iit jee 2012 paper 1 with solutions
TRANSCRIPT
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
1/36
IIT-JEE 2012 EXAMINATION
CODE : 9 08 / 04 / 2012
Part I : (PHYSICS)
SECTION I (Single Correct Answer Type)
This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which
ONLY ONE is correct.
Q.1 A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of
the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of
curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be
(A) 280.0 cm (B) 40.0 cm (C) 21.5 cm (D) 13.3 cm
Ans. [B]
Sol. For the combination( )
RRfeq
)1(11 21 +
=
20. =eqf
Here u = 40, f = 20
40=v
Q.2
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
2/36
Ans. [B]
Sol. Angular momentum about rotational axis
vx
x = vt L(t) = [I + m(vt)
2]
= tmv2dtdL2t ;
torque = (2mv2)t
Q.3
Ans. [C]
Sol. Under steady state ])T3(T[A]T)T2[(A44
14
14 =
(2T)4
41T =4
1T 34T
4
2 41T = (24
+ 34)T
4 ; 2
41T = (16 + 81)T
4
T1 = T2
974/1
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
3/36
Q.4
Ans. [D]
Sol.
O Rr
E(r)
r < R ; Er= 0
r > R ; Er 1/r2
r < R , V =R
KQ; r = R , V =
R
KQ; r > R , V =
R
KQ
r
(r)
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
4/36
Q.5
Ans. [A]
Sol.2d
MLg4Y
l= & % ymax = %M + %L + %l + 2%d
Least count of both instrument, l = d = 3105100
5.0 =
%l = %225.0
105100
3
==
l
l
%11005.0
105100
d
dd%
3
=
=
=
here we see that, contribution ofl, = 2%
contribution of d = 2% d = 2 1 = 2%
hence both terms l and d contribute equally.
Q.6
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
5/36
Ans. [A]
Sol.
4.9 m y = 0 0.2 m
10 m
45
B A
The block is released from A.
x = 4.9 m + (0.2 m) sin (t +2
)
at t = 15 ; x = 5 m
So range of projectile will be 5 m
Nowg
v =
90sin5
2
v2 = 50
v = 50
Q.7
Ans. [D]
Sol. As,d
D=
R> G > B
So R> G > B
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
6/36
Q.8
Ans. [C]
Sol. prL 00rrr
=
0Lr
is always directed along the axis & its magnitude is constant.
PLr
P'Lr
P
Q.9
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
7/36
Ans. [D]
Sol.m
KTvrms
3=
16.3104
40
(Ar)
(He) ===rms
rms
v
v
Q.10
Ans. [C]
Sol.
e
mg
qE
45
qE = mg
E =q
mg
q
mg
101
V2
=
19
312
106.1
1001830101.910V
=
= 1012
9.1 114
= 1000 1012
V = 1 109
V
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
8/36
SECTION II (Multiple Correct Answer(s) Type)
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which
ONE or MORE arecorrect.
Q.11
Ans. [A, C, D]
Sol. out =00
in qq
=
By symmetry
Ans is (A, C, D)
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
9/36
Q.12
Ans. [A,B,C,D]
Sol. The circuit can be simplified as
12V
4 4 4
2 2 2
11
P
Q T
S
because of symmetry no current passes through 1 resistor.
12V
4 4 4
2 2 2P
Q T
S
I1 I1 12V
12
6I2
Req =612
612
+
= 4
I1 =4
12= 3A ; I2 = =3
3
22A
Here we have
4VV QS = i.e., QS VV <
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
10/36
Q.13
Ans. [A, C]
Sol.
1N
1N
1 cos
1 sin
P
Q
If = 45 then cos = sin hence block will be at rest.If plane is rough & > 45 then sin > cos so friction will act up the planeIf plane is rough & < 45 then cos > sin so friction will act down the plane so (A, C) are correct
Q.14
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
11/36
Ans. [C, D]
Sol.
v
X
Y
E0, B0
* If = 0 or 10
then particle moves in helical path with increasing pitch along Y-axis.
