CLASS : XI (J-Batch) TIME : 40 Min. DPP. NO.-37
Select the correct alternative : (Only one is correct)
Q.145/st.line The lines y= mx + b and y= bx + m intersect at the point (m – b, 9). The sum of the x-intercepts of
thelinesis
(A) 9 (B*) –
20
41
(C)
20
41
(D)cannot be computed as data is insufficient
[Sol. mx + b = bx + m
(m – b)x = m – b  x = 1 (m  b since for m = b, x coordinate will become zero)
 m – b = 1 (as x coordinate of the point of intersection of the lines is m – b)
x = 1  y = b + m = 9
b + 1 + b = 9 (m = 1 + b)
 b = 4 and m = 5
sumofx-interceptsof linesis
– 






b
m
m
b
= – 






4
5
5
4
= –
20
41
Ans. ]
Q.213/st.line TS is the perpendicular bisector ofABwith coordinate ofA(0, 4) and B(p, 6) and the point S lies on
the x-axis. Ifx-coordinate of S is aninteger then the number ofintegral values of 'p'is
(A) 0 (B) 1 (C) 2 (D*) 4
[Hint: mAB =
p
2
; mST = –
2
p
; mid point ofAB =
2
p
, 5
equationofTS y – 5 = –
2
p







2
p
x
put y = 0 x =
p
10
+
2
p
Hence p = ± 2, ± 10 for x to be an integer ]
Q.375/st.line The line x = c cuts the triangle with corners (0, 0); (1, 1) and (9, 1) into two regions. For the area of
the two regions to be the same c must be equal to
(A) 5/2 (B*) 3 (C) 7/2 (D) 3 or 15
[Sol. 2 ·
2
1
·
1
9
c
c
1
1
c
1
1
9
=
2
1
1
0
0
1
1
1
1
1
9
 c = 3 Ans. ]
SUBJECTIVE:
Q.4 Find the roots of the equation (x – 3)3 + (x – 7)3 = (2x – 10)3.
[Sol. Let A = x – 3; B = x – 7; C = 10 – 2x [Ans. x = 3 or 5 or 7]
the equation canbe written as
(x – 3)3 + (x – 7)3 + (10 – 2x)3 = 0
 A + B + C = 0
 A3 + B3 + C3 = 3 ABC
 3(x – 3)(x – 7)(10 – 2x) = 0
x = 3 or 5 or 7 Ans. ]

Q.5174/5 Suppose that r1  r2 and r1r2 = 2 (r1 , r2 need not be real). If r1 and r2 are the roots of the biquadratic
x4 – x3 + ax2 – 8x – 8 = 0 find r1, r2 and a. [Ans. r1, r2 = 






 

2
7
i
1
and a = – 4]
[Sol. Since r1r2 = 2,
 x2 + px + 2 = 0
2
1
r
r
and r1r2r3r4 = – 8  r3r4 = – 4
 x4 – x3 + ax2 – 8x – 8 = (x2 + px + 2)(x2 + qx – 4)
comparecoefficient of x3 and x
 p + q = – 1 .....(1)
and 2q – 4p = – 8  q – 2p = – 4 ....(2)
 p = 1 and q = – 2
on comparing coefficient of x2; a = – 4
p = 1  x2 + x + 2 = 0
 r1, 2 =
2
7
i
1

Ans. ]
Q.6 If tan  =
2
11
where   (0, /2). Find the value of cos 
3
 and sin 
3
 .
[Sol. Given tan  =
2
11
;   




 
2
,
0 , 0 <
3

<
6

[Ans. sin
3

=
5
1
; cos  =
5
2
]
hence
)
3
(
tan
3
1
)
3
(
tan
)
3
tan(
3
2
3





=
2
11
let tan )
3
( = t, we get the cubic as
2t3 – 33t2 – 6t + 11 = 0
t3(2t – 1) – 16t(2t – 1)) – 11(2t – 1) = 0
(2t – 1)(t2 – 16t – 11) = 0
t = 1/2 or t = 8 ± 10 3 (rejected as the valueof  would liein the 2nd quadrant)
 tan )
3
( = 1/2
 sin
3

=
5
1
and cos  =
5
2
Ans. ]

CLASS : XI (J-Batch) TIME : 45 Min. DPP. NO.-38
Q.1102/2 The interior angle bisector of angle A for the triangle ABC whose coordinates of the vertices are
A (–8, 5) ; B(–15, –19) and C(1, – 7) has the equation ax + 2y + c = 0. Find 'a' and 'c'.
[Ans. a = 11 , c = 78]
[Sol. AB = 576
49  = 25
AC = 144
81 = 15
coordinate of the point D
8
)
5
45
( 

