that atleast one red ball is drawn is 0.6. 5. If a, b, c are non-zero real numbers such that

bc ca ab ca ab bc ab bc ca = 0, then (a) 1 1 1 0 a b c+ + = (b) 1 1 1 ₀ 2 a b+ w +cw = (c) 1 1₂ 1 0 aw+bw + =c (d) None of these 6. 2007201 _{+ 2019}201 _{– 1982}201 _{– 2044}201 _{is divisible by} (a) 74 (b) 50 (c) 1850 (d) 2013

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(a) 1 22 (b) 1 23 (c) 2 33 (d) 225

Paragraph for Questions 13 and 14

Consider a function f defined by f (x) = sin–1 _sin Êx+ x ËÁ sin2 ˆ¯˜,

" x Π[0, p], which satisfies f (x) + f (2p Рx) = p, " x Π[p, 2p] and f (x) = f (4p Рx) for all x Π[2p, 4p], then.

13. If a is the length of the largest interval on which f (x) is increasing, then

a-(a) p/2 (b) p (c) 2p (d) 4p 14. If f (x) is symmetric about x = b, then b =

(a) a/2 (b) a (c) a/4 (d) 2a

Paragraph for Questions 15 and 16

Straight line x + y = 18 and common tangents of the curve S1 = x2 + y2 – 28x + 160 = 0 and S2 = x y

2 2

64 36+ – 1 = 0

form a triangle ABC in the first quadrant and S3 = 0, the circumcircle of DABC.

15. The coordinates of the center of circle inscribed in triangle ABC are

(a) [12 2 2 10 2 2- , - ] (b) [12- 2 10, - 2] (c) [12 2 2 10 2 2+ , + ] (d) [12+ 2 10, + 2] 16. Equation of S3 is (a) x2 _{+ y}2 _{– 16x – 20x + 156 = 0} (b) x2 _{+ y}2 _{– 10x – 8x + 156 = 0} (c) x2 _{+ y}2 _{– 8x – 10x + 156 = 0} (d) x2 _{+ y}2 _{– 20x – 16x + 156 = 0}

Section 3: Matching List Type

17. The graphs of f and g are given in Fig. B.1. Use them to evaluate each limit.

-1 -1 -2 1 2 1 2 x y O y f x= ( )

7. If the equation sin2 _{x – a sin x + b = 0 has only one} solution in (0, p), then which of the following state-ments are correct?

(a) a Œ (– •, 1) » (2, •) (b) b Œ (– •, 0) » (1, •) (c) a = 1 + b

(d) None of these

8. Let a, b, and g be some angles in the 1st quadrant satisfying tan (a + b) = 15/8 and cosec g = 17/8, then which of the following holds good?

(a) a + b + g = p

(b) cot a cot b cot g = cot a + cot b + cot g (c) tan a + tan b + tan g = tan a tan b tan g (d) tan a tan b + tan b tan g + tan g tan a = 1

Section B: Paragraph Type

This section contains 4 paragraphs each describing theory, experiment, date etc. Eight questions relate to four para-graphs with two questions on each paragraph. Each question of paragraph has only one correct answer along the four choice (a), (b), (c) and (d).

Paragraph for Questions 9 and 10

A perpendicular is drawn from a fixed point (3, 4 ) to a variable line having x-intercept unity. P (x, y) = 0 represents the locus of the foot of perpendicular drawn from (3, 4) to the variable line, which is a circle. Then

9. The equation of the circle is: (a) x2 _{+ y}2 _{= 4}

(b) (x – 2)2 _{+ (y – 2)}2 _{= 5} (c) (x – 3)2 _{+ (y – 3)}2 _{= 4} (d) (x – 3)2 _{+ (y + 3)}2 _{= 5}

10. A tangent is drawn to P(x, y) = 0 from the origin, then the length of tangent is

(a) 2 (b) 1 (c) 3 (d) 2

Paragraph for Questions 11 and 12

Let a point P whose position vector is r xi y j zk= + + 

is called lattice point if x, y, z ΠN. If atleast two of x, y, z are equal then this lattice point is called isosceles lattice point. If all x, y, z are equal, then this lattice point is called equilateral lattice point.

11. The number of lattice points on the plane

   

r i j k.( + + )= 10 are

(a) 36 (b) 45 (c) 84 (d) 120 12. If a lattice point is selected at random from lattice

points which satisfy r i j k   .( + + )£ 11, then the probability that the selected lattice point is equilateral given that it is isosceles lattice point is

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-1 -1 -2 1 2 1 2 x y O y g x= ( ) Fig. B.1 Column I Column II (a) lim ( ( )) xÆ1f g x (p) 1 (b) lim ( )

xÆ2 3f x -2 (q) does not exist

(c) lim ( ) ( ) ( ) ( ) x f x g x f x g x Æ + Ê ËÁ ˆ ¯˜ 0 (r) 0 (d) lim _{( )}( ) _{( )}( ) x f x g x f x g x Æ+ -+ 1 3 _{(s) 2}

18. Match the locus of z given by equation in Column I with Column II.

Column I (a) arg z z 2 2-₊1₁ Ê ËÁ ˆ ¯˜ = 0; z π ± i, ± 1

(b) || z – cos–1 _{cos 12 | – | z – sin}–1 _{sin 12 || = 8 (p – 3)} (c) z2 _{+ k} 1 = i | z1 |2 + k2 ; k1 π k2 Œ R – {0} and z1 is fixed π 0 (d) z- -₁ - 1 + +z - - = 3 1 3 2 1 1 1 sin cos p Column II (p) Portions of a line

(q) Point of intersection of hyperbola (r) Pair of open rays

(s) line segment

19. The angles of a triangle are in the ratio 2 : 3 : 7.

Column I Column II

(a) If smallest side = 2, then (p) largest side = 3 1+ (b) If smallest side = 2,then (q) largest side = 6+ 2 (c) If s = +3 3+ 2,then (r) largest side = 1 (d) If D =( 3 1 4- )/ ,then (s) largest side = 2 2 3+

20. A bag contains 14 balls which are either white or black balls. (all number of white and black balls are equally likely). Five balls are drawn at random from the bag without replacement

Column I Column II

(a) Probability that all the five balls are

black is equal to (p) 3/13

(b) If the bag contains 11 black and 3 white balls, then the probability that

all five balls are black is equal to

(q) 1/6 (c) If all the five balls are black then

the probability that the bag contains 11 black and 3 white balls is equal to

(r) 6/65

(d) Probability that three balls are black and two are white is equal to (s) 3/65

Mock Test 2

PaPer 1

Section a: Only One Option Correct Type

1. If f (x) = ( ) , ( ) ( ) ( ) x n f f n n n -¢ = -=

’

51 1 50 51 51 then (a) 5050 (b) 1 1275 (c) 1 5050 (d) 1275

2. Through the focus of the parabola y2 _{= 2px (p > 0), a} line is drawn which intersects the curve at A(x1, y1) and B(x2, y2). The ratio y y_{x x}1 2

1 2 equals (a) 2 (b) –1 (c) –4 (d) some function of p 3. Lim x k n _{k k k k} k k Æ • -= + - + + + Ê Ë Á ˆ ¯ ˜

Â

cos ( ) ( ) ( ) ( ) 1 2 1 1 1 2 1 is equal to (a) p 6 (b) p 4 (c) p 3 (d) p 2

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12. Consider the binomial expansion of x x n + Ê ËÁ ˆ ¯˜ 1 24 ,

n ΠN, where the terms of the expansion are written in decreasing powers of x. If the coefficients of the first three terms form an arithmetic progression, then which of the following is/are true?

(a) Total number of terms in the expansion of the binomial is 8.

(b) Number of terms in the expansion with integral power of x is 3.

(c) There is no term in the expansion which is independent of x.

