### Sample Book - Career Point Kota

```LAWS OF MOTION
21 21
Contents
TOPIC
PAGE NO.
1.
Trigonometric Ratios ............................................................................................................................. 01
2.
Trigonometrical Equations ................................................................................................................... 11
3.
Properties of Triangle ............................................................................................................................. 18
4.
5.
Logarithm & Modulus Function ........................................................................................................... 34
6.
7.
Progressions ............................................................................................................................................ 49
8.
Binomial Theorem .................................................................................................................................. 56
9.
Permutation & Combination ................................................................................................................. 65
10.
Complex Number .................................................................................................................................... 75
11.
Point & Straight Line ............................................................................................................................. 84
12.
Circle ........................................................................................................................................................ 88
13.
Parabola ................................................................................................................................................... 99
14.
Ellipse ....................................................................................................................................................... 104
15.
Hyperbola................................................................................................................................................. 111
2
TRIGONOMETRICAL EQUATIONS
SINGLE CHOICE CORRECT TYPE QUESTIONS
1.
If the equation 4cos2 x sin x – 2sin2x = 3sin x, then x is equal to –
 3 

 10 
 3 

 10 
(B) n + (–1)n+1 
(A) n + 
 3 

 10 
(C) n + 
(D) None of these
2
2
2.
The equation 3sin 2x  2 cos x + 31 sin 2 x  2 sin
(A) cos x = 0, tan x = –1
(C) tan x = 1
x
3.
The solution of the trigonometric equation cos2  cos x 
= 28 is satisfied for the values of x given by –
(B) tan x = 0
(D) None of these

3
(A) cos–1(3k+8)
(C) 2n
4
(B) cos–1(3k–8)
(D) None of these
If tan m + cot n = 0, then the general value of  is(A)
(2r  1)
2 (m  n )
(B)
(2r  1)
2 (m  n )
(C)
1
5.
8 
 = 1 must be3 
The most general solution of 2sin x + 2cosx = 2


(A) n +
(B) n –
4
4
1
2
r
mn
(D)
r
mn
is(C) n +(–1)n

4
(D) 2n +

4
9
PERMUTATION & COMBINATION
SINGLE CHOICE CORRECT TYPE QUESTIONS
1.
A shopkeeper has 10 copies of each of nine different books, then number of ways in which atleast one book
can be selected is(A) 911–1
(B) 1010–1
(C) 119–1
(D) 109
2.
Number of quadrilaterals that can be constructed from the vertices of a decagon (10 sided polygon) such that
none of the sides of the quadrilateral is also a side of the decagon, is(A) 25
(B) 50
(C) 100
(D) 200
3.
In a conference 10 speakers are present. If S1 wants to speak before S2 and S2 wants to speak after S3, then
the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining
seven speakers have no objection to speak at any number is(A) 10C3
4.
(B) 10P8
(C) 10P3
(D)
10 !
3
There are n straight lines in a plane, no two of which are parallel, and no three pass through the same point.
Their points of intersection are joined. The number of fresh lines thus obtained is(A)
n (n  1) (n  2)
8
(B)
n (n  1) (n  2) (n  3)
6
(C)
n (n  1) (n  2) (n  3)
8
(D) None of these
Solutions
With
2 TRIGONOMETRICAL EQUATIONS
SINGLE CHOICE CORRECT TYPE QUESTIONS
1.[B]
We have [4 cos2 x – 2sin x – 3] sin x = 0.
Either sin x = 0;
 x = n;
or 4(1 – sin2 x) – 2sinx – 3 = 0  4sin2x + 2sinx – 1 = 0
sin x =
 1 5
 2  4  16
=
=
8
4
sin x = sin


 x = n + (–1)n
10
10
or sin x = – cos
2.[A]
5  1 1 5
,–
4
4

 3 
 3 
= sin     x = n (–1)n+1  
5
 10 
 10 
2
The given equation is 3sin 2x  2 cos x + 31 sin 2 x  2 sin
2
or 3sin 2x  2 cos x + 33 (sin 2 x  2 cos
y +
2
x)
2
x
= 28
= 28
2
27
= 28, where y = 3sin 2x  2 cos x y2 – 28y + 27 = 0
y
y = 27 or 1
2
If y = 27, then 3sin 2x  2 cos x = 33
sin2x + 2cos2 x = 3
sin2x + cos2x = 2 sin 2x = cos2x = 1
Which is not possible for any value of x and so y  27.
Also we have y = 1
 3sin 2x  2 cos
2
x
= 1 = 3º
sin 2x + 2 cos2x = 0
2 cosx (sin x + cos x) = 0
Either cos x = 0 or tan x = – 1
3.[C]
The equation is same as 2 cos2(t) = 2 where t =

 1 + cos 2t = 2


8
cos x –
3
3
 cos 2t = 1
 2t = 2k


 t = k


8
cos x –
= k
3
3
 cos x = 3k + 8
Since |cos x|  1, we must have |3k + 8|  1


 k = – 1  cos x = 1 x = 2k
Thus choices (A), (B) are ruled out and (C) is correct.



   then m = r + + n
2
2


4.[A]
Write tan m= tan 
5.[A]
 A.M.  G.M. 
 2sinx + 2cosx  2.
2
sinx
+2
cosx
 2.
2sin x  2 cos x

2
2sin x.2 cos x
2 sin x  cos x . But minimum value of cos x + sin x is – 2
2
 2
1
= 2
But the given equation is 2
sinx
1
2
+2
cos x
1
=2
1
2
, which can hold only if 2sinx = 2cosx
sin x = cos x  tan x = 1  x = n +

4
9 PERMUTATION & COMBINATION
SINGLE CHOICE CORRECT TYPE QUESTIONS
1.[C]
………. B
B1 B 2 B
9
  3
10
10
10
10
Selection of atleast one book (10  1) (10  1)......(10  1) – 1 = 119 –1

9 times
2.[A]
First vertex can be selected in 10C1 ways.
10
•1
9
×
2
8
×
3
×
×
4
7
6
×
First vertex
5
Now, two neighbouring vertices are not to be selected.
Among the remaining 7 vertices 3 are to be selected and 4 are not to be selected.
Now mark the 4 vertices not to be selected by × sign and then these can be partitioned by 5 partitions from
these partitions we are to select 3. This can be done in 5C3 ways and these are the places where the remaining
3 vertices can be chosen.
10
 number of ways =
3.[D]
If the order of speakers is S1, S3, S2 (need not be consecutive) treat S1, S3, S2 as identical and arrange. We
obtain
10 !
10 !
. Similarly if the order of speakers is S3, S1, S2 we obtain
.
3!
3!
Required = 2
4.[C]
C1  5C 3
= 25
4
10 ! 10 !
=
3!
3
Explanation:
(a) is wrong since the correct answer must become 0 for n = 3.
(b) Is also wrong since the correct answer must become 3 for n = 4
(c) Seems to be correct.
Alternative Solution : n lines will give nC2 points of intersections. These points can be again joined to get
nC

2
C 2 line segments which will contain n.
 The number of fresh lines =
nC
2
C 2 – n.
n–1
C2 old lines.
n–1
C2 =
n (n  1) (n  2) (n  3)
8
```