JEE Main 2016 Solved Paper PDF Download

 Page 1


1
1. A value of ? for which 
? ?
? ?
2 3isin
1 2isin
 is purely
imaginary, is :
(1) sin
–1
? ?
? ?
? ?
1
3
(2) 
?
3
(3) 
?
6
(4) sin
–1
? ?
? ?
? ?
? ?
3
4
Ans. (1)
Sol. Z = 
? ?
? ?
2 3isin
1 2isin
? Z = 
? ? ? ? ? ? ? ?
? ?
2
2 3isin 1 2isin
1 4sin
  = 
? ?
? ? ? ?
? ?
2
2
2 6sin 7isin
1 4sin
for purely imaginary Z, Re(Z) = 0
? 2 – 6sin
2
? = 0 ? sin ? = 
?
1
3
?? ? = ±sin
–1
? ?
? ?
? ?
1
3
2. The system of linear equations
x + ?y – z = 0
?x – y – z = 0
x + y – ?z = 0
has a non-trivial solution for :
(1) exactly three values of ?.
(2) infinitely many values of ?.
(3) exactly one value of ?.
(4) exactly two values of ?.
Ans. (1)
Sol.
? ?
? ? ?
? ?
1 1
1 1
1 1
 = 0    ?? ? ? ? ? = 0, 1, –1
PART A – MATHEMATICS
JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 03
th
 APRIL, 2016)
3. A wire of length 2 units is cut into two parts which
are bent respectively to form a square of side = x
units and a circle of radius = r units. If the sum
of the areas of the square and the circle so formed
is minimum, then :
(1) 2x = r
(2) 2x = ( ? + 4)r
(3) (4 – ?)x = ?r
(4) x = 2r
Ans. (4)
Sol.
Square
x
x
r
x
x
given that     4x + 2 ?r = 2
i.e. 
 
2x + ?r = 1
?   r = 
?
?
1 2x
..... (i)
Area A = x
2
 + ?r
2
     = x
2
 + 
?
1
(2x – 1)
2
for min value of area A
dA
dx
 = 0 gives x = 
? ?
2
4
..... (ii)
from (i) & (ii)
r = 
? ?
1
4
..... (iii)
? x = 2r
Page 2


1
1. A value of ? for which 
? ?
? ?
2 3isin
1 2isin
 is purely
imaginary, is :
(1) sin
–1
? ?
? ?
? ?
1
3
(2) 
?
3
(3) 
?
6
(4) sin
–1
? ?
? ?
? ?
? ?
3
4
Ans. (1)
Sol. Z = 
? ?
? ?
2 3isin
1 2isin
? Z = 
? ? ? ? ? ? ? ?
? ?
2
2 3isin 1 2isin
1 4sin
  = 
? ?
? ? ? ?
? ?
2
2
2 6sin 7isin
1 4sin
for purely imaginary Z, Re(Z) = 0
? 2 – 6sin
2
? = 0 ? sin ? = 
?
1
3
?? ? = ±sin
–1
? ?
? ?
? ?
1
3
2. The system of linear equations
x + ?y – z = 0
?x – y – z = 0
x + y – ?z = 0
has a non-trivial solution for :
(1) exactly three values of ?.
(2) infinitely many values of ?.
(3) exactly one value of ?.
(4) exactly two values of ?.
Ans. (1)
Sol.
? ?
? ? ?
? ?
1 1
1 1
1 1
 = 0    ?? ? ? ? ? = 0, 1, –1
PART A – MATHEMATICS
JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 03
th
 APRIL, 2016)
3. A wire of length 2 units is cut into two parts which
are bent respectively to form a square of side = x
units and a circle of radius = r units. If the sum
of the areas of the square and the circle so formed
is minimum, then :
(1) 2x = r
(2) 2x = ( ? + 4)r
(3) (4 – ?)x = ?r
(4) x = 2r
Ans. (4)
Sol.
Square
x
x
r
x
x
given that     4x + 2 ?r = 2
i.e. 
 
2x + ?r = 1
?   r = 
?
?
1 2x
..... (i)
Area A = x
2
 + ?r
2
     = x
2
 + 
?
1
(2x – 1)
2
for min value of area A
dA
dx
 = 0 gives x = 
? ?
2
4
..... (ii)
from (i) & (ii)
r = 
? ?
1
4
..... (iii)
? x = 2r
J E E ( M A I N ) - 2 0 1 6
2
4. A man is walking towards a vertical pillar in a
straight path, at a uniform speed. At a certain point
A on the path, he observes that the angle of
elevation of the top of the pillar is 30°. After
walking for 10 minutes from A in the same
direction, at a point B, he observes that the angle
of elevation of the top of the pillar is 60°. Then the
time taken (in minutes) by him, form B to reach
the pillar, is :
(1) 5 (2) 6 (3) 10 (4) 20
Ans. (1)
Sol.
30° 60°
A
x
B
y
P
pillar
Q
h
?QPA : 
?
h
x y
 = tan30° ? 
3
h = x + y....(i)
?QPB : 
h
y
 = tan60° ? h = 
3
y .... (ii)
By (i) and (ii) : 3y = x + y ? y = 
x
2
? speed is uniform
Distance x in 10 mins
? Distance 
x
2
 in 5 mins
5. Let two fair six-faced dice A and B be thrown
simultaneously. If E
1
 is the event that die A shows
up four, E
2
 is the event that die B shows up two
and E
3
 is the event that the sum of numbers on
both dice is odd, then which of the following
statements is NOT true ?
