Appendix B Mock Test 1 PAPER 1 Section A: Only One Option Correct Type 1. For q p Œ Ê Ë Á ˆ ¯ ˜ 0 2 , , the value of definite integral ln ( tan tan ) 1 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 1 u v + Ê Ë Á ˆ ¯ ˜ (c) a 1 1 u v - Ê Ë Á ˆ ¯ ˜ (d) None of these 3. If the function f (x) = 2 tan x + (2a +1) log e | 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 x x 2 4 2 - + - p sin = | x 2 – 2 | + | sin x | + p 4 , then (a) x Œ(, ) 0 2 (b) x Œ- ( , ) 2 2 (c) x ŒR (d) x Œ- ( , ) 20 7. If A and B are different matrices satisfying A 3 = B 3 and A 2 B = B 2 A, then (a) det (A 2 + B 2 ) must be zero. (b) det (A – B) must be zero. (c) det (A 2 + B 2 ) as well as det (A – B) must be zero (d) At least one of det (A 2 + 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 =- - Cengage Learning India Pvt. Ltd.
32
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Appendix BMock Test 1
PaPer 1Section a: Only One Option Correct Type
1. For qp
ŒÊËÁ
ˆ¯̃
02
, , the value of definite integral
ln ( tan tan )10
+Ú qq
x dx is equal to
(a) q ln (sec q) (b) q ln(cosec q)
(c) q ln 22
(d) 2q ln sec q
2. The focal length of a mirror is given by 1 1 2v u f
- = .
If errors made in measuring u and v are a, then the relative error in f is
(a) 2a
(b) a1 1u v
+ÊËÁ
ˆ¯̃
(c) a1 1u v
-ÊËÁ
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) 43
(b) 73
(c) 53
(d) 54
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 x x2
42- + -
p sin = | x2 – 2 | + | sin x | + p4
, 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 = B3 and A2B = B2A, then
(a) det (A2 + B2) must be zero. (b) det (A – B) must be zero. (c) det (A2 + B2) as well as det (A – B) must be zero (d) At least one of det (A2 + B2) 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) 124
(b) 136
(c) 132
(d) 18
9. The number of real solution of equation 16 sin–1 x tan–1 x cosec–1 x = p3 is/are (a) 0 (b) 1 (c) 2 (d) infinite 10. Let a, b, c are non-zero constant numbers. Then
Limr Æ •
cos cos cos
sin sin
ar
br
cr
br
cr
- equals
(a) a b cbc
2 2 2
2+ -
(b) c a bbc
2 2 2
2+ -
(c) b c abc
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
5Ú Ú= -
(d) f t dt g t dt( ) ( )0
5
0
5
5Ú Ú= - -
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B.2 Mathematics
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
– p), where n Œ Z, 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 + b2 = 6 (b) a/b = 1 (c) a – b = 4 (d) ab = –3
Section C: Integer Value Correct Type
16. The value of Limn
k n
n
kÆ •=
+
 1
2
21( )
is
17. If the slope of the curve y = axb 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 = z2 | z |n–2 + z | z |n–2 + 1, (n Œ N, n > 1) is
20. If x, y, z are distinct positive numbers such that
xy
yz
zx
+ = + = +1 1 1 , then the value of xyz is
PaPer 2Section 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 y
2 + 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 + 6x2 + 25 and 3x4 + 4x2 + 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 Œ Z (b) x > np – p/6, n Œ Z (c) x < np – p/4, n Œ Z (d) x < np + p/4, n Œ Z
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 abca ab bcab bc ca
= 0, then
(a) 1 1 1 0a b c
+ + = (b) 1 1 1 02a b c+ + =
w w
(c) 1 1 1 02a b cw w+ + = (d) None of these
6. 2007201 + 2019201 – 1982201 – 2044201 is divisible by (a) 74 (b) 50 (c) 1850 (d) 2013
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Appendix B: Mock Tests B.3
(a) 122
(b) 123
(c) 233
(d) 522
Paragraph for Questions 13 and 14
Consider a function f defined by f (x) = sin–1 sin x x+ÊËÁ
ˆ¯̃
sin ,2
" x Œ [0, p], which satisfies f (x) + f (2p – x) = p, " x Œ [p, 2p] and f (x) = f (4p – x) for all x Œ [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) 2aParagraph for Questions 15 and 16Straight line x + y = 18 and common tangents of the curve
S1 = x2 + y2 – 28x + 160 = 0 and S2 = x y2 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
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 TypeThis 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 10A 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 + y2 = 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) 2Paragraph 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 Œ N. 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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B.4 Mathematics
-1
-1-2 1 2
1
2
x
y
O
y g x= ( )
Fig. B.1
Column I Column II(a) lim ( ( ))
xf g x
Æ1(p) 1
(b) lim ( )x
f xÆ
-2
3 2 (q) does not exist
(c) lim ( )( )
( ) ( )x
f xg x
f x g xÆ
+ÊËÁ
ˆ¯̃0
(r) 0
(d) lim ( ) ( )( ) ( )x
f x g xf x g xÆ +
-+1
3 (s) 2
18. Match the locus of z given by equation in Column I with Column II.
Column I
(a) arg zz
2
211
-+
Ê
ËÁˆ
¯̃ = 0; z π ± i, ± 1
(b) || z – cos–1 cos 12 | – | z – sin–1 sin 12 || = 8 (p – 3) (c) z2 + k1 = i | z1 |
2 + k2 ; k1 π k2 Œ R – {0} and z1 is fixed π 0
(d) z z- - + + - =- -1 13
13 2
11 1sin 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 1Section a: Only One Option Correct Type
1. If f (x) = ( ) , ( )( )
( )x n ff
n n
n
-¢
=-
=’ 51
1
50 5151
then
(a) 5050 (b) 11275
(c) 15050
(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 yx x
1 2
1 2 equals
(a) 2 (b) –1 (c) –4 (d) some function of p
3. Limx
k
n k k k kk kÆ •
-
=
+ - + ++
Ê
ËÁ
ˆ
¯˜Âcos
( ) ( ) ( )( )
1
2
1 1 1 21
is
equal to
(a) p6
(b) p4
(c) p3
(d) p2
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Appendix B: Mock Tests B.5
12. Consider the binomial expansion of xx
n
+Ê
ËÁˆ
¯̃1
2 4,
n Œ N, 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 xa
yb
xc
yd
+ = + =1 1and intersect the axes at four
concyclic points and a2 + c2 = b2 + d2, 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 Limt
x t xt
dxÆ
+ -Ú00
2sin ( ) 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) pp
2,Ê
ËÁˆ¯̃
(c) 34p
p,ÊËÁ
ˆ¯̃
(d) 0 34
, pÊËÁ
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 xx
33
100
1 1+ +ÊËÁ
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) =
sin cos2 2a ax x+ is 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 ( ) ( )
( )
24
dx equals
(a) x2 f (x) + c (b) | x | f (x) + c
(c) 2 f xx
c( )+ (d)
2 f xx
c( )
