Title JEE Mathematics CENGAGE LEARNING Tangent Triangle Trigonometric Functions Sine Lattice (Group) 1.4 MB 32
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Page 1

Appendix B
Mock Test 1

PaPer 1
Section a: Only One Option Correct Type

1. For q
p

Œ
Ê
ËÁ

ˆ
¯̃

0
2

, , the value of definite integral

ln ( tan tan )1
0

+Ú q
q

x dx is equal to

(a) q ln (sec q) (b) q ln(cosec q)

(c)
q ln 2

2
(d) 2q ln sec q

2. The focal length of a mirror is given by
1 1 2
v u f

- = .

If errors made in measuring u and v are a, then the
relative error in f is

(a)
2
a

(b) a
1 1
u v

+
Ê
ËÁ

ˆ
¯̃

(c) a
1 1
u v

-
Ê
ËÁ

ˆ
¯̃

(d) None of these

3. If the function f (x) = 2 tan x + (2a +1) loge | sec x | +
(a –2) x is increasing on R, then

(a) a Œ (1/2, •) (b) a Œ (–1/2, 1/2)
(c) a = 1/2 (d) a Œ R
4. 3x + 4y – 7 = 0 and 3x – 4y –7 = 0 are equations of

asymptotes of a hyperbola H. From a point P(3, 4),
pair of tangents are drawn to hyperbola H in such
a way that both tangents touch the same branch of
hyperbola H. Then its eccentricity is

(a)
4
3

(b)
7
3

(c)
5
3

(d)
5
4

5. A parabola touches two given straight lines originat-
ing from a given point. The locus of the mid point
of the portion of any tangent, which is intercepted
between the given straight lines, is a/an

(a) parabola (b) ellipse
(c) straight line (d) hyperbola

6. If x x2
4

2- + -
p

sin = | x2 – 2 | + | sin x | +
p
4

, then

(a) x Œ( , )0 2 (b) x Œ -( , )2 2

(c) x ŒR (d) x Œ -( , )2 0

7. If A and B are different matrices satisfying A3 = B3
and A2B = B2A, then

(a) det (A2 + B2) must be zero.
(b) det (A – B) must be zero.
(c) det (A2 + B2) as well as det (A – B) must be zero
(d) At least one of det (A2 + B2) or det (A – B) must

be zero.
8. A fair dice is thrown 3 times. The probability that the

product of the three outcomes is a prime number is

(a)
1
24

(b)
1
36

(c)
1
32

(d)
1
8

9. The number of real solution of equation
16 sin–1 x tan–1 x cosec–1 x = p3 is/are
(a) 0 (b) 1 (c) 2 (d) infinite
10. Let a, b, c are non-zero constant numbers. Then

Lim
r Æ •

cos cos cos

sin sin

a
r

b
r

c
r

b
r

c
r

-
equals

(a)
a b c

bc

2 2 2

2
+ -

(b)
c a b

bc

2 2 2

2
+ -

(c)
b c a

bc

2 2 2

2
+ -

(d) Independent of a, b and c

Section B: One or More Options Correct
Type

11. Given 2 functions f and g which are integrable on
every interval and satisfy

(i) f is odd, g is even
(ii) g(x) = f (x + 5), then
(a) f (x – 5) = g(x)
(b) f (x – 5) = – g(x)

(c) f t dt g t dt( ) ( )
0

5

0

5

5Ú Ú= -

(d) f t dt g t dt( ) ( )
0

5

0

5

5Ú Ú= - -
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B.2 Mathematics

12. A continuous function f (x) on R Æ R satisfies the
relation f (x) + f (2x + y) + 5xy = f (3x – y) + 2x2 + 1
for " x, y Œ R, then which of the following hold(s)
good?

