Fiitjee Mock Paper
Fiitjee Mock Paper
Fiitjee Mock Paper
-1
-1 2 1 1 -1 sin x + + cos x - dx 2 2
where [ ] stands for the greatest integer function is (a) (c) Q.2 0 (b) (d)
p 2
2
If Z is a complex number satisfying |z| - z = |z+8|-|z-4|=0 then the locus of z is (a) (c) Circle A straight line (b) (d) An ellipse A single point
Q.3
A man takes a step forward with probability 0.5 and backward with probability 0.7. The probability that at the end of 13 steps he is one step away from the starting point is (a) (c) 0.7853 0.4231 (b) (d) 0.4251 0.3785
Q.4
The number of point at which the function f ( x) =| x - 0.7 | + | x 2 - 1| + | x - z | does not have a derivative in the interval (-z, z) is (a) (c) 1 3 (b) (d) 2 4
Q.5
e
1
dx is
e3 (b) 3e2 (c) 4e2 (d) 4e3
(a) Q.6
The sum of the squares of lengths of the chords intercepted by the lines x+y =n, nN on the circle (x-1)2+(y+1)2=9 is (a) (c) 84 44 (b) (d) 42 24
Q.7
p 2
(b)
p 4
(c)
p 3
(d)
p 6
Q.8
(a) Q.9
3 2
10
(b)
9 4
10
(c)
3 2
20
(d)
9 4
1 + x 2 + 2 x sin(cos -1 y ) = 0 are
(a) Q.10 0 (b) 1 (c) 2 (d) 3
In a ABC, tan A + tan B + tan C = 6 tan A tan B = 2 then (a) triangle is acute angles at C (b) (c) (d) triangle is obtuse angled at C right angled at C None of these SECTION II One or more options are correct
Q.11
(a,b) is a variable point on y=3x lying between the straight lines 2x+y+4=0, x+4y-8=0, Then (a)
-4 8 a , 5 13 -4 6 a , 5 13
n
(b)
-12 24 b , 5 13 -12 12 b , 5 13
(c)
(d)
Q.12
20 x 2 + 5 x
(b) (d)
25 x 2 + 5 xy 20 x 2 + 4 xy
25 x 2 + 5 x
Q.13
(a) (c)
g (o) = 0
g (e) = 9e2 - 4e
(b) (d)
4 g (1) + g = 0 9
g ' (e) = 9e - 8
Q.14
A factor of 25 (1+x) 10 15 (a) x2 10 4(1+x) 6 (b) 15 6 9 (1+x) x+2 (c) x+1 (d) x+3
Q.15
If the angles of a triangle are in the ratio 3:4:5 the sides opposite to these angles are in the ratio (a) (c)
3: 4:5
(b) (d)
1 3 3 +1 : : 2 2 2 2 1: 3 : 3 + 1
2 : 6 : 3 +1
SECTION III Single digit integer question. Q.16 Ans. Q.17 Let f ( x) - (sin x) x - 2 x , x 1 The slope of the tangent to a curve y = f ( x) at a point (x,y) in 2x+1 and the curve passes through the point (1,2). Find the area of the region bounded by the curve, the x-axis and the line x=1 1 Let K R . if the origin and non real roots of 3z2+3z+k=0 from the vertices of an equilateral triangle, then k is 1
1
p p p , If f ( x) is continuous at x = then f = is 2 2 2
Q.19
from a point on x2-y2=9 tangents are dawn to 16x2+25y2=400 If 1 and 2 are the inclinations of the tangents to the x-axis (anticlockwise direction), then sin2(1 + 2) is 1 The unit digit of the number 271968 + 431366 171049 is 3 If f ( x) is continuous and 2 A differentiable function satisfies f ' ( x) = f ( x) + 2e x with initial condition
f ( x )dx is
2 0
1 = cos -1 x 2 + - 1 2
1 When x 2 + = 0 , sin-1 (0) = 0 2 1 Then cos -1 x 2 - = cos -1 (-1) = p 2
1 When x 2 + = 1 2
p 2
I = p dx = 2p
-1
( z + 8) = z - 4 represents the perpendicular bisector of the line joining (-8,0) and (4,0)
where x=2 Solving it (1) \ locus in the single point (-2,0) Sol 3 Ans. (D) The man is one step away from the starting point is two case (i) He takes 7 steps forward and 6 steps backward (ii) He takes 6 steps forward and 7 steps backward
= 2 ye y dy
1
3 = 2[ ye y - e y ]1
= 4e3
Sol 6 Ans.