(d) Fourth and fifth are middle terms of the expan-sion. 13. If x a y b x c y d

+ =1and + =1 intersect the axes at four concyclic points and a2 _{+ c}2 _{= b}2 _{+ d}2_{, then these lines}

can intersect at (a, b, c, d > 0) (a) (1, 1) (b) (1, –1) (c) (2, –2) (d) (3, 3)

14. A forecast is to be made of the results of five cricket matches, each of which can be a win or a draw or a loss for the Indian team.

Let p = number of forecasts with exactly 1 error q = number of forecasts with exactly 3 errors r = number of forecasts with all 5 errors then the incorrect statement is

(a) 2q = 5r (b) 8p = q (c) 8p = 5r (d) 2(p + r) > q

15. A tangent drawn to the curve y = f (x) at P(x, y) cuts the x-axis at A. Now a perpendicular drawn from P(x, y) to the x-axis meets the x-axis at B. If B is the midpoint of OA (O being the origin), and f (1) = 1, then

(a) General point on the curve is (t, 1/t), where t is a real parameter.

(b) y = f (x) is a circle.

(c) Tangent at P(1, 1) is x + y = 2. (d) f ¢(2) = –1/4.

Section C: Integer Type

16. The value of Lim

t x t x t dx Æ +

-Ú

0 0 2 sin ( ) sin p is 4. If f ≤(x) > 0 and f ¢(1) = 0 such that g(x) = f (cot2 _x

+ 2 cot x + 2), where 0 < x < p, then the interval in which g(x) is decreasing is (a) (0, p) (b) p p 2, Ê ËÁ ˆ¯˜ (c) 3 4 p_,p Ê ËÁ ˆ¯˜ (d) 0 3, 4 p Ê ËÁ ˆ¯˜

5. A continuous and differentiable function y = f (x) is such that its graph cuts line y = mx + c at n distinct points. Then the minimum number of points at which f ≤(x) = 0 is/are

(a) n – 1 (b) n – 3

(c) n – 2 (d) Cannot say 6. The number of terms in x

x 3 3 100 1 1 + + Ê ËÁ ˆ¯˜ is (a) 300 (b) 200 (c) 100 (d) 201 7. The set of all values of a for which ax2 _{+ (a – 2)x – 2}

is negative for exactly two integral x is (a) (0, 2) (b) (1, 2) (c) (1, 2) (d) (0, 2)

8. If the fundamental period of the function f (x) =

sin2_ax+ cos2_{axis p/8, then the value of a is} (a) ± 4 (b) ± 2 (c) ± 8 (d) ± 1 9. x f x f x x f x ◊ ¢ -◊

Ú

( )4 _{( )}2 ( ) dx equals (a) x2 _{f (x) + c (b) | x | f (x) + c} (c) 2 f x x( ) +c (d) 2 f x x c ( ) +

10. In which one of the following intervals, the inequality sin x < cos x < tan x < cot x can hold good?

(a) (0, p/8) (b) (3p/4, p) (c) (5p/4, 3p/2) (d) (7p/4, 2p)

Section B: One or More Options Correct

Type

11. Let f (x) = et t dt x x - _>

Ú

[ ] _{( 0),} 0

where [x] denotes est integer less than or equal to x is

(a) continuous and differentiable " x Π(0, 3) (b) continuous but not differentiable " x Π(0, 3) (c) f (1) = e (d) f (2) = 2(e Р1)

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19. If 2 1 1 0 3 4 1 8 10 1 2 5 9 22 15 -È Î Í Í Í ˘ ˚ ˙ ˙ ˙ = - - -- -È Î Í Í Í ˘ ˚ ˙ ˙ ˙

A ,then the sum of

all the elements of matrix A is

20. Let f (x) = sin23 _{x – cos}22 _{x and g(x) = 1 1} 2

+ tan–1 _{| x |,} then the number of values of x in interval [–10p, 8p]

satisfying the equation f (x) = sgn(g(x)) is 17. If a, b and c are distinct positive real numbers such

that a + b + c = 1, then the least integral value of

( )( )( ) ( )( )( ) 1 1 1 1 1 1 + + + - - -a b c a b c is

18. The number of elements in the domain of the func-tion f (x) = sin- Ê - [ ] [ ] ËÁ ˆ ¯˜+ + -1 2 2 3 x x _{x x , where [◊]}

represents greatest integer function, is

PaPer 2

Section a: One or More Options Correct

Type

1. Let z be a complex number satisfying equation zp ₌ zq_{,where n, m Œ N, then}

(a) If p = q, then the number of solutions of equa-tion will be infinite.

(b) If p = q, then the number of solution of equation will be finite.

(c) If p π q, then the number of solution of equation will be p + q + 1.

(d) If p π q, then the number of solution of equation will be p + q. 2. If x x dx f x g x c 4 6 1 1 1 1 2 3 + + = - = - +

Ú

tan ( ) tan ( ) , then

(a) both f (x) and g(x) are odd functions. (b) g(x) is monotonic function. (c) f (x) = g(x) has no real roots. (d) f x g x dx x x c ( ) ( ) = + +

Ú

1 3 3

3. A tangent is drawn at any point (x1, y1) other than vertex on the parabola y2 _{= 4ax. If tangents are drawn} from any point on this tangent to the circle x2 _{+ y}2 _{= a}2 such that all the chords of contact pass through a fixed point (x2, y2), then

(a) x1, a, x2 are in GP. (b) y1/2, a, y2 are in GP. (c) –4, y1/y2, x1/x2 are in GP. (d) x1x2 + y1 y2 = a2

4. Let PM be the perpendicular from the point P(1, 2, 3) to x–y plane. If OP makes an angle q with the positive direction of the z-axis and OM makes an angle f with the positive direction of x-axis, where O is the positive direction of x-axis, where O is the origin, then (q and f are acute angles)

(a) tanq = 5 3/ (b) sin sinq f = 2 14/

(c) tan f = 2 (d) cos cosq f = 1 14/

5. If the side  AB of an equilateral triangle ABC lying in the x–y plane is 3i,then the side CB  can be

(a) - 3 -2(i 3j) (b) 3 2(i- 3j) (c) - 3 + 2(i 3j) (d) 3 2(i+ 3j)

6. In a precision bombing attack, there is a 50% chance that any one bomb will strike the target. Two direct hits are required to destroy the target completely. The number of bombs which should be dropped to give a 99% chance of better of completely destroying the target can be

(a) 12 (b) 11 (c) 10 (d) 13 7. The value(s) of p for which the equations

ax2 _{– px + ab = 0 and x}2 _{– ax – bx + ab = 0 may have} a common root, given a, b are non-zero real num-bers, is(are) (a) a + b2 _{(b) a}2 _{+ b} (c) a(1 + b) (d) b(1 + a) 8. Let f (x, t) = x t x t t x t x ( ), ( ), , - £ - < Ï Ì Ó 1 1 where t is a ous function of x in [0, 1]. Let g x( )=

Ú

f t( ) ( , ) .f x t dt 0 1 Then (a) g(0) = 1 (b) g(0) = 0 (c) g(1) = 1 (d) g(x) = f (x)

Section B: Paragraph Type

Paragraph for Questions 9 and 10

The curve y = (x – a)(x2 _{+ bx + c) meets the x-axis at a} point P(p, 0). A straight line through P meets the curve at two more points A and B. Tangents drawn to the curve at P meet the curves at C. (b2 _{– 4c < 0)}

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9. The abscissa of C is

(a) – a – b (b) – a + b

(c) a – b (d) a + b

10. Number of points on the curve where the slope of the tangent is same as slope of AB is

(a) 0 (b) 1 (c) 2 (d) 3

Paragraph for Questions 11 and 12

Let f (x) be one-one and onto function, such that | f (x) – f –1_{(x) | > 0 for x Œ (0, 2) » (2, 4) given f (2) = 2, f (4) = 0} and f (0) = 4. Let g(x) = 16_-x2 _{. Given f (x) < g(x) " x} Œ (0, 4) and graph of y = f (x) and y = f –1_{(x) are symmetrical} about the line x + y = 4.

