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Vidyamandir Classes
VMC | JEE Main-2020 1 Solutions |7th January Morning
SOLUTIONS
JEE Main – 2020 | 7th January 2020 (Morning)
PHYSICS
SECTION – 1
1.(2) A logic gate is reversible if we can recover input data from the output.
2.(1) 2000kgm=
4000rf N=
?=v
4000 20000T = + , 24000T N=
.P T v= ;
60 746 24000 u = , 1.9 m/su =
3.(1) 20 0I I t I t= −
0 IA = = BA n
0 0 (1 2 )
= − = − −R
dV nAI t
dt
1
0 at sec2
RV t= =
2
0 0
0 0
2(1 2 )
2
RR
nI RVI t
R R
− = = −
4.(4) 0 0 0
ˆ ˆcos60 sin 602 2 2
E x y −
= + −
0
3 1ˆ ˆ1
2 2 2y x
= − −
5.(4) 86000 10 cm− =
For 2nd minimum 2sin 2d = 3
4d
=
So, for 1st minimum, 1sind = 1
3sin
4d
= =
1 25.65 (from sin table) = , 1 25
6.(1) Energy conservation :
2 21 1
2 2 2
GMm GMmmu mv
R R
−− + = +
2 GM
V uR
= − …(i)
Momentum conservation:
9
10 10 2 2T orbital
m m GM GMV V
R R
= =
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2 [By (1)]10
r
m GMV m u
R= −
Kinetic energy ( )2 2 21 81100 100 (1)
2 10 20 2T r
m m GM GMV V u
R R
= + = + − =
7.(4) 1 2
1 2
1 2mixmix
mix 1 2
P Pp
V V V
n C n CC
C n C n C
+ = =
+
1 21 2
1 2mix
1 1 2
1 2
1 1
1 1
R Rn n
n R n R
+
− − =
+ − −
On rearranging, we get, 1 2 1 2
mix 1 21 1 1
n n n n+= +
− − −
mix
5 3 2
1 1/ 3 2 / 3= +
− mix mix
5 171 1.42
12 12 − = = =
8.(1) Time-period 2 r
Tv
= =
2
0
1
2 a n nT
V Z Z
= ;
30
21
2 a nT
V Z
=
3T n
3
1 12 13
2 2
18
8
T nT T
T n= = = ; 16
2 12.8 10 secT −=
142 16
17.8 10
12.8 10f
−=
9.(2) Loss in P.E. = Gain in K.E.
2 21 1
2 2mgh mv Iw= +
Put, (no slipping)v wr=
2
2 2 21 1
2 2 2
mrmgh mw r w= +
2 23 1 4
4 3= =
ghmgh mw r w
r
10.(3) Equal number of magnetic field lines enters the circular region & comes out of infinite plane excluding
circular area. So, magnetic flux are equal in magnitude but opposite in direction.
0i = −
11.(1) For damped oscillation : 0ma bv kx+ + =
2
20
d x dxm b kx
dtdt+ + = …(i)
For LCR series circuit 0di q
iR Ldt C
− − − =
2
2
10
d q dqL R q
dt Cdt+ + = …(ii)
Comparing equation (i) and (ii), 1
, ,L m C R bk
12.(4) 0
1e
L Dm
f f
= +
, if final image is least distance of distinct vision
150 25
375 15 ef
= +
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750
2.17 cm 21.7 mm 22 mm345
ef = = =
Also, 0 e
L Dm
f f
=
if final image is at infinity.