+
10. In which one of the following intervals, the inequality sin x < cos x < tan x < cot x can hold good?
est integer less than or equal to x is (a) continuous and differentiable " x Œ (0, 3) (b) continuous but not differentiable " x Œ (0, 3) (c) f (1) = e (d) f (2) = 2(e – 1)
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B.6 Mathematics
19. If 2 11 03 4
1 8 101 2 59 22 15
-
-
È
Î
ÍÍÍ
˘
˚
˙˙˙
=- - -
- -È
Î
ÍÍÍ
˘
˚
˙˙˙
A , then the sum of
all the elements of matrix A is
20. Let f (x) = sin23 x – cos22 x and g(x) = 1 12
+ 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 11 1 1
+ + +- - -
a b ca b c
is
18. The number of elements in the domain of the func-
tion f (x) = sin [ ] [ ]- -Ê
ËÁˆ
¯̃+ + -1
2 23
x x x x , where [◊]
represents greatest integer function, is
PaPer 2Section a: One or More Options Correct Type
1. Let z be a complex number satisfying equation zp = z q, 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 xx
dx f x g x c4
61 11
123
++
= = +- -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 xg x
dxx x
c( )( )
= + +Ú 1 33
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 + y2 = a2 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) tan /q = 5 3 (b) sin sin /q f = 2 14
(c) tan f = 2 (d) cos cos /q 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) - -32
3( )i j (b) 32
3( )i j -
(c) - +32
3( )i j (d) 32
3( )i 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 x2 – ax – bx + ab = 0 may have a common root, given a, b are non-zero real num-bers, is(are)
Paragraph for Questions 9 and 10The 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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Appendix B: Mock Tests B.7
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) 3Paragraph for Questions 11 and 12Let 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 2- x . 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 f x dx( ) ,=Ú 50
2
then f x dx-Ú 1
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, f x dx-Ú 1
2
4
( ) = 4, then
g x f x f x dx( ) max ( ( ), ( ))- -Ú 1
0
4
is
(a) 4p – 12 (b) 8p – 1 (c) 4p – 10 (d) 8Paragraph for Questions 13 and 14A 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 – •, lim
x Æ -•f (x) = 3, lim
x Æ •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) > 6Paragraph for Questions 15 and 16There are three curve defined here in Argand plane:
C z z195
15
2: - + - =
C z i2 24
: arg ( )- = -p
C z i3 1 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 z z495
15
2: - + - =
(b) C z i5 3 34
: arg ( )- =p
(c) C z i6 1 2: - - = (d) None of these
Section C: Matching List Type
17. Column I Column II
(a) Sum of coefficients in expansion of (px + qy – 5rz + 6t)3
(p) 0
(b) Coefficient of x103 in (1 + x + x2 + x3 + x4)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 -43
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 -43
, 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
(c) If f (x) = ex " x Œ [0, 1] and f (1) – f (0) = f ¢(c). where c Œ (0, 1) then ln (ec + 1) is equal to
(r) 3
(d) Number of values of x satisfying the equa-
tion tan tan- --ÊËÁ
ˆ¯̃
+ =1 12 110
12 4
xx
p(s) 4
20. Match the following:
Column I Column II
(a) x x
xdxln
( )1 2 20 +
•
Ú (p) 0
(b) ln (tan ( ))/
x dx0
2p
Ú (q) –2/5
(c) x dx
x x3 61
1
1+ +-Ú (r) –1
(d) [ ]x dx-Ú 1
1, where [◊] represents
greatest integer function
(s) –4/5
Mock Test 3
PaPer 1Section 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 = a3 (b) (x – a)2 y = a3
(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 – sin2t) dt. Then f (x) is/are
(a) xx1- sin
(b) sinsin
xx1-
(c) 1- coscos
xx
(d) tansin
xx1-
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 z z z12
22
32+ + 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 xa
yb
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) pp-1 (b) 2 1
2p
p-
(c) pp-1
2 (d) p
p-1
4 6. If x, y Œ R and satisfy the equation xy(x2 – y2) = x2 + y2
where x π 0, then the minimum possible value of x2 + y2 is
(a) 1 (b) 2 (c) 4 (d) 8
7. If f (x) = [ ] sin[ ]
[ ][ ]
[ ]
x xx
x
x
2
0
0 0
+π
=
ÏÌÔ
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) 17
176 3+ (b) 1
717
179 3 6+ -
(c) 17
173 9+ (d) 1
717
173 6 9+ -
9. Limn
k
n
n k nƕ
-
=
ÊËÁ
ˆ¯̃ + ( )
Ê
ËÁˆ
¯̃Âtan
tan1
1
1 11
has the value equal
to
(a) 1 12
+ ln(cos )
(b) 1 12
+ ln(sin )
(c) 1 1 12
- +ln(sin cos )
(d) 1 1 12
+ +ln(sin cos )
10. If f (x, y) = x2 + y2 – 2xy (x, y Œ R) and the matrix A is given by
A = f x y f x y f x yf x y f x y f x yf x y f
( , ) ( , ) ( , )( , ) ( , ) ( , )( , ) (
1 1 1 2 1 3
2 1 2 2 2 3
3 1 xx y f x y3 2 3 3, ) ( , )
È
Î
ÍÍÍ
˘
˚
˙˙˙
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)
= 21
t f t dtx
( )Ú , then which of the following does not
hold good? (a) f (p) = ep 2
(b) f (1) = e (c) f (0) = 1 (d) f (2) = 2 12. Let (a – 1)(x2 + 3x + 1)2 – (a + 1)(x4 – x2 + 1) £ 0
" x Œ R. Then which of the following is/are correct?
(a) a Œ-È
ÎÍ
˘
˚˙
13
43
,
(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 i a j k q( ) , ( )+ + = + + =1 22 and r ai a j k q( )+ + =2
3 intersect in a line, then the value of a is
(a) 1/4 (b) 1/2 (c) 1 (d) 2
14. If A = 13
1 2 22 1 2
2
1-È
Î
ÍÍÍ
˘
˚
˙˙˙
= -
a bA ATand then,
(a) a = –2 (b) a = 2 (c) b = –1 (d) b = 1
15. The value of a for which x x xx x x
a3 2
3 26 11 6
10 8 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)) = x2ya(a Œ R),
then find a. 17. If a, b, and c are real numbers such that a2 + 2b = 7,
b2 + 4c = –7, and c2 + 6a = –14, then find the value of (a2 + b2 + c2)/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
nn =
•Â 1
4.
19. Find the distance of the point P(3, 8, 2) from the line 12
1 14
3 13
2( ) ( ) ( )x y z- = - = - measured parallel
to the plane 3x + 2y – 2z + 15 = 0. 20. Consider the equation x2 + 2x – n = 0, where n Œ N and
n Œ [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 xx a x b x c
( )( ) ( ) ( )
,- - -
where f (x) is a
polynomial of degree < 3, then
(a) g x dxa f a x ab f b x bc f c x c
( )( ) log( ) log( ) log
Ú =---
111
∏ +
1
1
1
2
2
2
a a
b b
c c
k
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B.10 Mathematics
(b) dg xdx
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 xdx
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 dxa f a x ab f b x bc f c x c
( )( ) log( ) log( ) log
Ú =---
111
∏ +
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 dydx
yx
= 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) A2B = A2 (b) B2A = B2
(c) ABA = A (d) BAB = B 6. If the parabola y = (x2 + 4(b – c)x + 4a2)/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) = 31 2
1
tx
tdt
+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 TypeParagraph for Questions 9 and 10Let angles a, b, g of a triangle satisfy the relation,
sin sin sina b a g a-ÊËÁ
ˆ¯̃
+-Ê
ËÁˆ¯̃
+ ÊËÁ
ˆ¯̃
=2 2
32
32
.
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) scaleneParagraph for Questions 11 and 12
Let b xx
b xx x x
b Rcoscos
sin(cos sin ) tan
,2 2 1 32 2-
=+
-Œ
11. Equation has solutions if
(a) b Œ - •ÊËÁ
ˆ¯̃
- -ÏÌÓ
¸˝˛
, , ,12
1 0 13
(b) b Œ - •( ) - -ÏÌÓ
¸˝˛
, , ,1 1 0 13
(c) b Œ R – -ÏÌÓ
¸˝˛
1 0 13
, ,
(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 theseParagraph for Questions 13 and 14The 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 4 263
103
, ,- -ÊËÁ
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) 30Paragraph for Questions 15 and 16Consider two circles C1 : x
2 + y2 = r12 and C2 : x
2 + 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) 4r1
2 (b) 4r22
(c) 4r22 + 4r1
2 (d) None 16. The value of OA2 + OB2 + BC2 is (a) [5r2
2 – 3r12, 5r2
2 + r12]
(b) [4r22 – 4r1
2, 4r22]
(c) [4r12, 4r2
2] (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 y2 = 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
e xx
+ <≥
ÏÌÔ
ÓÔ
,,
00
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 – 3x2 – 12x in [–2, 5/2] is
(r) 4
(d) If lim ( ) ( )n
n n
n
Æ•
-+
-+
+
Ê
Ë
ÁÁÁÁ
ˆ
¯
˜˜˜˜
12 1
14 2
1
2 2
L
is
equal to k.p4
, then k is
(s) 16
20. Match the following:
Column I Column II
(a) If 1, a, a2, L a19 are 20th roots of unity, then 1 + a10 + a20 + L + a190 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) = x2e–2x, x > 0, then 12
ÊËÁ
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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B.12 Mathematics
Solutions of Mock Test 1
fi 2a –1 = 0 fi a = ½
4. Ans (c): m m1 234
34
= - =,
Since tangents are drawn on same branch and point P(3, 4) lies in obtuse angle between the asymptotes, therefore
tan q2
34
=
\ e ba
2 12
43
- = = =cot q
\ e =53
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 = bb
+4ah
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 – 2 < 0 and sin x < 0 fi x Œ -( , )2 0 7. Ans (d): A3 = B3 ...(1) and A2B = B2A ...(2) (1) – (2) gives A3 – A2B = B3 – B2A A2(A – B) = – B2(A – B) fi (A2 + B2) (A – B) = 0 det (A2 + B2) (A – B) = 0 det (A2 + B2) . det (A – B) = 0 fi (d) 8. Ans (a): Possible outcomes are 11 2 (C);113 (C);115 (C)
2 3 5p p p= = =
Hence, P(A) = 9216
124
=
9. Ans (b): Domain is x = ± 1. However, only x = 1 satisfies.
PaPer 1Section a
1. Ans (a): I = ln ( tan tan ( ))10
+ -Ú q qq
x dx
fi I = ln tan (tan tan )tan tan
11
0
+-
+ÊËÁ
ˆ¯̃Ú q q
q
qx
xdx
= ln tantan tan
11
2
0
++
Ê
ËÁˆ
¯̃Ú qq
q
xdx
fi I = ln ( tan ) ln ( tan tan )1 12
0 0
+ - +Ú Úq qq 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 2v u f
- =
fi dv u
df
1 1 2-Ê
ËÁˆ¯̃
=ÊËÁ
ˆ¯̃
fi - + = -1 1 22 2 2v
dvu
duf
df
fi 1 1 22 2 2v u f
df-ÊËÁ
ˆ¯̃
=a [Q du = dv = a]
fi a1 1 1 1 2
2v u v u fdf+Ê
ËÁˆ¯̃
-ÊËÁ
ˆ¯̃
=
fi a1 1 2 2
2v u f fdf+Ê
ËÁˆ¯̃
¥ = Q1 1 2v u f
- =È
ÎÍ
˘
˚˙
fi dff u v
= +ÊËÁ
ˆ¯̃
a1 1
Thus, relative error in f = a1 1u v
+ÊËÁ
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.
