For iit students

1. ANSWER KEY
12 th ABCD (Date: 14-06-2009) Review Test-2
Code-B
Paper-2
CHEMISTRY PHYSICS MATHS
SECTION-2 SECTION-3
SECTION-1
PART-A PART-A
PART-A
Q.13 C Q.1 D
Q.1 C
Q.24 D Q.2 A
Q.2 D
Q.35 C Q.3 A
Q.3 B
Q.41 A Q.4 B
Q.4 A
Q.52 D Q.5 B
Q.5 D
Q.66 B Q.6 B
Q.6 C
Q.77 A Q.7 C
Q.7 B
Q.88 A Q.8 C
Q.8 A
Q.99 B Q.9 D
Q.9 D
Q.1010 B Q.10 C
Q.10 B
Q.1111 A Q.11 B
Q.11 B
Q.1215 B Q.12 A, B, C, D
Q.12 A, D
Q.1312 A, C Q.13 A, D
Q.13 B, C, D
Q.1413 A, D Q.14 B, C
Q.14 A, B
Q.1514 A, B, C Q.15 C, D
Q.15 A,B,C,D
PART-B PART-B PART-B
Q.1 (A) T (B) R Q.1 (A) P, R,S (B) P,Q,S Q.1 (A) T ; (B) Q;
(C) P (D) P (C) P,Q,S (D) P,R,S (C) Q, R, S, T ; (D) R, S
PART-C PART-C PART-C
Q.1 0008 Q.1 0005 Q.1 0010
Q.2 0004 Q.2 0024 Q.2 0022
Q.3 0036 Q.3 8400 Q.3 0006
Q.4 0002 Q.4 0024 Q.4 0100

2. PHYSICS
PART-A
Q.1
[Sol. n11 = n22
1 4000Å
 n11 = n22  n2 = n1  = 60
6000Å
= 40 ]
2
Q.2
(I) O
[Sol.
D = 25cm
v = –25 cm
f = +10 cm
1 1 1  25.10  50
– = 
 25 u 10 25  10
u= = ]
7
Q.3
Com
[Sol. x

Greater the value of , lesser the value of x. hence for the sector of angle  position of COM will be
farther than the semicircle (sector of angle ) ]
Q.4
[Sol. Fringe width does not depend on distance between the source and plane of slits. ]
Q.5
  
[Sol. Angular spacing between two consecutive maxima = = =2× ]
D d 180
Q.6
[Sol. Luminous flux per unit area × area of hole
  d d 2
2
= ds = ×   = ]
4r 2 4r 2 2 16r 2
Q.7

T.I.R.
[Sol. S 
sin i = µ sin r
 1
sin = n sin   sin  = ]
2 n
Code-B Page # 1

3. PHYSICS
Q.8
[Sol. For TIR >c
90 –  > c  cos  > sin c
n 2 1 1
>  n> 2 ]
n n
Q.9
[Sol. Chemical energy of explosive is converted to heat which is connected to K.E. of the cannon-ball ]
Q.10
[Sol. Force on cannon ball = PA – Patm A
( P  Patm ) A
acceleration
m
displacement inside ballerl = length of barrel = 
v2 = 2as
 (P  Patm )A  2  (51  1)  105  8 10 3  2.5
 2  =
 
v= = 100 m/s ]
m 20
Q.11
[Sol. PV = nRT
dn dV
= R' & = A.V.
.
dt dt
dV dn
 P = RT
T
dt dt
 PAv = R' RT  v  R' ]
Q.12
[Sol. We know VIBGYOR
Since in option V & Y are not given so possible are BG YOR
R > B
R < B By cauchy’s relation
1
sin C =

and by T.I.R.
so (C)B < (C)R
so there is probability of reflection of Blue colour so D must be blue and C nearest to blue which is green
R > y so R < y i.e. A is red B is yellow.
Q.13
R
r
Fv FR
[Sol.
10cm
fR
Code-B Page # 2

4. PHYSICS
From similar triangles,
r fR  fv
= f
R R
1 f R  10 100
= f  fR =
10 R 9
1 2
f R = (µR – 1) R
2 9
 (µR – 1) =  µR = 1.27
6 100
fR  fv fR  fv 100 / 9  10 1
= fy  fR = = ~ 0.1 ]
100 / 9 10
Q.14
[Sol. We know
 O2 =
5 7
He = ,
3 5
PV = const.
L 
5/3
PHe (LA) = 5/3   A
2 
PHe

PHe = PHe (2)5/3 ... (1)

