IIT ADVANCED 2017 (PRACTICE PAPER I, FRIDAY 28-04-17)
Section I Single Answer Objective
1.
Let f : (1,1)  (1,1) be continuous, f ( x)  f ( x 2 ) for every x and f (0) 
(a)
2.
is
1
;
16
(b)
is
1
;
2
(c)
is
1
;
4
(d)
1
1
. Then f  
2
4
cannot be determined.
Let f r ( x)  x( x  1)( x  2)( x  3)  r ; where r is a real number. Then
(a)
f r ( x) has a real root only for finitely many values of r ;
(b)
f r ( x) can never have a repeated root; (c) f r ( x) has a real root for every value of r ;
(d)
f r ( x) can have a repeated root only for finitely many values of r .
3. Let X  1, 2,3, 4. The number of functions f : X  X satisfying f ( f (i ))  i for all 1  i  4 is
(a)
4.
1;
(b)
6;
(c)
10.
(d)
9;
Tangents are drawn to a circular arc at its middle point and its extremities ;  is the area of the
triangle formed by the chord of the arc and the two tangents at the extremities , and   the
area of triangle that formed by the three tangents . As the length of the arc tends to zero .
(a)
5.
 /   2
(b)
 /   4
(c)
 /   8
(d)
 /  1
Let pn ( x), n  0 ,1, 2,... be polynomials defined by p0 ( x) 1, p1 ( x)  x and
pn ( x)  xpn1 ( x)  pn2 ( x) then p10 (0) must be equal to
(a)
6.
0
(b)
10
(c)
1
(d)
-1
Let P( x)  x 4  ax3  bx 2  cx  d where a , b, c, d are integers. The sum of pairs of
1
2
roots of P(x) are given by 1,2,5,6,9,10 then P    must be
(a)
7.
33
(b)
28
(c)
-28
(d)
none of these.
The number of sequences of length five with 0 and 1 as terms which contain at least two
consecutive 0’s is.
(a)
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4.23
(b)
5
 
2
(c)
20
(d)
19
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Let X be a point on a straight line segment AB such that AB.BX  AX 2 . Let C be a point
on the circle with centre at A and radius AB such that BC  X . Then  BAC must be
(a)
(b)
(c)
30
60
72 (d)
36
8.
Section II Multiple Answer Objective
9
From a point P on a radius OA of a circle (radius r) produced beyond the circle a tangent PT is
drawn to the circle touching it at T . Draw TN  OA
AN
AN
1
NA
NA
2
tends to 1 (b)
tends to ½(c)
tends to
(d)
tents to
2
2
2r
AP
AP
r
(arcAT )
(arcAT )
(a)
10.
If P tends A then
Which of the following functions are necessarily periodic?
(a)
A function satisfying f(x + 8) + f(x) = 0 for all x
(b) A function satisfying f(x + 8) - f(x) = 0 for all x
(c) A function satisfying f(x + 1) + f(x – 1) =
3 f  x  for all x
(d) A function satisfying f(x + y) = f(x) + f(y) for all x, y
11.
Over ( 0,2 ) the function [ x 2  x  1] (where [ x ] denotes greatest integer  x) is discontinuous at
(a)
12.
1 2
2
  1 
  
(b)
Let f ( x)   x    , x  0
x
1 3
2
(c)
1 5
2
(d)
 .  denotes the greatest integer function . Then
1
f ( x) is
discontinuous
(a)
f ( x ) is discontinuous when x  (0,1).
(b)
f ( x ) is discontinuous when x is an irrational number in (0 , 1)
(c)
f ( x ) is discontinuous when x is a rational number in (0 , 1)
(d)
13.
at an infinite number of points and
The sides of a triangle are given by x 2  x  1, 2 x  1 and x 2  1 then
(a)
The largest side is x2 + x + 1,
(c)
The smallest side is x2 – 1 for all x in (1,1  3)
(d)
The smallest side is 2x + 1 for all x .
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(b)
The largest angle is 120
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14.
In a triangle ABC the equation of side BC is 2x – y = 3. The co-ordinates of orthocenter and
circumcentre of the triangle are (1, 2) and (2, 4) respectively then
(a)
15.
R
6
61
9
(b) sin B sin C =
(c) 2 R cos A 
7
5
2 61
(d) sin B sin C =
7
2 61
If n and r are positive integers (n  2r  1) , a  n  r 1 Cr , b  n  r 1Cr 1 then
(a)
a is number of selections of r objects from n objects of which no two are
consecutive .
(b)
a is number of selections of r objects from n  r  1 objects of which no
two are consecutive .
(c)
b is number of selections of n objects from r varieties of objects
( objects of one variety are identical )
(d)
b is number of selections of r objects from n varieties of objects
Section III (INTEGER ANSWERS)
16.
When 0  x  2 and  x  denotes greatest integer  x, then
sin x  cos x  sin x  cos x takes exactly k
integer values. Then k must be
17.
The maximum value of f ( x ); f ( x) 16 x C2 x 1  203 x C4 x 5 is
18.
The function f ( x) 
19.
The function f ( x) 
x5
x2  1
takes exactly k integer values, then k must be
x3
 (m  1) x 2  (m  5) x  11 is invertible only for k integer values
3
of m . Then k must be
20.
If β is a seventh root of unity β  1 and the value of 1  3 β  5 β 2  7 β 3  .....  13 β 6 is

