SS - FTS - 88 (Online) P1 - (Adv) A - 2021-08-06 - 2020 - A
SS - FTS - 88 (Online) P1 - (Adv) A - 2021-08-06 - 2020 - A
SS - FTS - 88 (Online) P1 - (Adv) A - 2021-08-06 - 2020 - A
Phase-I
A
CODE
ANSWERS
7. Answer (07)
PART – I : PHYSICS
1. Answer (B, C)
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Final Test Series for JEE (Advanced)-2021_PHASE-I Test-88_Paper-1_(Code-A)_(Answers and Solutions)
x2 – xR = 2R2
P = (ln 2)v 2
2 2
R 3R
x – 2 = 2
Fv = (ln 2)v 2
dv x = 2R
m = ln 2 v
dt
120
VBD = = 60 V
2
2v 0 t
ln v v = ln 2
0 m After every section voltage gets halved
t=2m VBC = 30 V
t = 2 × 4 = 8 sec
12. Answer (05)
10. Answer (09)
Resistance of heater
V 2 100 × 100
(RH=
) = = 10 Ω. Current in the
P 1000
circuit is
100
I=
10 R
10 +
(10 + R )
R 100 5R
IH =
I× = =
(10 + R ) 10 R (5 + R )
10 + 10 + R
Therefore, Q= IH2 × 10
Which gives R = 5 Ω.
Hence the correct choice is (D)
144
=
t2 = 16 sec
9
So, they finish the race simultaneously together
Let RAB = x Hence, their separation is zero.
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Test-88_Paper-1_(Code-A)_(Answers and Solutions) Final Test Series for JEE (Advanced)-2021_PHASE-I
∴ NH
=+
4 3.6 × 10 –5 M
13 1
⇒ x= = = 0.2 M
65 5
26. Answer (25)
32. Answer (20)
W = – ∫ Pdv
v2
k P2V2 – PV R (T2 – T1 )
∫ vn dv = n –1
1 1
=
n –1
v1
a=n–1
27. Answer (08)
3 stereo centres
C C
28. Answer (07) λm =λm –b C
1 1000 π 3π
λm= × × Also sec2x > cosec2x ∀ x ∈ ,
a R M 2 4
1000 π 3π
g′(x) > 0 ⇒ g(x) is increasing in ,
λm = (Cell constant] × 2 4
1000 × 25 × 10 –4
3π
35. Answer (A) Similarly g(x) is decreasing , π
4
is anti to –OH therefore it will 3π
Also g(x) has local maximum at x = .
4
41. Answer (A, C)
Let g(t) = |t – 1| – |t| + |t + 1| =
migrate.
−t , t < −1
t + 2, −1 ≤ t < 0
36. Answer (C)
2 − t , 0 ≤ t <1
t , t ≥1
1
2, −2 ≤ x ≤ −
2
1
∴ f(x) = 2 + x, <x≤0
2
PART – III : MATHEMATICS
2, 0<x≤2
0 3
37. Answer (A, C, D) 31
We have
Calculate ∫
−2
f ( x ) dx =
8
and ∫ f ( x ) dx = 6 .
0
42. Answer (A, D)
log a log a 2log a log a
+ = ⋅
log ( b + c ) log ( c − b ) log ( c + b ) log ( c − b ) C and D are corresponding point of A and B on
∑a =
k =0
k
n +1
C1 + n +1C2 + ... + n +1Cn +1= 2n +1 − 1
g′(x) = f′′ (tanx) sec2x − f′ (cotx) cosec2x For maximum area of ∆AOB, area of ∆COD
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Final Test Series for JEE (Advanced)-2021_PHASE-I Test-88_Paper-1_(Code-A)_(Answers and Solutions)
1 1
= × ×4×4=4
2 2 46. Answer (99)
Let slope of line CD = m 100
∏(x − i )
i (101 − i )
F(x) =
equation of line CD is mx − y − m 264 = 0 i =1
OM = 2 2 100
m 264
ln(f(x)) = ∑ i (101 − i ) ln ( x − i )
2 2 i =1
=2 2 ⇒ m= −
1+ m 2 16 100
1 i (101 − i )
1
⇒
(
f x )
⋅ f ′( x ) =
i =1
∑
(x − i )
Slope of AB = m/2 = −
8 2 100
f ′ (101) 101 − i
43. Answer (89)
⇒k=
f (101)
= ∑ i 101 − i = 5050
i =1
sin r º sin ( r + 1) º
=Tr + 1− 1 k
cos r º cos (1 + r ) º ⇒ − 2 = 99.
50
cos1º
= −1 47. Answer (15)
cos r º cos(r + 1)º
y
(0,2)
cos1º sin {(r + 1)º −r °} y=x
= −1 (0,3/2)
sin1º cos r º cos(r + 1)º B
C
= cot1º {tan ( r + 1) º − tan r º} − 1 A
min(α1, α2, α3) = 1, and max(α1, α2, α3) = 2 A = area OAB = 2[area OCM + area CMNA – area
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Test-88_Paper-1_(Code-A)_(Answers and Solutions) Final Test Series for JEE (Advanced)-2021_PHASE-I
1 ⇒ λ = 4.
X2 +Y 2 − Y =
0
2
sin–1sin4 = sin–1sin(π – 4) = π – 4.
y
So all points lies on circle x2 + y2 – =0
2 52. Answer (C)
1 Equation of plane contains both lines
r =
4
x − 3 y − 2 z −1
1 2π
Area of regular octagon = 8 × r 2 × sin 2 3 4 =0
2 8
3 2 3
1 1 1 1
= 8× × × = . (x – 3)(1) + (y – 2)(12 – 6) + (z – 1)(4 – 9) = 0
2 16 2 4 2
49. Answer (01) x + 6y – 5z = 10.
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