Limx
ax bx cxbx
bxcx
cxbc xÆ
-0 2
cos cos .cossin . sin . .
= 10 2bc
ax bx cxxx
LimÆ
-cos cos .cos
= 120bc
a ax b bx cxc cx bx
xxlim
sin sin cossin cos
Æ
- ++ ◊
= 12
2 2 2
bxb c a( )+ -
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 = g t dt( )-Ú 50
5
(Q f (t) = g(t – 5))
= g t dt( )50
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) + 2x2 + 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 tan sinx xx n
xx n2 2 2 2- -
=- -p p p p
for all value of x π np
+ p2
(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 ax bx dx= + =Ú ( )/1 2
0
4
8
fi 23
82
16 8a b◊ + ◊ =
fi 23
1a b+ = (2)
Solving (1) and (2), we get a = 3, b = –1.
Section C
16. Ans (2):
f(n) = 1 1
2
1
22 2 2n n n+
++
++
( ) ( )...
++ +
1
2 12n n
The terms of the sequence are decreasing and the number of terms are (2n + 2).
2 2
2 1
2 22 2
n
n nf n n
n
+
+ +£ £
+( )
Now Lim Limn n
n
n n
nn
nn n
Æ • Æ •
+
+ +=
+ÊËÁ
ˆ¯̃
+ +
2 1
2 1
2 1 1
1 2 12
2
( ) = 2
Similarly Lim Limn n
n
n
nnÆ • Æ •
+=
+=
2 1 2 1 22
( ) ( )
\ Limn
f xÆ •
=( ) 2
17. Ans (1): We have y = ax/(b – x)
fi dydx
b x a axb x
abb x
= - - --
=-
( ) ( )( ) ( )
12 2
dydx
abb
ÈÎÍ
˘˚̇
=-
=( , ) ( )1 1
212 (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 bb
--
11 2 = 2 fi b = 2 fi a = 2 – 1 = 1
Hence, a = 1, b = 2
18. Ans (5): AB i j k
= - = - - -b a 2 3 6 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 62 3 6 28
7
= 5 units
19. Ans (1): The given equation is | z |n = (z2 + z) | z |n – 2 + 1 fi z2 + z is real
fi z z z z2 2+ = +
fi ( ) ( )z z z z- + + =1 0
fi z z x z z x= = + + π π -as 1 0 1 2( / )
fi The given equation reduces to xn = xn + 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 zyz
y z z xzx
z x x yxy
- = - - = - - = -, ,
\ ( ) ( )( ) ( )( )( )( )
x y y z z x x y y z z xxyz
- - - = - - -2
fi xyz = 1
PaPer 2Section a 1. Ans (b, c): (1 – y)m (1 + y)n
= (1 – mC1 y + mC2 y2 – L) (1 + nC1 y + nC2 y
2 + L)
= + - +-
+-
-ÏÌÓ
¸˝˛
+1 12
12
2( ) ( ) ( )n m m m n n mn y L
Given: a1 = 10 fi a1 = n – m = 10 (1)
a2 = m n m n mn2 2 22
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 + 4x2 + 28x + 5) – 3(x4 + 6x2 + 25) = – 14x2 + 28x – 70 = – 14(x2 – 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 < 12
fi - < <12
12
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 13
24
35
0 9- ¥ ¥ = .
(b) P(E2) = 3P(BRR) = 3 23
14
25
0 2¥ ¥ ¥ = .
(c) P(E3) = P(RRR/(RRR » BBB))
= P RRRP RRR P BBB
( )( ) ( )+
= 0 1
0 1 23
34
45
.
. + ¥ ¥
= 0 10 1 0 4
0 2.. .
.+
=
(d) P(E4) = 1 – P(BBB) = 1 25
- = 0.6
5. Ans (a, b, c): We have bc ca abca ab bcab 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 02a b c c a b+ + = + + =, ,
w w
1 1 1 02c a b+ + =
w w 6. Ans (a, b, c, d): Odd + Odd – Even – Even = Even So the given number is divisible by 2. Also 2007201 – 1982201 is divisible by 2007 – 1982
= 25 2044201 – 2019201 is divisible by 2044 – 2019 = 25 Hence, the given number is divisible by 25. Also 2007201 – 2044201 is divisible by 2007 – 2044 = –37
and 2019201 – 1982201 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 + 2019201 is divisible by 2007 + 2019 = 4026 1982201 + 2044201 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
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 = 9C2 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 72 2 54 4 1
¸˝Ô
˛Ô
9 cases
1 1 62 2 43 3 2
¸˝Ô
˛Ô
9 cases
1 1 52 2 33 3 1
¸˝Ô
˛Ô
9 cases
1 1 4} 3 cases
1 1 32 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 = 369
123
=
Ans 13. (c), 14. (b):
g(x) = x x+ sin2
increasing function of x
range 02
, pÈÎÍ
˘˚̇
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\ f (x) = x x+ sin ,2
x Œ [0, p]
p £ t £ 2p, then f (t) + f (2p – t) = p
f (t) + 2 22
p p- + -t tsin ( ) = p
f (t) + p – t t2 2
-sin = p
f (t) = t t+ sin2
f (x) = x x+ sin2
p £ x £ 2p
f (x) = x x+ sin2
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 rs
= =+
=+
= -D 8
4 4 22
1 22 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 + y2 – 20x – 16y + 156 = 0
Section C 17. Ans: (a Æ q); (b Æ s); (c Æ r); (d Æ p) (a) lim
x Æ -1 f (g(x)) = f (2–) = 2
limx Æ +1
f (g(x)) = f (1–) = 1
Therefore, limit does not exist.
(b) limx Æ 2
f (x) = 2 fi lim ( )x
f xÆ
- =2
3 2 2
(c) limx Æ 0
f (x) = 0 and limx Æ 0
g(x) = finite quantity
\ lim ( )( )
( ) ( )x
f xg x
f x g xÆ
+ÊËÁ
ˆ¯̃
=0
0
(d) lim ( ) ( )( ) ( )
( )x
f x g xf x g xÆ +
-+
=-
+= =
1
3 3 1 11 1
22
1
18. Ans: (a Æ r), (b Æ r), (c Æ q), (d Æ s)
(a) arg zz
2
211
-+
Ê
ËÁˆ
¯̃ = 0 ; z π ± i
zz
zz
z z z z2
2
2
211
11
0 0-+
=-+
fi - = + =,
y = 0, x = 0 Locus of z is portion of pair of lines xy = 0.
Q zz
2
211
0-+
Ê
ËÁˆ
¯̃>
È
ÎÍÍ
˘
˚˙˙
(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 | z1 |2 = k2 – k1
x2 – y2 + 2 ixy – il1 = l2
x2 – y2 = l2 and xy = l1
2 The locus of z is the point of intersection of
hyperbola. (d) Given,
z z- - + + - =- -1 13
13 2
11 1sin cos p
Since 1 13
13 2
11 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 csin sin sin30 45 105∞
=∞
=∞
fi a b c1 2 1 2 3 1 2 2/ / ( )/
= =+
fi a b c k2 2 3 1
= =+
= (say)
Smallest side, a = 2 k and largest side, c = ( )3 1+ k
If a k c= = = +2 2 6 2, ,
If a k c= = = +2 1 3 1, ,
If s k c= + + = = +3 2 3 2 2 2 3, ,
If D =-3 1
4, then from D = 1
2 bc sin A, we have
3 14
12
2 3 1 30-= ¥ ¥ + ∞k k( ) sin
fi k k223 1
2 3 13 1
43 12
=-+
=-
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) = 115
(i = 0, 1, ... 14)
P(A/Ei) = 0 for i = 0, 1, 2, 3, 4
P(A/Ei) = i C
C5
145
for i ≥ 5
(a) P(A) = P P( ) ( / )E A Ei ii =Â
0
14
= 115
114
5C (5C5 + 6C5 + L + 14C5)
= 115
16
156
145
CC
=
(b) Clearly P(A/E11) = 11
514
5
313
CC
=
(c) From Baye’s theorem,
P(A/E11) = P PP
( ) ( / )( )
.E A EA
11 11
115
313
16
665
= =
(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 iC C
C3
142
145
-
for i = 3, 4, ..., 12
\ P(B) = P P( ) ( / )E B Ei ii =Â
0
14
= 115
114
5.
