JEE MAIN 2016
ONLINE EXAMINAT ION
DAT E : 09-04-2016
SUBJECT : MAT HEMAT ICS
T EST PAPER
WIT H SOLUT IONS &
ANSWER KEY
1.
If A and B are any two events such that P(A) =2/5 and P(A ! B) = 3/20, then the conditional probability,
P(A/(A' " B')), where A' denotes the complement of A, is equal to :
(1)
8
17
Ans.
(3)
Sol.
P(A) =
(2)
1
4
(3)
5
17
(4)
11
20
2
5
P(A ! B) =
3
20
P(A/(A' " B')) = ?
P(A/(A' " B')) =
P(A ! (A '" B'))
P(# " (A ! B'))
P((A ! A ') " (A ! B'))
=
=
P(A ! B)'
P(A '! B')
1 $ P(A ! B)
2 3
5
$
5
P(A ! B')
P(A) $ p(A ! B)
=
=
= 5 20 = 20 =
17
3
17
17
17
1$
20
20
20
20
2.
For x % R, x & 0, x & 1, let f0(x)=
1
and fn
1$ x
+ 1(x)
= f0 ' f(n (X) ( , n = 0, 1, 2,……… T hen the value of
. 2+
. 3+
f100(3) + f1 , ) + f2 , ) is equal to :
- 3*
- 2*
(1)
4
3
(2)
Ans.
(3)
Sol.
f0(x) =
1
1$ x
f1(x) = f0(f0(x)) =
=
=
1
1$
1
3
1
1$ x
1
; f0(x) & 1
1 $ f0 (x)
x&0
1$ x
$x
f2(x) = f0(f1(x)) =
1
; f1(x) & 1
1 $ f1(x)
(3)
5
3
(4)
8
3
1
=x
1$ x
1/
x
similarly
f3(x) = f0(x)
f4(x) = f1(x) .....
=
. 2+
. 3+
. 2+ 3
f100(3) + f1 , ) + f2 , ) = f1(3) + f1 , ) +
2
- 3*
- 2*
- 3*
=1–
3.
Ans.
Sol.
3
3
1
5
+1–
+
=
2
2
3
3
The distance of the point (1, –2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to
the planes x – y + 2z = 3 and 2x – 2y + z + 12 = 0, is
1
(2) 2
(3) 2
(4) 2 2
(1)
2
(4)
Equation of plane 0 to the planes.
x – y + 2z = 3 & 2x – 2y + z + 12 = 0
and passes through (1, 2, 2) is
x $1 y $ 2 z $ 2
1
2
$1
$2
2
1
=0
3(x – 1) + 3(y – 2) = 0
x+y=3
..... (1)
distance of plane x + y – 3 = 0 from (1, – 2, 4) is
=
4.
Ans.
Sol.
1$ 2 $ 3
1/ 1
= 2 2
2
2
If the equations x + bx – 1 = 0 and x + x + b = 0 have a common root different from – 1, then | b | is equal
to
(1) 2
(2) 2
(3) 3
(3)
2
2
x + bx – 1 = 0 & x + x + b = 0 have common root 1 .
2
2
1 + b1 – 1 = 0
2
1 +1 +b=0
12
2
2
=
(4) 3
1
1
=
$(b / 1)
(1 $ b)
2
(b + 1) = (b + 1) (1 – b)
3i
2
|b| =
2
b /1
2
2
3
3
2
b + 2b + 1 = b – b + 1 – b
2
b + 3b = 0
2
2
b = 0 or
b = –3
when b = 0 then common roots is (–1) hence b = 0 rejected.
2
so b = –3
2
b=±
3
2
1
5.
3
If 2 tan $1 xdx =
1
3
cot $1(1 $ x / x 2 ) dx then
0
0
(2)
(1)
Sol.
2 tan$1 xdx =
1
3
0
1
1
3cot
$1
4
/ log2
2
2
tan$1 (1 – x + x )dx =
2
1
54
379 2 $ cot
$1
0
4x
=
2
Sol.
(1 $ x / x 2 ) dx is equal to :
(3) log4
(1 – x + x )dx
0
Ans.