(1) E
1
, E
2
 and E
3
 are independent.
(2) E
1
 and E
2
 are independent.
(3) E
2
 and E
3
 are independent.
(4) E
1
 and E
3
 are independent.
Ans. (1)
Sol. E
1
 ? A shows up 4
E
2
 ? B shows up 2
E
3
 ? Sum is odd (i.e. even + odd or odd + even)
P(E
1
) = 
6
6.6
 = 
1
6
P(E
2
) = 
6
6.6
 = 
1
6
P(E
3
) = 
? ? 3 3 2
6.6
 = 
1
2
P(E
1
 ? E
2
) = 
1
6.6
 = P(E
1
) . P(E
2
)
? E
1
 & E
2
 are independent
P(E
1
 ? E
3
) = 
1.3
6.6
 = P(E
1
) . P(E
3
)
? E
1
 & E
3
 are independent
P(E
2
 ? E
3
) = 
1.3
6.6
 = 
1
12
 = P(E
2
) . P(E
3
)
? E
2
 & E
3
 are independent
P(E
1
 ? E
2 
? E
3
) = 0 ie imposible event.
6. If the standard deviation of the numbers 2, 3, a and
11 is 3.5, then which of the following is true ?
(1) 3a
2
 – 23a + 44 = 0
(2) 3a
2
 – 26a + 55 = 0
(3) 3a
2
 – 32a + 84 = 0
(4) 3a
2
 – 34a + 91 = 0
Ans. (3)
Sol. ?
2
2
i i
x x
S.D. –
n n
? ? ? ?
?
? ?
? ?
?
2
2
49 4 9 a 121 16 a
4 4 4
? ? ? ? ? ?
? ?
? ?
? ?
? 3a
2
 – 32a + 84 = 0
7. For x ? R, f(x) = |log2 – sinx| and g(x) = f(f(x)),
then :
(1) g is differentiable at x = 0 and g'(0) = –sin(log2)
(2) g is not differentiable at x = 0
(3) g'(0) = cos(log2)
(4) g'(0) = –cos(log2)
Ans. (3)
Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx
? g(x) = f(f(x)) = log2 – sin(f(x))
      = log2 – sin(log2 – sinx)
It is differentiable at x = 0, so
? g?(x) = –cos(log2 – sinx) (–cosx)
? g?(0) = cos(log2)
Page 3


1
1. A value of ? for which 
? ?
? ?
2 3isin
1 2isin
 is purely
imaginary, is :
(1) sin
–1
? ?
? ?
? ?
1
3
(2) 
?
3
(3) 
?
6
(4) sin
–1
? ?
? ?
? ?
? ?
3
4
Ans. (1)
Sol. Z = 
? ?
? ?
2 3isin
1 2isin
? Z = 
? ? ? ? ? ? ? ?
? ?
2
2 3isin 1 2isin
1 4sin
  = 
? ?
? ? ? ?
? ?
2
2
2 6sin 7isin
1 4sin
for purely imaginary Z, Re(Z) = 0
? 2 – 6sin
2
? = 0 ? sin ? = 
?
1
3
?? ? = ±sin
–1
? ?
? ?
? ?
1
3
2. The system of linear equations
x + ?y – z = 0
?x – y – z = 0
x + y – ?z = 0
has a non-trivial solution for :
(1) exactly three values of ?.
(2) infinitely many values of ?.
(3) exactly one value of ?.
(4) exactly two values of ?.
Ans. (1)
Sol.
? ?
? ? ?
? ?
1 1
1 1
1 1
 = 0    ?? ? ? ? ? = 0, 1, –1
PART A – MATHEMATICS
JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 03
th
 APRIL, 2016)
3. A wire of length 2 units is cut into two parts which
are bent respectively to form a square of side = x
units and a circle of radius = r units. If the sum
of the areas of the square and the circle so formed
is minimum, then :
(1) 2x = r
(2) 2x = ( ? + 4)r
(3) (4 – ?)x = ?r
(4) x = 2r
Ans. (4)
Sol.
Square
x
x
r
x
x
given that     4x + 2 ?r = 2
i.e. 
 
2x + ?r = 1
?   r = 
?
?
1 2x
..... (i)
Area A = x
2
 + ?r
2
     = x
2
 + 
?
1
(2x – 1)
2
for min value of area A
dA
dx
 = 0 gives x = 
? ?
2
4
..... (ii)
from (i) & (ii)
r = 
? ?
1
4
..... (iii)
? x = 2r
J E E ( M A I N ) - 2 0 1 6
2
4. A man is walking towards a vertical pillar in a
straight path, at a uniform speed. At a certain point
A on the path, he observes that the angle of
elevation of the top of the pillar is 30°. After
walking for 10 minutes from A in the same
direction, at a point B, he observes that the angle
of elevation of the top of the pillar is 60°. Then the
time taken (in minutes) by him, form B to reach
the pillar, is :
(1) 5 (2) 6 (3) 10 (4) 20
Ans. (1)
Sol.
30° 60°
A
x
B
y
P
pillar
Q
h
?QPA : 
?
h
x y
 = tan30° ? 