C[3C3.
11C2 + 4C3.10C2 + L+ 12C3.
2C2]
= 500515
1614
5. C=
Solutions of Mock Test 2
PaPer 1Section a
1. Ans (b): f (x) = ( ) ( )x n n n
n
- -
=’ 51
1
50
ln f (x) = n n x nn
( ) ln ( )511
50
- -=
Â
Differentiating both sides with respect to x, we get
¢ = --
=Âf x
f xn n
x nn
( )( )
( )51
1
50
¢ = --
=Âf
fn n
nn
( )( )
( )5151
5151
1
50
=1 + 2 + 3 + L + 50 = 1275
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or ff
( )( )5151
11275¢
=
2. Ans (c): y2 = 4ax, 4a = 2p > 0 and t1t2 = –1.
Ratio = 4 21 2
212
22
a t ta t t
= – 4
3. Ans (a):
Tk = cos .( )
( ) ( )( )( )
-
++
- + ++
Ê
ËÁ
ˆ
¯˜
1 1 11
1 1 21k k
k k k kk k
Let x = 1k
and y = 1
1k +
1 1 11
22- = -
+y
k( )
=+ -+
=++
( ) ( )kk
k kk
1 11
21
2
Tk is in the form of
cos .- + - -( )1 2 21 1xy x y = cos–1(y) – cos–1(x) (Q y < x)
Tk = cos cos- -
+ÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃
1 111
1k k
Substituting k = 2, 3, 4, ...
Sum = Limn nÆ •
- -
+ÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃
cos cos1 111
12
= cos ( ) cos- -- ÊËÁ
ˆ¯̃
1 10 12
Sum = p p p2 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 cosec2 x} for g(x) to be decreasing, g¢(x) < 0 fi f ¢{(cot x + 1)2 + 1} ◊ (–2 cosec2 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
" ŒÊËÁ
ˆ¯̃
» ÊËÁ
ˆ¯̃
x 0 34
34
, ,p pp
Thus, equation (i) holds, if cot x + 1 > 0
fi cot ,x x> - " ŒÊËÁ
ˆ¯̃
1 0 34p
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 1 1 133
100100
13
3+ +ÊËÁ
ˆ¯̃
È
ÎÍ
˘
˚˙ = + +Ê
ËÁˆ¯̃
xx
C xx
+ +ÊËÁ
ˆ¯̃
+ + +ÊËÁ
ˆ¯̃
1002
33
2100
1003
3
1001 1C xx
C xx
L
= (1 + r) + A1x3 + A2x
6 + L + A100(x3)100 + B1
13x
+ L + B100 13
100
xÊËÁ
ˆ¯̃
All other terms obtained by the combination of x3 and 1/x3 well get converted into a term involving x3 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 22a
Fig. B.4
So 2 2 1a
a£ fi ≥
and 2 1 2a
a> fi <
\ a Œ [1, 2) 8. Ans (a): f (x) = | sin a x | + | cos a x |
Period of f (x) = pa
p2 8
= (given)
fi a = ± 4 9. Ans (d): Given integral
= xf x f xx f x
dx x f x xf x x
f x xdx¢ -
= ¢ -ÚÚ ( ) ( )( )
( ( ) ( ))/
( )/
2 22
2 4
2
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Appendix B: Mock Tests B.19
Let t = f xx
dtdx
x f x x f xx
( ) ( ) ( )2
2
42
fi = ¢ -
Therefore, the required integral is
= 1 221 2
tdt t c
f xx
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 negative.
In 3rd quadrant, i.e., 54
32
p p,ÊËÁ
ˆ¯̃
, if tan x < cot x,
fi tan2x < 1, which is false. Hence, (a) can be correct.
Now sin x < cos x is true in 04
, pÊËÁ
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) = e dt e dtt tx
tx
-Ú Ú=[ ] { } ,0 0
So f (x) =
e dt x
e dt e dt x
e dt e dt
tx
t tx
t t
0
0
11
1
0
11
1
2
0 1
1 2
Ú
Ú Ú
Ú
Œ
+ Œ
+
-
-
if
if
( , )
( , )
ÚÚ Ú+ Œ
È
Î
ÍÍÍÍÍÍÍÍÍÍÍ
-e dt xtx
2
2
2 3if ( , )
⇒ 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
ÌÔÔ
ÓÔÔ
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): x xn
1 2 1 412
/ /+ÊËÁ
ˆ¯̃
-
T C x xrn
r r
n r r
+ = ¥ ¥- -
12 4
12
Coefficient of the first three terms are nC0, nC1
12
¥ ,
nC2 212
¥
\ n n nC C C0 2 114
2 12
+ ¥ = ¥ ¥
fi 1 18
+-
=n n n( )
fi n n n( ) ( )-= -
18
1
fi n = 8 (as n π 1)
\ T C x xr r r
r r
+-= ¥ ¥
-
18
82 4
12
= ¥ ¥-( )8 41
2
34C xr r
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 b(0, )
D d(0, )
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 dbc ad
bd c abc ad
( ) , ( )--
--
ÊËÁ
ˆ¯̃
Let the co-ordinates of the point of intersection be (h, k). Then
h ac b dbc ad
k bd c abc ad
=-
-=
--
( ) , ( )
Given c2 + a2 = b2 + d2
(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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14. Ans (a, b, d): p = 5C4 ◊ 2C1 = 10
q = 5C2(2C1)
3 = 80 r = 5C0(
2C1)5 = 32
fi 2q = 5r, 8p = q, and 2(p + r) > q 15. Ans (a, c, d): From Fig. B.6,
Ay x dy
dxdydx
=- +Ê
Ë
ÁÁÁ
ˆ
¯
˜˜˜
, 0
O B x( , 0) A
P x y( , )
Fig. B.6
As per the given condition,
2x dydx
y x dydx
= - +
fi x dydx
y= -
fi dxx
dyy
= -
fi ln x = –ln y + ln c fi xy = c Also, f (1) = 1 fi c = 1 fi xy = 1
Section C
16. Ans (4): I = Limt
x t xt
dxÆ
+ -Ú00
2sin ( ) sin
p
= Limt
x t t
tdt
Æ
+ÊËÁ
ˆ¯̃
¥Ê
Ë
ÁÁÁÁ
ˆ
¯
˜˜˜˜
Ú 00
2 22 2
cos sinp
= cos x dx0
2p
Ú = 4
17. Ans (9): a = 1 – b – c fi 1 + a = (1 – b) + (1 – c) > 2 ( )( )1 1- -b c
Similarly, 1 + b > 2 ( )( )1 1- -c a
and 1 + c > 2 ( )( )1 1- -a b Required expression > 8.
18. Ans (5): f (x) = sin [ ] [ ]- -Ê
ËÁˆ
¯̃+ + -1
2 23
x x x x
sin- -Ê
ËÁˆ
¯̃1
2 23
x x is defined for
- £-
£ + -1 23
12x x x xand [ ] [ ] defined only
for integral values of x. fi x = – 1, 0, 1, 2, 3 Therefore, total number of element in domain is 5. 19. Ans: Since the produce matrix is 3 ¥ 3 matrix and
the premultiplier of A is a 3 ¥ 2 matrix, A is a 2 ¥ 3 matrix.
Let A =1 m nx y z
È
ÎÍ
˘
˚˙ , then the given equation becomes
2 11 03 4
11 8 101 2 59 22 15
-
-
È
Î
ÍÍÍ
˘
˚
˙˙˙
È
ÎÍ
˘
˚˙ =
- - -- -
È
Î
ÍÍÍ
˘
˚
˙˙˙
m nx y z
fi 2 3 2
3 4 3 4 3 4
l x m y n zl m x
l x m y n z
- - -
- + - + - +
È
Î
ÍÍÍ
˘
˚
˙˙˙
=- - -
- -È
Î
ÍÍÍ
˘
˚
˙˙˙
1 8 101 2 59 22 15
fi 2l – x = –1, 2m – y = –8, 2n – z = –10, l = 1, m = –2, n = –5 fi x = 3, y = 4, z = 0, 1 = 1, m = –2, n = –5
fi Am n
x y z=
È
ÎÍ
˘
˚˙ =
- -È
ÎÍ
˘
˚˙
1 1 2 53 4 0
20. Ans (9): g(x) = 12
tan–1 | x | + 1 fi sgn (g(x)) = 1
\ sin23 x – cos22 x = 1 fi sin23 x = 1 + cos22 x, which is possible if sin x = 1
and cos x = 0.
fi x n= +22
pp
Hence, - £ + £10 22
8p pp
pn
fi - £ £214
154
n fi –5 £ n £ 3
Hence, the number of values of x = 9.
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Appendix B: Mock Tests B.21
Also, g(x) is monotonic.