150 25
3755 ef
=
, 22 mmef =
13.(4) 0 0 ,E B C C= = speed of light in vacuum
8 80 3 10 3 10 V/mE −=
0 9 V/mE = 3 10 ˆ9sin(1.6 10 48 10 ) V/mE x t k= +
14.(1) 2AB CDI I md= +
2 2
12 16
ml ml= +
27
48
ml=
Radius of gyration 27 7
48 48
ABI ll
m= = =
15.(4) 1
0.42.5
I A A= =
1 0.22
II A= =
16.(4) Take origin at 1 kg mass y = vertical, x = horizontal
1 0 1.5 3 2.5 0
cm 0.9 cm5
X +
= = ; 1 0 1.5 0 2.5 4
2cm5
Ycm + +
= =
17.(1) 20 cosI I= ,
200 cos
10
II=
1
cos 0.31 0.70710
= =
45 &90 45 − ; 71.6 = Angle rotated 90 71.6 18.4= − =
18.(4) All 'dc s are in series
1 1 1 1
....c dc dc dc= + + +
1 1
c dc=
0k Adc
dx
=
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0 00
1 1ln |1 |
(1 )
ddx
d dc x AK K A= = +
+
1d
2
ln|1 |2
dd d
+ = −
0
11
2
d d
c K A
= −
; 0 1
2
= +
K A dc
d
19.(1) 6.0m g= , 60 cml = , 21.0 mmA = , 190 msv −= , 11 216 10 NmY −=
2Tv T v= =
;
2v lY
A l
=
2 2 3
6 11
6 10 8100
10 16 10
v l mvl m
AY AY
−
−
= = =
5 33 10 m = 0.03 10 m = 0.03 mm− −=
20.(1) 11 1 2 2 2 1
2
VPV PV P P
V
= =
1.41
13
atm
=
Work
3 5
1 1 2 2
310 1.01 10 1
4.65
1 0.4
PV PVJ
− − −
= = −
Work 90J
Closest answer is 90.5 J
SECTION – 2
21.(600) 300 2
1900 3
= = − =in
W
Q ;
31800
2inQ W J= = ; 600low inQ Q Q J= − =
22.(60) 6 5 62 12 2 5 10 5 10 60 10− − − = + = + = 60C =
23.(10) = −A PKE PE PE A Bmgh mgh= − (2 1) 1 10 1 10mg J= − = =
24.(175) x yz z xyB A B A = + 3 25 4 25 175Wb= + =
25.(11) 1240
4310
photonE ev= =
No. of electrons emitted . .
photon
I An
E=
53 11
19
6.4 10 110 1 10
4 1.6 10
−−
−
= =
11x =
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CHEMISTRY
SECTION – 1
1.(2) It is Gay Lussac law of gaseous volume
2.(4) 2 3 3CS CH COCH+
A B
A……..A (non polar –non polar)
B……..B (polar –polar)
A……B (non polar…..polar)
So
A......AA......B
B......B
So, it is non ideal solution showing positive deviation.
So, volume should be greater than 200ml
3.(1) E Cu / Cu 0.34V++ = E Cu / Cu 0.522V+ =
Cu 2e Cu++ + → E 0.34V = -(1) oCuE / Cu ?++ + =
Cu e Cu+ → E 0.522V = -(2)
Cu e Cu++ ++ → oE ?=
(1)-(2)
Cu 2e Cu++ + → E 0.34V = -(3)
Cu Cu e++→ + oE 0.522V= − -(4)
Cu e Cu++ ++ → E ? = -(5)
3 4 5G G G + =
G nfE = −
3G 2 F .34 = − ( )4G 1 F 522 = − −
4G 1 f E = −
( ) ( ) ( )2f .34 1 .522 1 f E − + − − = −
E 0.158V =
4.(2) Vitamins Deficiency
Vitamin 2B -(Riboflavin) Cheilosis
Vitamin 1B - (Thiamine) Beriberi
Vitamin 6B - (Pyridoxine) Convulsions
Vitamin C - (Ascorbic acid) scurvy)
5.(2) Diammine chloride (methanamine) (Platinum II) chloride
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6.(2)
7.(1)
8.(4) n 5= Ss,Sp,Sd,Sf ,Sg
2 2n 5 25= =
No. of orbitals =1 3 5 7 9 25+ + + + =
Each orbital has one electron with 5
1m
2= +
9.(4) Potassium is alkali metal
It always show +1 oxidation state
22