3
h = x + y....(i)
?QPB : 
h
y
 = tan60° ? h = 
3
y .... (ii)
By (i) and (ii) : 3y = x + y ? y = 
x
2
? speed is uniform
Distance x in 10 mins
? Distance 
x
2
 in 5 mins
5. Let two fair six-faced dice A and B be thrown
simultaneously. If E
1
 is the event that die A shows
up four, E
2
 is the event that die B shows up two
and E
3
 is the event that the sum of numbers on
both dice is odd, then which of the following
statements is NOT true ?
(1) E
1
, E
2
 and E
3
 are independent.
(2) E
1
 and E
2
 are independent.
(3) E
2
 and E
3
 are independent.
(4) E
1
 and E
3
 are independent.
Ans. (1)
Sol. E
1
 ? A shows up 4
E
2
 ? B shows up 2
E
3
 ? Sum is odd (i.e. even + odd or odd + even)
P(E
1
) = 
6
6.6
 = 
1
6
P(E
2
) = 
6
6.6
 = 
1
6
P(E
3
) = 
? ? 3 3 2
6.6
 = 
1
2
P(E
1
 ? E
2
) = 
1
6.6
 = P(E
1
) . P(E
2
)
? E
1
 & E
2
 are independent
P(E
1
 ? E
3
) = 
1.3
6.6
 = P(E
1
) . P(E
3
)
? E
1
 & E
3
 are independent
P(E
2
 ? E
3
) = 
1.3
6.6
 = 
1
12
 = P(E
2
) . P(E
3
)
? E
2
 & E
3
 are independent
P(E
1
 ? E
2 
? E
3
) = 0 ie imposible event.
6. If the standard deviation of the numbers 2, 3, a and
11 is 3.5, then which of the following is true ?
(1) 3a
2
 – 23a + 44 = 0
(2) 3a
2
 – 26a + 55 = 0
(3) 3a
2
 – 32a + 84 = 0
(4) 3a
2
 – 34a + 91 = 0
Ans. (3)
Sol. ?
2
2
i i
x x
S.D. –
n n
? ? ? ?
?
? ?
? ?
?
2
2
49 4 9 a 121 16 a
4 4 4
? ? ? ? ? ?
? ?
? ?
? ?
? 3a
2
 – 32a + 84 = 0
7. For x ? R, f(x) = |log2 – sinx| and g(x) = f(f(x)),
then :
(1) g is differentiable at x = 0 and g'(0) = –sin(log2)
(2) g is not differentiable at x = 0
(3) g'(0) = cos(log2)
(4) g'(0) = –cos(log2)
Ans. (3)
Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx
? g(x) = f(f(x)) = log2 – sin(f(x))
      = log2 – sin(log2 – sinx)
It is differentiable at x = 0, so
? g?(x) = –cos(log2 – sinx) (–cosx)
? g?(0) = cos(log2)
C O D E - G
3
8. The distance of the point (1, –5, 9) from the plane
x – y + z = 5 measured along the line x = y = z
is :
(1) 
20
3
(2) 3 10 (3) 10 3 (4) 
10
3
Ans. (3)
Sol. Equation of line parallel to x = y = z through
(1, –5, 9) is 
x 1 y 5 z 9
1 1 1
? ? ?
? ? ? ?
If P( ? + 1, ? ?– 5, ? ?+ 9) be point of intesection
of line and plane.
P
(1, –5, 9)
? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ?? ?? ?? ?
? ? ? ?? ? ??
? Coordinates point are (–9, –15, –1)
? Required distance = 
10 3
9. The eccentricity of the hyperbola whose length of
the latus rectum is equal to 8 and the length of its
conjugate axis is equal to half of the distance
between its foci, is :
(1) 3 (2) 
4
3
(3) 
4
3
(4) 
2
3
Ans. (4)
Sol. Given
2
2b
8
a
? .... (1)
2b = ae .... (2)
we know
  b
2
 = a
2
(e
2
 – 1) .... (3)
substitute 
b
a
 = 
e
2
 from (2) in (3)
? 
2
e
4
 = e
2
 – 1
??4 = 3e
2
??e = 
2
3
10. Let P be the point on the parabola, y
2
 = 8x which
is at a minimum distance from the cente C of the
circle, x
2
 + (y + 6)
2
 = 1. Then the equation of the
circle, passing through C and having its centre at
P is :
(1) x
2
 + y
2
 – 4x + 9y + 18 = 0
(2) x
2
 + y
2
 – 4x + 8y + 12 = 0
(3) x
2
 + y
2
 – x + 4y – 12 = 0
(4) x
2
 + y
2
 – 
x
4
 + 2y – 24 = 0
Ans. (2)
Sol. Circle and parabola are as shown :
(0, -6)C
(2t , 4t)
2
O
y = 8x
a =2
2
P
Minimum distance occurs along common normal.
Let normal to parabola be
y + tx = 2.2.t + 2t
3
pass through (0, –6) :
–6 = 4t + 2t
3
 ??t
3
 + 2t + 3 = 0
? t = –1(only real value)
? P(2, –4)
? CP = 4 4 2 2 ? ?
? equation of circle
(x – 2)
2
 + (y + 4)
2
 = 
? ?