* If = 90 then magnetic force on the particle is zero and particle moves along Y-axis with constantacceleration.
Q.15 A person blows into open end of a long pipe. As a result, a high pressure pulse of air travels down the pipe.
When this pulse reaches the other end of the pipe,
(A) a high pressure pulse starts traveling up the pipe, if the other end of the pipe is open.
(B) a low pressure pulse starts traveling up the pipe, if the other end of the pipe is open
(C) a low pressure pulse starts traveling up the pipe, if the other end of the pipe is closed
(B) a high pressure pulse starts traveling up the pipe, if the other end of the pipe is closed
Ans. [B, D]
Sol.
At rigid end there is no phase difference in pressure wave, when end is open phase difference of come inpressure wave.
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
12/36
SECTION III (Integer Answer Type)
This section contains 5 questions. The answer to each question is a SINGLE DIGIT INTEGER, ranging from 0 to 9
(both inclusive).
Q.16
Ans. [6]
Sol.
2
1
(1) + (2) = Complete cylinder
E1 + E2 = E
E =)R2(2
R
0
2
=04
R
E2 =3
4
3
2
R
)R4(4
12
0=
0424
R
E1 = E E24
R
24
11 =
4
R
64
23
=
616
R23
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
13/36
Q.17
Ans. [5]
Sol.
OP
a
a/2
I = J a2
B =
a2
aJ 20
4
a
2a32
J 20
B = 0Ja
12
1
2
1
B = 0Ja 12
5
Q.18
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
14/36
Ans. [3]
Sol.
O
R
2R
P
Let mass of original disc = m
The mass of disc removed =)R4(
m2
R2 =4
m
So M.O.I of remaining section about axis passing
through "O" I0 =2
)R2(m 2
+ 2
2
R4
m
)2(
R
4
m
2mR2
+8
mR2mR 22
83
2 mR2
813
mR2
MOI of remaining section about "P"
IP =
+ 2
2
)R2(m2
)R2(m
+ 2
2
R54
m
)2(
R
4
m
[2mR2 + 4mR2]
+
4
mR5
8
mR 22
6mR2 8
11mR
2
8
37mR
2
O
P
I
I=
8
37
13
8 3
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
15/36
Q.19
Ans. [7]
Sol. Assume circular wire loop as primary and square loop as secondary coil
secondary = 2/322
20
)RR3(2iR2
+ a2 cos 45
=2
2a
R82
iR 23
20
M =i
ondarysec=
R22
a2/13
20
M =
R2
a
2
7
20
Q.20
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
16/36
Ans. [7]
Sol.
10 fme
v
Ui = 0
Uf= 15
2
1010
e120k
Ki =2
mv2
1and Kf= 0
From Energy conservation,
Kf+ Uf= Ki + Ui
2
15
2
mv2
1
1010
e120K=
v = 5.76 107
727
34
1076.5103
5
1067.6
mv
h
==
= 7 fm (approx)
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
17/36
Part II : (CHEMISTRY)
SECTION I SINGLE CORRECT ANSWER TYPE
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
Q.21 As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is -
(A) Tetraaquadiaminecobalt (III) chloride
(B) Tetraaquadiamminecobalt (III) chloride
(C) Diaminetetraaquacobalt (III) chloride
(D) Diamminetetraaquacobalt (III) chloride
Ans. [D]
Sol. Factual
Q.22 In allene (C3H4), the type (s) of hybridization of the carbon atoms is (are)(A) sp and sp
3(B) sp and sp
2(C) only sp
2(D) sp
2and sp
3
Ans. [B]
Sol. CH2=C=CH2
sp2
spsp
2
Q.23 For one mole of a van der Waals gas when b = 0 and T= 300 K, the PVvs. 1/Vplot is shown below. The
value of the van der Waals constant a (atm. liter2 mol2)
20.1
3.00
21.6
24.6
23.1
2.0
PV(liter-atmm
ol1)
V
1(mol liter
1)
[Graph not to scale ]
(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0
Ans. [C]
Sol.
