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Memory Based Questions & Solutions
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_______________________________________________________________________________________________________________ Resonance Eduventures Ltd.
Registered & Corporate Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005
Tel.No.: 0744-6607777, 3012100, 3012222, 6635555 | Toll Free: 1800 258 5555 | Fax: +91-022-39167222 | 08003 444 888
Website: www.resonance.ac.in | E-mail: [email protected] | CIN: U80302RJ2007PLC024029
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph.No. : +91-744-3012222, 6635555 | To Know more : sms RESO at 56677
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PAGE # 1
PART : MATHEMATICS
Straight Objective Type
This section contains 30 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
1. There are 5 girls and 7 boys. A team of 3 boys and 2 girls is to be formed such that no two specific boys
are in the same team. Number of way to do so
(1) 400 (2) 250 (3) 200 (4) 300
Ans. (4)
Sol. Without any restriction, total number of ways of forming team is 7C3 × 5C2 = 350 If two specific boys B1,
B2 are in same team then total number of ways of forming team equals to 5C1 × 5C2 = 50 ways
total ways = 350 – 50 = 300 ways
2. The equation x2 + 2x + 2 = 0 has roots and . Then value of 15 + 15 is
(1) 512 (2) 256 (3) –512 (4) –256
Ans. (4)
Sol. roots are –1 + i and –1 – i
Let = –1 + i and b = –1 – i
then 15 + 15 = (–1 + i)15 + (–1 – i)15 = 16 16(–1 i) ( 1 i)
3 1 i 1 i
=
168
2 2
( 1 i 1 i)2
1 i
= 128(–2) = –256
3. 3
0| cos x |
dx is equal to
(1) 4
3 (2)
2
3 (3) 0 (4)
8
3
Ans. (1)
Sol. / 2
3 3
0 0
| cos x | dx 2 | cos x | dx
= / 2
3
0
12 cos xdx
2
2
.13
= 4
3 (by walli's formula)
4. If x2 n + 1, n N then 2 2
2 2
2sin(x 1) sin2(x 1)x
2sin(x 1) sin2(x 1)
dx is equal to
(1) 2x 1
lncos2
+ c (2) 21 x 1
lncos2 2
+ c
(3) ln2x 1
sec2
+ c (4) 21 x 1
lnsec2 2
+ c
Ans. (4)
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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PAGE # 2
Sol. 2
2
(1 cos(x 1))x dx
(1 cos(x 1))
=
2
2
x 1sin
2x dx
x 1cos
2
= 2x 1
x.tan dx2
2x 1
t2
2xdx = 2dt
= dt
tan t2
= 1
2nsect + c =
1
2nsec
2x 1c
2
5. If A = i – j , B = i + j + k are two vectors and C is another vector such that A × C + B = 0 and A . C = 0
then | C |2 =
(1) 9 (2) 8 (3) 19
2 (4)
17
2
Ans. (3)
Sol. a × ( a × c ) + a × b = 0
( a × c ) a – ( a . a ) c + a × b = 0
4 a – 2 c + a × b = 0
a × b =
ˆ ˆ ˆi j k
1 1 0
1 1 1
= i (–1) – j (1) + k(2) = – i – j + 2 k
2 c = 4( ˆ ˆi j ) + (– i – j +2 k )
2 c = 3 i – 5 j + 2 k
2|c| = a 25 4
2|c| = 38
4|c|2 = 38 |c| = 19/2
6. f(x) =
5 ; x 1
a bx ; 1 x 3
b 5x ; 3 x 5
30 ; x 5