$1
(4)
4
$ log4
2
...(1)
0
3
6.
3tan
0
(1) log2
Ans.
1
@ 3
>
If P = > 2
> 1
$
>
? 2
1 =
;
2 ; ,A=
3;
2 ;<
1 =
@2015
(1) >
;
0
2015
?
<
(2)
@ 3
>
T
PP = >2
> 1
>$
? 2
T
New P Q
1
3
$ 2 tan$1 xdx =
0
0
4
. 4 1
+
– 2 , $ ! n2 ) = ! n2
2
-4 2
*
@1 1=
T
T
2015
P is
>
; and Q = PAP , then P Q
0
1
?
<
@1 2015 =
(2) >
;
1 <
?0
1 =@ 3
; >
2 ; >2
3; > 1
; >
2 <? 2
2015
1
6
(1 $ x / x 2 )dx 8
:
@0 2015 =
(3) >
;
0 <
?0
0 =
@2015
(4) >
;
1
2015
?
<
1=
$ ;
2 ; = @1 0 = = PT P
>
;
3;
?0 1<
;
2 <
T
P = P PAPT
PAPT
.......PAPT
P
" # # # # # # $ # # # # # # %
2015 times
2015
because = A
2
Now A – 2A + A= 0
n
A = nA – (n – 1) I
2
7.
If
3cos
(1)
Ans.
(1)
16
5
dx
3
x 2 sin 2 x
2
A
2015
A
@1 1=
@1 0 = @1 2015 =
= 2015 >
; – (2014) >
; = >
;
1 <
?0 1<
?0 1< ?0
B
= (tan x) + C(tan x) + k, where k is a constant of integration, then A + B + C equals
(2)
21
5
(3)
7
10
(4)
27
10
Sol.
dx
3
A=
3
cos
1
x sin 2
1
x cos 2
=
x
1
2
(tan2 x / 1)sec 2 x
3
1
(tan x) 2
dx
tanx = t
1
2
A=
3
3
t 2 dt +
1
2
3
t
$
1
2
5
dt
5
t2
(tan x) 2
1/2
1/2
=
+t +c=
+ (tanx)
5
5
A=
1
5
1
,B= ,C=
2
2
5
A+B+C=
8.
16
5
The point (2, 1) is translated parallel to the line L : x – y = 4 by 2 3 units. If the new point Q lies in the
third quadrant, then the equation of the line passing through Q and perpendicular to L is :
6
(1) 2x + 2y = 1 –
(3) x + y = 2 –
6
Ans.
(4)
Sol.
Slopes of x – y = 4
6
(4) x + y = 3 – 2
6
.
1
1 +
,cos B C
, sin B C
)
2
2*
-
2
tanB = 1
or
.
1
1 +
cos B C $
, sin B C $
)
2,
2*
-
2
(2) x = y = 3 – 3
.
.
.
1 +
1 ++
Q is , 2 / 2 3 , $
) ,1 / 2 3 , $
))
,
2*
2 * )*
(2 $ 6, 1 $ 6 )
equation of required line is x + y = 3 – 2 6
9.
xF1
5
a
D$ x,
If the function f(x) = 7
is differentiable at x = 1, then
is equal to :
$1
b
D
/
/
E
E
a
cos
(
x
b
),
1
x
2
9
(1)
Ans.
(3)
$4 $ 2
2
–1
(2) – 1 – cos (2)
(3)
4/ 2
2
(4)
4$2
2
Sol.
L.H.L. at x = 1 is – 1
–1
R.H.L at x = 1 is a + cos (1 + b)
–1
2
–1 = a + cos (1 + b)
now
cos (1 + b) = – 1 – a
L.H.D. at x = 1 is – 1
–1
R.H.D at x = 1 is
....(i)
$1
1 $ (1 / b)2
2
2
(1 + b) = 0
2
–1
Now cos (1 – 1) = – 1 – a
a=–1–
b=–1
4
2
$(2 / 4 )
2/ 4
a
=
=
2( $ 1)
2
b
10.
The value of
15
Gr
r C1
Ans.
Sol.
11.