Also f xg x
dx x xx
dx( )( )
/Ú Ú=- 1
3
= -ÊËÁ
ˆ¯̃
= - + +Ú 1 1 1 32 4 3x x
dxx x
c
3. Ans (b, c, d): Let (x1, y1) ∫ (at2, 2at) Tangent at this point is ty = x + at2.
Any point on this tangent is h h att
, +Ê
ËÁˆ
¯̃
2
.
Chord of contact of this point with respect to the circle
x2 + y2 = a2 is hx h att
y a++Ê
ËÁˆ
¯̃=
22
or (aty – a2) + +ÊËÁ
ˆ¯̃
=h x yt
0
which is a family of straight lines passing through
the point of intersection of ty – a = 0 and x yt
+ = 0
So, the fixed point is -ÊËÁ
ˆ¯̃
at
at2 , . Therefore,
x at
y at2 2 2= - =,
Clearly, x1x2 = – a2, y1 y2 = 2a2
Also, xx
t yy
t1
2
4 1
2
22= - =,
fi 4 01
2
1
2
2xx
yy
+ÊËÁ
ˆ¯̃
=
4. Ans (a, b, c): Here, let P be (x, y, z). Then, x = r sin q cos f, y = r sin q sin f, z = r cos q fi 1 = r sin q cos f, 2 = r sin q sin f, 3 = r cos q (i) fi 12 + 22 + 32
1. Ans (a, c): If p = q, then the equation becomes zp = z p and it has infinite solution because any z Œ real will satisfy it.
If p π q, let p > q, then zp = z q
fi | z |p = | z |q
fi | z |p (| z |p – q – 1) = 0 fi | z | = 0 or | z | = 1 | z | = 0 fi z = 0 + i0 | z | = 1 fi z = eiq
fi e(p + q)qi = 1 fi z = 11/(p + q)
Hence, the number of solution is p + q + 1. 2. Ans (a, c, d):
Let I = ( )( )xx
dx4
611
++Ú
= ( )( )( )
x xx x x
dx2 2 2
2 4 21 2
1 1+ -
+ - +Ú = ( )
( ) ( )x dx
x xx dxx
2
4 2
2
61
12
1+
- +-
+ÚÚ
= 1 1
1 12
1
2
22
2
3 2
+ÊËÁ
ˆ¯̃
- +ÊËÁ
ˆ¯̃
-+ÚÚ x
dx
xx
x dxx( )
In first integral, put xx
t- =1
Therefore, 1 12+Ê
ËÁˆ¯̃
=x
dx dt and in second integral,
put x3 = u.
\ x dx du2
3= , then
I = dtt
duu1
23 12 2+
-+ÚÚ
= tan tan- -- +1 123
t u c
= tan tan ( )- --ÊËÁ
ˆ¯̃
- +1 1 31 23
xx
x c
Here f x xx
g x x( ) ( )= - =1 3and
Both the functions are odd.
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B.22 Mathematics
x
y
M
O
Z
rq
f
P(1, 2, 3)
Fig. B.7
5. Ans (b, d):
Let a i j b i j= - - = -
32
3 32
3( ), ( )
c i j d i j= - + = +
32
3 32
3( ), ( )
AB i
= =3 a (say)
Clearly, a b c d= = = = 3
If a makes angle q and f, y and b with
a b c, , and d , respectively, then
cos /.
qa
a=
◊=
-= -
aa
9 23 3
12
cos /.
fa
a=
◊= =
bb
9 23 3
12
\ f = 60°
cos /.
ya
a=
◊=
-= -
cc
9 23 3
12
cos /.
ba
a=
◊= =
dd
9 23 3
12
\ b = 60° 6. Ans (a, b, d): We have P = probability that the bomb
strikes the target = 1/2. Let n be the number of bombs which should be
dropped to ensure 99% chance or better of completely destroying the target. Then the probability that out of n bombs at least two strike the target is greater than 0.99.
Let X denote the number of bombs striking the target.
Then P(X = r) = nr
r n rn
r
n
C C r12
12
12
ÊËÁ
ˆ¯̃
ÊËÁ
ˆ¯̃
= ÊËÁ
ˆ¯̃
-
;
= nr
r n rn
r
n
C C r12
12
12
ÊËÁ
ˆ¯̃
ÊËÁ
ˆ¯̃
= ÊËÁ
ˆ¯̃
-
; = 0, 1, 2, ..., n
We should have P(X ≥ 2) ≥ 0.99
fi {1 – P(X < 2)} ≥ 0.99 fi 1 – {P(X = 0) + P(X = 1)} ≥ 0.99
fi 1 1 12
0 99- +ÏÌÓ
¸˝˛
≥( ) .n n
fi 0 001 12
. ≥+ n
n
fi 2n > 100 + 100 n fi n ≥ 11 Thus, the minimum number of bombs is 11. 7. Ans (b, c): x2 – (a + b)x + ab = 0 or (x – a)(x – b) = 0 fi x = a or b If x = a is the root of other equation, then a3 – ap + ab = 0 fi p = a2 + b If x = b is the root of the other equation, then ab2 – pb + ab = 0 or p = a(1 + b) 8. Ans (b, d): f (0, t) = 0 for t ≥ 0, so
g(0) = f t dt( )◊ ◊ =Ú 0 00
1
f (1, t) = t (1 – 1) = 0 for t < 1; so g(1) = 0
Also g(x) = f t t x dt f t x t dtx
x( ) ( ) ( ) ( )- + -Ú Ú1 1
0
1
= ( ) ( ) ( ) ( )x t f t dt x f t t dtx
x- + -Ú Ú1 1
0
1
\ g¢(x) = t f t dt f t dtx
( ) ( )0
1 1
Ú Ú-
g≤(x) = f (x)
Section B
9. Ans (a) 10. Ans (c) y = (x – a) (x2 + bx + c) for y = 0, x = a \ p = a fi The point P is (a, 0). Let the slope of the line through P be m. Therefore, equation of the line is y = m (x – a). It meets the curve in A and B. Therefore, m (x – a) = (x – a) (x2 + bx + c)
dydx
= x2 + bx + c + (x – a) (2x + b)
= x2 + bx + c + 2x2 – 2ax + bx – ab = 3x2 + (2b – 2a) x + c – ab Therefore, slope of the tangent at P(a, 0) is a2 + ab + c. Therefore, equation of the tangent y – 0 = (a2 + ab + c) (x – a)
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It meets the curve at C. \ (a2 + ab + c) (x – a) = (x – a) (x2 + bx + c) i.e., a2 + ab + c = x2 + bx + c i.e., x2 + bx – a2 – ab = 0 i.e., (x – a) (x + a) + b (x – a) = 0 i.e., (x – a) (x + a + b) = 0 Therefore, absicca of C is – a – b. Since the line through P intersects at two different
points A and B, therefore, b2 – 4 (c – m) > 0 ...(i) Now if the slope of any tangent is m, then 3x2 + (2b – 2a) x + c – ab = m Its D = 4(b – a)2 – 12(c – ab – m) = 4[a2 + b2 – 2ab – 3c + 3ab + 3m] = 4[a2 + b2 + ab + 3m – 3c] ...(ii) From (i) and (ii), we get
D a b ab b> + + -ÈÎÍ
˘˚̇
4 34
2 2 2
= 44
22
a b ab+ +È
ÎÍÍ
˘
˚˙˙
= 4a2 + b2 + 4ab = (2a + b)2 ≥ 0 \ D > 0 Hence, there are 2 points.
11. Ans (b): f x dx( ) =Ú 50
2
= shaded region PSTR
y f x= ( )–1
T M
A3
A2
A4
y f x= ( )
RA1
Q
S
P
Fig. B.8
Area of PQR = 5 – Area QRST = 5 – 4 = 1
= Area RMT = f x dx-Ú 1
2
4
( ) = 1
12. Ans (a): ( ( ) ( ))f x f x dx A A- = fi = =-Ú 11 2
Center is the mid point of foci which is (1, 0). Therefore, vertices are (0, 0) and (2, 0).
C2 : Arg (z – 2i) = -p4
represents a ray originating
from (0, 2) and making an angle, if 135° with the positive x-axis.
C3 : | z – 1 – i | = 2 which is the circle having center at (1, 1) and radius
2 .
-1 1 2 3
–1
1
2
3
C1 C3
C2
Fig. B.11
From Fig. B.11, there is only one point common to all curves which is (2, 0).
16. Ans (a): From the above diagram, C1 is symmetrical about x-axis.
So, C4 will be the same as C1.