then
(1) f(x)is discontinuous a R, b R
(2) f(x) is continuous if a = 0 & b = 5
(3) f(x) is continuous if a = 5 & b = 0
(4) f(x) is continuous if a = – 5 & b = 10
Ans. (1)
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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PAGE # 3
Sol. f(x) =
5 ; x 1
a bx ; 1 x 3
b 5x ; 3 x 5
30 ; x 5
for function to be continuous at x = 1
f(1) = h 0lim
f(1 + h) = h 0lim
f(1 – h)
f(1) = 5
h 0lim
f(1 + h) = h 0lim
a + b(1 + h) = a + b
h 0lim
f(1 – h) = h 0lim
5 = 5
hence a + b = 5 ..…(1)
for function to be continuous at x = 3
f(3) = h 0lim
f(3 + h) = h 0lim
f(3 – h)
f(3) = b + 15
h 0lim
f(3 + h) = h 0lim
b + 5(3 + h) = b + 15
h 0lim
f(3 – h) = h 0lim
a + b(3 – h) = a + 3b
hence b + 15 = a + 3b a + 2b = 15 ………(2)
for function to be continuous at x = 5
f(5) = h 0lim
f(5 + h) = h 0lim
f(5 – h)
f(5) = 30
h 0lim
f(5 + h) = 30
h 0lim
f(5 – h) = h 0lim
b + 5(5 – h) = b + 25
hence b + 25 = 30 b = 5 ………..(3)
from equation (2) a = 5
from equation (1) a = 0
hence f(x) is discontinuous for a, bR s
7. Average height & variance of 5 students in a class is 150 and 18 respectively. A new student whose
height is 156 cm is added to the group. Find new variance.
(1) 20 (2) 22 (3) 16 (4) 14
Ans. (1)
Sol. Let 5 students are x1, x2, x3, x4, x5
Given x = ix150
5
5
i
i 1
x 750
…..(1)
2
2ix(x) 18
5
22ix
(150) 185
2
ix = (22500 + 18)× 5 5
2
i
i 1
x
= 112590 …… (2)
Height of new student = 156 (Let x6)
now x1 + x2 + x3 + x4 + x5 + x6 = 750 + 156
newx = 1 2 3 4 5 6x x x x x x 906
6 6
= 151 ……..(3)
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
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PAGE # 4
Variance (new) = 2
2ix(x)
6
= 2 2 2 2 2 2
21 2 3 4 5 6x x x x x x(151)
6
from equation (2) and (3)
var (new) = 2
2112590 (156)(151)
6
= 22 821 – 22801 = 20
8. a, b, c are in G.P. a + b + c = x b , x can not be
(1) 2 (2) –2 (3) 3 (4) 4
Ans. (1)
Sol. a + ar + ar2 = xar
since a 0 so 2r r 1
xr
; 1 +
1r x
r
r + 1
r (–, –2] [2, ) x (–, –1] [3, )
9. 4032
15
= k
15then find k.
(1) 2 (2) 8 (3) 1 (4) 4
Ans. (2)
Sol. 2032
15
8.2200 8.1650 = 8(1 + 15)50 = 8(1 + 15)
hence remainder is 8.
10. y 0lim
4
4
1 1 y 2
y
=
(1) 1
4 2 (2)
1
2 2 (3)
1
2 2(1 2) (4) does not exist
Ans. (1)
Sol. Using rationalization
y 0lim
4 4
44
1 1 y 2 1 1 y 2
y 1 1 y 2
= 4 4
4 4y 0 4
1 y 1 1 y 11lim
y 1 y 11 1 y 2
= y 0lim 44
1 1
1 y 11 1 y 2
by putting value of limit
= 1 1 1
2 2 2 4 2
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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PAGE # 5
11. There is a parabola having axis as x -axis, vertex is at a distance of 2 unit from origin & focus is at
(4, 0). Which of the following point does not lie on the parabola.