11. If f (x) – f –1_{(x) < 0 for x Œ (0, 2) and}

Ú

_{f x dx( ) =5,} 0 2 then

Ú

f-1 x dx 2 4 ( ) is (a) 2 (b) 1 (c) 5 (d) 7 12. If f (x) – f –1_{(x) > 0 for x Œ (0, 2) and} ( ( )f x - f-( ))x dx

Ú

1 0 2 = 2,

Ú

f-1x dx 2 4 ( ) = 4, then g x( ) max ( ( ),- f x f-( ))x dx

Ú

1 0 4 is (a) 4p – 12 (b) 8p – 1 (c) 4p – 10 (d) 8

Paragraph for Questions 13 and 14

A function f (x) having the following properties: (i) f (x) is continuous except at x = 3.

(ii) f (x) is differentiable except at x = –2 and x = 3 (iii) f (x) = 0,lim x Æ 3f (x) Æ – •,x Æ - •lim f (x) = 3,x Æ •lim f (x) = 0 (iv) f ¢(x) > 0 " x Œ (– •, – 2) » (3, •) and f ¢(x) £ 0 " x Œ (– 2, 3) (v) f ≤(x) > 0 " x Œ (– •, – 2) » (– 2, 0) and f ≤(x) < 0 " x Œ (0, 3) » (3, •)

13. Maximum possible number of solutions of f (x) = | x | is

(a) 2 (b) 1 (c) 3 (d) 4 14. f (x) + 3x = 0 has five solution if

(a) f (– 2) > 6 (b) f ¢(0) < – 3 and f (– 2) > 6 (c) f ¢(0) > – 3 (d) f ¢(0) > – 3 and f (– 2) > 6

Paragraph for Questions 15 and 16

There are three curve defined here in Argand plane:

C z1: -9₅ + z-1₅ =2

C2: arg(z-2i)= -p₄ C z3: - - =1 i 2

15. The number of points common to curves C1, C2 and C3 is/are

(a) 0 (b) 1

(c) 2 (d) None of these

16. Curves C4, C5, C6 are the reflections of the curves C1, C2 and C3, respectively, about the real axis. Then

(a) C z4: -9₅ + z-1₅ =2 (b) C₅ z 3i 3 4 : arg( - )= p (c) C z6: - - =1 i 2 (d) None of these

Section C: Matching List Type

17.

Column I Column II

(a) Sum of coefficients in expansion of (px _{+ q}y _{– 5r}z _{+ 6t)}3 (p) 0

(b) Coefficient of x103 _{in (1 + x + x}2 ₊

x3 _{+ x}4₎199_{(x – 1)}201 _is (q) 100

(c) Remainder when 22003 _{is divided by}

17 is

(r) 27 (d) Remainder when (11111 … 1001

times) is divided by 1001 is

(s) 8

18. A tangent having slope - 4

3 touches the ellipse x2

18

+ y2

32 = 1 at point P and intersects the major and

minor axes at A & B respectively, O is the centre of the ellipse

Column I Column II

(a) Distance between the parallel tangents having slopes - 4

3, is

(p) 24

(b) Area of DAOB is (q) 7/24

(c) If coordinates of p are (l, m), then the value

of l ¥ m is (r) 48/5

(d) If equation of the tangent intersecting posi-tive axes is lx + my = 1, then l + m is equal to (s) 12

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19.

Column I Column II

(a) The number of integer lying in the domain of the function f (x) = log0 5. Ê_ËÁ5 2-_x xˆ_¯˜

(p) 1

(b) Number of positive roots of the equation (x – 1) (x – 2) (x – 3) + (x – 1) (x – 2) (x – 4) + (x – 2) (x – 3) (x – 4) + (x – 1) (x – 3) (x – 4) = 0 is/are

(q) 2

(c) If f (x) = ex _{" x Œ [0, 1] and f (1) – f (0) = f ¢(c).}

where c Π(0, 1) then ln (ec _{+ 1) is equal to} (r) 3

(d) Number of values of x satisfying the equa-tion tan- Ê - tan

-ËÁ ˆ¯˜+ = 1 2 1 1 10 1 2 4 x x p (s) 4

20. Match the following:

Column I Column II (a) x x x dx ln (1 2 2) 0 + •

Ú

(p) 0 (b) / ln (tan ( ))x dx 0 2 p

Ú

(q) –2/5 (c) x dx x3 x6 1 1 1 + +

-Ú

(r) –1 (d) [ ]x dx

-Ú

1 1 , where [◊] represents greatest integer function

(s) –4/5

Mock Test 3

PaPer 1

Section a: Only One Option Correct Type

1. A curve passes through the point (2a, a) and is such

that the sum of subtangent and abscissa is a. Its equa-tion is (a) (x – a)y2 _{= a}3 _{(b) (x – a)}2 _{y = a}3 (c) (x – a) y = a2 _{(d) none of these} 2. A function f(x) satisfies f (x) = sin x + 0 x

Ú

f ¢(t) (2 sin t – sin2_{t) dt. Then f (x) is/are}

(a) x x 1- sin (b) sin sin x x 1-(c) 1- cos cos x x (d) tan sin x x

3. The number 916238457 is an example of nine-digit

number which contains each of the digit 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. Number of such numbers are

(a) 2268 (b) 2520 (c) 2975 (d) 1560

4. Let z1, z2, z3 be complex numbers such that z1 + z2 + z3 = 0 and | z1 | = | z2 | = | z3 | = 1. Then z12+z22+z32 is

(a) greater than zero (b) equal to 3 (c) equal to zero (d) equal to 1

5. P and Q are two points on the upper half of the

ell-ipse x a y b 2 2 2 2

+ = 1. The center of the ellipse is at the origin O and PQ is parallel to the x-axis such that

the triangle OPQ has the maximum possible area. A point is randomly selected from inside of the upper half of the ellipse. The probability that it lies outside the triangle is (a) p p -1 _(b) 2 1 2 p p -(c) p p -1 2 (d) p p -1 4

6. If x, y Œ R and satisfy the equation xy(x2 _{– y}2_{) = x}2 _{+ y}2 where x π 0, then the minimum possible value of x2 _{+ y}2 _is (a) 1 (b) 2 (c) 4 (d) 8 7. If f (x) = [ ] sin[ ] [ ] [ ] [ ] x x x x x 2 0 0 0 + _π = Ï Ì Ô ÓÔ for for

where [x] denotes the greatest integer less than or equal to x, then lim_xÆ0 f (x) equals:

(a) 1 (b) 0

(c) –1 (d) None of these

8. Matrices of order 3 ¥ 3 are formed by using the

ele-ments of the set A = {–3, –2, –1, 0, 1, 2, 3}. Then the

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probability that matrix is either symmetric or skew symmetric is (a) 1 7 1 7 6 + 3 (b) 1 7 1 7 1 7 9 + 3 - 6 (c) 1 7 1 7 3 + 9 (d) 1 7 1 7 1 7 3 + 6 - 9 9. Lim n _k n n k n Æ• -= Ê ËÁ ˆ¯˜ +

( )

Ê ËÁ ˆ ¯˜

Â

tan tan 1 1 1 1

1 has the value equal

to (a) 1 1 2 + ln(cos ) (b) 1 1 2 + ln(sin ) (c) 1 1 1 2 -ln(sin +cos ) (d) 1 1 1 2 +ln(sin +cos )

10. If f (x, y) = x2 _{+ y}2 _{– 2xy (x, y Œ R) and the matrix A} is given by A = f x y f x y f x y f x y f x y f x y f x y f ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( 1 1 1 2 1 3 2 1 2 2 2 3 3 1 xx y3, )2 f x y( , )3 3 È Î Í Í Í ˘ ˚ ˙ ˙ ˙ such that trace (A) = 0, then

(a) det(A) ≥ 0 (b) det(A) £ 0 (c) det(A) = 0 (d) None of these

Section B: One or More Options Correct

Type

11. If f : R Æ R be a continuous function such that f (x) = 2

1

t f t dt x

( )

Ú

, then which of the following does not hold good?