.
,
-
2,
15
15
(1) 1085
(3)
15
.
r2 ,
,
r C1
G
Cr +
) is equal to :
Cr – 1 )*
(2) 560
15
Cr +
) =
15
Cr $1 )*
(3) 680
15
. 15 $ r / 1 +
r2 ,
) =
r
*
r C1
G
15
(4) 1240
. 15 H16 + 15 H16 H 31 15 H16
–
=
(17) = 680.
2 )*
6
6
G r (16 – r) = 16 ,r C1
In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively
3 î / ĵ – k̂ , – î / 3ˆj / pk̂ and 5 î / q ĵ $ 4k̂ , then the point (p, q) lies on a line
Ans.
Sol.
(1) parallel to y-axis
(2) making an acute angle with the positive direction of x-axis
(3) parallel to x-axis
(4) making an obtuse angle with the position direction of x-axis.
(2)
&&&'
AB = – 4i + 2j + (p + 1) k̂
&&&'
AC = 2 î + (q – 1) ĵ – 3 k̂
&&&' &&&'
2
– 8 + 2(9 – 1) – 3 (p + 1) = 0
2
– 3p + 2q – 13 = 0
AB.AC = 0
2
(p, q) lies on line
3x – 2y + 13 = 0
slope =
3
2
2x
12.
a 4 +
.
If Lim , 1 / $ 2 )
xJ I x x *
(1)
3
= e , then ‘a’ is equal to :
2
3
(2)
Ans.
(2)
Sol.
a 4 +
.
L = Lim , 1 / $ 2 )
xJ I x x *
2x
L=
K
(4)
1
2
must be of the from 1I
. a 4 +
lim , $ 2 ) 2x
e -x x * K
lim
=e
(3) 2
xJ I
L = e xJ
2
3
2
2a
2' ax $ 4 (
x
I
3
=e
3
2
a=
2 sin4 x / 18 cos2 x $ 2 cos 4 x / 18 sin2 x = 1 is
13.
The number of x % [0, 24] for which
Ans.
(1) 6
(3)
Sol.
2sin x + 18cos x = 1 + 2cos x + 18sin x + 2 2cos4 x / 18 sin2 x
(2) 4
4
2
2
(3) 8
4
2
2
(4) 2
2
2
2(sin x – cos x) + 18(cos x – sin x) = 1 + 2 2cos4 x / 18 sin2 x
2
2
2
16(cos x – sin x) = 1 + 2 2cos4 x / 18 sin2 x
2
. 1 / cos 2x +
16cos2x – 1 = 2 2 ,
) / 9(1 $ cos 2x)
2
*
2
. 1 / 2cos 2x / cos2 2x
+
2
/ 9(1 $ cos 2x) )
256 cos 2x + 1 – 32 cos2x = 4 ,
,
)
2
*
2
256 cos 2x + 1 – 32 cos2x = 2(19 – 16cos2x + cos 2x)
2
254 cos 2x = 37
2
cos 2x =
2
2
2
2
2
clearly 8 solutions
37
254
2
cos2x = ±
37
% [–1, 1]
254
14.
If m and M are the minimum and the maximum values of 4 +
1
2
4
sin 2x – 2 cos x, x % R, then M – m is
2
equal to
(1)
Ans.
(3)
Sol.
4+
7
4
(2)
15
4
(3)
9
4
(4)
1
4
1
1
2
2 2
sin 2x – (2cos x)
2
2
2
=4+
1
1
17
1+
.
2
2
2
2
sin 2x – (1 + cos2x) = – cos 2x – cos2x + 4 = –[cos 2x + cos2x – 4] =
– , cos 2x / )
2
2
4
2
*
17
4
m = minimum value = 2
M = maximum value =
M–m=
15.
17
9
–2= .
4
4
If a variable line drawn through the intersection of the lines
x y
x y
= 1 and /
= 1, meets the coordinate
/
3 4
4 3
axes at A and B, (A & B), then the locus of the midpoint of AB is
(1) 7xy = 6(x + y)
(2) 6xy = 7(x + y)
2
2
Ans.