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2k
then k must be .
1 β
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SOLUTIONS PTS I (IIT ADVANCED 2017)
= .......... f (0)
=
1
 1
f   f  
4
 16 

f ( x)  f ( x 2 )
1. .(b) :
1
2
 (b) is true .
(d): x( x  1) ( x  2) ( x  3) r  0
2.
 1 
 1 

f



  256 2 
 256 


= f
Put t  x 2  3x  t (t  2)  r  0

( x 2  3x ( x 2  3x  2)  r  0
t 2  2t  1  r  1
(t  1)2  r  1  0 ( x 2  3x  1)2  r  1 0
 ( x 2  3x  1)2 1  r
If r 1 then no real roots  choices (a),(c) are wrong
if r  1 then the original equation has two repeated roots
3. (c) :
 (d) is true .
Such a function has to be one-one . Now for a one-one such function four cases arise
Case (i) all four fixed points ie f (1)  1, f (2)  2, f (3)  3, f (4)  4
 one function of our interest i e f ( f (i )  i )
 Case (i) only
If two are fixed e.g f (1) 1, f (2)  2 then f (3) and f (4) may be
Case (ii)
If three points are fixed then four th is
Case (iii)
interchanging
also fixed
i .e f (3)  4, f (4)  3  4 C 2 such function
(since two fixed points may be chosen is 4 C 2 ways )
Case (iv)
one fix point wont lead to such a function
but if we divide 1, 2,3, 4 in two
Indinguishable boxes and cross the images for example
then 3 more such functions will come
 total = 10
4.
(b) : Consider the are PAQ of the
circle
x 2  y 2  r 2 whose mid point is A.
Then P is (r cos θ , r sin θ ), Q is
(r cos θ  r sin θ )
 chord PQ  2r sin θ
Now tangents at P and Q are
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x cos θ  y sin θ  r  0
(i)
x cos θ  y sin θ  r  0
(ii)
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
y  0, x  r sec θ
 Tangents at P and Q intersect at R(r sec θ ,0)
   Area of triangle RPQ . =
1
(2r sin θ ) ( r sec θ  r cos θ ) θ
2
sin θ
r 2 sin θ
(1  cos 2 θ ) =
(1  cos θ )(1  cos θ )
cos θ
cos θ
Again equation of tangent at middle point of arc is x  r
r (1  cos θ )
2r (1  cos θ )
 LM 
( on solving with (i) )
 L is ( r ,
sin θ
sin θ
=
r2

5.
1 2r (1  cos θ )
r 2 (1  cos θ ) 2
  .
.(r sec θ  r ) 
2
sin θ
sin θ cos θ
1

sin 2 θ (1  cos θ )


 (1  cos θ )2  4 as θ  0
1

1  cos θ
(d) : On putting x = 0 in the recursion relation we get pn (0)   pn 2 (0)

p10 (0)   p8 (0)  () () p6 (0)  (1)3 p4 (0)
 (1) 4 (1)   1 (
6.
(b):
 (1) 4 p2 (0)
p2 (0)   p0  1sin ce p0 ( x) 1  p0 (0)  1)
If the roots be x1, x2, x3, x4 then x1 + x2, x2 + x3, x3 + x4, x4 + x5, x1 + x4,
x1 + x3 have values 1, 2, 5, 6, 9, 10 (Not necessarily in this order)
We will accept the choice for which a, b, c, d become integers. Taking the choice
x1 + x2 = 1, x2 + x4 = 2, x1 + x3 = 9, x3 + x4 = 10, x2 + x3 = 6, x1 + x4 = 5
Now x1 + x2 + x3 + x4 = -a  1 + 10 = -a  a = -11
From the above equations we easily get x1 = 2, x2 = -1, x3 = 7, x4 = 3
 p(x) = x4 – 11x3 + 29x2 + x – 42
p"(x) = 12x2 – 66x + 58
1
 p     28
2
7. (d ): The sequence will contain either 5 zeros or 3 zeros or 2 zeros. A sequence having a single zero
cannot satisfy the condition of the problem. If the sequence contains 5 zeros then it will satisfy
the condition of the problem. If it contains exactly four zeros then five sequences are possible
and all of them satisfy the condition of the problem.
(00001, 00010, 001000, 0100, 10000 or
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5!
 5)
4!
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If it contains 3 zeros then they are either together or two of them are together. If all of them are
together then treat 000 as one object β and arrange β 11 in
3!
 3 ways. If two zeros are
2!
together and one is separated then consider
1
1
. There are three empty spaces. We
can put 00 in 3 ways and 0 in 2 ways  total ways. = 3x2 = 6 ways and total ways in this
case = 9.
Finally when only two zeros are there then treat 00 as one object β and arrange β 111 in
4!
 4 ways. Thus final answer = 1 + 5 + 9 +4 = 19.
3!
8. ( d ) : If ABC   then chord BC  2r sin
 AX  2r sin