+
2V
aP V = RT
PV = RT V
a
Slope = a
=1
6.211.20= 1.5
a = 1.5
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
18/36
Q.24 The number of optically active products obtained from the complete ozonolysis of the given compound is
CH3CH=CHCCH=CHCCH=CHCH3
CH3 H
H CH3 (A) 0 (B) 1 (C) 2 (D) 4
Ans. [A]
Sol. CH3CH=CHCCH=CHCCH=CHCH3O3, Zn, H2O
O O O O OO
CH3
H
H
CH3
2CH3.CHO + OHCCCHO + OHCCCHO
CH3
H
H
CH3
Q.25 A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below.
The empirical formula of the compound is -
(A) MX (B) MX2 (C) M2X (D) M5X14
Ans. [B]
Sol.
8
18
X
+ 6 2
1= X4
4
14
M
+ 1 = M2
Simplest ration = X2M
Q.26 The number of aldol reaction (s) that occurs in the given transformation is -
(A) 1 (B) 2 (C) 3 (D) 4Ans. [C]
Sol. CH3CHO + 3HCHO
nrexaldol3
NaOH.aq
.conc HOCH2CCHO + HCHO
CH2OH
CH2OH
nrex
Cannizaro HOCH2CCH2OH + HCOONa
CH2OH
CH2OH
Q.27 The colour of light absorbed by an aqueous solution of CuSO4 is -
(A) orange-red (B) blue-green (C) yellow (D) violet
Ans. [A]
Sol. Factual.
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
19/36
Q.28 The carboxyl functional group (COOH) is present in -
(A) picric acid (B) barbituric acid (C) ascorbic acid (D) aspirin
Ans. [D]
Sol.
OCOCH3
COOH
Aspirin
Q.29 The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius]
(A)20
2
2
4
h
ma(B)
20
2
2
16
h
ma(C)
20
2
2
32
h
ma(D)
20
2
2
64
h
ma
Ans. [C]
Sol. mvr =2
nh
mv =r2
nh
m2v
2=
22
22
r4
hn
2
1m
2v
2=
22
22
r8
hn
2
1
m
vm 22= K.E =
20
22
2
22
z
nm8
hn
a
=20
22
2
nm8
h
a
==
1z2n
=20
2
2
32
h
ma
Q.30 Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen ?
(A) HNO3, NO, NH4Cl, N2 (B) HNO3, NO, N2, NH4Cl
(C) HNO3, NH4Cl, NO, N2 (D) NO, HNO3, NH4Cl, N2
Ans. [B]
Sol. 3
5
ONH
+
ON
2+
2
0
N ClHN 4
3
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
20/36
SECTION II : MULTIPLE CORRECT ANSWER(S) TYPE
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which
ONE or MORE are correct.
Q.31 Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction, as
shown in the given scheme.
(A) C6H5OH and C6H5COOH (B) C6H5COOH and C6H5CH2OH
(C) C6H5CH2OH and C6H5OH (D) C6H5CH2OH and C6H5CH2COOHAns. [B,D]
Sol. Factual
Q.32 Choose the correct reason(s) for the stability of the lyophobic colloidal particles.
(A) Preferential adsorption of ions of on their surface from the solution
(B) Preferential adsorption of solvent on their surface from the solution
(C) Attraction between different particles having opposite charges on their surface
(D) Potential difference between the fixed layer and the diffused layer of opposite charges around the
colloidal particles
Ans. [A,D]Sol. Factual.
Q.33 Which of the following molecules, in pure form, is (are) unstable at room temperature ?
(A) (B) (C)
O
(D)
O
Ans. [B,C]
Sol. Antiaromaticity
Q.34 Which of the following hydrogen halides react(s) with AgNO3(aq) to give a precipitate that dissolves in
Na2S2O3(aq) ?