(3) 4(x + y) – 28(x + y) + 49 = 0
(1)
Sol.
4x + 3y = 12
....(1)
3x + 4y = 12
....(2)
(4) 14(x + y) – 97 (x + y) + 168 = 0
equation of lines passing through the intersection of the lines
B
p(! , k)
A
4x + 3y – 12 + L(3x + 4y – 12) = 0
. 12(1 / L )
+
A=C ,
, 0)
- 4 / 3L
*
12(1 / L ) +
.
B = , 0,
)
3 / 4L *
-
6(1 / L )
4 / 3L
6(1 / L )
k=
3 / 4L
from (3) & (4)
3k $ 4h
put in (1)
L=
3h $ 4k
7hk = 6(h + k)
hence locus is 7xy = 6(x + y)
!n =
.....(3)
... (4)
16.
If f(x) is a differentiable function in the interval (0, I ) such that f(1) = 1 and Lim
Ans.
Sol.
. 3+
x > 0, then f , ) is equal to :
- 2*
13
23
(1)
(2)
6
18
(4)
Differentiate w.r.t. t
tJ x
lim
tJ x
2
2t f(x) $ x 2 f M
(t)
=1
1
2x f(x) – x2fM
(x) = 1
2x f(x) $ 1
fM
(x) =
x2
2y
dy
1
=
– 2
x
dx
x
I.F. = e
$
2
3x dx
= e $2! nx =
1
x2
1
. 1 +
y , 2 ) = $ 4 dx
x
-x *
1
y
=
/ c
2
3x 3
x
at x = 1, y = 1
2
2
c=
3
3
1 2x 2
/
3x
3
. 3 + 31
f, ) =
18
- 2*
f(x) =
(3)
25
9
(4)
31
18
t 2 f ( x) $ x 2f (t )
= 1, for each
t$x
17.
Ans.
Sol.
2
3
If the tangent at a point P, with parameter t, on the curve x = 4t + 3, y = 8t – 1, t % R, meets the curve
again at a point Q, then the coordinates of Q are :
2
3
2
3
2
3
2
3
(1) (t + 3, – t – 1)
(2) (t + 3, t – 1)
(3) (16t + 3, – 64t – 1) (4) (4t + 3, – 8t – 1)
(1)
2
2
P(x = 4t + 3, y = 8t – 1)
2
3
let Q(4t1 + 3, 8y1 – 1)
dy
dy / dt
24t 2
=
=
= 3t
dx
dx / dt
8t
3
2
N
tangent at P is y – 8t + 1 = 3t (x – 4t – 3)
Q will satisfy it
3
2
3
2
N
8t1 – 8t = 3t (4t1 – 4t )
2
2
8(t1 – t)(t1 + t1t + t ) = 3t. 4(t1 – t) (t1 + t)
2
2
2(t1 + t1t + t ) = 3t(t1 + t)
2
2
2
2t1 + 2t1t + 2t = 3t t1 + 3t
2
2
2t1 – t1t – t = 0
(t1 – t) (2t1 + t) = 0
t
t1 = –
2
2
3
N
Q(t + 3, –t – 1) Ans. (1)
at P,
Ans.
x2 y2
/
= 1 meets the coordinate axes at A and B, and O is the
27 3
origin, then the minimum area (in sq. units) of the triangle OAB is :
9
(1) 9
(2)
(3) 9 3
(4) 3 3
2
(1)
Sol.
Let P (3 3 cosB,
18.
If the tangent at a point on the ellipse
N tangent is
x
3 3
3 sinB)
cosB +
2 A (3 3 secB, 0)
N Area of OOAB =
y
3
sinB = 1
B (0, 3 cosecB)
1
OA. OB
2
1
(3 3 secB. 3 cosecB
2
9
9
=
=
2sin B cos B
sin2B
=
N minimum area of OOAB =
9
=9
1
19.
The point represented by 2+i in the Argand plane moves 1 unit eastwards, then 2 units northwards and
finally from there 2 2 units in the south-westwards direction. Then its new position in the Argand plane is
Ans.