(a) f is many-one
(b) f has no minima
(c) f is neither odd nor even
(d) f is bounded
13. The equation x2 – 4x + a sin a = 0 has real roots
(a) for all values of a
(b) for all values of a provided –p/4 < a < p/4
(c) for all values of a ≥ 4 provided p £ a £ 2p
(d) for all a provided | a | £ 4
14. Which of the following statement(s) is/are not correct?
(a) If the roots of a quadratic equation are imagi-

nary, then these are conjugates of each other.
(b) If a continuous function is strictly monotonic,

then it is differentiable.
(c) If f (x) is periodic, then | f (x) | is also periodic.
(d) sin x/(2x – 2np – p) and cos x tan x/(2x – 2np

– p), where n Œ Z, are identical functions.
15. If the curve y = ax1/2 + bx passes through the point

(1, 2) and lies above the x-axis for 0 £ x £ 9 and the

area enclosed by the curve, the x-axis and the line
x = 4 is 8 sq. units, then

(a) a2 + b2 = 6 (b) a/b = 1
(c) a – b = 4 (d) ab = –3

Section C: Integer Value Correct Type

16. The value of Lim
n

k n

n

kÆ •
=

+

Â 1
2

21( )

is

17. If the slope of the curve y =
ax

b x-
at the point (1, 1)

is 2, then the value of a + b is
18. Given

a = 3i + j + 2k,

b = i – 2j – 4k are the posi-

tion vectors of point A and B. Then the distance of
point –i + j + k from the plane passing through B and
perpendicular to AB is

19. If z = x + iy (x, y Œ R, x π –1/2), then the number of
values of z satisfying | z |n = z2 | z |n–2 + z | z |n–2 + 1,
(n Œ N, n > 1) is

20. If x, y, z are distinct positive numbers such that

x
y

y
z

z
x

+ = + = +
1 1 1

, then the value of xyz is

PaPer 2
Section a: One or More Options Correct
Type

1. For natural numbers m and n, if (1 – y)m (1 + y)n
= 1 + a1 y + a2 y

2 + L, and a1 = a2 = 10, then
(a) m < n (b) m > n
(c) m + n = 80 (d) m – n = 20
2. Let P(x) = x2 + bx + c, where b and c are integer. If

P(x) is a factor of both x4 + 6x2 + 25 and 3x4 + 4x2 +
28x + 5, then

(a) P(x) = 0 has imaginary roots
(b) P(x) = 0 has roots of opposite sign
(c) P(1) = 4
(d) P(1) = 6
3. Let tan x – tan2 x > 0 and | 2 sin x | < 1. Then the int-

ersection of which of the following two sets satisfies
both inequalities?

(a) x > np + p/6, n Œ Z
(b) x > np – p/6, n Œ Z
(c) x < np – p/4, n Œ Z
(d) x < np + p/4, n Œ Z

4. A bag initially contains one red and two blue balls. An
experiment consisting of selecting a ball at random,
noting its colour and replacing it together with an
additional ball of the same colour. If three such trials

(a) probability that atleast one blue ball is drawn
is 0.9.

(b) probability that exactly one blue ball is drawn
is 0.2.

(c) probability that all the drawn balls are red given
that all the drawn balls are of same colour is 0.2.

(d) probability that atleast one red ball is drawn is 0.6.
5. If a, b, c are non-zero real numbers such that

bc ca ab
ca ab bc
ab bc ca

= 0, then

(a)
1 1 1

0
a b c

+ + = (b)
1 1 1

02a b c
+ + =

w w

(c)
1 1 1

02a b cw w
+ + = (d) None of these

6. 2007201 + 2019201 – 1982201 – 2044201 is divisible by
(a) 74 (b) 50 (c) 1850 (d) 2013

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B.16 Mathematics

\ f (x) =
x x+ sin

,
2

x Œ [0, p]

p £ t £ 2p, then
f (t) + f (2p – t) = p

f (t) +
2 2

2
p p- + -t tsin ( )

= p

f (t) + p –
t t
2 2

-
sin

= p

f (t) =
t t+ sin

2

f (x) =
x x+ sin

2
p £ x £ 2p

f (x) =
x x+ sin

2
for 0 £ x < 2p

f (x) = f (4p – x) for x Œ [2p, 4p]
f (x) is symmetric about x = 2p.

O 2p 4p

Fig. B.2

a = 2p – 0 = 2p from graph
b = a
Ans 15. (a), 16. (d):
From the graph (Fig. B.3), y = 6 and x = 8 are the

common tangents.