(A) HCl (B) HF (C) HBr (D) HI
Ans. [A,C,D]
Sol. AgBr + 2Na2S2O3 Na3[Ag(S2O3)2] + NaBr
Sodium argentothio sulphate
colourless soluble
Similar reaction observed in AgCl, AgI
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
21/36
Q.35 For an ideal gas, consider onlyP-Vwork in going from an initial state Xto the final stateZ. The final stateZ
can be reached by either of the two paths shown in the figure. Which of the following choices(s) is (are)
correct ? [take Sas change in entropy and w as work done]
(A) Sxz = Sxy + Sy z (B) Wxz = Wxy + Wy z
(C) Wxyz = Wxy (D) Sxyz = SxyAns. [A,C]
Sol. Sx z = Sx y + Sy zState
Wx y z = Wx y + Wy z 0
SECTION III : INTEGER ANSWER TYPE
This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both
inclusive)
Q.36 The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides
are positively charged at pH = 7.0 ?
H3NCHCONHCHCONHCHCONHCHCOO
H R1 R2 H
Peptide R1 R2
I H H
II H CH3
III CH2COOH H
IV CH2CONH2 (CH2)4NH2
V CH2CONH2 CH2CONH2
VI (CH2)4NH2 (CH2)4NH2VII CH2 COOH CH2CONH2
VIII CH2 OH (CH2)4NH2
IX (CH2)4 NH2 CH3
Ans. [4]
Sol. IV, VI, VIII, IX
Four amides are positively charged at pH = 7.0
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
22/36
Q.37 The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a
nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic
table ?
X++++ H2n6HCu1
1
1
0
1
1
63
29 Ans. [8]
Sol. Z = 29 + 1 2 2 = 26
Fe 8th groups.
Q.38 When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its
pyranose form is -
CHO
CH2
CHOHCHOH
CHOH
CH2OH
Ans. [8]
Sol. 8 stereoisomers possible for given compound in pyranose form of D-configuration.
Q.39 29.2% (w/w) HCl stock solution has a density of 1.25 g mL1
. The molecular weight of HCl is 36.5 g mol1
.
The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is.Ans. [8]
Sol.wt.mol
10dx V = 200 0.4
5.36
V1025.12.29 = 200 0.4
V = 8 ml.
Q.40 An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and
1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the value of]t[]t[
10/1
8/1 10 ? (take log102 = 0.3)
Ans. [9]
Sol. =10t
t
10/1
8/1
10/a
alog
K
303.2
8/a
alog
K
303.2
10 = 9
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
23/36
Part III : (MATHEMATICS)
SECTION I (Single Correct Answer Type)
This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.
Q.41 The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line
4x 5y = 20 to the circle x2
+ y2
= 9 is
(A) 20(x2
+ y2) 36x + 45y = 0 (B) 20(x
2+ y
2) + 36x 45y = 0
(C) 36(x2
+ y2) 20x + 45y = 0 (D) 36(x
2+ y
2) + 20x 45y = 0
Ans. [A]
Sol.
x2 + y2 = 9(h, k)P
5
204,
4x 5y = 20
A
B
Equation of chord AB is T = 0
x +
5
204y = 9 (i)
& hx + ky 9 = h2
+ k2
9 (ii)
Q Equation (i) & (ii) both represent the same line
Soh
=
k
5
204
=22 kh
9
+
=22 kh
h9
+=
)kh(4
)kh(20k4522
22
+
++
36h = 45k + 20(h2
+ k2)
20(x2
+ y2) 36x + 45y = 0
Q.42 The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that
each person gets at least one ball is
(A) 75 (B) 150 (C) 210 (D) 243Ans. [B]
Sol. G1 G2 G3
1 1 3
1 2 2
+
!2!2!2!1
!5
!2!3!1!1
!53!