2
. Now AB.BX  AX 2
2

 r (r  2r sin )  4r 2 sin 2
2
2
 4sin 2

 2sin

1 0 
2
2
 1  5


( sin  0) 
18    36
sin 
 (d) is correct
2
4
2
2
9 (a,c) : Let POT  x then TN  r sin x , ON  r cos x , AN  r  r cos x
OP  r sec x

 AP  r sec x  r
AN
r (1  cos x)

 cos x
AP
r (sec x  1)
Now arcAT  rx

lim
x 0
when P tends to A , x  0  limit = 1
AN
r (1  cos x)
1
 lim

2
2
2
x 0
(arcAT )
r x
2r -
Thus (a) and (c) are correct .
10. (b) ,(c) : b is correct since f ( x ) is periodic with period 8.c is also correct since on
Replacing x by x + 1 and x -1 we can show that
f ( x  2)  f ( x  2)  f ( x)
From which we can conclude that f ( x  12)  f ( x) for all x .
11.
12.
(c) ,(d) : On ( 0 ,2) x 2  x  1 becomes an integer at x  1,
(a) , (d):
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1 5
2
Observe the table
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x
1 1
 , 
5 4
1 1
 , 
 4 3
1 1
 , 
3 2
1
1
x
1
 x 
 4,5
4
 3, 4 
3
 2,3
2
 1  1 
But f     

 2   2 1/ 2  
  1 
x  x 
  
1
x 
x
4 
 ,1
5 
1 
 ,1
4 
0
0
2 
 ,1
3 
0
1 
   2  1
2 
 1 1
,  then f ( x)  0 but f
 n 1 n 
Thus f is discontinuous at infinite number of points in (0,1)
In general if n  N and x  
 1  1 
    n  1
 n  n 
 (a) and (d) are correct .
We must have x 2  x  1  0, 2 x  1  0, x 2  1 0
13. (a) , (b) ,(c) :
We must also have x 2  x  1  2 x  1  x 2  1, 2 x  1  x 2  1 x 2  x  1
x 2  x  1  x 2  1  2 x  1 . Solving all six inequations we get x  1
Now x2 + x + 1 > 2x + 1 since x2 – x > 0 (as x > 1) and x2 + x + 1 > x2 – 1 since x > 1
 x2 + x + 1 is largest side.
 choice (a) is correct.
Since x2 + x + 1 is largest side cosine of largest angle is given by
 2 x  1
2
  x 2  1   x 2  x  1
2
2  2 x  1  x  1
2
Finally smallest side is x2 – 1
2

1
 largest angle is 120
2
 x2 – 1 < 2x + 1
But x > 1 is an essential conditions.
 x (1, 1 +
 x  (1 -
3 ,1 +
3)
3)
14. (b) , (c): the distance between the circumcentre O and orthocenter H is given by OH2 = R2 (1 – 8 cos
A cos B cos C) ………..(*)
A
We know that OM = R cos A
H
HD = 2R cos B cos C
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O
B
M D'
C
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7
Now equation of BC is 2x - y – 3 = 0
 OM =
From above relations we easily determine R 
sin B sin C =
15.
9
2 61
3
5
, HD  
3
5
OH2 = (1 – 2)2 + (2 – a)2 = 5
61
6
, 2 R cos A 
5
5
Thus choices (b) and (c) are correct.
(a,c) : (a) is correct since the selections are equivalent to selecting r empty spacesfrom
n – r + 1 empty spaces.Since (a) is correct (b) is wrong.
(c) is correct since number of selections in this case are number of non negative integer
solutions: of x1 + x2 + ……… + xr = n which is n + r - 1Cr - 1
(d) is wrong since number of these selections = r + n - 1Cr or n + r - 1Cr .
16.
π π 3π 5π 3π 7π
, , , , ,
it takes 5 values
4 2 4 4 2 4
f ( x ) is defined only for x  2,3 compare f (2) and f (3) we note that f (3) is greatest
calculate f ( x ) at 0,
17.
1  5x
1
, f    26 , f ()  1, f ()  1
2 3/ 2
(1  x )
5
values attained are 0,1, 2,3, 4,5
f ( x) 
18.

f ( x)  x 2  2(m  1) x  m  5
19.
f is invertible if f is one-one
 4(m  1)2  4(m  5)  0  1  m  4)  m takes 6 integer values
S  1  3 β  5 β 2  7 β 3  ....13 β 6 then βS  β  3 β 2  5 β 3  ....11β 6  13 β 7
20.
on subtracting we get

S 1  β   1  2 β  2 β 2  .....  2 β 6 13

β 7  1

= 1  2 β 1  β  β 2  ....  β 5  13
1 β6
= 2 β.
 12
1 β
 S
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14
1 β
β  β7
 2.
 12
1 β
 k 7
 2.
β 1
 12  14
1 β
Ans …7
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