PO 2 (LA)7/5 = PO 2 (LA)7/5
L
7/5

 
 L 
PO 2 = PO 2 ... (2)
Now PO 2  PHe   
PO 2  PHe
 L 
PO 2 PHe 7/5
1
  =     =
L
PO 2 PHe 5/3
2
7/5
L7 / 5
(L) =
25 / 3
L L
L = 5/3 5/ 7 =
2 x 2 25 / 21
LA
so volume of O2 = LA = ]
2 25 / 21
Code-B Page # 3

5. PHYSICS
Q.15
[Sol. (A) For convex mirror object is not at focus (focus is on other side). The mirror will cause the
diverging incident beam to diverge even more.
(B) The rays emerging towards left from the lens must be parallel to P.A.
(C) Object is between optic center and focus. Image will be diverging (virtual)
(D) Initial object is at O whose image is formed at pole of mirror by the lens. Hence find image will
also be at O. ]
PART-B
Q.1
[Sol. (1) y1 = n11 = n22 = LCM of 1 and 2
2y1 = 2n11 = 2n22
Hence at this point both maxima again coincide.
 1  1
y2 =  n1   1 =  n 2   2
 2  2
n2 
1
1 2
2 =
n1 
1
2
1 2n 2  1 1 odd
 2 = 2n  1 which will have a solution. Iff  expressed as a proper fraction will be of form .
1 2 odd
1 odd
Option B & C :  is of form . Hence no solution i.e. the two minima will never coincide.
2 even
1 odd
For option A& D :  is of form . Hence at some finite y2 the two minima will coincide.
2 odd
At 2y2 the two maxima (and not minima) will coincide.
 y = 3y2 is the next nearest point where minima coincide. ]
PART-C
Q.1
[Sol. New volume of flask = 2(1 + 3Glass T) litres (1dm3 = Litre)
New volume of water = 1(1 + Ywater T) litres
New volume of iron = 1(1 + 3iron T) dm3
Volume of water overflow = Vwater + Viron – Vflask
= (Ywater + 3iron – 6Glass)T
= [(180 + 3 × 12 – 6 × 8) × 10–6] × 50
= 168 × 50 × 10–6 = 8400 × 10–6 litres = 8400 mm3 ]
Q.2
[Sol. Heat released by water = Heat absorbed by ice
(15 × 103(gm))(cal/gm) × (80 – 20)°C = ms (20°C – 0°C) + mL
= (100 cal/g) m(gram)
Code-B Page # 4

6. PHYSICS
15 60
 m= × 103 gm = 9kg
100
total water = 15 + 9 = 24 kg
 minimum volume of vessel = 24 litres. ]
Q.3
dv Adx
[Sol. Temperature is constant  E = 0, dW = nRT = nRT
v AL / 2
Q = E + W
dW nRT dx
 =
dt L / 2 dt
1 T 2nRT  dx 
  
dQ dW

L  dt 
k =
dt dt 900 L
dx k  27 415.5  27  2 1

900  0.5  8.31 300 200
= = = m/s = 5 mm/s ]
dt 900.nRT
Q.4
[Sol. Heat supplied (1  2 & 2  3)
= nCv9T0 + nCp90T0
Heat rejected (3  4 & 4  1)
= nCv90T0 + nCp9T0
 10C v  C p 
10  
 = 1  C  10C  = 1 –
11.4 3.6 12
 =1–
 p 1  10
= =
 V  15 15 50
 %efficiency = 24% ]
Code-B Page # 5

7. CHEMISTRY
PART-A
Q.1
[Sol. H = V + (PV)
V = W = – 4 × – 10 = 40 l-bar
(PV) = P2V2 – P1V1 = 4 × 30 – 2 × 40 = 40 l – bar ]
Q.2
[Sol. (A) nBeH2(g) Cool. (BeH2)n(s)
 H H
H –– Be –– H Be Be Be
Hyb. of Be : sp H H
(gas) 3
.. Hyb. of Be : sp
.. (solid)
..
..O O. .
(B) .. [NO2]+ [NO3]+
N–O–N
.. Hyb. of N in NO2+ : sp
..
Hyb. of N in NO3¯ : sp2
. .
O. . O:
..
2 (N2O5 in solid state)
Hyb. of each N:sp
(gaseous state)
(C) XeF6 [XeF5]+ F¯
Gaseous state Solid state
Hyb. sp3d3 Hyb. sp3d2
(D) PF5 PF5
Gaseous state Solid state
Hyb. : sp3d Hyb.: sp3d ]
Q.3
[Sol. q = – E
I law
E = –  + 
 2E = 
2CVdT = – PdV
RT
2CVdT = – dV
V
 
