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Mathematics
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Q . Tangents draw from the point P (1, 8) to the circle x2 + y2 – 6x – 4y – 11 = 0
touch the circle at points A and B. the equation of the circumcircle of the triangle
PAB is
[IIT-JEE 2009]
2
2
2
2
(a) x + y + 4x – 6y + 19 = 0
(b) x + y – 4x – 10y + 19 = 0
2
2
(c) x + y – 2x + 6y – 29 = 0
(d) x2 + y2 – 6x – 4y + 19 = 0
BOOK & COACHING SOLUTION
Equation of chord of contact i.e. AB is
x.1 + y.8 – 3 (x + 1) – 2 (y + 8) – 11 = 0
or x – 3y + 15 = 0
Now equation of circle passing through
intersection point of circle
x2 + y2 – 6x – 4y – 11 = 0
and of line x – 3y + 15 = 0 is given by
S+
L = 0 i.e.
(x2 + y2 – 6x – 4y – 11) +
(x – 3y + 15) = 0
At this circle passes through P(1, 8) also,
we get ,
(1 + 64 – 6 – 32 – 11)+
(1 – 24 + 15) = 0
=2
The required circle is
(x2 + y2 – 6x – 4y – 11)+ 2(x – 3y + 15) = 0
or x2 + y2 -4x– 10y + 19 = 0
PIONEER’S SMART SOLUTION
Tangents PA and PB are drawn from the
point P(1, 8) to circle x2 + y2 – 6y – 4y –
11 = 0 centre C(3, 2).Clearly the circum
circle of PAB will pass through C and as
A = 900,
PC must be a diameter of the circle
(x – 1) (x – 3) + (y – 8) (y – 2) = 0
x2 + y2 – 4x – 10y + 19 = 0
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3 i
2
Q: If z
(a)Re z =0
6
3 i
2
6
then
(b) Re z, Im z >0
(c) Im (z) = 0
BOOK & COACHING SOLUTION
6
3 i
2
z
PIONEER’S SMART SOLUTION
ei
for which :
|a:b|= 1: 3 or 3 :1 can be expressed in terms
of i, , 2 .
3 i
2
6
1 i 3
2
6
1 i 3
2
6
(
i
6
i
12
2
2 6
)
2 0i
Im(z) 0
6
e
6
i
2cos
z
6
3 1
2
z
6
3 i
2
Fact: Using fact every complex number a +ib
3 i
2
(d) Re z > 0, Im (z)< 0
2
2 0i
3 i
2
6
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1 1 0
Q: For the matrix A = 1 2 1 which of following is correct? [IIT-JEE 2005]
2 1 0
(a) A3 + 3A2 – I = 0
(b) A3 – 3A2 – I = 0 (c) A3 + 2A2 – I = 0
BOOK & COACHING SOLUTION:
A
1 1 0
1 2 1
2 1 0
2
1 1 0
1 2 1
2 1 0
1 1 0 1 2 0 0 1 0
= 1 2 2 1 4 1 0 2 0
2 1 0 2 2 0 0 1 0
2 3 1
= 5 6 2
3 4 1
A
1 1 0
1 2 1
2 1 0
2 3 2 2 6 1 0 3 0
= 5 6 4 5 12 2 0 6 0
3 4 2 3 8 1 0 4 0
7 9 3
= 15 19 6
9 12 4
A
3
3A
2
7
9 3
15 19 6
9 12 4
7
9
9
12 4
3
2 3 1
3 5 6 2
3 4 1
6 9 3
15 18 6
9 12 3
= 15 19 6
1 0 0
= 0 1 0
0 0 1
PIONEER’S SMART SOLUTIONS:
Using Cayley-Hamilton theorem
A λI 0
1 λ
1
2
1
I
A3 3A2 I 0
0
1
λ
2 λ
1
1 λ
1 λ
2 3 1
5 6 2
3 4 1
3
(d) A3 – A2 + I = 0
2 λ
1
1
λ
0
1
2
1
2λ λ2 1
λ3 3λ2 I 0
A3 3A2 I 0
1
λ
0 0
λ 2
0
λ3 3λ2 I 0
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Q: Determine the ratio in which y – x + 2 = 0 divide the line joining
(3 – 1) and (8, 9).
BOOK & COACHING SOLUTION:
Pioneer Smart Solution:
Suppose the line Y – X + 2 = 0 divided the
line segment joining A (3, –1) and B (8.9) in
the ratio λ : 1 at the point P,
then the coordinate of the point P are
8λ 3 9λ 1
,
.