= 150
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
24/36
Q.43 Let f(x) =
=
0x,0
0x,x
cosx2, x IR,
then f is(A) differentiable both at x = 0 and at x = 2
(B) differentiable at x = 0 but not differentiable at x = 2
(C) not differentiable at x = 0 but differentiable at x = 2
(D) differentiable neither at x = 0 nor at x = 2
Ans. [B]
Sol. f(0 + h) =0h
lim
0h
0h
cosh2
= 0
f(0 h) =0h
lim h
0h
cosh2
= 0
Q f(0+) = f(0) = 0 = finite
So f(x) is differentiable at x = 0
f(2 + h) =0h
lim
h
0h2
cos)h2( 2
+
+
=
f(2 h) =0h
lim h
0h2cos)h2(2
=
Qf(2+) f(2) but both are finite so f(x) is not differentiable at x = 2 but continuous at x = 2
Q.44 The function f : [0, 3] [1, 29], defined by f(x) = 2x3 15x2 + 36x + 1, is
(A) one-one and onto. (B) onto but not one-one.
(C) one-one but not onto. (D) neither one-one nor onto.
Ans. [B]
Sol. Given f : [0, 3] [1, 29]f(x) = 2x3 15x2 + 36x + 1
f'(x) = 6x2 30x + 36
= 6(x 2) (x 3)
f'(x) > 0 if x(0, 2)& f '(x) < 0 if x(2, 3) Function is many one & continuousNow f(0) = 1
f(2) = 29
Range = co-domain
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
25/36
0 2 3
(2,29)
(3,28)
(0,1)
Hence function is onto.
Q.45 If
+++
bax
1x
1xxlim
2
x= 4, then
(A) a = 1, b = 4 (B) a = 1, b = 4 (C) a = 2, b = 3 (D) a = 2, b = 3
Ans. [B]
Sol. xlim
+++
1x
bbxaxax1xx 22
= 4
xlim
+++
1x
b1)ba1(x)a1(x 2= 4
As limit is finite so 1 a = 0
a = 1
Nowx
lim
+
+
x
11
x
b1)ba1(
= 4
1 a b = 4
as a = 1 b = 4
Q.46 Let z be a complex number such that the imaginary part of z is nonzero and a = z 2 + z + 1 is real. Then a
cannot take the value
(A) 1 (B)3
1(C)
2
1(D)
4
3
Ans. [D]
Sol. As a is real
So a = a
z2 + z + 1 = 2z + z + 1
(z z ) (z + z +1) = 0
As z is imaginary
So z z 0
z + z + 1 = 0
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
26/36
z + z = 1 z = x + iy
x =2
1
so a = (x + iy)
2
+ (x + iy) + 1
= (x2 + x + 1 y2) + (2x + 1)yi x = 2
1
a =4
3 y2
so a 0
So a 4
3
Q.47 The ellipse E1 :4y
9x 22 + = 1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes.
Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the
ellipse E2 is
(A)2
2(B)
2
3(C)
2
1(D)
4
3
Ans. [C]
Sol.