22 2 2
2 2
K O 2K O
K O 2K O
KO K O
+ −
+ −
+ −
⎯⎯→ +
⎯⎯→ +
⎯⎯→ +
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VMC | JEE Main-2020 7 Solutions |7th January Morning
10.(3) (A)
3 3
H3 3 3 3
33 2
CH CH| |
CH C C CH CH C C C CH| | | |
CHCH OHon
+
+
− − − ⎯⎯⎯→ − − − −
3 3
33
H|
CH C C CH| |
CHCH
+
− − − 1,2methyl
shift⎯⎯⎯⎯⎯
3
3 3
3
CH|
CH C — C CH| |
CH H
+
− −
(B) ( )
alcohol3 3 3 3E2
3 3
CH CH CH CH CH C C CH| | | |
BrCH CH H
⎯⎯⎯⎯→− − − − = −
(C) ( )CH O K3 3
3 3 3
3 3
CH CH CH CH CH CH CH CH| | |
BrCH CH
− +
− − − ⎯⎯⎯⎯⎯⎯→ − − =
( )3 3CH O K− + ⎯⎯→Sterically hindered base (Hoffmann elimination)
(give Hoffmann major product)
(D)
11.(1) Due to lanthenoid contraction, size of 4d size of 5d series except La
12.(2) Fact
13.(2) Fact
14.(4) Dipole moment of 4CH & CC 4 0=
Dipole moment of 3CHc 0
15.(4) ( )c f Br I
16.(2)
17.(2) Fact
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18.(2)
19.(4) B→ Guanidine Type strongest organic base (conjugate acid is stablised by equivalent Resonance)
(three resonating structure)
A→ 2NH CH NH− = (also guanidine type but two equivalent resonating structure of conjugate
acid) (so less base then (B)
C→ 3 3CH NHCH (Aliphatic amine)
Order of basic nature B A C
Order of bkP B A C
20.(4) MOT can explain the nature of bonding due to synergic bonding 4[Ni(CO) ] .
SECTION – 2
21.(2)
22.(10.60)
2 4NaOH H SO
(A) (B)
+
Molarity of 34
NaOH 1040 100
−= =
Molarity of 3
2 4
9.8H SO 10
98 100
−= =
2NaOH + 2 4 2 4 2H SO Na SO H O→ +
310 M−
310 M−
40 10
Moles 340 10−= moles 310 10−=
Limiting Reagent is 2 4H SO
So, moles of NaOH left ( ) ( )3 340 10 10 10 2− −= −
320 10−
Molarity of 3
420 10OH 4 10
50
−− −
= =
OHP 4 log4 3.398= − =
HP 14 3.398 10.602= − =
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23.(23.03)
2yt 6.93years=
2.303 100t log
K 10=
0.693
K 0.16.93
= =
2.303
t 1 23.03.1
= = years
24.(–2.70) ( ) ( )gA 2B→
U 2.1kcal, s 20cal / k = = , T 300K=
G H T S = −
H U ngRT = + ng 2 =
( )( )g 3300 300 20 = −
3300 6000= −
2700cal= −
2.7kcal= −
25.(1.67)( ) ( ) ( )
2 3 2Hot &conc x y
3Cl 6NaOH 5NaCl NaClO 3H O+ → + +
3 3NaCl AgNO AgCl NaNO+ →
Y is N 3NaClO 3Na ClO+ −
O Cl 0
O
− − =
Bond order =
(in Resonance)
noof band
1no.of bond
= +
2
13
= +
5
1.673
= =
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MATHEMATICS
SECTION – 1
1.(1) 1 x ydye e
dx
−− =
y y xdy
e e edx
− =
y x ydy
e e edx = +
ye t=
y dy dt
edx dt
=
xdt
e tdx
= +
xdt
t edx
− =
I.F. 1dx xe e
− −=
x x xt e e e dx− − =
xt e x c− = +
x xt xe ce= +
y x xe xe ce= +
0, 0x y= =
1 0 c= + 1c =
At 1x =
( 1)y xe e x= +
(2)ye e=
log log 2ye ee e=
log 2 1ey = +
2.(4) Given that plane passes through the point 1 1 1 2 2 2
(2, 1, 0) (4, 1, 1),
, , , ,x y z x y z and
3 3 3
(5, 0, 1).