(a) f (p) = ep2

(b) f (1) = e (c) f (0) = 1 (d) f (2) = 2

12. Let (a – 1)(x2 _{+ 3x+ 1)}2 _{– (a + 1)(x}4 _{– x}2 _{+ 1) £ 0} " x Œ R. Then which of the following is/are correct?

(a) a ŒÈ -Î Í ˘ ˚ ˙ 1 3 4 3 ,

(b) Largest possible value of a is 3.

(c) Number of possible integral values of a is 3. (d) Sum of all possible values of a is 0.

13. If planes r i j k   ( + + )=q r i1, ( +2a j k + )=q2 and

   

r ai a j k( + 2 + )=q

3 intersect in a line, then the value of a is (a) 1/4 (b) 1/2 (c) 1 (d) 2 14. If A = 1 3 1 2 2 2 1 2 2 1 -È Î Í Í Í ˘ ˚ ˙ ˙ ˙ = -a b AT A and ,then (a) a = –2 (b) a = 2 (c) b = –1 (d) b = 1 15. The value of a for which x x x

x x x

a

3 2

3-₊62_-+₁₀11 ₊-₈6+₃₀ = 0 does not have real solution is

(a) –10 (b) 12 (c) 5 (d) –30

Section C: Integer Value Correct Type

16. If f : (0, •) Æ (0, •) satisfies f (x f (y)) = x2_ya_{(a Œ R),}

then find a.

17. If a, b, and c are real numbers such that a2 _{+ 2b = 7,} b2 _{+ 4c = –7, and c}2 _{+ 6a = –14, then find the value} of (a2 _{+ b}2 _{+ c}2_)/2.

18. Let an = 16, 4, 1, ... be a geometric sequence. Define Pn as the product of the first n terms. Then find the

value of pn n n = •

Â

1 4 .

19. Find the distance of the point P(3, 8, 2) from the line

1 2 1 1 4 3 1 3 2 (x- =) (y- =) (z- ) measured parallel to the plane 3x + 2y – 2z + 15 = 0.

20. Consider the equation x2 _{+ 2x Рn = 0, where n ΠN and} n Π[5, 100]. Find the total number of different values of n so that the given equation has integral roots.

PaPer 2

Section a:

One or More Options Correct

Type

1. If g(x) = f x

x a x b x c ( )

( - ) ( - ) ( - ), where f (x) is a

polynomial of degree < 3, then

(a) g x dx a f a x a b f b x b c f c x c ( ) ( ) log ( ) log ( ) log

Ú

= -1 1 1 ∏ + 1 1 1 2 2 2 a a b b c c k

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(b) dg x dx a f a x a b f b x b c f c x c a a b b c ( ) ( ) (_{( ) ( )}) ( ) ( ) = -∏ -1 1 1 1 1 2 2 2 2 2 2 _{cc 1} (c) dg x dx a f a x a b f b x b c f c x c a a b b c ( ) ( ) (_{( ) ( )}) ( ) ( ) = -∏ -1 1 1 1 1 1 2 2 2 2 2 cc2 (d) g x dx a f a x a b f b x b c f c x c ( ) ( ) log ( ) log ( ) log

Ú

= -1 1 1 ∏ + a a b b c c k 2 2 2 1 1 1

2. If A and B are two events such that P(A) = 3/4 and P(B) = 5/8, then

(a) P(A » B) ≥ 3/4 (b) P(A¢ « B) £ 1/4 (c) 3/8 £ P(A « B) £ 5/8 (d) 1/8 £ P(A « B¢) £ 3/8

3. If (5, 12) and (24, 77) are the focii of a hyperbola passing through the origin, then

(a) e = 386 12/ (b) e = 386 13/ (c) LR = 121/6 (d) LR = 121/3 4. y = ae–1/x _{+ b is a solution of} dy dx y x = ₂. Then (a) a Œ R (b) b = 0

(c) b = 1 (d) A takes finite number of values

5. If AB = A and BA = B, then

(a) A2_{B = A}2 _{(b) B}2_{A = B}2 (c) ABA = A (d) BAB = B

6. If the parabola y = (x2 _{+ 4(b – c)x + 4a}2_{)/4 touches} the x-axis, then the line ax + by + c = 0 always passes through the fixed point/points

(a) (1, 1) (b) (–1, –1) (c) (–1, 1) (d) (1, –1) 7. Let f (x) = 3 1 2 1 t x t dt +

Ú

,x > 0, then (a) for 0 < a < b, f (a) < f (b) (b) for 0 < a < b, f (a) > f (b) (c) f (x) + p/4 < tan–1 _{x, "x ≥ 1} (d) f (x) + p/4 > tan–1 _{x, "x ≥ 1} 8. If 10! = 2p₃q₅r₇s_{, then} (a) 2q = p (b) pqrs = 64 (c) number of divisors of 10! is 270.

(d) number of ways of putting 10! as a product of two natural numbers is 135.

Section B: Paragraph Type

Paragraph for Questions 9 and 10

Let angles a, b, g of a triangle satisfy the relation,

sinÊ_ËÁa b- _¯˜ˆ+sinÊ_ËÁa g- ˆ_¯˜+sinÊ_ËÁ aˆ_¯˜=

2 2

3 2

3 2.

9. The largest angle of triangle is

(a) 70° (b) 100° (c) 110° (d) 130° 10. The triangle is (a) acute-angled (b) right-angled (c) isosceles (d) scalene

Paragraph for Questions 11 and 12

Let b x x b x x x x b R cos cos sin

(cos sin ) tan ,

2 2 _-1= 2 _-+3 2 Œ

11. Equation has solutions if (a) b Œ Ê_ËÁ- • ˆ_¯˜- -ÏÌ Ó ¸ ˝ ˛ ,1 , , 2 1 0 13 (b) b Œ

(

- •

)

- -Ï_Ì Ó ¸ ˝ ˛ ,1 1 0 1, , 3 (c) b Œ R – ÏÌ -Ó ¸ ˝ ˛ 1 0 1 3 , , (d) None of these

12. For any value of b for which the equation has solution, then the number of solutions for x Œ(0, 2p) are always

(a) infinite

(b) depends upon the value of b (c) 4

(d) none of these

Paragraph for Questions 13 and 14

The base of pyramid is rectangular, three of its vertices of the base are A(2, 2, –1), B(3, 1, 2) and C(1,1,1) (points

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may or may not be in order). Its vertex at the top is P 4 26 3 10 3 ,- , -Ê

ËÁ ˆ¯˜ and fourth vertex of the base is D.

13. Coordinates of D are

(a) (4, 0, 2) (b) (4, 2, 0) (c) (2, 0, 4) (d) (0, 2, 4)

14. Volume of the pyramid is (in cubic units)

(a) 20 (b) 10

(c) 40 (d) 30

Paragraph for Questions 15 and 16

Consider two circles C1 : x2 + y2 = r12 and C2 : x2 + y2 = r22 (r1 > r2). Let A be a fixed point on circle C1 say (r1, 0) and B be a variable point on circle C2. The line BA meets circle C2 again at C. Then

15. The maximum value of BC2 _is (a) 4r12 (b) 4r22 (c) 4r22 + 4r12 (d) None 16. The value of OA2 _{+ OB}2 _{+ BC}2 _is (a) [5r22 – 3r12, 5r22 + r12] (b) [4r22 – 4r12, 4r22] (c) [4r12, 4r22] (d) None of these

Section C: Matching List Type

17. Match the following:

Column I Column II

(a) Common normals to the parabola y2

= 4ax and x2 _{= 4ay is/are} (p) x = a

(b) The locus of P point is, if tangents

from P to the parabola y2 _{= 4ax}

intersects the co-ordinate axes in concyclic points

(q) x + y = 3a

(c) The locus of P, if tangents from it to the parabolas y2 _{= 4a (x + a) and y}2

= 8a (x + 2a) are perpendicular is

(r) x + 3a = 0 (d) The chord of contact of a point P

w.r.t. the parabola y2 _{+ 4ax = 0}

subtends right angle at the vertex. The locus of P is

(s) x = 4a

18. Match the following:

Column I Column II

(a) If the vectors a, b, c form sides BA , CA ,  AB of DABC, then

(p) a c b c c a     . = . = .