Section C
17. Ans: (a Æ r); (b Æ p); (c Æ s); (d Æ q) (a) Put px = qy = rz = t =1
(b) Coefficient of x103 in 11
5 199--
Ê
ËÁˆ
¯̃xx
(x – 1)201
= Coefficient of x103 in – (1 – x)2 (1 – x5)199
= Coefficient of x103 in – (1 + x2 – 2x) (1 – 199C1x
5 + 199C2x10 + L) = 0
(c) 22003 = 23 (17 – 1)500 = 8 (17k + 1); k ŒI \ Remainder = 8 (d) Therefore, 111111 is divisible by 1001 com-
pletely (6 one’s) So 6 ¥ 166 = 996 ones are divisible. Now remaining ones are 11111 when divisible
by 1001 gives 100 as remainder 18. Ans: (a Æ r), (b Æ p), (c Æ s), (d Æ q)
(a) y mx a m b= ± +2 2 2
y x= - ± ¥ +43
18 169
32 y x= - ±43
8
0
A
Bx
yP
Fig. B.12
Distance between tangent = 16
1 169
+
= 16 35
485
¥=
(b) y x= - +43
8 A(6, 0) and B(0, 8)
Area of DAOB = 12
6 8 24¥ ¥ =
(c) Point of contact
-+ +
Ê
ËÁÁ
ˆ
¯˜˜
a m
a m b
b
a m b
2
2 2 2
2
2 2 2,
Product of coordinates
= -+
a b ma m b
2 2
2 2 2 = -¥ ¥ -Ê
ËÁˆ¯̃
18 32 43
64 = 12
(d) 4x + 3y = 24 \ l = 424
m = 324
\ l m+ =724
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19. Ans: (a Æ p); (b Æ r); (c Æ p); (d Æ q)
(a) log .0 55 2 0-Ê
ËÁˆ¯̃
≥x
x
0 5 2 1 53
52
<-
£ fi ŒÊËÁ
ˆ¯̃
xx
x ,
2 is the only integer in the domain. (b) L.H.S. of given equation is f ¢(x) where f (x) = (x – 1) (x – 2) (x – 3)(x – 4) Because f (x) is continuous and derivable and f (1) = f (2) = 0 Therefore, f ¢(x) = 0 has atleast one root in (1, 2). Similarly f ¢(x) = 0 has atleast one root in (2, 3)
and (3, 4). Because f ¢(x) is a cubic function, hence it has
exactly one root in (1, 2), (2, 3), (3, 4). (c) Applying LMVT in [0, 1]
e --
11 0
= ec for some c Œ (0, 1)
ec + 1 = e fi ln (ec + 1) = 1
(d)
2 110
12
1 2 110
12
xx
xx
- +
- - . = 1
fi 4x2 – 20x + 9 = 0 or (2x – 9) (2x – 1) = 0
or x = 12
92
,
At x = 12
, tan–1(0) + tan–11 = p4
(true)
At x = 92
,
tan tan tan- - -ÊËÁ
ˆ¯̃
+ = =1 1 145
19
14p (true)
So two values of x satisfy the above equation. 20. Ans: (a Æ p); (b Æ p); (c Æ q); (d Æ r)
(a) I = x xx
dxln | |( )1 2 2
0 +
•
Ú
xz
I zz
zz
dz= =+Ê
ËÁˆ¯̃
¥-
-
•Ú1
1
1 11
1
2
2 2
0
,log ( )
I = log ( )( )
zz
z dz1 2 2
0
+¥
•Ú = x x
xdxln
( )1 2 2
0
+•Ú = -
+
•
Ú x xx
dxln( )1 2 2
0
I + I = 0 or 2I = 0 or I = 0
(b) I = ln tan/
x dx0
2p
Ú = ln cot
/x dx
0
2p
Ú 2I = ln (tan cot )
/x x dx¥Ú0
2p
= 0 or I = 0
(c) I = xdx
x x3 61
1
1+ +-Ú = x x x dx1 6 3
1
1+ -( )
-Ú = x x dx x dx1 6 4
1
1
1
1+ -
Ø-- ÚÚ
Zero
I = - = -Ú2 2 54
0
1x dx /
(d) [ ] ( )x dx dx dx= - + ÚÚÚ --1 0
0
1
1
0
1
1
= – [x]0–1 = – [0 – (–1)] = –1
Solutions of Mock Test 3
PaPer 1Section a
1. Ans (c): Y – y = dydx
(X – x)
Therefore, the coordinates of T are x -ÊËÁ
ˆ¯̃
y dxdy
, 0 .
Therefore, the sum of subtangent and abscissa
= y dxdy
+ x = a
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\ y dxdy
= a – x
i.e., dxa x-
= dyy
\ –log (a – x) = log y + log C
or cy = 1a x-
Curve passes through (2a, a)
or c = – 12a
Therefore, equation of the curve is (x – a) y = a2. 2. Ans (b): Differnciate both sides w.r.t. x, f ¢(x) = cos x + f ¢(x) (2 sin x – sin2 x)
f ¢(x) = cossin sin
xx x1 22+ -
= cos( sin )
xx1 2-
Integrating,
f (x) = cos( sin )
xdxx1 2-Ú (Put 1 – sin x = t)
= - = + =-
+Údtt t
Cx
C2
1 11 sin
Also f (0) = 0, hence C = –1.
f (x) = 11- sin x
–1 = 1 11- +
-sin
sinx
x = sin
sinx
x1- 3. Ans (b): Six places can be selected in 9C2 ways and 6
can be placed only at 5 places except the right most of other 6 selected. Remaining numbers, i.e., 7, 8, 9 in 3! ways.
Hence, number of number 9C6 . 5 . 3! = 84 . 30 = 2520 4. Ans (c): z z z z z z1 1 2 2 3 31 1 1= = =; ; Given z1 + z2 + z3 = 0 fi z z z1 2 3+ + = 0
or (z1 + z2 + z3)2 = 0
z12Â + 2z1z2z3
1 1 1
1 2 3z z z+ +
È
ÎÍ
˘
˚˙ = 0
or z12Â + 2z1z2z3 [ z z z1 2 3+ + ] = 0
\ z12Â = 0
5. Ans (a):
A = 2a bcos sinq q2
= ab sinq cos q = absin 22
q
Hence, A is maximum if q = p/4 and Amax = ab/2. Also sample space = pab/2.
Therefore, favourable and come = pab2
– ab2
= ab2
(p – 1)
Probability = ab2
. ( )pp-1 2ab
= pp-1
6. Ans (c): If x = r cosq and y = r sin q, then E = x2 + y2 = r2. Hence, we have to minimize r2. Now in the given equation substituting x = r cosq and y = r sinq,
we get r2 = 4 cosec 4q fi r 2 4min = 7. Ans (d): As x Æ 0– (i.e., approaches 0 from the left),
[x] = –1
\ limxÆ -0
f (x) = limxÆ -0
1 11
+ --
sin ( ) = –1 + sin 1
whereas if x Æ 0+, we get [x] = 0. \ f (x) = 0 fi lim
xÆ +0 f (x) = 0
Thus, limxÆ0
f (x) does not exist.
8. Ans (d): For symmetric matrix, each place in upper triangle and leading diagonal can be filled in seven ways. Then the number of symmetric matrices are 76.
For skew symmetric matrix, leading diagonal ele-ments are zero.