2
2 2
? x
2
 + y
2
 – 4x + 8y + 12 = 0
Page 4


1
1. A value of ? for which 
? ?
? ?
2 3isin
1 2isin
 is purely
imaginary, is :
(1) sin
–1
? ?
? ?
? ?
1
3
(2) 
?
3
(3) 
?
6
(4) sin
–1
? ?
? ?
? ?
? ?
3
4
Ans. (1)
Sol. Z = 
? ?
? ?
2 3isin
1 2isin
? Z = 
? ? ? ? ? ? ? ?
? ?
2
2 3isin 1 2isin
1 4sin
  = 
? ?
? ? ? ?
? ?
2
2
2 6sin 7isin
1 4sin
for purely imaginary Z, Re(Z) = 0
? 2 – 6sin
2
? = 0 ? sin ? = 
?
1
3
?? ? = ±sin
–1
? ?
? ?
? ?
1
3
2. The system of linear equations
x + ?y – z = 0
?x – y – z = 0
x + y – ?z = 0
has a non-trivial solution for :
(1) exactly three values of ?.
(2) infinitely many values of ?.
(3) exactly one value of ?.
(4) exactly two values of ?.
Ans. (1)
Sol.
? ?
? ? ?
? ?
1 1
1 1
1 1
 = 0    ?? ? ? ? ? = 0, 1, –1
PART A – MATHEMATICS
JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 03
th
 APRIL, 2016)
3. A wire of length 2 units is cut into two parts which
are bent respectively to form a square of side = x
units and a circle of radius = r units. If the sum
of the areas of the square and the circle so formed
is minimum, then :
(1) 2x = r
(2) 2x = ( ? + 4)r
(3) (4 – ?)x = ?r
(4) x = 2r
Ans. (4)
Sol.
Square
x
x
r
x
x
given that     4x + 2 ?r = 2
i.e. 
 
2x + ?r = 1
?   r = 
?
?
1 2x
..... (i)
Area A = x
2
 + ?r
2
     = x
2
 + 
?
1
(2x – 1)
2
for min value of area A
dA
dx
 = 0 gives x = 
? ?
2
4
..... (ii)
from (i) & (ii)
r = 
? ?
1
4
..... (iii)
? x = 2r
J E E ( M A I N ) - 2 0 1 6
2
4. A man is walking towards a vertical pillar in a
straight path, at a uniform speed. At a certain point
A on the path, he observes that the angle of
elevation of the top of the pillar is 30°. After
walking for 10 minutes from A in the same
direction, at a point B, he observes that the angle
of elevation of the top of the pillar is 60°. Then the
time taken (in minutes) by him, form B to reach
the pillar, is :
(1) 5 (2) 6 (3) 10 (4) 20
Ans. (1)
Sol.
30° 60°
A
x
B
y
P
pillar
Q
h
?QPA : 
?
h
x y
 = tan30° ? 
3
h = x + y....(i)
?QPB : 
h
y
 = tan60° ? h = 
3
y .... (ii)
By (i) and (ii) : 3y = x + y ? y = 
x
2
? speed is uniform
Distance x in 10 mins
? Distance 
x
2
 in 5 mins
5. Let two fair six-faced dice A and B be thrown
simultaneously. If E
1
 is the event that die A shows
up four, E
2
 is the event that die B shows up two
and E
3
 is the event that the sum of numbers on
both dice is odd, then which of the following
statements is NOT true ?
(1) E
1
, E
2
 and E
3
 are independent.
(2) E
1
 and E
2
 are independent.
(3) E
2
 and E
3
 are independent.
(4) E
1
 and E
3
 are independent.
Ans. (1)
Sol. E
1
 ? A shows up 4
E
2
 ? B shows up 2
E
3
 ? Sum is odd (i.e. even + odd or odd + even)
P(E
1
) = 
6
6.6
 = 
1
6
P(E
2
) = 
6
6.6
 = 
1
6
P(E
3
) = 
? ? 3 3 2
6.6
 = 
1
2
P(E
1
 ? E
2
) = 
1
6.6
 = P(E
1
) . P(E
2
)
? E
1
 & E
2
 are independent
P(E
1
 ? E
3
) = 
1.3
6.6
 = P(E
1
) . P(E
3
)
? E
1
 & E
3
 are independent
P(E
2
 ? E
3
) = 
1.3
6.6
 = 
1
12
 = P(E
2
) . P(E
3
)
? E
2
 & E
3
 are independent
P(E
1
 ? E
2 
? E
3
) = 0 ie imposible event.
6. If the standard deviation of the numbers 2, 3, a and
11 is 3.5, then which of the following is true ?
(1) 3a
2
 – 23a + 44 = 0
(2) 3a
2
 – 26a + 55 = 0
(3) 3a
2
 – 32a + 84 = 0
(4) 3a
2
 – 34a + 91 = 0
Ans. (3)
Sol. ?
2
2
i i
x x
S.D. –
n n
? ? ? ?
?
? ?
? ?
?
2
2
49 4 9 a 121 16 a
4 4 4
? ? ? ? ? ?
? ?
? ?
? ?