λ 1 λ 1
The straight line ax + by + c = 0 divides
the joint of points A (x 1, y1) and B Q: (x 2,
y2) in the ratio
But P lies on y – x + 2 = 0
λ
therefore,
9λ 1
λ 1
8λ 3
λ 1
2 0
9λ 1 8λ 3 2λ 0
3λ 2 0
2
3
So, the required ratio is
2
: 1, i.e. 2 : 3
3
(internally) since here λ is positive.
or λ
1 3 2
9 8 2
2 or
3
λ: 1 = 2 : 3
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Q: If Tm = n, Tn = m, then Tm+n=0
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to
zero is
(a) 501th
(b) 502th
(c) 508th
(d) none of these
BOOK & COACHING SOLUTION:
Pioneer Smart solution:
T9 = 449
T9
449 a 8d. …….. (i)
T449 = 9
T449
9 a 448d ……… (ii)
T9+449 = 0
Tn 0
By Using : If Tm = n, Tn = m, then
n=?
Subtract (ii) from (i)
440 =
440 d
d
1
Put in (1)
449 = a 8
a
Tn
457
449 8
a
457
n
457
n 1 d 0
n 1
T458 = 0, so (d) is correct option
1
0
n 1
458 , So (d) is the correct option.
Tm+n=0
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Q: If two vertices of an equilateral triangle are (0,0) and (3, 3 ) then find the
co-ordinates of the third vertex?
BOOK & COACHING SOLUTION:
Let O = (0, 0) and A = 3, 3 be the given
Pioneer Smart Solution
x1
3 y 2 y1
2
points and let B = (x, y) be the required
point. Then,
x2
i.e.,
0 3
3
,
3 0
2
y1
0,2 2
OA = OB = AB
(OA)2 = (OB)2 = (AB)2
(3 – 0)2 + ( 3 –0)2 = (x – 0)2 + (y – 0)2 =
(x – 3)2 + (y – 3 )2
12 = x2 + y2 = x2 + y2 – 6x – 2 3 y + 12
Taking first two members then x 2 + y2 = 12
……(1)
and taking last two members then
6x + 2 3 y = 12 or y = 3 (2 - x) …..(2)
(2) From (1) and (2), we get
x2 + 3 (2 – x)2 = 12 or 4x 2 – 12x = 0
x
= 0, 3
Putting x = 0, in (2), we get y = 2 3 – 3
Hence the co-ordinates of the third vertex
B are (0, 2 3 ) or (3, – 3 ).
3 x 2 x1
2
,
0
or 3 3 , 3
2
y2
3
3 3
2
3 3
2
or 3,
3
0
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Physics
Q: A car is fitted with a convex side-view mirror of focal length 20 cm. A second
car 2.8m behind the first car is overtaking the first car at a relative of 15 m/s.
The speed of the image of the second car as seen in the mirror of the first one
is:
1
1
(1)
m/s
(2)
m/s
(3) 10 m/s
(4) 15 m/s
10
15
BOOK & COACHING SOLUTION:
Focal length of mirror
f = + 20 cm
Given that speed of object with respect to
du
mirror i.e.
= 15 m/s.
dt
dv
Speed of image with respect to mirror,
=?
dt
v = position of image w. r. t. mirror
u = position of object w. r. t. mirror
By mirror formula,
1
1
1
u
v
f
Differentiate w. r. t. time t
1 dv 1 du
0
v 2 dt u2 dv
dv
v2 du
…(1)
dt
u2 dt
1 1 1
from mirror formula:
v u f
1
1
1
v
2.8
0.2
1 15
2.8
v
m
v 2.8
15
2
dv
2.8 1
15
from (1)
dt
15 2.8
dv 1
Speed of image,
m/s.
dt 15
PIONEER’S SMART SOLUTION
By technique of Longitudinal
magnification,
dv
du
2
f
u f
dv
dt
du
dt
f
2
u f
object speed du = 15 m/s, f = + 20
dt
cm, u = – 280 cm
dv
dt
15
speed,
2
20
280
dv
dt
20
1
m/s.