2C V dT dV
– = ]
R T V
Q.4
 a 
 P  2  (V – b) = RT
 V 
[Sol. T
RT a
– 2
Vb V
P=
PV V a
–
V  b VRT
Z= =
RT
Code-B Page # 1

8. CHEMISTRY
 a 
T  
V b
–
Vb V  Rb 
=
b2
V ( V  b)
=1+ ]
Q.5
[Sol. Tertiary alkyl halides can take part in the reaction but prefers disproportionation so the yield of dimerized
product will be less but it can show Wurtz reaction. ]
Paragraph for question nos. 6 to 8
Me CH2 CH3
Me Me
[Sol.  + +
Me Me CH3 CH3 CH3 CH3
Me
(X) (R) 3° Free radical (P) 1° Free radical (Q) 2° Free radical
]
CH2–Cl CH3
Me Me Cl
H
Cl CH3 CH3 CH3 CH3
Me
Dia +
Dia + CH3
Me Me H
Cl
Me CH3 CH3
Cl
9
9×1 =9 1°Product = × 100 = 19.86%
45.3
22.8
6 × 3.8 = 22.8 2° Product = × 100 = 50.33%
45.3
13.5
3 × 4.5 = 13.5 3° Product = × 100 = 29.80%
45.3
45.3
Code-B Page # 2

9. CHEMISTRY
Q.9
F
N
[Sol. (A) F F
F
Hybridisation of N : sp3
 FNF = 109°
As all substituents are same
Cl
C
(B) Cl Cl
Cl
Hybridisation of c : sp3
 l CC = 109°
As all substituents are same.
O¯ O O O
(C) Cl Cl Cl Cl
O O O¯ O O O O O¯
O O O¯ O
Hybridisation of Cl : sp3
 OClO : 109° , as well Cl–O bonds are of equal length. ]
Q.10
F F
Br —F
< 90°
[Sol. (A) B
F F F
Hybridisation of B : sp2 Hybridisation of Br : sp3d
Shape : Trigonal planar Shape : Bent 'T' shape.
Cl
I 180°
Cl — Be — Cl
(B)
Cl
Hybridisation : sp3d Hybridisation : sp
Shape : linear Shape : linear
Code-B Page # 3

10. CHEMISTRY
Cl
(C) B P
Cl Cl Cl Cl
Cl
Hybridisation of N : sp3 Hybridisation of P : sp3
Shape : Trigonal planar Shape : Triangular pyramidal
N P
(D) Cl Cl
Cl Cl
Cl Cl
Hybridisation of N : sp3 Hybridisation of P : sp3
Shape : Triangular pyramidal Shape : Triangular pyramidal. ]
Q.11
H C
[Sol. (A) C H (B) H F
H F
Triangular pryamidal
(Planar)
(Nonplanar)
•
C N
(C) F F (D)
O O
Planar (singlet carbene) (Planar)
Q.12
[ B]t k1
[Sol.(A) [C] = 2k = 1
t 2
 k1 
(B) ln    E a for same temp. change
k
 
0.693
(C) t1/2 = k  k = 2.772 sec. ]
1 2
Code-B Page # 4

11. CHEMISTRY
Q.13
[Sol. NCl3 undergoes hydrolysis and H2O attacks at Cl-atoms.
H 2SO3  4HF
SF4 + 3H2O  

R .T.
 
 
Pr oton donor acids
+
Cl OH2
B B¯
H 2O 
 
Cl Cl Cl Cl
2 Cl
Hyb. sp
3
Hyb. sp
(Transition state)
Q.15
COOH COOH
NO2
[Sol. (A) Ethanol can solvate anion of as compared to anion of
O2N NO2
(B) H H
N N N N
No flipping 2°amine 3°amine l.p.is present in more EN orbital
(C) L.G. tendency does not depend on solvent nature.
(D) Polar Aprotic solvent does not solvate anion. ]
PART-B
Q.1
Potential
Energy
[Sol. (A) As after rearrangement two product are formed.
Reaction progress
(B) No rearrangement
(C) Two products formed.
(D) Two products formed. ]
Code-B Page # 5

12. CHEMISTRY
PART-C
Q.1
[Sol. WAB = WCD = 0 D C
WBC = – nRT = – 1 × R × 300 = 300 R
40
WDA = – nRT = – 1 × R × – 200 = 200 R
V
(L)
Wtotal = – 100 R 20 lit.
A
B
 |W| = 8 atm. lit. ]
200 300 400 600
T (K)
Q.2
[Sol. AsCl3 + 3H2O  H 3AsO3  3HCl ]
Basicity  3 Basicity  1
   