B M 1,-1 C A
x+y=
The given circle ( x - 1) 2 + ( y + 1) 2 = 9 has centre at (1,-1) and radius 3, if AB is the chord x+y=n n Then cm = 2
\ AB 2 = (2 AM ) 2
= 4 AM 2
n2 = 4 9 - 2
= 2[72 - 30]
2 cos q b - a + c = 0 2 cos q b - a + c = 0
Where q in the angle between a and b
a - 2 cos q b = c
4 - 4cos2 q - 8cos2 q = 1
3 3 cos 2 q = , cos q = 4 2 As q in acute p q= 6
Sol. 9 1 + x 2 + 2 x sin(cos -1 y ) = 0
cos -1 y = sin -1 ( 1 - y 2 ) 1 + x 2 + 2 x sin(sin -1 1 + y 2 ) = 0
1 + x2 + 2 x 1 - y 2 ) = 0
Q.10 Ans. (A) In a DABC tan A + tan B + tan C = tan A tan B tan C Hence 6 = 2 tan c
tan c = 3 Now tan A + tan B = 3 and tan A tan B = 2 \ either tan A = 1 tan B = 2 Or tan A = 2 and tan B = 1
Section II
Sol 11 Ans.(A,B)
-4 -12 solving y = 3x and 2 x + y + 4 = 0 we get (a,b) as , 5 5 8 24 Solving y = 3x and x + 4 y - 8 = 0 we get (a,b) as , 13 13 Since (a,b) lies between the two lines -4 8 a , 5 13 -12 24 b , 5 13
= 20 x ( x + 1 - 1) + 5 x( x + 1 - x)
2 3
= 20 x 2 + 5 x
Also 5 x(1 - x) + 25 x 2
= 5 xy + 25 x 2
Using leibnitz rule g ( x) = (9 x 2 - 4 x) log x When x = 0, log x is not defined is the defined
(A) Is not correct
3 3 3(1 + x )
2 1 1 1+ x
1
1+ x 1 1 = 900 - x -x 1+ x = 900 x
2
x 0 0 1 1 x 1 -1 0 1 0 -1
1+ x 2 + x = 900 x 2 1 0 1 1
1 0 -1
Sol 15 . Ans.(B,C)
1 3 3 +1 : : 2 2 2 2
1 3 3 +1 : : 2 2 2 2
= 2 : 6 : 3 +1
p = f 2
log(cosh)
lim(cosh)
b 0
1 -2b
=e
b0 ( -2 b )
lim
(1)
Hence (1) becomes y = x 2 + x This curve cut the x-axis at (0,0) and (-1,0) also it cut y axis at (0,0) If meets the line x=1 at the print (1,2) Hence the curve is as Required area
(1,2) x=1
= =
-1 0
ydx + ydx
0 2
X
1
-1
(x
+ x)dx + ( x 2 + x)dx
0
(-1,0 ) 00
o
(0,0)
X
(1,0)
x3 x 2 x3 x2 = + + + 3 2 -1 3 2 0
-1 1 1 1 = - + + + 3 2 3 2 1 5 = + =1 6 6
Sol 18 Ans. (1) If z1 , z2 , z3 are the vertices of an equilateral triangle then we prove
Z12 + Z 22 + Z 32 = Z1Z 2 + Z 2 Z 3 + Z 3 Z1
Here Z 3 = 0
Z12 + Z 22 = Z1Z 2
( Z1 + Z 2 ) 2 = 3Z1Z 2 ,
y = mx + 25m 2 + 16
Let (x1y1) be a point on x 2 - y 2 = 9 Then m 2 (25 - x12 ) + 2mx1 y1 + 16 - y = 0
If m1 + m2 =
q1 + q 2 =
p 2 2 sin (q1 + q 2 ) = 1
271968 + 431366 - 171049 = The unit digit in 27 4492 + 434341.432 - 174262.17 = so the required unit digit = 1+ 9 - 7 = 3
Sol 21 Ans. [2] Let I = x f ( x 2 )dx
0 3
2x - dx = dt
(put x 2 = t )
1 f (t )dt 2 0
1 = .4 = 2 2
Sol 22 Ans.(2) Let f ( x) = y
dy - y = 2e x dx
y.e - x = 2e x e - x dx
If = e- x
= 2x + c y =0=0C =0
f ( x) = y = 2 xe x
dy = 2[ xe x + e x ] = 0 dx X= -1 0 0 x 0 x A = 2 xe dx = 2 ( xe ) - - e x dx - -