C(8,10)
A(8,6) B(12,6)

y

x

Fig. B.3

15. Area of DABC = 8 sq. unit
2s = 4 + 4 + 4 2

s

= =
+

=
+

= -
D 8

4 4 2
2

1 2
2 2 1

Co-ordinate of centre are ( , )7 2 2 5 2 2+ +

16. Centre of circumcircle is (10, 8) and radius 2 2 .
Therefore, equation of circle is
x2 + y2 – 20x – 16y + 156 = 0

Section C
17. Ans: (a Æ q); (b Æ s); (c Æ r); (d Æ p)
(a) lim

x Æ -1
f (g(x)) = f (2–) = 2

lim
x Æ +1

f (g(x)) = f (1–) = 1

Therefore, limit does not exist.

(b) lim
x Æ 2

f (x) = 2 fi lim ( )
x

f x
Æ

- =
2

3 2 2

(c) lim
x Æ 0

f (x) = 0 and lim
x Æ 0

g(x) = finite quantity

\ lim
( )
( )

( ) ( )
x

f x
g x

f x g x
Æ

+
Ê
ËÁ

ˆ
¯̃

=
0

0

(d) lim
( ) ( )
( ) ( )

( )
x

f x g x
f x g xÆ +

-
+

=
-

+
= =

1

3 3 1 1
1 1

2
2

1

18. Ans: (a Æ r), (b Æ r), (c Æ q), (d Æ s)

(a) arg
z
z

2

2
1
1

-
+

Ê

Ë
Á

ˆ

¯̃
= 0 ; z π ± i

z
z

z
z

z z z z
2

2

2

2
1
1

1
1

0 0
-
+

=
-
+

fi - = + =,

y = 0, x = 0
Locus of z is portion of pair of lines xy = 0.

Q
z
z

2

2
1
1

0
-
+

Ê

Ë
Á

ˆ

¯̃
>

È

Î
Í
Í

˘

˚
˙
˙

(b) Given
|| z – cos–1 cos 12 | – | z – sin–1 sin 12 || = 8 (p – 3)
Since, | cos–1 cos 12 – sin–1 sin 12 | = 8 (p – 3)
Therefore, locus of z is the portion of a line

joining z1 and z2 except the segment between
z1 and z2.

(c) z2 – i | z1 |
2 = k2 – k1

x2 – y2 + 2 ixy – il1 = l2

x2 – y2 = l2 and xy =
l1
2

The locus of z is the point of intersection of
hyperbola.

(d) Given,

z z- - + + - =- -1
1
3

1
3 2

11 1sin cos
p

Since 1
1
3

1
3 2

11 1+ + - =- -sin cos
p

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Appendix B: Mock Tests B.17

| z – z1 | + | z – z2 | = | z1 + z2 |
Thus, the locus of z is line segment joining z1

and z2.
19. Ans: (a Æ q), (b Æ p), (c Æ s), (d Æ r)
Angles of the triangle are 30º, 45°, 105°.

Then
a b c

sin sin sin30 45 105∞
=

=

fi
a b c

1 2 1 2 3 1 2 2/ / ( )/
= =

+

fi
a b c

k
2 2 3 1

= =
+

= (say)

Smallest side, a = 2 k and largest side, c = ( )3 1+ k

If a k c= = = +2 2 6 2, ,

If a k c= = = +2 1 3 1, ,

If s k c= + + = = +3 2 3 2 2 2 3, ,

If D =
-3 1

4
, then from D =

1
2

bc sin A, we have

3 1
4

1
2

2 3 1 30
-

= ¥ ¥ + ∞k k( ) sin

fi k k2
23 1

2 3 1
3 1

4
3 1
2

=
-
+

=
-

fi =
-

( )
( )

and hence c = 1
20. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ q)
Let Ei denotes the event that the bag contains i black

and (14 – i) white balls (i = 0, 1, 2, ... 14) and A
denotes the event that five balls drawn are all black.
Then

P(Ei) =
1

15
(i = 0, 1, ... 14)

P(A/Ei) = 0 for i = 0, 1, 2, 3, 4

P(A/Ei) =
i C

C
5

14
5

for i ≥ 5

(a) P(A) = P P( ) ( / )E A Ei i
i =
Â

0

14

=
1

15
1

14
5C

(5C5 +
6C5 + L +

14C5)

=
1

15
1
6

15
6

14
5

C
C

=

(b) Clearly P(A/E11) =
11

5
14

5

3
13

C
C

=

(c) From Baye’s theorem,

P(A/E11) =
P P

P
( ) ( / )