Total basicity of products  4
Q.3
[Sol. A(g)  B(g) + 2C(g)
P0 – –
P0 – x x 2x
– P0 2P0
if N is pressure of inert gas
P0 + N = 200
3P0 + N = 500
 P0 = 150 ; N = 50
at t = 70 sec.
Pt = 350 = P0 + 2x + N
 x = 75
 t = t1/2 = 70 sec.
0.693
k (hr–1) = t sec × 3600 ]
1/ 2
Q.4
[Sol. Cl –– Be –– Cl
No. of 2C – 2e– bonds = 2   Total no. of 2C – 2e– &
No. of 3C – 4e– bonds = 0  3C – 2e– bonds = 2
 ]
Code-B Page # 6

13. MATHEMATICS
PART-A
Q.1
x2 t2 x2
[Sol. 2 f ' (2x) = x2  f ' (2x) = ; put 2x = t  f ' (t) =  f ' (x) = Ans.]
2 8 8
Q.2
dx
Let cos–1x = t  –
1 x2
[Sol. = dt
(1  cos t )  sin t  1 sin t 
I = – e
t
dt = –  e t   
 1  cos t (1  cos t ) 2  dt
(1  cos t ) 2
 
1
et e cos x
=C– =C–
1  cos t
Ans. ]
1 x
Q.3
[Sol. Equation of line : y – 1 = m(x – 7)
mx – y + 1 – 7m = 0
r | 7m  1 | r

1 m2
Perpendicular distancer from (0, 0)= = =5
2 2
(7m – 1) 2 = 25(1 + m2)
49m2 – 14m + 1 = 25 + 25m2
24m2 – 14m – 24 = 0  m1m2 = – 1 Ans. ]
Q.4
2 x (5 x  1)  (5 x  1) ( 2 x  1) (5 x  1) Lim ( 2  1) (5  1)
x x
l = Lim = Lim
x · x ln (sec x )1 x
[Sol. = x 0 2
x 0 ln (sec x ) x 0 ln (sec x )
x 2 ln 2 · ln 5 (ln 2 · ln 5) x 2
= Lim
x 0 ln (1  (sec x  1))
= Lim = 2 (ln 2 · ln 5) = (ln 4) (ln 5) Ans.]
x 0 (1  cos x )
Q.5
a a
[Sol. 1  r1 = r1 and 1  r2 = r2
a
 r2 – r + a = 0 
1 r
hence r1 and r2 are the roots of =r r1 + r2 = 1 Ans.]
Code-B Page # 1

14. MATHEMATICS
Paragraph for question nos. 6 to 8
[Sol. y = f (x) = x 2 – 4ax + a2 + 5a + 2
4a
vertex = = 2a
2
(i) f (2a) = 4a2 – 8a2 + a2 + 5a + 2
= – 3a2 + 5a + 2
 2 5a 2   5  25 24  49 
2
5
2
= – 3 a    = – 3  a      =  3 a  
 3 3 
 6  36 35  12 
 6
5
f (2a) is maximum if a = Ans.(i)
6
(ii) f (x)  0  x  R
x2 – 4ax + a2 + 5a + 2  0
D0
16a2 – 4(a2 + 5a + 2)  0  4a2 – a2 – 5a – 2  0  3a2 – 6a + a – 2  0
(3a + 1)(a – 2)  0  – 1/3  a  2  amax = 2 Ans.(ii)
(iii) Given minimum value of f (x) is 2, for x  0
Case-1: when vertex is  0
i.e. 2a  0  a0
minimum f (x) occurs at x = 2a
f (2a) = 4a2 – 8a2 + a2 + 5a + 2 = – 3a2 + 5a + 2
f (2a) = 2  a(5 – 3a) = 0  a = 0 or a = 5/3
Case-2: when vertex is < 0
i.e. a<0
in this case, minimum occurs when x = 0
f (0) = a2 + 5a + 2 = a(a + 5) = 0  a = 0 or a = – 5
hence a  {0, – 5, 5/3}  3 values of 'a' Ans.(iii) ]
Paragraph for question nos. 9 to 11
Sol. We have
 r    
sin   x     x  (r  1)  
 
n
6  6 
   r 
f1(x) = 2
r 1
cos  x  (r  1)  · cos  x  
 6  6
      2       3   2  
= 2 ·  tan  x    tan x    tan  x    tan x      tan  x    tan  x   
    