Column I Column II

(b) If a, b, c are forming three adjacent sides of regular tetrahedron, then

(q) a b b c c a     . = . = . = 0 (c) If a b ¥ =c, and b c ¥ =a,

then (r)

      a b b c c a¥ = ¥ = ¥ (d) If a, b, c are unit vectors

and a b c  + + = 0, then (s)

      a b b c c a. + . + . = - 3

2

19. Match the following:

Column I Column II

(a) If Langrange’s mean value theorem is applicable in [–2, 2] for the function

f (x) = mx c x ex x + < ≥ Ï Ì Ô ÓÔ , , 0

0 then the value of

m + 3c is

(p) 2

(b) If the ends of latus rectum of parabola are (2, 6) and (6, 2) and the equation of the possible directrix is x + y = li where i = 1, 2.

Then the value of l1 + l2 is

(q) 8

(c) The maximum value of f(x) = 2x3 _{– 3x}2 _{– 12x}

in [–2, 5/2] is (r) 4 (d) If lim_n ( n ) ( n ) n Æ• - + - + + Ê Ë Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ 1 2 1 1 4 2 1 2 2 L is equal to k.p 4, then k is (s) 16

20. Match the following:

Column I Column II

(a) If 1, a, a2_{, L a}19 _{are 20th roots of unity,}

then 1 + a10 _{+ a}20 _{+ L + a}190 _{is equal to} (p) 4

(b) Number of integral values of x satisfying the

inequality (log2x)2 + log2 0.03125x + 3 < 0

(q) 2 (c) If l is the maximum value of the function

f(x) = x2_e–2x_{, x > 0, then} 1 2 Ê ËÁ ˆ¯˜ ln ( )l is (r) 0 (d) Number of points on x2 9 – y2 = 1 from

which pair of perpendicular tangents can

be drawn to parabola y2 _{= – 12x is} (s) 1 (Contd.)

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Solutions of Mock Test 1

fi 2a –1 = 0

fi a = ½

4. Ans (c): m1= - 3₄,m2=3₄

Since tangents are drawn on same branch and point P(3, 4) lies in obtuse angle between the asymptotes, therefore tanq 2 3 4 = \ e b a 2 ₁ 2 4 3 - = =cotq = \ e = 5 3

5. Ans (c): Let the parabola is y2 _{= 4ax.} a = at1t2, b = a(t1 + t2), Q ∫ (at1t3, a(t1 + t3)), R ∫ (at2t3, a(t2 + t3))

Let T ∫ (h, k). Then h = at3

2 (t1 + t2) = bt3

2 and

2k = a(t1 + t2) + 2at3 = b+ 4ah_b

Therefore, the locus is 4ax – 2by + b2 _{= 0, which is} a straight line.

6. Ans (d): | x + y + z | = | x | + | y | + | z | fi x, y, z are of same sign.

Thus, x2 _{Р2 < 0 and sin x < 0} fi x Π-( 2 0, ) 7. Ans (d): A3 _{= B}3 _...(1) and A2_{B = B}2_{A ...(2)} (1) Р(2) gives A3 _{РA}2_{B = B}3 _{РB}2_A A2_{(A РB) = РB}2_{(A РB)} fi (A2 _{+ B}2_{) (A РB) = 0} det (A2 _{+ B}2_{) (A РB) = 0} det (A2 _{+ B}2_{) . det (A РB) = 0 fi (d)} 8. Ans (a): Possible outcomes are

11 2 (C);113 (C);115 (C) 2 3 5 p= p= p=    Hence, P(A) = 9 216 1 24 =

9. Ans (b): Domain is x = ± 1. However, only x = 1 satisfies.

PaPer 1

Section a

1. Ans (a): I = ln (1 tan tan ( ))

0

+

-Ú

q q

q

x dx

fi I = ln tan (tan tan ) tan tan 1 1 0 + -+ Ê ËÁ ˆ ¯˜

Ú

q qq q x x dx = ln tan tan tan 1 1 2 0 + + Ê ËÁ ˆ ¯˜

Ú

q q q x dx

fi I = ln (1 tan )2 ln (1 tan tan )

0 0 + - +

Ú

q

Ú

q q q dx x dx fi I = 2q ln sec q – I fi 2I = 2q ln sec q fi I = q ln sec q 2. Ans (b): We have 1 1 2 v u- = f fi d v u d f 1 1_- 2 Ê ËÁ ˆ¯˜= Ê ËÁ ˆ ¯˜ fi - 1₂ + 1₂ = - 2₂ v dv u du f df fi 1₂ 1₂ 2₂ v -u f df Ê ËÁ ˆ¯˜a= [Q du = dv = a] fi a 1 1 1 1 2₂ v u+ v u f df Ê ËÁ ˆ¯˜ÊËÁ - ˆ¯˜= fi a 1 1 2 2₂ v u+ f f df Ê ËÁ ˆ¯˜¥ = Q 1 1v u- = 2f È Î Í ˘ ˚ ˙ fi df f = u v+ Ê ËÁ ˆ¯˜ a 1 1

Thus, relative error in f = a 1 1 u v+ Ê ËÁ ˆ¯˜.

Hence, (b) is the correct answer.

3. Ans (c): f (x) is strictly increasing ⇒ f ¢(x) ≥ 0 fi 2 sec2 _{x + (2a +1) tan x + (a – 2) ≥ 0, " x Œ R} fi 2 tan2 _{x + (2a +1) tan x + a ≥ 0, " x Œ R} fi (2a +1)2 _{– 8a £ 0} fi (2a –1)2 _{£ 0}

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10. Ans (c): Let 1/r = x so that as r Æ •, x Æ 0. Lim x ax bx cx bx bx cxcx bc x Æ -0 2

cos cos .cos

sin _{. sin . .} = 1 0 2 bc ax bx cx x x Lim Æ

-cos cos .cos

= 1 2 0 bc a ax b bx cx c cx bx x xlim

sin sin cos

sin cos Æ - + + ◊ = 1 2bx(b2+c2-a2)

Section B

11. Ans (b, c): g(x) = f (x + 5) from (ii) fi g(–x) = f (–x + 5) fi g(–x) = – f (x – 5) from (i)

Thus, choice (b) is true. Further I = f t dt( ) 0 5

Ú

\ I =

Ú

g t( -5)dt 0 5 (Q f (t) = g(t – 5)) = g(5 t dt) 0 5

-Ú

(Q g is even)

Therefore, choice (c) is correct.

12. Ans (a, b): Let 2x + y = 3x – y fi 2y = x fi y = x/2 Put y = x/2

f (x) + f (5x/2) + 5x2_{/2 = f (5x/2) + 2x}2 _{+ 1} \ f (x) = 1 – (x2_/2)

13. Ans (c, d): x2 _{– 4x + a sin a ∫ (x – 2)}2 _{+ 4 – a sin a} The roots are real if a sin a £ 4.

As | sin a | £ 1, the roots are real for all values of a if | a | £ 4.

For p £ a £ 2p, sin a £ 0. Even if a ≥ 4, a sin p £ 4 and the roots are real.