, ,x y z
Equation of plane passes through 3 non-collinear point
1 1 1
2 1 2 1 2 1
3 1 3 1 3 1
0
x x y y z z
x x y y z z
x x y y z z
− − −
− − − =
− − −
2 1 0
4 2 1 1 1 0 0
5 2 0 1 1 0
x y z− − −
− − − =
− − −
2 1
2 0 1 0
3 1 1
x y z− −
=
−
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( 2)(0 1) ( 1)( 1) ( 2) 0x y z− + − − − + − = 2 1 2 0x y z− + − − =
2 3x y z− − =
Now image of (2, 1, 6)R w.r.t 2 3x y z+ − = is
2 2 22 1 6 2(2 1 12 3)
1 1 6
− − − − + − −= = =
−
x y z
x
2 2 22 1 64
1 1 2
x y z− − −= = =
−
2 6x = 2 5y = 2 2z = −
3.(4) Let x be a random variable and k be the value of assigned x for 3, 4, 5k =
k 0 1 2 3 4 5
( )P k 1
32
12
32
11
32
5
32
2
32
1
32
Now expected value ( )= xP k
1 12 11 5 2 1
1 ( 1) ( 1) 3 4 532 32 32 32 32 32
= − + − + − + + + 24 28 4 1
32 32 8
− += =
4.(4) 4y mx= +
Let 1
y mxm
= + is tangent to 2 4y x=
Now solving with 2 2x by=
2 1
2x b mxm
= +
2 2
2 0b
x bmxm
− − =
0x = (because line touches the curve)
2 2
( 2 ) 4 1 0b
bmm
− − − =
2 2 8
4 0b
b mm
+ = 2 2 8
4b
b mm
= −
3 2bm = − 3 2
mb
= −
1/ 32
mb
= −
Now
1/ 3
42
b − =
642
b − =
128b = −
2nd method
Given 4y mx= + ........ (i)
Let the equation of common tangent of 2 4y x= & 2 2x by= is
1
y mxm
= + ........ (ii)
(i) & (ii) are identical
1
4m =
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VMC | JEE Main-2020 12 Solutions |7th January Morning
So the line 1
44
y x= + is also common tangent of 2 2x by=
Solving 2 162
4
xx b
+ =
22 16x bx b= +
22 16 0x bx b− − =
0D =
2 4 2 ( 16 ) 0b b− − =
2 128 0b b+ =
128b = − & 0b = (which is not possible)
So 128b = −
5.(2) 125 124 249 49 ..... 49 49 1+ + + + +
Clearly given series in G.P.
126 126 63 631(49 1) 49 1 (49 1)(49 1)
49 1 48 49
− − + −= =
−
Greatest value of 63k =
6.(4) Given digits are 1, 3, 5, 7, 9
For digits to repeat we have 5 choices, hence total members are, 51
6! 5 6!
2! 2C
=
7.(3) Given equation, k k kx y a+ =
Differentiating w.r.t. x,
1 1 0k k dy
kx kydx
− −+ =
1kdy x
dx y
−
= −
........ (i)
& given that
1/ 3dy y
dx x
= −
........ (ii)
(i) & (ii) are tangent
1 1/ 3kx x
y y
− −
=
1
13
k − = − 2
3k =
8.(3) Given equation 2( 1) tan 2 tan 1k x x k+ − = −
2( 1) tan 2 tan ( 1) 0+ − + − =k x x k
Sum of roots 2
tan tan1
b
a k
+ = − =
+
Product of roots 1
tan tan1
c k
a k
− = =
+
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VMC | JEE Main-2020 13 Solutions |7th January Morning
2 2tan ( ) (tan( )) 50+ = + =
22
tan tan 150 5011 tan tan
11
kk
k
+ + = = = = −− − +
22
1 501 1
1
kk k
k
+ =+ − +
+
2
502
=
2 100 = 10 =
9.(2) Given equation,