11. Ans (a, b, c, d): Q f ¢(x) = 2x f (x) fi ln f (x) = x2 + c fi f (x) = ex2 f (x) = lex2
Q f (1) = 0 fi 0 = le fi l = 0 Hence, f (x) = 0, x Œ R 12. Ans (c, d): (a – 1)(x2 + 3x + 1)2
– (a + 1) [(x2 + 1)2 – ( ) ]x 3 2 ≤ 0 or (a – 1) (x2 + 3x + 1)2
– (a + 1) (x2 + x 3 + 1) (x2 – x 3 + 1) ≤ 0 (x2 + 3x + 1) [(a – 1) (x2 + 3x + 1) – (a + 1) (x2 – 3x + 1)] £ 0, " x Œ R or –2(x2 + 1) + 2 3a x £ 0 or x2 – a x3 + 1 ≥ 0, " x Œ R or 3a2 – 4 £ 0 (D £ 0)
or a Œ -È
ÎÍ
˘
˚˙
23
23
,
Therefore, number of possible integral value of a is {– 1, 0, 1} fi 3 fi (c) and the sum of all integral values of a is –1 + 0 + 1
= 0 fi (d) 13. Ans (b, c):
r n q r n q r n q◊ = ◊ = ◊ =1 1 2 2 3 3, , intersect in a line if [ ]
n n n1 2 3 0= . So,
1 1 11 2 1
1
02
a
a a
=
or 0 0 1
1 2 2 1 1
1 1
02 2
- -
- -
=a a
a a a or (1 – 2a) (a2 – 1) – (a – a2) (2a – 1) = 0
or a =12
1,
14. Ans (a, c): AAT = I
fi 13
1 2 22 1 2
2
13
1 22 1 22 2
1 0 00 1 00 0 1
-È
Î
ÍÍÍ
˘
˚
˙˙˙◊
-
È
Î
ÍÍÍ
˘
˚
˙˙˙
=È
Î
Í
a b
a
bÍÍÍ
˘
˚
˙˙˙
or 19
1 2 22 1 2
2
1 22 1 22 2
1 0 00 1 00 0 1
-È
Î
ÍÍÍ
˘
˚
˙˙˙ -
È
Î
ÍÍÍ
˘
˚
˙˙˙
=È
Î
ÍÍÍ
˘
a b
a
b ˚̊
˙˙˙
or 9 0 4 20 9 2 2 2
4 2 2 2 2 42 2
a ba b
a b a b a b
+ ++ -
+ + + - + +
È
Î
ÍÍÍ
˘
˚
˙˙˙
= 9 0 00 9 00 0 9
È
Î
ÍÍÍ
˘
˚
˙˙˙
or a + 4 + 2b = 0, 2a + 2 – 2b = 0 and a2 + 4 + b2 = 9 or a + 2b + 4 = 0, a – b + 1 = 0 and a2 + b2 = 5 or a = –2, b = –1
15. Ans (b, c, d): Let f x a( ) + =30
0
where
f (x) = x x xx x x
x x xx x x
3 2
3 26 11 6
10 81 2 31 2 4
- + -+ - +
=- - -- - +
( ) ( ) ( )( ) ( ) ( )
= ( ) , , ,xx
x-+
π -34
1 2 4
Range of f (x) = R - - -ÏÌÓ
¸˝˛
1 25
16
, ,
So (1) does not have solution if
a30
1 25
16
= - , ,
a = –30, 12, 5
Section C
16. Ans (4): f (0, •) Æ (0, •) f (x f (y)) = x2ya(a Œ R) Put x = 1, we get f ( f (y)) = ya
Put f (y) = 1x
, we get f (1) = 12( ( ))f y
ya
and put y = 1, we get ( f (1))3 = 1 f (1) = 1 For y = 1, we have f (x( f (1))) = x2
f (x) = x2
Thus, a = 4 17. Ans (7): a2 + 2b = 7 b2 + 4c = –7 c2 + 6a = –14 a2 + b2 + c2 + 2b + 4c – 6a = –14 (a + 3)2 + (b + 1)2 + (c + 2)2 – 14 = –14 fi a = –3, b = –1, and c = –2. or a2 + b2 + c2 = 14
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B.28 Mathematics
18. Ans (8): For the GP a, ar, ar2, ... Pn = a, (ar), (ar2), ..., (arn – 1) = anrn (n – 1)/2
S = P arnn
n
n
n=
•-
=
•
 Â=1
1 2
1
( ) /
Now,
ar a r r r rn
n
( ) /-
=
•= + + + + + •È
΢˚ 1 2
11 L
= ar1 -
Given a = 16 and r = 1/4
\ S = 161 1 2- ( / )
= 32
19. Ans (7): Let general point of line be A(2l + 1, 4l + 3, 3l + 2). Let this point lies at the same distance as point P(3,
8, 2) from the plane 3x + 2y – 2z + 15 = 0
Therefore, 3 3 2 8 2 2 1517
. . .+ - +
= 3 2 1 2 3 3 2 3 2 1517
( ) ( ) ( )l l l+ + + - + +
fi 36 = 8l + 20 fi l = 2
P(3, 8, 2)A
Fig. B.13
Therefore, A is (5, 11, 8)
PA = ( ) ( ) ( )5 3 11 8 8 22 2 2- + - + -
= 4 9 36 7+ + = .
20. Ans (8): x2 + 2x – n = 0 fi (x + 1)2 = n + 1 fi x = –1 ± n + 1 Thus, n + 1 should be a perfect square. Since m Œ [5, 100] fi n + 1 fi [6, 101] Perfect square values of n + 1 are 9, 16, 25, 36, 49,
64, 81, 100. Hence, the number of values is 8.
PaPer 2Section a
1. Ans (a, b): By partial fractions, we have
g x f ax a a b a c
( ) ( )( ) ( ) ( )
=- - -
+- - -
+- - -
f bb a x b b c
f cc a c b x c
( )( ) ( ) ( )
( )( ) ( ) ( )
fi g xa b b c c a
( )( ) ( ) ( )
=- - -
1
¥-
-+
--
+-
-È
ÎÍ
˘
˚˙
f a c bx a
f b a cx b
f c b ax c
( ) ( )( )
( ) ( )( )
( ) ( )( )
fi g xa f a x ab f b x bc f c x c
a a
b b
c c
( )( )/( )( )/( )( )/( )
=---
∏111
1
1
1
2
2
2
fi g x dxa f a x ab f b x bc f c x c
( )( ) log( ) log( ) log
Ú =---
111
∏ +
1
1
1
2
2
2
a a
b b
c c
k
and df xdx
a f a x a
b f b x a
c f c x a
a a
b b
c
( )( ) ( )
( ) ( )
( ) ( )
=
-
-
-
∏
-
-
-
1
1
1
1
1
1
2
2
2
2
2
cc2
=
1
1
1
1
1
1
2
2
2
2
2
2
a f a x a
b f b x a
c f c x a
a a
b b
c c
( ) ( )
( ) ( )
( ) ( )
-
-
-
∏
-
-
-
2. Ans (a, b, c, d): A Õ A » B,
fi P(A) £ P(A » B) fi P(A » B) ≥ 34
Also, P(A « B) = P(A) + P(B) – P(A » B) ≥ P(A) + P(B) – 1
= 34
58
1 38
+ - =
Now, A « B Õ B
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Appendix B: Mock Tests B.29
fi P(A « B) £ P(B) = 58
\ 38
58
£ « £P A B( )
Next, P(A « B¢) = P(A) – P(A « B)
fi 34
58
34
38
- £ « ¢ £ -P A B( )
fi P(A « B) = P(B) – P(A¢ « B) (Using (2))
\ 38
58
£ - ¢ « £P B P A B( ) ( )
fi 0 14
£ ¢ « £P A B( )
Hence, all the options are correct. 3. Ans (a, c): | PS1 – PS2 | = 2a
Comparing with y = ae– 1/x + b, a Œ R, b = 0. 5. Ans (a, b, c, d): We have A2B = A(AB) = AA = A2, B2A = B(BA) = BB = B2, ABA = A(BA) = AB = A, and BAB = B(AB) = BA = B 6. Ans (b, d): Since x-axis touches the given parabola, the equation
14
02 2x b c x a+ - + =( ) must have equal roots.
fi (b – c)2 – a2 = 0 fi (c – a – b)(c + a – b) = 0 Hence, the given line passes through the points
(–1, –1) and (1, –1). 7. Ans (a, d):
¢ =+
> " >f xx
xx
( ) 31
0 02
=+
>+
" >3
11
112 2
x
x xx,
fi ¢ >+Ú Úf x dx
xdx
x x
( )1
21
11
fi f (x) > tan–1 x – tan–1 1 fi f (x) + p/4 > tan–1 x 8. Ans (a, b, c, d):
Exponent of 2 = 102
102
1022 3
ÈÎÍ
˘˚̇
+ ÈÎÍ
˘˚̇
+ ÈÎÍ
˘˚̇
= 5 + 2 + 1 = 8
Exponent of 3 = 103
1032
ÈÎÍ
˘˚̇
+ ÈÎÍ
˘˚̇
= 3 + 1 = 4
Exponent of 5 = 105
ÈÎÍ
˘˚̇
= 2
Exponent of 7 = 107
ÈÎÍ
˘˚̇
= 1
Number of divisors of 10 is (8 + 1) (4 + 1) (2 + 1) (1 + 1) = 270.
Number of ways of putting N as a product of two natural numbers is 270/2 = 135.
Section B
Ans 9. (b), 10. (c): a + b + g = p
sin sin sina b a g a-ÊËÁ
ˆ¯̃
+-Ê
ËÁˆ¯̃
+ ÊËÁ
ˆ¯̃
=2 2
32
32
fi sin( )
sin( )
sin ( )
p b g b p g b g
p b g
- +( ) -ÊËÁ
ˆ¯̃
+- +( ) -Ê
ËÁˆ¯̃
+- +Ê
2 2
3 32ËËÁ
ˆ¯̃
=32
fi cos cos cos ( )22
22
32
32
b g g b b g+ÊËÁ
ˆ¯̃
++Ê
ËÁˆ¯̃
++Ê
ËÁˆ¯̃
=
fi 2 34 4
1 2 34
32
2
cos ( ) cos
cos ( )
b gb g
b g
+ÊËÁ
ˆ¯̃
-ÊËÁ
ˆ¯̃
+ - +ÊËÁ
ˆ¯̃
=
fi 2 34 4
2 34
12
2
cos ( ) cos
cos ( )
b gb g
b g
+ÊËÁ
ˆ¯̃
-ÊËÁ
ˆ¯̃
- +ÊËÁ
ˆ¯̃
=
fi 4 34
4 34 4
1 0
2cos ( )
cos ( ) cos
b g
b gb g
+ÊËÁ
ˆ¯̃
- +ÊËÁ
ˆ¯̃
-ÊËÁ
ˆ¯̃
+ = (1)
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B.30 Mathematics
Since cos ( )34
b g+ÊËÁ
ˆ¯̃
is a real number, D ≥ 0. Hence,
164
16 02cos b g-ÊËÁ
ˆ¯̃
- ≥
cos2
41b g-Ê
ËÁˆ¯̃
≥
cos2
41b g-Ê
ËÁˆ¯̃
=
or b = g (2) Hence, Expression (2) for b = g is
2 34
12
cos ( )b g+ÊËÁ
ˆ¯̃
-ÈÎÍ
˘˚˙ = 0
Hence 2 34
cos ( )b g+ = 1
cos ( )34
12
b g+ =
34
60( )b g+ = ∞
or b + g = 80° or a = 100° or b = 40°, g = 40°, and a = 100°.Ans 11. (a), 12. (c):
b xx
b xx x x
coscos
sin(cos sin ) tan2 2 1 32 2-
=+
-
(1) 2 cos 2x – 1 π 0 fi x π np ±p6
(2) tan x π 0 fi x π ± p2
(3) cos2 x – 3 sin2 x π 0 fi x π n p ± p6
Also 2 cos 2x – 1 = 2 (cos2 x – sin2x) – (cos2 x + sin2 x) = cos2x – 3 sin2 x Now, the given equation reduces to b sin x = b + sin x
fi sin x = bb - 1
\ – 1 £ sin x £ 1
or – 1 £ bb - 1
£ 1
or bb - 1
+ 1 ≥ 0 and bb - 1
– 1 £ 0
or 2 11
bb
--
≥ 0 and 11b-
£ 0
or b Œ - •ÊËÁ
˘˚̇
, 12
» [1, •) and b Œ (– •, 1)
when b = 12
fi sin x = 1 which is not possible.