-1
2e1
Hence, area is 2
Q.1
(b)
1 3 1 1 x + + In 2 x + 3 + 3 x + + k x x x
3
3 1 1 1 x + 3 + In x + + x3 + 3 + k x x x 3
(c)
(d)
x2 3 + I n x + k 2
Q.2
Given that x+ y = 5k where k is a constant and that all values of x between 0 and 9 5k are equally likely, then the probability that xy > k 2 is 4 (a)
3 5
(b)
4 5
(c)
2 5
(d)
1 2
Q.3
(a) Q.4
179
(b)
(c)
176
(d)
178
p If , , , 0, and cos( b + g ) cos( b - g ) = sin a (2sin g + sin a ) then the 2 straight line
(a)
(1, 1) (b)
(1, -1)
(c)
(-1,1)
(d)
(-1, -1)
Q.5
If z = 2a sin q (sin q + i cos q ) , then the locus of z is (a) (b) (c) (d) a circle whose centre is a a circle whose center is a a circle whose radius is a circle of radius 2a
a 2
Q.6
In a ABC, coordinates of A are (4, 6), AB=12 and BAC= AC is (4, 12) Then the radius of the in-circle of ABC is (a)
p . The midpoint of 3
(d)
3 3
(b)
2 3
(c)
4 3
Q.7
1 2
Q.8
2 If log3 2, log3 (3x - 1) and log 3 (3x + 3) be the 11th, 16th and 26th term 3 respectively of A.P. then
(a) x=1 (b) x = log 3 5 (c) x=2 (d) x=
log 3 3
SECTION II Each paragraph contain 2 Question The sides of a triangle a, b and c are natural numbers with the condition a b c Q.9 If c = 12, the number of triangle, that are possible is (a) Q.10 21 (b) 42 (c) 12 (d) 20
If c = 24, the number of isosceles or equilateral triangle possible is (a) 13 (b) 24 Paragraph 2 Let f ( x) = cos(1 + 2 x - 2 x ) where (y) = greatest y (a) (c) 11 (d) 12
Q.11
The graph of y = f(x) cuts the x-axis at the point with x coordinate.
1 2
(b)
p 1 + 4 2
(c)
Q.12
B where nI (d) 2
p 1 4 2
(1 - sin1)
1 (1 - sin1) 2
(b)
1 3 sin - sin1 2 2
(c)
(d)
Paragraph 3
(d) Q.14
Scalene
The area of the triangle formed by three tangents in Q15 is equal to (a) (c)
3 3 sq units 2
(b) (d)
6 3 sq units
3 3 sq units
3 sq units 4
SECTION 3 One or more options are correct Q.15 If z + 4i = min { z + 3i , z + 5i } , then z satisfies (a) (c) Q.16
Im z = Im z =
-7 2 7 2
(b) (d)
Im z = Re z =
-9 2 7 2
The value of a N such that the graph of the function y = x 2 + 2(a + 4) x - 5a + 64 is strictly above the x-axis is (a) 3 (b) 6 (c) 1 (d) 2 Let A be a square matrix of order 3 satisfying A3 = 0 . Then which of the following statements is true ? (a) (b) (c) (d)
Q.17
Q.18
Suppose {an }n 1 is a sequence satisfying a = 2009 and an +1 = Then which of the following statements is true ? (a) (c) a2009=2009 (b) a2011=2009 P, q N, ap = aq if p q is divisible by 64
(d) Q.19
Q.20
Let f ( x) = x 6 - 6 x5 + 15 x 4 - 20 x 3 + 15 x - 6 x + 1 then (a) (b) (c) (d) f has neither maximum nor minimum f decreases on [1, ] f increases on [1, ] f has a minimum at x = 1