( )

.E A E
A

11 11

1
15

3
13

1
6

6
65

= =

(d) Let B denotes the probability of 3 black and 2
white balls, then

P(B/Ei) = 0 if i = 0, 1, 2 or 13, 14

P(B/Ei) =
i iC C

C
3

14
2

14
5

-

for i = 3, 4, ..., 12

\ P(B) = P P( ) ( / )E B Ei i
i =
Â

0

14

=
1

15
1

14
5

.
C

[3C3.
11C2 +

4C3.
10C2 + L +

12C3.
2C2]

=
5005

15
1
614 5. C

=

Solutions of Mock Test 2

PaPer 1
Section a

1. Ans (b): f (x) = ( ) ( )x n n n

n

- -

=
’ 51

1

50

ln f (x) = n n x n
n

( ) ln ( )51
1

50

- -
=

Â

Differentiating both sides with respect to x, we get

¢

=
-

-
=

Âf xf x
n n

x n
n

( )
( )

( )51

1

50

¢

=
-

-
=

Âff
n n

n
n

( )
( )

( )51
51

51
51

1

50

=1 + 2 + 3 + L + 50
= 1275

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Appendix B: Mock Tests B.31

= -
- -Ê

ËÁ
ˆ
¯̃

-

ËÁ
ˆ
¯̃

+

+ +

Ê

Ë

Á
Á
Á
Á

ˆ

¯

˜
˜
˜̃

4 5
26
3

2
10
3

6

1 25 4

x y z-

=
+

-
=

+

-
= -

4
1

26
3

5

10
3

2
2

fi x = 2, y =
4
3

, z =
2
3

\ 2
4
3

2
3

, ,Ê
ËÁ

ˆ
¯̃

Volume of the pyramid

=
1
3

(Base area) ¥ Height

=
1
3

AB AC
   

¥ ¥ EP

=
1
3

5 2 120i j k  - - ¥

=
1
3

30 120 = 20 cubic units

15. Ans (b): Coordinate of any point on line AB can be
taken as

h = r1 + r cos q

C
B

O

C1

C2

A r( , 0)1

X

Y

Fig. B.15

k = 0 + r sin q
Therefore, it lies on C2.
\ (r1 + r cos q)

2 + r2 sin2q = r2
2

r2 + 2 r r1 cos q + r1
2 – r2

2 = 0
Let AB = rAB and AC = rAC. Therefore,
rAB + rAC = –2r1cos q, rAB.rAC = r1

2 – r2
2

So (BC) = | AC – AB |
= | rAC – rAB |

= ( )r r r rAC AB AC AB+ -
2 4

= 4 4 41
2 2

1
2

2
2r r rcos q - +

BC = - +4 41
2 2

2
2r rsin q

Therefore, for max sin q = 0
BC2max = 4r2

2

16. Ans (a):
Now OA2 + OB2 + BC2 = r1

2 + r2
2 + 4r2

2 – 4r1
2sin2q

= 5r2
2 + r1

2 – 4r1
2sin2q

\ 0 £ sin2q £ 1
So OA2 + OB2 + OC2 Œ [5r2

2 – 3r1
2, 5r2

2 + r1
2]

Section C

17. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ s)
(a) we get common normal perpendicular to y = x.

So, slope = – 1 fi x + y = 3a
(b) Tangent to the parabola y = mx + a/m passes

through the point P(h, k).
fi m2h – mk + a = 0. If its roots are m1 and m2,

then m1m2 = +1. Locus is x = a.
(c) The tangents are y = m(x + a) + a/m (1)

and y =
-1
m

(x + 2a) – 2am (2)

Subtracting from (1) to (2), we get x + 3 a = 0.
(d) If the chord joining t1 and t2 subtends 90º at

vertex, then t1t2 = –4. point of intersection of
tangents is (–at1t2, –a(t1 + t2)). So the locus is
x = 4a.

18. Ans: (a Æ r); (b Æ p); (c Æ q); (d Æ s)
(a) | |

 
a b¥ = | |

 
b c¥ = | |

 
c a¥ = 2DABC

Also directions of
 
a b¥ ,

 
b c¥ and

 
c a¥ are

the same.
Hence,

 
a b¥ =

 
b c¥ =

 
c a¥

(b) For regular tetrahedron all sides are of equal
length, hence, | | | | | |

  
a b c= = . Also, all the faces

are equilateral triangle.
Therefore, angle between

a and

b is 60º,

b

and

c is 60º, and between

a and

c is 60º.