fi b Œ - •ÊËÁ
ˆ¯̃
- -ÏÌÓ
¸˝˛
, , ,12
1 0 13
For any other value b, sin x takes two values for x Œ(0, p).
Hence, four solutions for x Œ(0, 2p). 13. Ans (b):
A
E
D( , , )x y z
C (1, 1, 1)
B (3, 1, 2)
(2, 2, – 1)
P(4, –26/3, –10/3)
Fig. B.14
A (2, 2, – 1), B(3, 1, 2), C(1, 1, 1)
AB i j k AC i j k BC i k = - + = - - + = - -3 2 2, ,
\ AC ^ BC Mid point of AB = Mid point of CD
x + =12
52
fi x = 4 fi y + =12
32
fi y = 2
z + =12
12
fi z = 0
Therefore, coordinates of D are (4, 2, 0). 14. Ans (a): Equation of the base a(x – 1) + b(y – 1) + C(z – 1) = 0 which also passes through A(2,2,–1) and B(3,1,2). \ a + b – 2c = 0 (i) and 2a + c = 0 (ii) Therefore, c = –2a, b = –5a a (x – 1) – 5a (y – 1) – 2a (z – 1) = 0 \ x – 5y – 2z + 6 = 0 Foot of the normal from P
x y z- =+
-=
+
-4
1
263
5
103
2
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Appendix B: Mock Tests B.31
= -- -Ê
ËÁˆ¯̃
- -ÊËÁ
ˆ¯̃
+
+ +
Ê
Ë
ÁÁÁÁ
ˆ
¯
˜˜˜̃
4 5 263
2 103
6
1 25 4
x y z- =+
-=
+
-= -4
1
263
5
103
22
fi x = 2, y = 43
, z = 23
\ 2 43
23
, ,ÊËÁ
ˆ¯̃
Volume of the pyramid
= 13
(Base area) ¥ Height
= 13
AB AC
¥ ¥ EP
= 13
5 2 120i j k - - ¥
= 13
30 120 = 20 cubic units
15. Ans (b): Coordinate of any point on line AB can be taken as
h = r1 + r cos q
CB
O
C1
C2
A r( , 0)1
X
Y
Fig. B.15
k = 0 + r sin q Therefore, it lies on C2. \ (r1 + r cos q)2 + r2 sin2q = r2
2
r2 + 2 r r1 cos q + r12 – r2
2 = 0 Let AB = rAB and AC = rAC. Therefore, rAB + rAC = –2r1cos q, rAB.rAC = r1
2 – r22
So (BC) = | AC – AB | = | rAC – rAB |
= ( )r r r rAC AB AC AB+ -2 4
= 4 4 412 2
12
22r r rcos q - +
BC = - +4 412 2
22r rsin q
Therefore, for max sin q = 0 BC2
max = 4r22
16. Ans (a): Now OA2 + OB2 + BC2 = r1
2 + r22 + 4r2
2 – 4r12sin2q
= 5r22 + r1
2 – 4r12sin2q
\ 0 £ sin2q £ 1 So OA2 + OB2 + OC2 Œ [5r2
2 – 3r12, 5r2
2 + r12]
Section C
17. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ s) (a) we get common normal perpendicular to y = x.
So, slope = – 1 fi x + y = 3a (b) Tangent to the parabola y = mx + a/m passes
through the point P(h, k). fi m2h – mk + a = 0. If its roots are m1 and m2,
then m1m2 = +1. Locus is x = a. (c) The tangents are y = m(x + a) + a/m (1)
and y = -1m
(x + 2a) – 2am (2)
Subtracting from (1) to (2), we get x + 3 a = 0. (d) If the chord joining t1 and t2 subtends 90º at
vertex, then t1t2 = –4. point of intersection of tangents is (–at1t2, –a(t1 + t2)). So the locus is x = 4a.
(b) For regular tetrahedron all sides are of equal length, hence, | | | | | | a b c= = . Also, all the faces are equilateral triangle.
Therefore, angle between a and b is 60º,
b
and c is 60º, and between a and c is 60º.
Hence, a b b c c a. . .= =
(c) Since
a b¥ = c fi a c^ and b c^ and
b c a¥ = fi
b a^ and c a^ .
Therefore, a ,b , c are mutually perpendicular.
(d) Since a b c+ + = 0
fi a b c a b b c c a2 2 2 2 0+ + + + + =( . . . )
fi a b b c c a. . .+ + = - 3
2
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B.32 Mathematics
19. Ans: (a Æ r); (b Æ s); (c Æ q); (d Æ p)
(a) f (x) = mx c x
e xx
+ <≥
ÏÌÔ
ÓÔ
00
For application of LMV function must be con-tinous and derivable in [–2, 2].
f (0+) = f (0–) fi c = 1 f ¢(0+) = f ¢(0–) fi m = 1 fi m + 3c = 4 (b)
(6, 2)
(4, 4)
(2, 6)
Fig. B.16
Length of latus rectum is 4 2 = 4a fi a = 2 Therefore, equation of directrix is x + y + l = 0 whose distance from (4, 4) is 2 2 .
\ 4 42
+ + l = 2 2 fi l = ± 4 – 8
fi l = –4 or –12 Therefore, directrix are x + y = 4 and x + y = 12. \ l1 + l2 = 4 + 12 = 16 (c) f ¢(x) = 6x2 – 6x – 12 = 0 fi (x – 2)(x + 1) = 0 fi x = –1, 2 f ¢¢(x) = 12x – 6 f ¢¢(–1) = –18 < 0 fi x = –1 is the point of maxima f ¢¢(2) = 18 > 0 fi x = 2 is the point of minima.
Now f (–1) = 8, f (2) = –19, f(5/2) = - 332
,
f (–2) = –5 \ maximum value of f(x) in [–2, 5/2] is 8.
(d) lim( ) ( )
....
n
n n
n n
Æ•
-+
-+
+-
Ê
Ë
ÁÁÁÁ
ˆ
¯
˜˜˜˜
12 1
14 2
12
2 2
2 2
= limn
r
n
rn rƕ= -
 12 2
1
= limn
r
n
n rn
rn
Æ•= -
 1
22
21
= dxx x2 2
0
1
-Ú
= dxx1 1 2
0
1
- -Ú ( )
= sin ( )- -ÈÎ ˘̊10
11x
= 02 2 4
2- -ÊËÁ
ˆ¯̃
= = fi =p p pk k
20. Ans: (a Æ r); (b Æ s); (c Æ p); (d Æ s) (a) 1 + a10 + a20 + L + a190 = 0 as 10 is not an integral multiple of 20 (b) (log2x)2 + log2(0.03125) + log2x + 3 < 0 fi (log2x)2 + log2(1/32) + log2x + 3 < 0 fi (log2x)2 + (log2x) – 2 < 0 fi (log2x + 2) (log2x – 1) < 0 fi –2 < log2x < 1
fi 14
< x < 2
Number of integral solution = 1 (c) f (x) = x2e–2x
f ¢(x) = x2(–2) e–2x + e–2x.2x = 2x e–2x(1 – x) f ¢(x) > 0 "x Œ (0, 1) f ¢(x) < 0 "x Œ (1, •) \ f (x)max = f (1) = e–2 = l
12
12
42
ÊËÁ
ˆ¯̃
= ÊËÁ
ˆ¯̃
=-ln ( ) ln ( )l e
(d)
y x2= – 12 x = 3
y
(3, 0)x
Fig. B.17
Perpendicular tangents can be drawn to the parabola from points which lies on directrix. There is only one point (3, 0) which lies on its directrix as well as on the hyperbola.