Sol.
at the point represented by :
(1) 2 + 2i
(2) – 2 – 2i
(3)
Let P(2 + i)
By rotation theorem
(3) 1 + i
(4) – 1 – i
y
2 2
S
1
45
Q(3,1)
P(2,1)
O
R(3,3)
x
2
z – (3 / 3i)
2 2 ( $ 4 / 4) i
=
e
3 / i – (3 / 3i)
2
z – 3 – 3i
=1–i
–2i
z – 3 – 3i = – 2i – 2
z=1+i
20.
A circle passes through (–2, 4). Which one of the following equations can represent a diameter of this
Ans.
Sol.
circle?
(1) 4x + 5y – 6 = 0
(3)
Required circle is
2
(2) 5x + 2y + 4 = 0
2
(x – 0) + (y – 2) + L(x) = 0
it passes (– 2, 4)
N 4 + 4 – 2L = 0
L=4
2
2
N circle is x + y – 4y + 4x + 4 = 0
centre (–2, 2) which satisfy
2x – 3y + 10 = 0 Ans. 3
(3) 2x – 3y + 10 = 0
(4) 3x + 4y – 3 = 0
cos x
21.
The number of distinct real roots of the equation,
(1) 4
Ans.
Sol.
(2) 1
sinx
sinx
sin x
sin x
@ 4
cos x sin x = 0 in the interval >$ ,
? 4
sin x cos x
(3) 2
4=
is :
4 ;<
(4) 3
(3)
cos x
sin x
sin x
sin x
cos x
sin x C 0
sin x
sin x
3
cos x
3
3
2
2 cos x + sin x + sin x – 3sin xcosx = 0
2
2
2
2
2 (cosx + sinx + sinx) (cos x + sin x + sin x – cosxsinx – cosxsinx – sin x) = 0
2 cosx = – 2sinx
tanx = –
x = – tan
22.
1
2
–1
or
cosx = sinx
tan = 1 2 x = 4/4
1
2
N two solutions
The shortest distance between the lines
(1) (2, 3]
(2) [0, 1)
(3) (3, 4]
Ans.
(1)
Sol.
x y z
x/ 2
z–5
y– 4
=
and
=
=
z z 1
–1
8
4
shortest distance
'
'
'
'
= ' a – a ( . b1 H b2
2
x y z
x/ 2 y$4 z$5
lies in the interval :
C C and
C
C
2 2 1
8
4
$1
1
'
(
'
'
here b1 – b2 = (2i + 2j + k) × (–i + 8j + 4k)
= – 9j + 18k
'
'
' b Hb (
1
2
= – j / 2k
5
'
'
a 2 – a 1 = – 2i + 4j + 5k
N S.D. (–2i + 4j + 6k).
' –j/
2k (
5
=
6
5
which lies in (2,3]
(4) [1, 2)
23.
If the four letter words (need not be meaningful) are to be formed using the letters from the word
“MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such
words is :
(1)
Ans.
Sol.
11 !
(2! )3
(2) 59
Ans.
Sol.
(4) 56
(1)
There are 1M, 3E, 1D, II, IT, 2R, 2A, 2N
R– –E – – – – – – – – – – – –
rest of 11 letters can be arranged in
24.
(3) 110
11!
(2 !)3
Let a and b respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity
2
satisfies the equation 9e- – 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of
2
2
hyperbola, then a – b is equal to
(1) – 7
(2) – 5
(3) 5
(4) 7
(1)
2
9e – 18e + 5 = 0
2 e=
5
3
b2
25
2
=e =
............. (i)
9
a2
Also distance between foci and directrix is
N 1+
a+
9
.
= , ae – ) = 5 –
e*
5
. 5 3 + 16
2 a , – ) =
2 a=3
5
- 3 5*
from (i)
1+
b2
25
2
= e2 C
2 b = 16
9
9
2
2
N a – b = 9 – 16 = – 7
25.
Ans.
Consider the following two statements :
P : If 7 is an odd number, then 7 is divisible by 2.
Q : If 7 is a prime number, then 7 is an odd number.