1 6 1 x + - x + 6 + 2 x x = 3 1 1 3 x+ +x + 3 x x
3 2 1 3 1 x + - x + 3 x x 2
1 3 1 x+ +x + 3 x x
3 3 1 3 1 1 3 1 x + + x + 3 x + - x + 3 x x x x = 3 1 3 1 x+ +x + 3 x x
x2 = 3 + ln x + k 2
9 x 2 - 5 xk + k 2 < 0 4 k 9 x - x - k < 0 2 2
k 9k x , 2 2 Faceable cases = Length of the above Interval = 4k Exhaustive case = length of the interval (o,5k) = 5k 4k 4 Probability = 5k 5
p 3 The DABC is equilateral p 1 + 2 x - [2 x] = The radius of the in-circle 2 1/3 of an altitude 1 (12sin 60) 3 2 3
AC=AB and BAC= Sol. 7 Ans. (c) Now x 2 + y 2 + z 2 - xy - yz - zx 0
tan
Sol. 8
a + 10d - log 3 2
a + 15d = log 3 (3x - 1) 2 3
Hence 5d = log
(3x - 1) 2 3 2
2 3
10d = log
(3x + 3) (3x + 1)
(3
- 1) 2
(3 (3
x
+ 3)
2 3
- 1)
3x - 1 = 4 or -2 (-2 is rejected)
3x = 5 x = log 3 5
SECTION-2 Sol 9 Ans. (B) C= 12 the possible values of a and b are below a b 1 12 2 11 to 12 3 10 to 12 : : : : 12 12 Total number of possible =2(1+2+3+4+5+6)=42 Sol 10 Ans. (D) C=24 the possible values of a and b are a b 13 13 14 14 : : : : 23 23
25 25 Total number of . = 12 Sol 11 Ans. (D) y = f ( x)cut the x - axis If cos(1 + 2 x - [2 x]) = 0
p 2 p 1 if 2x = 2 2 p Since [2 x ] = - 1 = 0 2 p 1 f x= 2 2
1 + 2 x - [2 x] =
cos(1 + 2 x - [2 x])dx
p 1 4 2
cos (1 + 2 x)dx
y = mx +
y2 = 4x
1 is a tangent to m
If it torches x 2 + y 2 + 2 x = 0 then
1 x 2 + mx + + 2 x = 0 m Must have equal roots 1 16 - 4(1 + m 2 ) 2 = 0 m
2
1 3 The two common tangents are 1 y= x+ 3 3 -1 y= x- 3 2 These meet the x-axis at A(-3,0) The third common tangent is x = 0 CAB = 60 and symmetry D is equilateral m=
So l 14 Ans. (c) one tangent meets the y-axis at (0, 3) Second at (0, - 3) The side of the equilateral
D=2 3
And its area
(2 3)2 3 =3 3 4
-7 -7 on Im Z = 2 2
Similarly if z + 5i < z + 3i Im z =
-9 2
Sol 16 Ans. (C.D) The graph of y = x 2 + 2(a + 4) x - 5a + 64 is strictly above the x-axis. if only the discriminate is negative (i.e.) 4(a + 4) 2 - 4(64 - 5a ) < 0
3a n -1 ,n 1 an + 3 The form brining to the mind the formula for difference of tangent function Let a1 = tan u u R an +1 =
an +1 = an 1 3
1 an + 1 3 a1 -
p = tan 4 - 6 Continuing the pattern are have p an = tan(4 - (n - 1) , n 1 6 As tangent is periodic with periodic II we obtain an = an +6 showing that the sequence has period 6
Sol 19 Ans. (A,B,C) The idea is to count in two different ways i.e. use a combinational arrangement Sol 20 (C,D) f '( x) = 6( x5 - 5 x 4 + 10 x3 - 10 x 2 + 5 x - 1)