? 3a
2
 – 32a + 84 = 0
7. For x ? R, f(x) = |log2 – sinx| and g(x) = f(f(x)),
then :
(1) g is differentiable at x = 0 and g'(0) = –sin(log2)
(2) g is not differentiable at x = 0
(3) g'(0) = cos(log2)
(4) g'(0) = –cos(log2)
Ans. (3)
Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx
? g(x) = f(f(x)) = log2 – sin(f(x))
      = log2 – sin(log2 – sinx)
It is differentiable at x = 0, so
? g?(x) = –cos(log2 – sinx) (–cosx)
? g?(0) = cos(log2)
C O D E - G
3
8. The distance of the point (1, –5, 9) from the plane
x – y + z = 5 measured along the line x = y = z
is :
(1) 
20
3
(2) 3 10 (3) 10 3 (4) 
10
3
Ans. (3)
Sol. Equation of line parallel to x = y = z through
(1, –5, 9) is 
x 1 y 5 z 9
1 1 1
? ? ?
? ? ? ?
If P( ? + 1, ? ?– 5, ? ?+ 9) be point of intesection
of line and plane.
P
(1, –5, 9)
? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ?? ?? ?? ?
? ? ? ?? ? ??
? Coordinates point are (–9, –15, –1)
? Required distance = 
10 3
9. The eccentricity of the hyperbola whose length of
the latus rectum is equal to 8 and the length of its
conjugate axis is equal to half of the distance
between its foci, is :
(1) 3 (2) 
4
3
(3) 
4
3
(4) 
2
3
Ans. (4)
Sol. Given
2
2b
8
a
? .... (1)
2b = ae .... (2)
we know
  b
2
 = a
2
(e
2
 – 1) .... (3)
substitute 
b
a
 = 
e
2
 from (2) in (3)
? 
2
e
4
 = e
2
 – 1
??4 = 3e
2
??e = 
2
3
10. Let P be the point on the parabola, y
2
 = 8x which
is at a minimum distance from the cente C of the
circle, x
2
 + (y + 6)
2
 = 1. Then the equation of the
circle, passing through C and having its centre at
P is :
(1) x
2
 + y
2
 – 4x + 9y + 18 = 0
(2) x
2
 + y
2
 – 4x + 8y + 12 = 0
(3) x
2
 + y
2
 – x + 4y – 12 = 0
(4) x
2
 + y
2
 – 
x
4
 + 2y – 24 = 0
Ans. (2)
Sol. Circle and parabola are as shown :
(0, -6)C
(2t , 4t)
2
O
y = 8x
a =2
2
P
Minimum distance occurs along common normal.
Let normal to parabola be
y + tx = 2.2.t + 2t
3
pass through (0, –6) :
–6 = 4t + 2t
3
 ??t
3
 + 2t + 3 = 0
? t = –1(only real value)
? P(2, –4)
? CP = 4 4 2 2 ? ?
? equation of circle
(x – 2)
2
 + (y + 4)
2
 = 
? ?
2
2 2
? x
2
 + y
2
 – 4x + 8y + 12 = 0
J E E ( M A I N ) - 2 0 1 6
4
11. If A = 
? ? ?
? ?
? ?
5a b
3 2
 and A adj A = A A
T
, then
5a + b is equal to :
(1) 13 (2) –1 (3) 5 (4) 4
Ans. (3)
Sol. A = 
5a –b
3 2
? ?
? ?
? ?
 and A
T
 = 
5a 3
–b 2
? ?
? ?
? ?
AA
T
 = 
2 2
25a b 15a – 2b
15a 2b 13
? ? ?
? ?
?
? ?
Now, A adj A = |A|I
2
 = 
10a 3b 0
0 10a 3b
? ? ?
? ?
?
? ?
Given AA
T
 = A. adj A
15a –2b = 0 ...(1)
10a + 3b = 13 ...(2)
Solving we get
5a = 2 and b = 3
? 5a + b = 5
12. Consider f(x) = tan
–1
? ?
?
? ?
? ?
?
? ?
1 sin x
1 sin x
, x ? 
? ? ?
? ?
? ?
0,
2
.
A normal to y = f(x) at x = 
?
6
 also passes through
the point :
(1) 
? ? ?
? ?
? ?
,0
4
(2) (0, 0)
(3) 
? ? ?
? ?
? ?
2
0,
3
(4) 
? ? ?
? ?
? ?
,0
6
Ans. (3)
Sol.
1
1 sin x
f (x) tan
1–sin x
?
? ?
?
?
? ?
? ?
? ?
 where 
x 0,
2
? ? ?
?
? ?
? ?
 = 
2
1
2
(1 sin x)
tan
1– sin x
?
? ?
?
? ?
? ?
? ?
     
1
1 sin x
tan
| cos x |
?
? ? ?
?
? ?
? ?
1
1 sin x
tan as x 0,
cos x 2
?
? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ? ?
2
1
2 2
x x
cos sin
2 2
tan
x x
cos sin
2 2
?
? ?
? ?
? ? ?
? ?
? ?
? ?
?
? ?
?
? ?
? ?
? ?
1
x
tan tan
4 2
?
? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
f(x) 
x
4 2
?
? ?
  as 
x 0,
2
? ? ?
?
? ?
? ?
??
1
f
6 2
? ? ?
? ?
? ?
? ?
? Equation of normal
y – 2 x
3 6
? ? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
which passes through 
2
0,
3
? ? ?
? ?
? ?
13. Two sides of a rhombus are along the lines,
x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals
intersect at (–1, –2), then which one of the
following is a vertex of this rhombus ?