If V1 is the truth value of contrapositive of P and V2 is the truth value of contrapositive of Q, then the
ordered pair (V1, V2) equals :
(1) (F, T)
(2) (T, F)
(3) (F, F)
(4) (T, T)
(1)
Sol.
Statement P is False
Statement Q is True.
V1 P F
V2 P T
Ans. 1
26.
The minimum distance of a point on the curve y = x – 4 from the origin is :
2
(1)
Ans.
Sol.
15
2
(2)
(1)
Let point at minimum distance from O is
2
(h, h – 4)
2
2
2
2
N OP = h + (h – 4)
2
d(OP )
2
= 2h + 2(h – 4)2h = 0
dh
2 h=±
7
2
Ans.
Sol.
15
2
(4)
19
2
R0
7
2
N OP is min at h = ±
OPmin =
(3)
7
,0
2
. d2 (OP2 ) +
,
)
2
- dh
*h CQ
27.
19
2
7
/
2
.7
,2–
-
2
+
4) =
*
15
2
3 4 5
3
3
3
3
Let x, y, z be positive real numbers such that x + y + z = 12 and x y z = (0.1) (600) . Then x + y + z is
equal to
(1) 270
(2) 258
(3) 216
(4) 342
(3)
x + y + z = 12
x3y4 z5 = (0.1) (600)3
. x+ . y+ . z+
1/12
3, ) / 4, ) / 5, )
5. x +3 . y +4 . z +5 D
6
- 3*
- 4*
- 5* S D
7, ) , ) , ) 8
12
D
9- 3 * - 4 * - 5 * D
:
x3 y 4 z5
(60)3 (4 H 25)
3 4 5
3
x y z E (0.1) (600)
3 4 5
3
But x y z = (10.1) (600)
Clearly AM = GM
x y z
Hence C C
2 x = 3, y = 4, z = 5
3 4 5
3
3
3
2 x + y + z = 27 + 64 + 125 = 216
1S
28.
If the mean deviation of the numbers 1, 1 + d, ….., 1 + 100d from their mean is 255, then a value of d is :
Ans.
(1) 10
(4)
Sol.
(2) 20.2
Mean is
101 /
(3) 5.05
(4) 10.1
100 H101
2
= 1 + 50 d
101
sum of deviation about mean is
50 d + 49d + ................... d + 0 + d + ........ + 50 d
= 50. 51 d
Mean deviation =
d=
29.
50 H 51d
= 255
101
255 H101
= 10.1
2550
2016
For x % R, x = – 1, if (1 + x)
2015
+ x (1 + x)
2
2014
+ x (1 + x)
+ …….. + x
2016
=
2016
G a x , then a
i
i
iC0
(1)
Ans.
Sol.
2016!
16!
(2)
2017!
2000!
(3)
2017!
17! 2000!
(3)
2016
G c .x
i C0
i
i
= (1 + x)
2016
+ x(1 +x)
2015
2
+ x (1 + x)
2017
.
. x + +
(1 / x)2016 , 1– ,
))
)
,
- 1/ x *
*
=
x
1–
1/ x
(1 / x)2016
x 2017
–
(1 / x)2017 – x 2017
1
(1 / x)
=
=
x / 1– x
1
1/ x
N Ka17 =
2017
C17 =
2007!
17! 2000!
2014
+ ......... + x
2016
(4)
2016!
17! 1999!
17
is equal to :
30.
2
The area (in sq. units) of the region described by A = {(x, y) | y S x – 5x + 4, x + y S 1, y E 0} is :
(1)
7
2
(2)
13
6
(3)
Ans.
(4)
Sol.
A = {(x,y)|y S x – 5x + 4, x + y S 1, y E 0}
2
2
Here y S x – 5x + 4 , x + y S 1 , y E 0
4
5/2 3
4
–9
4
Required area =
(3, –2)
1
.2 2+
2
4
3(5x – x
2
– 4)dx
3
4
@5x 2 x3
=
=2+ >
–
– 4x ;
2
3
?
<3
=2+
5
1
(16 – 9) – (64 – 27) – 4 (4 – 3)
2
3
-2+
35 37
19
–
–4=
2
3
6
17
6
(4*)
19
6