2 1 0x x+ + =
2, =
Taking =
2
2 4
1 1 11
13
1
A
=
2 2
2 2 2 4 4 2
2 3 3 4 3
1 1 1 1 1
1 1 1
1 1 1
A
+ + ++ + + = ++ + + + + + + + + + +
3 0 01
0 0 33
0 3 0
=
2A I= 4A I=
Now, 31 28 3 4 7 3( )A A A A A= = 3 3I A A =
10.(3) The system of linear equations
2 2 0x ay az+ + =
2 3 0x by bz+ + =
2 4 0x cy cz+ + = has non zero solution
Then 0,D =
2 2
2 3 0
2 4
a a
b b
c c
=
1 2
1 3 0
1 4
a a
b b
c c
=
1( 4 ) 2 ( ) (4 3 ) 0bc bc a c b a c b− − − + − =
3 4 2 2 4 3 0bc bc ac ab ac ab− − + + − =
2 2 3 0bc ac ab ab− + + − =
2bc ab ac− − = −
2bc ab ac+ =
2 1 1
b a c= +
1 1 1, ,
a b c are in AP
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VMC | JEE Main-2020 14 Solutions |7th January Morning
11.(2) Given statement
( ) ( )p q q p
p q p q p q p ( )( )p q p q −
T T T F F F
T F F F T F
F T T T T T
F F T T T T
( ) ( )p q p q p =
12.(3) 2
tan cot 1( ) 2
1 tan siny x
x
+ = +
+
2 2
2
sin cos
1cos sin2sin sin
1cos
x +
+
+
2 2
2 2 2
2
cos sin
sin cos 12
cos sin sin
cos
+
+ +
2
2
cos 12
sin cos sin x
+
22cot cosec +
21 cot 2cot+ + 1 cot+ (1 cot )− +
for 3
,4
2cosec
dy
d=
dy
d at
25 5cosec 4
6 6
= = =
13.(3) z x iy= +
1 ( 1)
2 2 (2 1)
z x iy
z i x y i
− − +=
+ + +( 1) (2 (2 1) )x iy x y i= − + − +
(2 (2 1) ) (2 (2 1) )x y i x y i= + + − +
2 2
2 2
2 2 2Re( ) 1
4 4 4 1
x x y yz
x y y
+ + += =
+ + +
2 2 2 22 2 2 4 4 4 1x x y y x y y− + + = + + +
2 22 2 2 3 1 0x y x y+ + + + =
2 2 3 1
02 2
+ + + + =x y x y
2 2 1 9 1 5
4 16 2 4r g f c= + − = + − =
Circle whose diameter is 5
.2
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VMC | JEE Main-2020 15 Solutions |7th January Morning
14.(1) Using LMVT for [ 7, 1]x − −
( 1) ( 7)
2( 1 7)
f f− − −
− +
( 1) 3
26
f − +
( 1) 9f −
Using LMVT for [ 7, 0]x −
(0) ( 7)
2(0 7)
f f− −
+
(0) 3
27
f + (0) 11f
(0) ( 1) 20f f+ −
15.(2)
Required area = Total area − shaded area
11 23/ 2
0 0
22 ( ) 2
3 2
xx x dx x
= − − = − −
2 1 12 (12 1)
3 2 6
= − − = −
16.(4) Distance between foci 2 6ae= =
3ae = ........ (i)
Distance between directrixes 2
a a
e
= − −
2
12a
e= 6
a
e= ........ (ii)
Multiply (i) and (ii)
2 18a =
Dividing (i) by (ii)
2 1
2e =
Now, 2
2
21
be
a= −
21
12 18
b= − 2 9b =
Length of 22 2 9
3 23 2
bLR
a
= = =
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VMC | JEE Main-2020 16 Solutions |7th January Morning
17.(4) Let the numbers 2 , , , , 2a d a d a a d a d− − + +
Sum 5 25a= = 5a =
Product 2 2 2 2( 4 )( ) 2520a d a d a= − − =
2 2(25 4 )(25 )5 2520d d− − =
2 2(4 25)( 25) 504d d− − =
Solving we get 1d = or 11
2d =
Now 11
,2
d = since 11 1
52 2