(0, 4)
(0, 2) (3, 2)
(3, 0)
(0,2)
(3,0)
Let equation of ellipse E2 is
1b
y
a
x2
2
2
2=+
it passes through (0, 4)
so b2 = 16
and also passes through (3, 2)
So 1b
4
a
922
=+ 14
1
a
92
=+ a2 = 12
as a < bso 12 = 16 (1 e
2)
e2 =4
1 e =
2
1
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
27/36
Q.48 Let P = [aij]be a 3 3 matrix and let Q = [b ij], where bij = 2i + j
aij for 1 i, j 3, If the determinant of P is 2,then the determinant of the matrix Q is
(A) 210
(B) 211
(C) 212
(D) 213
Ans. [D]
Sol. Let P =
333231
232221
131211
aaa
aaaaaa
So Q =
336
325
314
235
224
213
134
123
112
a2a2a2
a2a2a2
a2a2a2
Now det (Q) = 222
32
4
332
322
312
232221
131211
a2a2a2
a2a2a2
aaa
= 29
2 22
333231
232221
131211
aaa
aaa
aaa
= 212
det (P)
= 212
2 = 213
Ans. [C]
Sol. I = + dx)xtanx(secxsec
2/9
2
sex x + tan x = t
(sec x tan x + sec2 x) dx = dt
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
28/36
sec x dx =t
dt
also, sec x tan x =t
1
sec x =2
1
+t
1t
So, I =
+
2/11t
dtt
1t
2
1
=
+ dtt
1
t
1
2
12/132/9
= K
t
1
11
2
t
1
7
2
2
12/112/7
+
= K11
1
7
t
t
1 2
2/11+
+
=
+++
11
1)xtanx(sec
7
1
)xtanx(sec
1 22/11
+ K
Ans. [A]
Sol. Equation of line
=
=
=
1
5z
4
3y
1
2x
General points { + 2, 4 + 3, + 5}
Intersection point with plane
5( + 2) 4(4 + 3) ( + 5) = 1
5 + 10 16 12 5 = 1
12 8 = 0
= 12
8=
3
2
Point
+
++
5
3
2,3
3
8,2
3
2
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
29/36
P
3
13,
3
1,
3
4
T(2, 1, 4)
Dr's( , 4 + 2, + 1)
S( +2, 4 + 3, + 5)Dr's(1, 4, 1)
Now
+ 4 (4 + 2) + ( + 1) = 0
+ 16 + 8 + + 1 = 0
18 = 9
= 2
1
Points
+++
5
2
1,32,2
2
1
2
9,1,
2
3
Distance at PS =
222
2
9
3
131
3
1
2
3
3
4
+
+
PS =36
1
9
4
36
1++ =
36
1161 ++=
36
18=
2
1
SECTION II (Multiple Correct Answer(s) Type)
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which
ONE or MORE are correct.
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
30/36
Ans. [A, C, D]
Sol. tan (2 ) > 0
0 < 2 2
1ve ; x <
2
1+ve
x 1 f(x) = x2
x 1 f '(x) = 2x 1 ve
11/2 0 1/2 1
Ans. [4]
Sol. Let x = ...23
14
23
14
23
14
x2
= 4 23
1x
23 x2
+ x 12 2 = 0
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
35/36
x =26
212.23.411 ++
x =26
171+=
23
8
6 + log3/2
9
4= 6 + log3./2
2
2
3
= 6 2 = 4
Ans. [9]
Sol. P'(1) = 0, P'(3) = 0
P'(x) = K(x 1) (x 3)
= K(x2
4 x + 3) P'(0) = 3K
P(x) =3
Kx
3 2K x
2+ 3K x +
3
K 2K + 3K + = 6, 9K 18 K + 9K + = 2
34 K + = 6,
34 K = 4
K = 3
P'(0) = 9
Ans. [3]
Sol.2|ba|
rr + 2|cb|
rr + 2|ac|
rr = 9
a.b + b.c + c.a = 3/2 ....(1)
Q2|cba|
rrr++ 0 ....(2)
a.b + b.c + c.a 2
3....(3)
Q from (1) & (3)
-
8/2/2019 IIT JEE 2012 Paper 1 With Solutions
36/36
so |cba|rrr
++ = 0
cbarrr
++ = 0
cbarrr
=
on squaring
1 = 2 + 2 cos B
cos B = 2
1 B = b
r^ cr
.
Let T = |2 ar
+ 5br
+ 5 cr
|
= |3 br
+ 3 cr
|
= |cb|3rr
+
= Bcos223 +
= 3
Ans. [4]
Sol. (x 1)2
+ (y 2)2
= ( 5 )2
(2, 0)
y
SQ
P (2, 4)
y2
= 8x
x
2+ 8x 2x 4.2 x2 = 0
x2
+ 6x 8 x2 = 0
x3/2
+ 6x1/2
28 = 0
x1/2
= t
t3 + 6t 28 = 0
(t 2 ) (t2
2 t + 4) = 0
t = 2 x = 2
y = 4
P(2, 4) Q(0, 0) S(2, 0)
Area PQS =2
1 2 4 = 4