  6    6   6    6   6 
  n     
 ................   tan  x    tan  x  (n  1)  

  6   6  

  n  
 f1(x) = 2 ·  tan  x    tan x 
 
  6  
   
for n = 3, f1(x) = 2  tan   x   tan x  = 2  cot x  tan x  = – 2
1
 
 2   sin x cos x
Code-B Page # 2

15. MATHEMATICS
 n 
2 f2(x) = f1(x) – 2 · tan  x  
 6 
Now
 n   n 
= 2 tan  x   – 2 tan x – 2 tan  x  
 6   6 
 f2(x) = – tan x
 f3(x) = – f2(x), so
f3(x) = tan x
 e( e x 1) 1
 2·(e x 1)
e ; x 0

Now f 4 ( x )   k1 ; x0

1 | tan x | tan x ; x  0
k2


1
f (0¯) = e 2  e
lne
Clearly
f (0+) = e k 2 & f (0)  k1
(i) As f4(x) is continuous at x = 0, so by definition of continuity, we get
f (0¯ ) = f (0+) = f (0)
 e = e k 2 = k1
1
 k1  e & k2 =  (D)
2
(ii) As y = f3(x) = tan x
clearly f3(x) is continuous as well as derivable everywhere in 0,  2 .  (C)
(iii) for n = 3,
f1(x) = – 4

sin x cos x
 – 2(tan x + cot x) = – 4 cos x sin x = 2  1 = 2 sin x · cos x
  
  0,  
4  2
x= (B)]
Q.12
[Sol. (A) f (x) = x – cos x ; f (0) < 0, f (/2) > 0
(B) f (x) = x + sin x – 1
 1
f (0) = – 1 < 0; f (/6) =  1  0
6 2
(C) f (x) = a(x – 3) + b(x – 1) in [1, 3]
f (1) = – 2a < 0; f (3) = 2b > 0  f (x) = 0 in (1, 3)
(D) h (x) = f (x) – g (x)
h (a) = f (a) – g (a) > 0
h (b) = f (b) – g (b) < 0
hence using IVT all the four have at least one root in indicated interval. ]
Code-B Page # 3

16. MATHEMATICS
Q.13
[Sol. g ' (x) = aeax + f ' (x)
g ' (0) = a + f ' (0) = a – 5 ....(1)
g '' (x) = a2eax + f '' (x)
g '' (0) = a2 + f '' (0) = a2 + 3 ....(2)
g ' (0) + g '' (0) = 0
 a – 5 + a2 + 3 = 0  a2 + a – 2 = 0  (a + 2)(a – 1) = 0  a = 1, – 2 Ans.]
Q.14
             
= 1  cos  1  cos    1  cos    1  cos    
  
 4k    2 4k     2 4k     4k  

[Sol. P(k)
        
= 1  cos  1  sin  1  sin  1  cos 
 4k   4k   4k   4k 
 
4 sin 2 · cos 2
    
= 1  cos  1  sin  =
2 2 4k 4k
 4k   4k  4
1 2  
sin   · =
1 1 1

 2k 
P(k) = P(3) = ;
4 4 4 16
 2  1 2  1 
sin  sin  1  cos  = 2  2  (B)
8 8 4
P(4) =
4 2k 4 16
1 2  1 2   1 5 1 3 5
=  2 sin  = 1  cos 36  = 1  
1
sin  4  = 32  (C)
8 10  8
P(5) = 8
4 10 
1 2  1 2   1  1 3  2 3
=  2 sin  = 1  cos  = 1 


2  = 16
sin
8 12  8  6 8
P(6) = ]
4 12 
Q.15
[Sol. for continuity at x = 0
1
f (0) = 0 ; f (0–) = 0; f (0+) = Lim h sin 
n
=0 n>0
h0 h
for derivability at x = 0
1
h n sin
f ' (0–) = 0; f ' (0+) = Lim h for this limit not to exist n  1
h0 h
hence 0 < n  1
 n can not be 3/2 or 2  C, D ]
Code-B Page # 4