,
8
)
35
57
( 

 







2
23
,
5
equationofAD:
y – 5 =
 
5
8
2
23
5



(x + 8)  –3(y – 5) =
2
33
(x + 8)
– 6y + 30 = 33x + 264
33x + 6y + 234 = 0 ....(1)
given ax + 2y + c = 0 ....(2)
comparing(1)and (2)
33
a
=
6
2
=
234
c
 a = 11 ; c =
3
234
= 78
 a = 11 and c = 78 ]
Q.2110/2 Find the combined equation of the pair of lines through (2, 3) perpendicular to the lines
3x2 – 8xy + 5y2 = 0. [Ans. 5x2 + 8xy + 3y2 – 44x – 34y + 95 = 0]
[Sol. 3x2 – 8xy + 5y2 = 0 ....(1)
Let m
x
y

5m2 – 8m + 3 = 0
5m2 – 5m – 3m + 3 = 0
m = 1,
5
3
nowequationof straight lineperpendicularto (1) and passingthrough (2, 3)
y – 3 = – 1 (x – 2)  x + y – 5 = 0 ....(2)
y – 3 = –
3
5
(x – 2)  3y – 9 = – 5x + 10
5x + 3y – 19 = 0 ....(3)
thenequationofrequiredline
(x + y – 5) (5x + 3y – 19) = 0
 5x2 + 8xy + 3y2 – 44x – 34y + 95 = 0 ]

Q.3111/2 Each sideof a squareisof length 4. Thecentre of the squareis (3, 7) andone of its diagonals is parallel
to y= x. Find the coordinates of its vertices. [Ans. (1, 5) ; (1, 9) ; (5, 9) ; (5, 5) ]
[Sol. Let lineAC is parallel to y= x
AC2 = 16 + 16
AC = 2
4





sin
7
y
cos
3
x
= 2
2
x = 3 + 2
2 cos ( tan = 1, then sin =
2
1
, cos =
2
1
)
y= 7 + 2
2 sin
x = 3 + 2 = 5 ; y = 9  C (5, 9)
To findcoordinates ofA
x = 3 – 2
2 cos  x = 3 – 2 = 1
y= 7 – 2
2 sin  y = 7 – 2 = 5  A (1, 5)
Then B (5, 5) and D (1, 9) ]
Q.4116/2Avariable straight linewhoselengthis Cmoves in such awaythat one ofits end lies on the x-axis and
the otheron the y-axis. Show that the locus of the feet of theperpendicular from origin on the variable
linehastheequation
(x2 + y2)3 = C2x2y2
[Sol. equationofAB
y – k = )
h
x
(
k
h


hx + ky = h2 + k2 ........(1)
OA =
h
k
h 2
2

OB =
k
k
h 2
2

Hence
2
2
2
2
2
2
2
C
k
k
h
h
k
h








 








 
(h2 + K2)2 




 
2
2
2
2
k
h
k
h
= C2
(h2 + K2)3 = C2h2K2
or (x2 + y2)3 = C2x2y2 ]
Q.5120/2 Arayof light is sent along the line x  2y  3 = 0. Upon reaching the line 3x  2y  5 = 0, the rayis
reflectedfromit.Find theequationofthelinecontainingthereflected ray.
[Ans: 29x  2y= 31] [ REE ’90, 6+6 ]

[Sol: Equation of the reflected rayBP can be takens as
x – 2y – 3 +  (3x – 2y – 5) = 0 ....(1)
Now perpendicular from Q (x1, y1) are (1)
and x – 2y – 3 = 0 should be equal
2
2
1
1
1
1
1
1
)
1
(
4
)
1
3
(
)
5
y
2
x
3
(
3
y
2
x
5
3
y
2
x














or (3 + 1)2 + 4 ( + 1)2 = 5]
92 + 1 + 6 + 42 + 8 + 4 = 5
132 + 14 = 0   = –
13
14
(as  0)
 requiredequationis x – 2y – 3 –
13
14
(3x – 2y – 5) = 0
13 x – 26 y – 39 = 42 x – 28 y – 70  29 x – 2 y = 31 Ans.]
Q.6136/2 Find the equations of the sides of a square whose each side is of length 4 units and centre is (1, 1) .
Given that one pair of sides is parallel to 3 x  4y = 0 .
[ Ans. : 3 x  4y + 11 = 0 ; 3 x  4 y  9 = 0 ; 4 x + 3 y + 3 = 0 ; 4 x + 3 y  17 = 0 ]