14. Ans (a, b): If the coefficients are imaginary, then the imaginary roots may not be conjugate of each other. Hence, Statement (a) is false.

The continuous monotonic function may not be dif-ferentiable. Hence, Statement (b) is false.

Statement (c) is true

cos tanx x sin

x n

x

x n

2 -2 p p- =2 -2 p p- for all value of x π np

+ p

2 (which are not in the domains of both the sides).

Hence, Statement (d) is true.

15. Ans (c, d): Since the curve y = ax1/2 _{+ bx passes} through the point (1, 2),

2 = a + b (1)

By observation, the curve also passes through (0, 0). Therefore, the area enclosed by the curve, x-axis and x = 4 is given by A=

Ú

(ax1 2/ +bx dx) = 0 4 8 fi 2 3 8 2 16 8 a_{◊ + ◊ =}b fi 2 3a b+ =1 (2)

Solving (1) and (2), we get a = 3, b = –1.

Section C

16. Ans (2): f(n) = 1 1 2 1 2 2 2 2 n + (n + ) + (n + ) +... + + + 1 2 1 2 n n

The terms of the sequence are decreasing and the number of terms are (2n + 2).

2 2 2 1 2 2 2 2 n n n f n n n + + + £ £ + ( )

Now Lim Lim

n n n n n n n n n n Æ • Æ • + + + = + Ê ËÁ ˆ¯˜ + + 2 1 2 1 2 1 1 1 2 1 2 2 ( ) _{= 2}

Similarly Lim Lim

n n n n n n Æ • Æ • + ₌ + ₌ 2 1 2 1 ₂ 2 ( ) ( ) \ Lim nÆ •f x( ) 2=

17. Ans (1): We have y = ax/(b – x) fi dy dx b x a ax b x ab b x = - - -- = -( ) ( ) ( ) ( ) 1 2 2 dy dx ab b È ÎÍ ˘ ˚˙( , )1 1 =( -1)2 = 2 (given) (1)

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Since the curve passes through the point (1, 1), there-fore,

1 = a/(b – 1) fi a = b – 1 (2)

On putting a = b – 1 in Eq. (1), we get

( ) ( ) b b b -1 12 = 2 fi b = 2 fi a = 2 – 1 = 1 Hence, a = 1, b = 2 18. Ans (5):  AB = - = - -b a  2 3i j-6k

Equation of the plane passing through B and perpen-dicular to  AB is

(r OB AB  -   )◊ = 0

fi r ◊ (2i + 3 j + 6k) + 28 = 0

Hence, the required distance from c = – i + j + k is

c i j k i j k ◊ + + + + + = - + + + (2 3 6 ) 28 2 3 6 2 3 6 28 7 = 5 units

19. Ans (1): The given equation is | z |n _{= (z}2 _{+ z) | z |}n – 2 _{+ 1}

fi z2 _{+ z is real} fi z2_{+ =}z z2₊z

fi (z z z z- ) ( + + =1 0)

fi z z x z z= = as + + π1 0(xπ -1 2/ )

fi The given equation reduces to xn _{= x}n _{+ x | x |}n – 2 _{+ 1}

fi x | x |n – 2 _{= –1}

fi x = –1.

So, the number of solution is 1. 20. Ans (1): x y y z yz y z z xzx z x x yxy - = - , - = - , = \ ( )( )( ) ( )( )( ) ( ) x y y z z x x y y z z x xyz - - - = - - ₂ fi xyz = 1

PaPer 2

Section a

1. Ans (b, c): (1 – y)m _{(1 + y)}n = (1 – m_C 1 y + mC2 y2 – L) (1 + nC1 y + nC2 y2 + L) = + - +Ï_Ì - + - -Ó ¸ ˝ ˛ + 1 1 2 1 2 2 (n m) m m( ) n n( ) mn y L Given: a1 = 10 fi a1 = n – m = 10 (1) a2 = m n m n mn 2 2 ₂ 2 10 + - - - ₌ fi (m – n)2 _{– (m + n) = 20} fi m + n = 80 (2)

2. Ans (c): Since P(x) divides into both of them, hence P(x) also divides

(3x4 _{+ 4x}2 _{+ 28x + 5) – 3(x}4 _{+ 6x}2 _{+ 25)} = – 14x2 _{+ 28x – 70 = – 14(x}2 _{– 2x + 5} which is a quadratic.

Hence, P(x) = x2 _{– 2x + 5 fi P(1) = 4}

3. Ans (a, d): tan x – tan2 _{¥ > 0 fi tan x(tan x – 1) < 0} fi 0 < tan x < 1 or 0 < x < p/4 or np < x < np + p/4, n Œ Z (generalizing) sin x < 1 2 fi - <1 < 2 1 2 sin x or –p/6 < x < p/6 or –p/6 + np < x < p, n Œ Z (generalizing) Then the common values are np + p/6 < x < np + p/4. 4. Ans (a, b, c, d): (a) P(E1) = 1 – P(RRR) = 1 1 3 2 4 3 5 0 9 - ¥ ¥ = . (b) P(E2) = 3P(BRR) = 3 2¥ ¥ ¥ = .₃ 1₄ 2₅ 0 2 (c) P(E3) = P(RRR/(RRR » BBB)) = P RRR P RRR P BBB ( ) ( )+ ( ) = 0 1 0 1 2 3 3 4 4 5 . . + ¥ ¥ = 0 1 0 1 0 4 0 2 . . + . = . (d) P(E4) = 1 – P(BBB) = 1 2-₅ = 0.6 5. Ans (a, b, c): We have

bc ca ab ca ab bc ab bc ca

= 0 fi (ab)3 _{+ (bc)}3 _{+ (ca)}3 _{– 3(ab)(bc)(ca) = 0} or ab + bc + ca(abw + bcw2 _{+ ca)} (abw2 _{+ bcw + ca) = 0}

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or ab + bc + ca = 0, abw + bcw2 _{+ ca = 0,} abw2 _{+ bcw + ca = 0} or 1 1 1 0 1 1 1 0₂ a b c+ + = ,cw + +a bw = , 1 1 1₂ 0 c a+ w +bw =

6. Ans (a, b, c, d): Odd + Odd – Even – Even = Even So the given number is divisible by 2.

Also 2007201 _{– 1982}201 _{is divisible by 2007 – 1982} = 25

2044201 _{– 2019}201 _{is divisible by 2044 – 2019 = 25} Hence, the given number is divisible by 25. Also 2007201 _{– 2044}201 _{is divisible by 2007 – 2044 = –37}

and 2019201 _{– 1982}201 _{is divisible by 2019 – 1982 = 37} Hence, the Given number is divisible by 37. So the given number is divisible by 2 ¥ 37, 2 ¥ 25,

2 ¥ 25 ¥ 37 = 74, 50, 1850.

2007201 _{+ 2019}201 _{is divisible by 2007 + 2019 = 4026} 1982201 _{+ 2044}201 _{is divisible by 1982 + 2044 = 4026} So, the given number is divisible by 4026 = 2 ¥ 2013 [\ an + bn has a factor a + b if n is odd]

7. Ans (a, b, c): sin2 _{x – a sin x + b = 0 has only one} solution in (0, p).

So, sin x = 1 gives one solution and sin x = a gives other solution such that a > 1 or a £ 0.

Hence, (sin x – 1)(sin x – a) is the same equation as sin2 _{x – a sin x + b = 0}

fi 1 + a = a and a = b or 1 + b = a and b > 1 or b £ 0 or b Œ (–•, 0) » (1, •) and a Œ (–•, 1) » (2, •)

8. Ans (b, d): tan (a + b) = 15/8 and tan g = 8/15 \ a + b + g = p/2 fi (b) and (d)

Section B

Ans 9. (b), 10. (c):

Let the variable line AC be x + ay – 1 = 0 which passes through the point A(1, 0).