Hence,
     
a b b c c a. . .= =

(c) Since
 
a b¥ =

c fi

 
a c^ and

 
b c^ and

  
b c a¥ = fi

 
b a^ and

 
c a^ .

Therefore,

a ,

b ,

c are mutually perpendicular.

(d) Since
  
a b c+ + = 0

fi
        
a b c a b b c c a2 2 2 2 0+ + + + + =( . . . )

fi
     
a b b c c a. . .+ + = -

3
2

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B.32 Mathematics

19. Ans: (a Æ r); (b Æ s); (c Æ q); (d Æ p)

(a) f (x) =
mx c x

e xx
+ <

Ï
Ì
Ô

ÓÔ

0
0

For application of LMV function must be con-
tinous and derivable in [–2, 2].

f (0+) = f (0–) fi c = 1
f ¢(0+) = f ¢(0–) fi m = 1 fi m + 3c = 4
(b)

(6, 2)

(4, 4)

(2, 6)

Fig. B.16

Length of latus rectum is 4 2 = 4a
fi a = 2
Therefore, equation of directrix is x + y + l = 0
whose distance from (4, 4) is 2 2 .

\
4 4

2
+ + l

= 2 2 fi l = ± 4 – 8

fi l = –4 or –12
Therefore, directrix are x + y = 4 and x + y = 12.
\ l1 + l2 = 4 + 12 = 16
(c) f ¢(x) = 6x2 – 6x – 12 = 0 fi (x – 2)(x + 1) = 0
fi x = –1, 2
f ¢¢(x) = 12x – 6
f ¢¢(–1) = –18 < 0 fi x = –1 is the point of maxima
f ¢¢(2) = 18 > 0 fi x = 2 is the point of minima.

Now f (–1) = 8, f (2) = –19, f(5/2) = -
33
2

,

f (–2) = –5
\ maximum value of f(x) in [–2, 5/2] is 8.

(d) lim
( ) ( )

....

n

n n

n n

Æ•

-
+

-
+

+
-

Ê

Ë

Á
Á
Á
Á

ˆ

¯

˜
˜
˜
˜

1
2 1

1
4 2

1
2

2 2

2 2

= lim
n

r

n

rn rÆ• = -
Â 1

2 21

= lim
n

r

n

n
r
n

r
n

Æ•
=

-
Â 1

2
2

2
1

=
dx
x x2 20

1

-
Ú

=
dx
x1 1 20

1

- -
Ú

( )

= sin ( )- -ÈÎ ˘̊
1

0

1
1x

= 0
2 2 4

2- -Ê
ËÁ

ˆ
¯̃

= = fi =
p p p

k k

20. Ans: (a Æ r); (b Æ s); (c Æ p); (d Æ s)
(a) 1 + a10 + a20 + L + a190 = 0
as 10 is not an integral multiple of 20
(b) (log2x)

2 + log2(0.03125) + log2x + 3 < 0
fi (log2x)

2 + log2(1/32) + log2x + 3 < 0
fi (log2x)

2 + (log2x) – 2 < 0
fi (log2x + 2) (log2x – 1) < 0
fi –2 < log2x < 1

fi
1
4

< x < 2

Number of integral solution = 1
(c) f (x) = x2e–2x

f ¢(x) = x2(–2) e–2x + e–2x.2x
= 2x e–2x(1 – x)
f ¢(x) > 0 "x Œ (0, 1)
f ¢(x) < 0 "x Œ (1, •)
\ f (x)max = f (1) = e

–2 = l

1
2

1
2

4
2

Ê
ËÁ

ˆ
¯̃

=
Ê
ËÁ

ˆ
¯̃

=
-ln ( ) ln ( )l e

(d)

y x2 = – 12 x = 3

y

(3, 0)
x

Fig. B.17

Perpendicular tangents can be drawn to the
parabola from points which lies on directrix.
There is only one point (3, 0) which lies on its
directrix as well as on the hyperbola.

Ce
ng

ag
e L

ea
rn

ing
In

dia
P

vt.
Lt

d.