(1) 
? ?
? ?
? ?
? ?
10 7
,
3 3
(2) (–3, –9)
(3) (–3, –8) (4) 
? ?
?
? ?
? ?
1 8
,
3 3
Ans. (4)
Sol. Equation of angle bisector of the lines x – y + 1 = 0
and 7x – y – 5 = 0 is given by
x y 1 7x y 5
2 5 2
? ? ? ?
? ?
? ?5(x – y + 1) = 7x – y – 5
and
  5(x – y + 1) = –7x + y + 5
? 2x + 4y – 10 = 0 ? x + 2y – 5 = 0 and
12x – 6y = 0 ??2x – y = 0
Now equation of diagonals are
(x + 1) + 2(y + 2) = 0 ? ?x + 2y + 5 = 0 ...(1)
and
2(x + 1) – (y + 2) = 0 ? ?2x – y = 0      ...(2)
Clearly 
? ?
?
? ?
? ?
1 8
,
3 3
 lies on (1)
Page 5


1
1. A value of ? for which 
? ?
? ?
2 3isin
1 2isin
 is purely
imaginary, is :
(1) sin
–1
? ?
? ?
? ?
1
3
(2) 
?
3
(3) 
?
6
(4) sin
–1
? ?
? ?
? ?
? ?
3
4
Ans. (1)
Sol. Z = 
? ?
? ?
2 3isin
1 2isin
? Z = 
? ? ? ? ? ? ? ?
? ?
2
2 3isin 1 2isin
1 4sin
  = 
? ?
? ? ? ?
? ?
2
2
2 6sin 7isin
1 4sin
for purely imaginary Z, Re(Z) = 0
? 2 – 6sin
2
? = 0 ? sin ? = 
?
1
3
?? ? = ±sin
–1
? ?
? ?
? ?
1
3
2. The system of linear equations
x + ?y – z = 0
?x – y – z = 0
x + y – ?z = 0
has a non-trivial solution for :
(1) exactly three values of ?.
(2) infinitely many values of ?.
(3) exactly one value of ?.
(4) exactly two values of ?.
Ans. (1)
Sol.
? ?
? ? ?
? ?
1 1
1 1
1 1
 = 0    ?? ? ? ? ? = 0, 1, –1
PART A – MATHEMATICS
JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 03
th
 APRIL, 2016)
3. A wire of length 2 units is cut into two parts which
are bent respectively to form a square of side = x
units and a circle of radius = r units. If the sum
of the areas of the square and the circle so formed
is minimum, then :
(1) 2x = r
(2) 2x = ( ? + 4)r
(3) (4 – ?)x = ?r
(4) x = 2r
Ans. (4)
Sol.
Square
x
x
r
x
x
given that     4x + 2 ?r = 2
i.e. 
 
2x + ?r = 1
?   r = 
?
?
1 2x
..... (i)
Area A = x
2
 + ?r
2
     = x
2
 + 
?
1
(2x – 1)
2
for min value of area A
dA
dx
 = 0 gives x = 
? ?
2
4
..... (ii)
from (i) & (ii)
r = 
? ?
1
4
..... (iii)
? x = 2r
J E E ( M A I N ) - 2 0 1 6
2
4. A man is walking towards a vertical pillar in a
straight path, at a uniform speed. At a certain point
A on the path, he observes that the angle of
elevation of the top of the pillar is 30°. After
walking for 10 minutes from A in the same
direction, at a point B, he observes that the angle
of elevation of the top of the pillar is 60°. Then the
time taken (in minutes) by him, form B to reach
the pillar, is :
(1) 5 (2) 6 (3) 10 (4) 20
Ans. (1)
Sol.
30° 60°
A
x
B
y
P
pillar
Q
h
?QPA : 
?
h
x y
 = tan30° ? 
3
h = x + y....(i)
?QPB : 
h
y
 = tan60° ? h = 
3
y .... (ii)
By (i) and (ii) : 3y = x + y ? y = 
x
2
? speed is uniform
Distance x in 10 mins
? Distance 
x
2
 in 5 mins
5. Let two fair six-faced dice A and B be thrown
simultaneously. If E
1
 is the event that die A shows
up four, E
2
 is the event that die B shows up two
and E
3
 is the event that the sum of numbers on
both dice is odd, then which of the following
statements is NOT true ?
(1) E
1
, E
2
 and E
3
 are independent.
(2) E
1
 and E
2
 are independent.
(3) E
2
 and E
3
 are independent.
(4) E
1
 and E
3
 are independent.
Ans. (1)
Sol. E
1
 ? A shows up 4
E
2
 ? B shows up 2
E
3
 ? Sum is odd (i.e. even + odd or odd + even)
P(E
1
) = 
6
6.6
 = 
1
6
P(E
2
) = 
6
6.6
 = 
1
6
P(E
3
) = 
? ? 3 3 2
6.6
 = 
1
2
P(E
1
 ? E
2
) = 
1
6.6
 = P(E
1
) . P(E
2
)
? E
1
 & E
2
 are independent
P(E
1
 ? E
3
) = 
1.3
6.6
 = P(E
1
) . P(E
3
)
? E
1
 & E
3
 are independent
P(E
2
 ? E
3
) = 
1.3
6.6
 = 
1
12
 = P(E
2
) . P(E
3
)
? E
2
 & E
3
 are independent
P(E
1
 ? E
2 
? E
3
) = 0 ie imposible event.