Let the foot of perpendicular on variable line from B(3, 4) is C(h, k).

Now BC ^ AC

\ (slope of BC) ¥ (slope of AC) = –1 or k h k h -Ê ËÁ ˆ ¯˜ -Ê ËÁ ˆ ¯˜= -4 3 0 1 1 or x2 _{+ y}2 _{– 4x – 4y + 3 = 0} or (x – 2)2 _{+ (y – 2)}2 _{= 5}

Thus, the length of tangent from origin is S1= 3.

Ans 11. (a), 12. (b): Let r xi y j zk= + + 

r i j k   .( + + )= 10

\ x + y + z = 10 fi x ≥ 1, y ≥ 1, z ≥ 1

Therefore, the number of latice points = 9_C 2 x + y + z £ 11

For equilateral latice points, x = y = z = 1, 2, 3 So, three cases are possible. For isosceles lattice point, x = y π z 1 1 9 5 5 1  ¸ ˝ Ô ˛ Ô 15 cases 1 1 8 4 4 2   ¸ ˝ Ô ˛ Ô 12 cases 1 1 7 2 2 5 4 4 1 ¸ ˝ Ô ˛ Ô 9 cases 1 1 6 2 2 4 3 3 2 ¸ ˝ Ô ˛ Ô 9 cases 1 1 5 2 2 3 3 3 1 ¸ ˝ Ô ˛ Ô 9 cases 1 1 4

}

3 cases 1 1 3 2 2 1 ¸ ˝ ˛ 6 cases 1 1 2

}

3 cases

Total number of isosceles lattice points which are not equilateral points

= 15 + 12 + 9 + 9 + 9 + 3 + 6 + 3 = 66 Therefore, total number of isosceles latice point = 66 + 3 = 69

Therefore, required probability, P = 3

69 1 23 = Ans 13. (c), 14. (b): g(x) = x+ sinx 2 increasing function of x range 0 2 ,p È ÎÍ ˘ ˚˙

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\ f (x) = x+ sin ,x 2 x Œ [0, p] p £ t £ 2p, then f (t) + f (2p – t) = p f (t) + 2 2 2 p- +t sin ( p-t) _{= p} f (t) + p – t t 2- sin2 = p f (t) = t+ sint 2 f (x) = x+ sinx 2 p £ x £ 2p f (x) = x+ sinx 2 for 0 £ x < 2p f (x) = f (4p – x) for x Œ [2p, 4p] f (x) is symmetric about x = 2p. O 2p 4p Fig. B.2 a = 2p – 0 = 2p from graph b = a Ans 15. (a), 16. (d):

From the graph (Fig. B.3), y = 6 and x = 8 are the common tangents. C(8,10) A(8,6) B(12,6) y x Fig. B.3 15. Area of DABC = 8 sq. unit

2s = 4 + 4 + 4 2 Radius r s = = + = + = -D 8 4 4 2 2 1 2 2 2 1

Co-ordinate of centre are (7 2 2 5 2 2+ , + )

16. Centre of circumcircle is (10, 8) and radius 2 2. Therefore, equation of circle is

x2 _{+ y}2 _{– 20x – 16y + 156 = 0}

Section C

17. Ans: (a Æ q); (b Æ s); (c Æ r); (d Æ p) (a) lim x Æ1- f (g(x)) = f (2 –_{) = 2} lim x Æ1+ f (g(x)) = f (1 –_{) = 1} Therefore, limit does not exist. (b) lim

x Æ 2 f (x) = 2 fi xlimÆ2 3f x( )- =2 2 (c) lim

x Æ 0 f (x) = 0 and x Æ 0lim g(x) = finite quantity

\ lim ( ) ( ) ( ) ( ) x f x g x f x g x Æ + Ê ËÁ ˆ ¯˜= 0 0 (d) lim ( ) ( ) ( ) ( ) ( ) x f x g x f x g x Æ+ -+ = -+ = = 1 3 3 1 1 1 1 2 2 1 18. Ans: (a Æ r), (b Æ r), (c Æ q), (d Æ s) (a) arg z z 2 2-₊1₁ Ê ËÁ ˆ ¯˜ = 0 ; z π ± i z z z z z z z z 2 2 2 2 1 1 1 1 0 0 -+ = -+ fi - = , + = y = 0, x = 0

Locus of z is portion of pair of lines xy = 0.

Q z z 2 2-₊1₁ 0 Ê ËÁ ˆ ¯˜> È Î Í Í ˘ ˚ ˙ ˙ (b) Given

|| z – cos–1 _{cos 12 | – | z – sin}–1 _{sin 12 || = 8 (p – 3)} Since, | cos–1 _{cos 12 – sin}–1 _{sin 12 | = 8 (p – 3)} Therefore, locus of z is the portion of a line

joining z1 and z2 except the segment between z1 and z2. (c) z2 _{– i | z} 1 |2 = k2 – k1 x2 _{– y}2 _{+ 2 ixy – il} 1 = l2 x2 _{– y}2 _{= l} 2 and xy = l₂1

The locus of z is the point of intersection of hyperbola. (d) Given, z- -₁ - 1 + z+ - - = 3 1 3 2 1 1 1 sin cos p Since 1 1 3 1 3 2 1 1 1 +_sin- +_cos- -p =

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| z – z1 | + | z – z2 | = | z1 + z2 |

Thus, the locus of z is line segment joining z1 and z2.

19. Ans: (a Æ q), (b Æ p), (c Æ s), (d Æ r) Angles of the triangle are 30º, 45°, 105°.

Then a b c

sin30∞=sin45∞=sin105∞

fi a b c

1 2 1 2/ = / = ( 3 1 2 2+ )/

fi a b c k

2 = =2 3 1+ = (say)

Smallest side, a = 2 kand largest side, c =( 3 1+ )k

If a=2,k= 2,c= 6+ 2 If a= 2,k=1,c= 3 1+ If s= +3 2+ 3,k=2,c= +2 2 3 If D = 3 1 -4 , then from D = 1 2 bc sin A, we have 3 1 4 1 2 2 3 1 30 - _{= ¥ k¥( + ) sink ∞} fi k2 3 1 2 k 2 3 1 3 1 4 3 1 2 = -+ = - _{fi =} -( ) ( ) and hence c = 1 20. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ q)

Let Ei denotes the event that the bag contains i black

and (14 – i) white balls (i = 0, 1, 2, ... 14) and A denotes the event that five balls drawn are all black. Then P(Ei) = ₁₅1 (i = 0, 1, ... 14) P(A/E_i) = 0 for i = 0, 1, 2, 3, 4 P(A/E_i) = iC C5 14 5 for i ≥ 5 (a) P(A) = P( ) ( / )E_i P A E_i i =

Â

0 14 = 1 15 1 14 5 C ( 5_C 5 + 6C5 + L + 14C5) = 1 15 1 6 15 6 14 5 C C = (b) Clearly P(A/E11) = 11 5 14 5 3 13 C C =

(c) From Baye’s theorem, P(A/E11) = P( ) ( /_PP_{( )} ) . E A E A 11 11 1 15 3 13 1 6 6 65 = =

(d) Let B denotes the probability of 3 black and 2 white balls, then

P(B/Ei) = 0 if i = 0, 1, 2 or 13, 14 P(B/Ei) = i_C i_C C 314 2 14 5 for i = 3, 4, ..., 12 \ P(B) = P( ) ( / )Ei PB Ei i =

Â

0 14 =1 15 1 14 5 . C [3C3.11C2 + 4C3.10C2 + L + 12C3.2C2] = 5005 15 1 6 14 5 . C =

Solutions of Mock Test 2

PaPer 1

Section a

1. Ans (b): f (x) = (_{x n})n( n) n - -=

’