a d− = − = −
Therefore, largest number 11
2 5 2 162
a d
= + = + =
18.(1) 2( ) 1g x x x= + −
2( ( )) ( ) ( ) 1g f x f x f x= + −
24 5 51
5 4 4g f f f
= + −
Also, 2( ( )) 4 10 5g f x x x= − +
25 5 5 25 50 5
4 10 5 54 4 4 4 4 4
g f
= − + = − + = −
2 5 51
4 4f
= − = −
25 1 1 5
14 2 4 4
f
+ − − = −
25 1
04 2
f
+ =
5 1
4 2f
= −
19.(1 & 3) ( 1 ) ( )f a b x f x x R+ + − =
1( ( ) ( 1)
b
a
I x f x f x dxa b
= + ++ . . . (i)
1( ) ( ( ) ( 1 ))
b
a
I a b x f a b x f a b x dxa b
= + − + − + + + −+ . . . (ii)
1( ) ( ( 1) ( ))
b
aI a b x f x f x dx
a b= + − + +
+ . . . (iii)
(i) + (iii)
2 ( ( 1) ( )
b
a
a bI f x f x dx
a b
+= + ++
2 ( 1) ( )
b b
a a
I f x dx f x dx= + +
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VMC | JEE Main-2020 17 Solutions |7th January Morning
2 ( 1 ) ( )
b b
a a
I f a b x f x dx= + + − +
( ) ( )
b b
a a
I f x dx f a b x dx= = + −
( ) ( 1)
b b
a a
f x dx f x dx= +
1 1
1 1
( 1) ( )
b b
a a
I f x dt f x dx
− +
− +
= + =
20.(2) Since a bisects the angle between and ,b c
ˆˆ ˆ ˆ ˆ 4ˆ ˆ( )
2 3 2
i j i j ka b c
+ − += =
ˆˆ ˆ4 2 4
3 2
i j ka
+ +=
or
ˆˆ ˆ2 4 4
3 2
i j ka
+ −=
On comparing, we get
4, 4 = = or 1, 2 = = −
ˆˆ ˆ4 2 4a i j k= + + or ˆˆ ˆ2 2a i j k= + −
2 2 2 0a k + = − + =
SECTION – 2
21.(36) 3
/ 2 12
3 3 12lim
3 3
x x
x xx
−
− −→
+ −
−
2
2
273 12
3lim1 3
3 3
→
−
+ −
−
x
x
x
x x
2
/ 22
3 12 3 27lim
3 3
x x
xx→
− +
−
/ 22
(3 3)(3 9)lim
(3 3)
x x
xx→
− −
−
/ 2 2 2
/ 22
(3 3)((3 ) (3) )lim
3 3
x x
xx→
− −
−
/ 2 / 2
2
/ 22
(3 3)(3 3)(3 3)lim (3 3)(3 3) 36
(3 3)
x x x
xx→
− − += − + =
−
Vidyamandir Classes
VMC | JEE Main-2020 18 Solutions |7th January Morning
22.(30) Let 2 2 20 1 2(1 ......)(1 .......) .......x x x x a a x a x− + + + + = + + +
Put 1x =
0 1 2 21(2 1) ......+ = + + + + nn a a a a
Put 1x = −
0 1 2 3(2 1)(1) .......n a a a a+ = − + −
0 2 24 2 2( ...... )nn a a a+ = + + +
0 2 2...... 2 1na a a n+ + + = +
2 1 61n+ = 30n =
23.(18) (1, 2,...... ) 10Var n =
22 2 2 21 2 3 ...... 1 2 ......10
n n
n n
+ + + + + + + − =
2( 1)(2 1) ( 1)
106 2
n n n n n
n n
+ + + − =
2 1 120n − = 11n =
Now, (2, 4, 6......2 ) 16Var m =
(1, 2, 3...... ) 4Var m =
18m n+ = 2 1 48m − = 7m =
24.(5)
Since ,APC APB and BPC have equal area, P is a centroid of .ABC
1 2 3 1 2 3 17 8, ,
3 3 6 3
x x x y y yP
+ + + + =
Length of line segment
217 7 8 1
5units6 6 3 3
PQ
= − − − − =
25.(3) ( ) 2 | 3|f x x= − −
Clearly, ( )f x is non differentiable at points 1, 3x x= = and 5x =
( ( )) 2 || 2 | 3|| 3|f f x = − − − −
( ( )) ( ( 1)) ( (3)) ( (5)) 1 1 1 3
x S
f f x f f f f f f
= − + + = + + =