6. If the standard deviation of the numbers 2, 3, a and
11 is 3.5, then which of the following is true ?
(1) 3a
2
 – 23a + 44 = 0
(2) 3a
2
 – 26a + 55 = 0
(3) 3a
2
 – 32a + 84 = 0
(4) 3a
2
 – 34a + 91 = 0
Ans. (3)
Sol. ?
2
2
i i
x x
S.D. –
n n
? ? ? ?
?
? ?
? ?
?
2
2
49 4 9 a 121 16 a
4 4 4
? ? ? ? ? ?
? ?
? ?
? ?
? 3a
2
 – 32a + 84 = 0
7. For x ? R, f(x) = |log2 – sinx| and g(x) = f(f(x)),
then :
(1) g is differentiable at x = 0 and g'(0) = –sin(log2)
(2) g is not differentiable at x = 0
(3) g'(0) = cos(log2)
(4) g'(0) = –cos(log2)
Ans. (3)
Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx
? g(x) = f(f(x)) = log2 – sin(f(x))
      = log2 – sin(log2 – sinx)
It is differentiable at x = 0, so
? g?(x) = –cos(log2 – sinx) (–cosx)
? g?(0) = cos(log2)
C O D E - G
3
8. The distance of the point (1, –5, 9) from the plane
x – y + z = 5 measured along the line x = y = z
is :
(1) 
20
3
(2) 3 10 (3) 10 3 (4) 
10
3
Ans. (3)
Sol. Equation of line parallel to x = y = z through
(1, –5, 9) is 
x 1 y 5 z 9
1 1 1
? ? ?
? ? ? ?
If P( ? + 1, ? ?– 5, ? ?+ 9) be point of intesection
of line and plane.
P
(1, –5, 9)
? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ?? ?? ?? ?
? ? ? ?? ? ??
? Coordinates point are (–9, –15, –1)
? Required distance = 
10 3
9. The eccentricity of the hyperbola whose length of
the latus rectum is equal to 8 and the length of its
conjugate axis is equal to half of the distance
between its foci, is :
(1) 3 (2) 
4
3
(3) 
4
3
(4) 
2
3
Ans. (4)
Sol. Given
2
2b
8
a
? .... (1)
2b = ae .... (2)
we know
  b
2
 = a
2
(e
2
 – 1) .... (3)
substitute 
b
a
 = 
e
2
 from (2) in (3)
? 
2
e
4
 = e
2
 – 1
??4 = 3e
2
??e = 
2
3
10. Let P be the point on the parabola, y
2
 = 8x which
is at a minimum distance from the cente C of the
circle, x
2
 + (y + 6)
2
 = 1. Then the equation of the
circle, passing through C and having its centre at
P is :
(1) x
2
 + y
2
 – 4x + 9y + 18 = 0
(2) x
2
 + y
2
 – 4x + 8y + 12 = 0
(3) x
2
 + y
2
 – x + 4y – 12 = 0
(4) x
2
 + y
2
 – 
x
4
 + 2y – 24 = 0
Ans. (2)
Sol. Circle and parabola are as shown :
(0, -6)C
(2t , 4t)
2
O
y = 8x
a =2
2
P
Minimum distance occurs along common normal.
Let normal to parabola be
y + tx = 2.2.t + 2t
3
pass through (0, –6) :
–6 = 4t + 2t
3
 ??t
3
 + 2t + 3 = 0
? t = –1(only real value)
? P(2, –4)
? CP = 4 4 2 2 ? ?
? equation of circle
(x – 2)
2
 + (y + 4)
2
 = 
? ?
2
2 2
? x
2
 + y
2
 – 4x + 8y + 12 = 0
J E E ( M A I N ) - 2 0 1 6
4
11. If A = 
? ? ?
? ?
? ?
5a b
3 2
 and A adj A = A A
T
, then
5a + b is equal to :
(1) 13 (2) –1 (3) 5 (4) 4
Ans. (3)
Sol. A = 
5a –b
3 2
? ?
? ?
? ?
 and A
T
 = 
5a 3
–b 2
? ?
? ?
? ?
AA
T
 = 
2 2
25a b 15a – 2b
15a 2b 13
? ? ?
? ?
?
? ?
Now, A adj A = |A|I
2
 = 
10a 3b 0
0 10a 3b
? ? ?
? ?
?
? ?
Given AA
T
 = A. adj A
15a –2b = 0 ...(1)
10a + 3b = 13 ...(2)
Solving we get
5a = 2 and b = 3
? 5a + b = 5
12. Consider f(x) = tan
–1
? ?
?
? ?
? ?
?
? ?
1 sin x
1 sin x
, x ? 
? ? ?
? ?
? ?
0,
2
.
A normal to y = f(x) at x = 
?
6
 also passes through
the point :
(1) 
? ? ?
? ?
? ?
,0
4
(2) (0, 0)
(3) 
? ? ?
? ?
? ?
2
0,
3
(4) 
? ? ?
? ?
? ?
,0
6
Ans. (3)
Sol.
1
1 sin x
f (x) tan
1–sin x
?
? ?
?
?
? ?
? ?
? ?
 where 
x 0,
2
? ? ?
?
