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MATHEMATICS DPP
TARGET : JEE (Advanced) 2015
TEST IN F OR M AT ION

Course : VIJETA & VIJAY (ADP & ADR)

Date : 08-04-2015

DP P

NO
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04
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Time : 151 min
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1 to 10
Multiple choice objective (no negative marking) Q
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33 to 37
Match the Following (no negative marking) Q
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39,40

1
...


3
...

5
...
)
(5 marks, 4 min
...
)
(8 marks, 8 min
...
)

(C) (1, )

[30, 30]
[110, 88]
[15, 15]
[8, 8]
[8, 10]

(D) [1, )

 1 
7
2  
cos–1 
 cos 5 – sin 5   is equal to
2


23
13
3
17
(A)
(B)
(C)
(D)
20
20
20
20
{x}) + x, x [–1, 4] where [x] and {x} denote integral
Number of solutions of equation 3 + [x] = log2(9 – 2
and fractional part of x respectively, is
(A) 6
(B) 12
(C) 2
(D) 1
If f(x) = x + sinx then all points of intersection of y = f(x) and y = f–1(x) lie on the line
(A) y = x
(B) y = –x
(C) y = 2x
(D) y = –2x


1
Range of f() = tan  cos ec 1 
 2 sin

(A) (–, ) – {n}
(C) [0, )


  is


6
...
99
...


For each positive integer n, let f(n + 1) = n(–1)n + 1 – 2f(n) and f(1) = f(2010)
...


(A) 335
(B) 336
(C) 331
If f(x) = x + tanx and f(x) is inverse of g(x), then g'(x) is equal to
1
1
1
(A)
(B)
(C)
2
2
1 g x  x
1 g x  x
2  g x  x

(D) 333

2

1

(D)

2

2  g x  x
















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10
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12
...


14
...


(B) (x + y) (x2 – y2)

 12 – 2 x 2 

For f(x) = tan  4
 x  2x 2  3 





(A) fmax =
(B) fmin = 0
12

1
x3
x  y3
y
1
cosec2  tan –1  
sec 2  tan–1  is equal to
2
2
y 2
x
2

(C) (x + y) (x2 + y2)
(D) (x – y) (x2 – y2)

–1





(C) fmin does not exist

(D) fmax =


2

 x 2  1, x  1
 x  1, x  0

If f(x) = 
and g(x) = 
then
2  x, x  0
2x  3, x  1


(A) Range of gof (x) is (–, –1)  [2, 5]
(B) Range of gof (x) is (–, –1)  [2, 5)
(C) gof (x) is one-one for x[0, 1]
(D) gof (x) is many one for x[0, 1]
If f(x) is identity function, g(x) is absolute value function and h(x) is reciprocal function then
(A) fogoh(x) = hogof(x)
(B) hog(x) = hogof(x)
(C) gofofofohogof(x) = gohog(x)
(D) hohohoh(x) = f(x)
x
The function y =
: R  R is
1 | x |
(A) one-one
(B) onto
(C) odd
(D) into

 1
If , ,  are roots of equation tan–1 (|x2 + 2x| + |x + 3| – ||x2 + 2x| – |x + 3||) + cot–1    =  in
 2
ascending order ( <  < ) then
(A) sin–1 is defined
(B) sec–1 is defined
(C)  –  = 2
(D) || > ||

16
...


1 + [sin–1x] > [cos–1x] where [
...


(D) [cos1, 1]
2

4   sin cos1 x  cos sin1 x  is a, then




3

(A) sin–1a + cos–1a =
(B) 2sin–1a + cos–1a = (C) sin–1a + 3cos–1a =
(D) tan–1a + cos–1a =
2
2
2
2
x
2   1
If f(x) =
then (where {x} represent fractional part of x)
x
2   1
1
(A) Df R
(B) Rf  [0, )
(C) period of f(x) is 1
(D) f(x) is even function
3
If the solution of equation sin(tan–1x) =









19
...


Which of the following is true for f(x) = (cosx)cosx, x   cos 1 1 , cos 1 1 


e

e



(A) Rf   1 
 
 e
 

21
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22
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(B) many-one

(C) into

(D) onto

Let f(x) = ([a] – 5[a] + 4)x + (6{a} – 5{a} + 1)x – tanx
...
]
and {
...


 
Let f(x) = cot–1(x2 + 4x + 2 – 3) be a function defined on R   0,  , is an onto function then
 2
(A)   [–1, 4]
(B) f'(0) = –4/17
(C) f(x) is one-one
(D) f(x) is many-one

25
...
] is greatest integer function then

2
2
 n(x  e)  1  x

26
...


28
...
h  x 
is
3





  



h repeated n times

(A) identity function

(B) one-one

(C) odd
x

29
...


31
...



...
If one of the solutions of the
equation f(x) = f–1(x) is 2014, then the other solution may be
(A) 2013
(B) 2015
(C) 2016
(D) 2012

3
x + 1 and fn + 1(x) = f(fn(x)) n  1, nN
...

(B)  is a linear polynomial in x
...

(D) line 4y =  touches a circle of unit radius with centre at origin
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Comprehension # 1 (Q
...
33 to 35)
Let f : [2, )  [1, ) defined by f(x) = 2 x

4

 4x 2

sin x  4
 
and g :  ,   A, defined by g(x) =
be two
2 
sin x  2


invertible functions, then
33
...


35
...
no
...

36
...


2

(C) (–, 0]   , 3 
3


(D) None of these

Area bounded between the curves y = f(x) and y = g(x) is
(A)

38
...

Column – I

Column-II

(A) Domain of f(x) is

(p)

  2n  2 ,2n  2 



(B) Domain of g(x) is

(q)

R –  2n  1  









n

2



(C) If fundamental period of g(x) is k then k is element of set

(r)

(D) gog–1 is an identity for x 

(s)

e1 x + 1 + x +

  3 
 2, 2 


 3 5 
 2 , 2 



x 2 x3

...

 6

39
...


1 1 
Let f : R+  R+ be a function which satisfies the relation f(x)
...

2

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