17. MATHEMATICS
PART-B
Q.1
[Sol.
e ln x ! x!
= 20  = 20  x(x – 1) = 20  x2 – x – 20 = 0  (x – 5)(x + 4) = 0
( x  2)! ( x  2)!
(A)
 x = 5 Ans.  (T)
(B)  4{x} = x + [x] ....(1)
2x 2x 5
= x + x – {x}  5{x} = 2x  {x} =  0 <1  0x<
5 5 2
hence [x] = 0, 1, 2
Again from (1) 4x – 4[x] = x + [x]
3x = 5[x]
Case-I: If x  [0, 1)  [x] = 0
 3x = 0  x0
Case-II: If x  [1, 2)  [x] = 1
x
5
 3x = 5 
3
Case-III: If x  (2, 5/2]  [x] = 2
10
 3x = 10  x= (reject)
3
 number of solutions = 2 Ans.
Alternatively :
Let x = I + f
2I 2 4
 4f = I + f + I  f=  I = 0, 1  x=0&x=1+ =
3 3 3
(C) Given equation is
x2 + 2x(y + g) + y2 + 2fy + 4 = 0
2 x = – 2 (y + g) ±
  4( y  g) 2  4( y 2  2fy  4)
 
x = – (y + g) ± (g 2  4)  2 y(g  f ) ....(1)
(1) will represent a pair of lines if its discriminant is zero
 + 4(g – f) = 0  g=f
 x = – (y + g) ± g 2  4 . For two real lines
g2  4  g  2 or g  – 2 Ans.  Q, R, S, T
(D) If d = 6; A = (0, 0); B(6, 0)
Consider a circle with centre A are radius 2(r1) and a
circle with centre B and radius 3(r2). The circles will be
separated. There will 4 common tangents at a distance
of 2 from A and 3 from B  4 lines
|||ly if d = 5; r1 + r2 = 5  circles touches externally
 3 common tangents  3 lines  R, S
NOTE: |||ly for two intersecting circles
r2 – r1 < d < r1 + r2 i.e. 1 < d < 5  2 common tangents  2 lines
if d = r2 – r1 i.e. if d = 1  circles touches internally  1 lines ]
Code-B Page # 5

18. MATHEMATICS
PART-C
Q.1
[Sol. AAAAA | B B B B B B M
         
Middle digit must be A (think !)
so that even number of A's and B's are available
Take AABBB on one side of M (6th place) and then their image about M in a unique way
5!
 Number of ways = 2!· 3! = 10 Ans.]
Q.2
[Sol. Using L'Hospital's Rule using f g h ' + g h f ' + h f g'
5 cos 4 x (sin x ) cos 3 2x cos 3 3x  3 cos 2 2 x 2 sin 2x  cos 5 x cos 3 3x  3 cos 2 3x 3 sin 3x  cos 5 x cos 3 2 x
Lim
x 0 2x
 6   3 = 22 Ans.
5 9
=
2 2
Alternatively: Using expansion
 x2   (2x ) 2   9x 2 
5 3 3
5x · cos32x · cos33x = 1   ....  · 1   ....  · 1   .... 
cos      
 2!   2!   2! 
 5x 2
1 
=


 ....  1  6 x 2  ....
   27 x 2
1 


 .... 

 2   2 
  5 2   27 x 2 
1   6   x  ....  1 
=   .... 

  2  2 
27 x 2 5x 2
= 1  6x 2   ......... = 1 – 22x2 + .....
2 2
1  cos5 x · cos3 2 x ·cos3 3x 1  (1  22x 2  ....)
 Lim = Lim = 22 Ans. ]
x 0 x2 x 0 x2
Q.3
[Sol. Let the roots are a – 3d, a – d, a + d, a + 3d
sum of roots = 4a = 0  a=0
hence roots are – 3d, – d, d, 3d
m
product of roots = 9d4 = m2  d2 = ....(1)
3
again  x 1x 2 = 3d2 – 3d2 – 9d2 – d2 – 3d2 + 3d2
= – 10d2 = – (3m + 2)
= 10d2 = 3m + 2
10m
= = 3m + 2
3
= 10m = 9m + 6
Code-B m = 6 Ans. ] Page # 6

19. MATHEMATICS
Q.4
[Sol. Let P(x) = k(x – 1)(x – 3)(x – 5)(x – r) where r  1, 3, 5
ln P(x) = ln k + ln (x – 1) + ln (x – 3) + ln (x – 5) + ln (x – r)
differentiate w.r.t. x
P' ( x )
  
1 1 1 1
x 1 x  3 x  5 x  r
=
P( x )
put x = 7, as P'(7) = 0  P(7)  0
23 6
    
1 1 1 1 1 1 11
 =0  =0 
6 4 2 7r 12 7r r  7 12
12 89 p
 r–7=  r= =  p + q = 100 Ans.]
11 11 q
Code-B Page # 7