? ?
? ?
 = 
2
1
2
(1 sin x)
tan
1– sin x
?
? ?
?
? ?
? ?
? ?
     
1
1 sin x
tan
| cos x |
?
? ? ?
?
? ?
? ?
1
1 sin x
tan as x 0,
cos x 2
?
? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ? ?
2
1
2 2
x x
cos sin
2 2
tan
x x
cos sin
2 2
?
? ?
? ?
? ? ?
? ?
? ?
? ?
?
? ?
?
? ?
? ?
? ?
1
x
tan tan
4 2
?
? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
f(x) 
x
4 2
?
? ?
  as 
x 0,
2
? ? ?
?
? ?
? ?
??
1
f
6 2
? ? ?
? ?
? ?
? ?
? Equation of normal
y – 2 x
3 6
? ? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
which passes through 
2
0,
3
? ? ?
? ?
? ?
13. Two sides of a rhombus are along the lines,
x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals
intersect at (–1, –2), then which one of the
following is a vertex of this rhombus ?
(1) 
? ?
? ?
? ?
? ?
10 7
,
3 3
(2) (–3, –9)
(3) (–3, –8) (4) 
? ?
?
? ?
? ?
1 8
,
3 3
Ans. (4)
Sol. Equation of angle bisector of the lines x – y + 1 = 0
and 7x – y – 5 = 0 is given by
x y 1 7x y 5
2 5 2
? ? ? ?
? ?
? ?5(x – y + 1) = 7x – y – 5
and
  5(x – y + 1) = –7x + y + 5
? 2x + 4y – 10 = 0 ? x + 2y – 5 = 0 and
12x – 6y = 0 ??2x – y = 0
Now equation of diagonals are
(x + 1) + 2(y + 2) = 0 ? ?x + 2y + 5 = 0 ...(1)
and
2(x + 1) – (y + 2) = 0 ? ?2x – y = 0      ...(2)
Clearly 
? ?
?
? ?
? ?
1 8
,
3 3
 lies on (1)
C O D E - G
5
14. If a curve y = f(x) passes through the point (1, –1) and
satisfies the differential equation, y(1 + xy) dx = x dy,
then f
? ?
?
? ?
? ?
1
2
 is equal to :
(1) 
4
5
(2) 
?
2
5
(3) 
?
4
5
(4) 
2
5
Ans. (1)
Sol. Given differential equation
ydx + xy
2
dx = xdy
?
2
xdy – ydx
y
= xdx
2
x x
d d
y 2
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
Integrating we get
2
x x
– C
y 2
? ?
??It passes through (1, –1)
?
1 1
1 C C
2 2
? ? ? ?
?
2
2
2x 2x
x 1 0 y
y x 1
?
? ? ? ? ?
?
?
1 4
f –
2 5
? ?
?
? ?
? ?
15. If all the words (with or without meaning) having
five letters, formed using the letters of the word
SMALL and arranged as in a dictionary; then the
position of the word SMALL is :
(1) 58
th
(2) 46
th
(3) 59
th
(4) 52
nd
Ans. (1)
Sol. Total number of words which can be formed using
all the letters of the word 'SMALL'
= 
5!
60
2!
?
Now, 60
th
 word is ??SMLLA
59
th
 word is ??SMLAL
58
th 
word is ??SMALL
16. If the 2
nd
, 5
th
 and 9
th
 terms of a non-constant A.P.
are in G.P., then the common ratio of this G.P. is :-
(1) 
7
4
(2) 
8
5
(3) 
4
3
(4) 1
Ans. (3)
Sol. Let 'a' be the first term and d be the common
difference
2
nd
 term = a + d, 5
th
 term = a + 4d,
9
th
 term = 4 + 8d
? ? ? ? ? ? ? ? ??Common ratio = 
a 4d a 8d
a d a 4d
? ?
?
? ?
 = 
4d 4
3d 3
?
17. If the number of terms in the expansion of
n
2
2 4
1 ,x 0
x x
? ?
? ? ?
? ?
? ?
, is 28, then the sum of the
coefficients of all the terms in this expansion, is :-
(1) 729 (2) 64 (3) 2187 (4) 243
Ans. (1 or Bonus)
Sol. Number of terms in the expansion of
n
2
2 4
1–
x x
? ?
?
? ?
? ?
 is 
n + 2
C
2
 (considering 
1
x
 and 
2
1
x
distinct)
?
n + 2
C
2
 = 28 ??n = 6
? Sum of coefficients = (1 – 2 + 4)
6
 = 729
But number of dissimilar terms actually will be
2n + 1 (as 
1
x
 and 
2
1
x
 are functions as same
variable)
Hence it contains error, so a bonus can be
expected.
18. If the sum of the first ten terms of the series
2 2 2 2
2
3 2 1 4
1 2 3 4 4 ...,
5 5 5 5
? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
is 
16
m,
5
then m is equal to :-
(1) 99 (2)  102
(3) 101 (4) 100
Ans. (3)
Sol. Given series is
2 2
2 2 2
8 12 16
S ...10 terms
5 5 5
?
? ? ? ?
  = 
2
2 2 2
2
4
(2 3 4 ....10 terms)
5
? ? ?
  = 
16 11.12.23
1
25 6
? ?
?
? ?
? ?
= 
16
505
25
?
? m = 101