51 1 50 ln f (x) = n n x n n (51 ) ln ( ) 1 50 - -=

Â

Differentiating both sides with respect to x, we get

¢ = -=

Â

f x f x n n x n n ( ) ( ) (51 ) 1 50 ¢ = -=

Â

f f n n n n ( ) ( ) ( ) 51 51 51 51 1 50 =1 + 2 + 3 + L + 50 = 1275

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or f f ( ) ( ) 51 51 1 1275 ¢ = 2. Ans (c): y2 _{= 4ax, 4a = 2p > 0 and t} 1t2 = –1. Ratio = 4 21 2 2 1222 a t t a t t = – 4 3. Ans (a): Tk = cos- ._{( + )}+ ( - ) (_{( +}+₎)( + ) Ê Ë Á ˆ ¯ ˜ 1 1 1 1 1 1 2 1 k k k k k k k k Let x = 1 k and y = 1 1 k + 1 1 1 1 2 2 - = -+ y k ( ) = + -+ = + + (k ) ( ) k k k k 1 1 1 2 1 2 Tk is in the form of

cos-1

(

_{xy+ 1-x}2. _1-y2

)

_{= cos}–1_{(y) – cos}–1_(x)

(Q y < x) T_k = cos- cos -+ Ê ËÁ ˆ ¯˜ -Ê ËÁ ˆ¯˜ 1 1 1 1 1 k k Substituting k = 2, 3, 4, ... Sum = Lim nÆ • n - -+ Ê ËÁ ˆ ¯˜ -Ê ËÁ ˆ¯˜ cos1 1 cos 1 1 1 2 = cos ( ) cos- _- - Ê ËÁ ˆ¯˜ 1₀ 1 1 2 Sum = p p p 2- =3 6

4. Ans (d): g(x) = f (cot2 _{x + 2 cot x + 2)} fi g¢(x) = f ¢(cot2 _{x + 2 cot x + 2)}

◊ {–2 cot x cosec2 _{x – 2 cosec}2 _x} for g(x) to be decreasing, g¢(x) < 0

fi f ¢{(cot x + 1)2 _{+ 1} ◊ (–2 cosec}2 _{x) (cot x + 1) < 0} fi f ¢{(cot x + 1)2 _{+ 1} ◊ (cot x + 1) > 0 …(i)} {as f ≤(x) > 0 fi f ¢(x) is increasing, then f ≤(cot x + 1)2 _{+ 1} > f ¢(1) = 0} " ŒÊ ËÁ ˆ¯˜» ÊËÁ ˆ¯˜ x 0 3 4 3 4 , p p,p

Thus, equation (i) holds, if cot x + 1 > 0 fi cotx> - " ŒÊ1 x _ËÁ0 3, ˆ_¯˜

4 p

5. Ans (c): From LMVT, there exists atleast (n – 1) point where f ¢(x) = m.

fi $ atleast (n – 2) points where f ≤(x) = 0 (using Rolle’s theorem) 6. Ans (d): 1 3 1 1 1 3 100 100 1 3 3 +Ê_ËÁ + ˆ_¯˜ È Î Í ˘ ˚ ˙ = + Ê_ËÁ + ˆ_¯˜ x x C x x +100 Ê_ËÁ + ˆ_¯˜ + + Ê_ËÁ + ˆ_¯˜ 2 3 3 2 100 100 3 3 100 1 1 C x x L C x x = (1 + r) + A1x3 + A2x6 + L + A100(x3)100 + B1 _x13 + L + B100 13 100 x Ê ËÁ ˆ¯˜

All other terms obtained by the combination of x3 _and 1/x3 _{well get converted into a term involving x}3 _or 1/x3 _{and hence it will be present among above terms.} So number of terms = 1 + 100 + 100 = 201 7. Ans (b): f(x) = ax2 _{+ (a – 2) x – 2 = (ax – 2) (x + 1)}

f (0) = – 2 and f (–1) = 0

If a is negative, then the expression becomes negative for infinite values of x. Therefore, it must be positive. Expression to be negative for exactly two integral values of x O 1 -1 2 2 a Fig. B.4 So 2 2 1 a£ fi a≥ and 2 1 2 a> fi a< \ a Œ [1, 2)

8. Ans (a): f (x) = | sin a x | + | cos a x | Period of f (x) = p

a p 2 = 8 (given)

fi a = ± 4

9. Ans (d): Given integral = xf x f x x f x dx x f x xf x x f x x dx ¢ - ₌ ¢

-Ú

Ú

( ) ( )( ) ( ( ) ( ))/ ( )/ 2 2 2 2 4 2

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Let t = f x x dt dx x f x x f x x ( ) ( ) ( ) 2 2 42 fi = ¢

Therefore, the required integral is

= 1 21 2 2 t dt t c f x x c = + = +

Ú

/ ( )

10. Ans (a): In 2nd quadrant, sin x < cos x is false as sin x is positive and cos x in negative.

In 4th quadrant, cos x < tan x is false as cos x is posi-tive and tan x in negaposi-tive.

In 3rd quadrant, i.e., 5 4 3 2 p_, p Ê ËÁ ˆ¯˜, if tan x < cot x,

fi tan2_{x < 1, which is false.} Hence, (a) can be correct. Now sin x < cos x is true in 0

4 ,p Ê

ËÁ ˆ¯˜ and tan x < cot x

is also true.

Further, cos x < tan x as tan x = (sin x)/(cos x) and cos x < 1.

Section B

11. Ans (b, d): We have f (x) = et t dt e dt x t x

-Ú

[ ] =

Ú

{ } _, 0 0 So f (x) = e dt x e dt e dt x e dt e dt t x t t x t t 0 0 1 1 1 0 1 1 1 2 0 1 1 2

Ú

Ú

Ú

Ú

Π+ Π+ -if if ( , ) ( , )

ÚÚ

+

Ú

Œ È Î Í Í Í Í Í Í Í Í Í Í Í -e dtt x x 2 2 2 3 if ( , ) ⇒ f (x) = e x e e x e e x x x x - Œ - + - Œ - + - Œ -1 0 1 1 1 1 2 2 1 1 2 1 2 if if if ( , ) ( ) ( ) ( , ) ( ) ( ) ( , 33) Ï Ì ÔÔ Ó Ô Ô

Clearly f (x) is continuous " x > 0 but not differen-tiable " x > N.

Also f (2) = 2(e – 1) = 0 = 2(e – 1) 12. Ans (b, c): x1 2 1x 1 4 n 2 / ₊ / Ê ËÁ - ˆ¯˜ Tr nC xr r x n r r + = ¥ ¥ - -1 2 ₂1 4

Coefficient of the first three terms are nC0, nC1¥ ,1₂

n_C 2 2 1 2 ¥ \ n_C n_C n_C 0+ 2¥ = ¥1₄ 2 1¥ 1₂ fi 1 1 8 +n n( - )=n fi n n( -1) ( )= n -8 1 fi n = 8 (as n π 1) \ Tr C xr r x r r + = ¥ ¥ -1 8 8 2 1 4 2 =8 ¥ 1 ¥

(

4-

)

2 3 4 Cr r x r

Terms of x with integer power occur when r = 0, 4, 8. Thus, three terms.

Hence, (b) and (c) are correct. 13. Ans (a, b, c, d): O(0,0) y x B(0, )b D(0, )d A a( , 0) C c( , 0) Fig. B.5

Points A, B, C and D are concyclic, then ac = bd. The co-ordinates of the points of intersection of lines

are ac b d bc ad bd c a bc ad ( - _{) , (} ) -Ê ËÁ ˆ ¯˜

Let the co-ordinates of the point of intersection be (h, k). Then h ac b d bc ad k bd c abc ad = -- = -( _{) ,} ( ) Given c2 _{+ a}2 _{= b}2 _{+ d}2 (Q ac = bd) fi (c – a)2 _{= (b – d)}2 _{or (c – a) = ± (b – d)} Then the locus of the points of intersection is y = ± x.

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