15
image
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Q: A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v0
in the x-y plane as shown in the figure. At a time t <
v0 sin θ
, the angular momentum of the
g
particle is :
i , j and k are unit vectors along x, y and z-axis respectively. [ AIEEE – 2010]
where
BOOK & COACHING SOLUTION:
PIONEER’S SMART SOLUTION
Angular momentum L being “Axial vector” is
always normal to the plane of rotation and its
direction is determine by “Right hand thumb
rule”. Since plane of rotation is xy-plane and
rotation is clockwise hence L is directed
along negative z-axis i.e. - k hence option (4)
is correct.
v o sin θ – (given)
g
t
It means the projectile doesn’t attain max. height. By
concept of torque
τ
r
τ
F
0
τ
0 i
0
0 j
mgx
mg x k
since τ
dL or dL
dt
τ dt or dL
(since
x
dL
mg v o cos θt k dt
vo cos θ t ).
L
dL
L
L
L
L
0
t
mg v o cos θ K
0
tdt
t2
2
mg v o cos θ k
1
mg v o cos θt 2 k .
2
1
mv o cos θt 2
2
mg x k dt
VISIT-COMPARE-DECIDE
Integrating both sides.
0
0 k
k
Hence option (4) is correct.
t
0
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2
. It is
3
surrounded by air. A light ray is incident at the mid-point of one end of the rod
as shown in the figure. The incident angle θ for which the light ray grazes along
the wall of the rod is :
[AIIEEE – 2009]
Q: A transparent solid cylindrical rod has a refractive index of
(1) sin
1
2
1
(2) sin
1
3
2
BOOK & COACHING SOLUTION:
μ
cos α
1
1
μ2
μ2 1
..(i)
μ
Applying Snell’s law at P
μ
sin θ
sin α
sin θ μ sin α
2
3
sin θ
or θ
sin
1
μ
sin θ
1 cos2 α
2
1
.
3
1
(4) sin
1
1
3
By technique of T. I. R. in optical fibre
1
cos α
1
μ
sin α
2
3
1
PIONEER’S SMART SOLUTION
Applying Snell’s law at Q
sin 900
sin 900 α
(3) sin
4
1
3
μ2 1 from(i)
1
3
θ
sin
2
1
μ
2
0
μ0
1
1 .
3
2
3
2
1
1
2
1
3
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Q: A point is subjected to two simultaneous sinusoidal displacements in xdirection.
2π
x1 (t) = A sin ω t and x2 (t) = A sin ωt
. Adding a third sinusoidal
3
displacement x3 (t) = B sin wt
brings the mass to a complete rest. The
values of B and are :
4π
3π
(1) 2 A,
(2) A,
3
4
BOOK & COACHING SOLUTION:
The particle comes to complete rest. If
the resultant of x1(t), x2(t) and x3(t)
becomes zero i.e. x(t) = 0
Bu superposition of waves
5π
π
(4) A,
6
3
PIONEER’S SMART SOLUTION
(3) 3 A,
The particle comes to complete rest. If resultant of
three waves becomes zero i.e.
xt(1) + x2(t) + x3(t) = 0
By technique of vectors: This is possible only if
three waves (i.e. vectors) have equal magnitude
and equally inclined.
x(t) = x1(t) + x2(t) + x3(t)
0 = x12(t) + x3(t)
x3(t) = –x12(t)
hence x3 t
x12 t
B=A
π
4π
3
α
π
π
3
Hence B = A &
2π
3
2π
3
4π
.
3
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CHEMISTRY
Q. Increasing order of bond strength of O2 , O2 , O22 and O2 is
(a) O2 O2 O2 O22 (b) O2 O2 O2 O22
(c) O2 O22
O2 O2 (d) O22
[AIEEE-2002]
O2
O2 O2
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Q: The electrons identified by quantum Numbers n and l : [AIEEE – 2012]
(a) n = 4, l = 1
(b) n = 4, l = 0
(c) n = 3, l = 2
(d) n = 3, l = 1
Can be placed in order of increasing energy as :
(1) (c) < (d) < (b) < (a)
(2) (d) < (b) < (c) < (a)
(3) (b) < (d) < (a) < (c)
(4) (a) < (c) < (b) < (d)
BOOK & COACHING SOLUTION
Orbitals
PIONEER SMART SOLUTION
As per (n + l) Rule
(a) n = 4, l = 1
4p
the increasing energy is :
(b) n = 4, l = 0
4s
(d) < (b) < (c) < (a)
(c) n = 3, l = 2
(d) n = 3, l = 1
3d
3p
Greater the (n + l) value
According to Aufbau principle :
greater will be the energy.
Increasing order of energy of
orbitals
1s < 2s < 2p < 3s < 3p < 4s <
3d < 3d < 4p < 5s < 4d < 5p <
6s -----So According to Aufbau
principle the increasing order
of energy should be :
(d) < (b) < (c) < (a)
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