(1) (6, 8) (2) (5, 2 6 ) (3) (8, 4 3 ) (4) (4, –4)
Ans. (1)
Sol. As per the given data diagram of parabola must be as below
(2, 0)
(4, 0)
then equation of parabola will be
(y – 0)2 = 4.2(x – 2)
y2 = 8(x – 2)
now check all options (C)
12. Find sum of all possible values of in the interval ,2
for which 3 2isin
1 2isin
is purely imaginary
(1) 3
(2) (3)
2
3
(4)
2
Ans. (3)
Sol. Let z = 3 2isin
1 2isin
,
2
for z to be purely imaginary
z + z = 0
3 2isin 3 2isin
01 2isin 1 2isin
2 2
2
3 6isin 2isin 4sin 3 6isin 2isin 4sin
1 4sin
3 – 4sin2 = 0 sin2 = 3
4
= n ± 3
Given ,2
hence = 2
, ,3 3 3
sum of all possible value of = 2
3
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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PAGE # 6
13. Let A = cos sin
sin cos
Find the value of A–50 at = 12
(1)
3 1
2 2
1 3
2 2
(2)
3 1
2 2
1 3
2 2
(3)
3 1
2 2
1 3
2 2
(4)
1 3
2 2
3 1
2 2
Ans. (2)
Sol. A = cos sin
sin cos
A–50 = cos( 50 ) sin( 50 )
sin( 50 ) cos( 50 )
A–50 at = 12
is
25 25cos sin
6 6
25 25sin cos
6 6
=
cos sin6 6
sin cos6 6
=
3 1
2 2
1 3
2 2
14. If (A B) ( A B) = A B what should be proper symbol in place of and to hold the equation
(1) and (2) and (3) and (4) and
Ans. (1)
Sol. Check all option repeatedly
(i) (A B) (~A B) A(B (~ A B))
A (B) A B
(i) is correct
(ii) (A B) (~A B) (A~A) B
fB f
(iii) (A B) (~A B) B
(iv) (A B) (~A B)
B (A ~A) = B f f
only (1) is correct
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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PAGE # 7
15. If y(x) is solution of x dy
dx + 2y = x2 , y(1) = 1 then value of y
1
2
=
(1) –49
16 (2)
49
16 (3)
45
8 (4)
45
8
Ans. (2)
Sol. xdy
dx+2y = x2
dy
dx+
2
xy = x
This is linear differential equation
I.F. = 2
dxxe = 2 nxe = x2
Solution of differential equation is
y . x2 = x . x2 dx + C
yx2 = 4x
4+ C
y = 4
2
x C
4 x
given y (1) = 1
1 = 1
4+ C C =
3
4
Hence y(x) = 2x
4+
2
3
4x
y1
2
= 1
16+
3
14
4
y1
2
= 49
16
16. From a well shuffled deck of cards, 2 cards are drawn with replacement. If x represent numbers of
times ace coming, then value of P(x = 1) + P (x = 2) is
(1) 25
169 (2)
24
169 (3)
49
169 (4)
23
169
Ans. (1)
Sol. P(x = 1) = 4
52 ×
48
52× 2 =
24
169
P(x = 2) = 4
52 ×
4
52=
1
169
P(x = 1) + P (x = 2) = 25
169
17. If eccentricity of the hyperbola 2
2
x
cos –
2
2
y
sin = 1 is more than 2 when 0,
2
Find the possible
values of length of latus rectum
(1) (3, ) (2) (1, 3/2) (3) (2, 3) (4) (–3, –2)
Ans. (1)
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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PAGE # 8
Sol. 2
2
x
cos –
2
2
y
sin = 1
e > 2 (given)
e2 > 4 1 + 2
2
sin
cos
> 4
1 + tan2 > 4 tan2 > 3
0,2
hence ,3 2
Latus rectum = 22sin
cos
= 2 tan sin
for ,3 2
, 2 tan sin is
An increasing function
Hence latus rectum (3, ) Ans(1)
18. If slant height of a right circular cone is 3 cm then the maximum volume of cone is
(1) 2 3 cm3 (2) 4 3 cm3 (3) (2 + 3 ) cm3 (4) (2 – 3 ) cm3
Ans. (1)
Sol. V = 1
3 r2 h , r2 + h2 = 9
V = 1
3 h (9 – h2)
dv
dh=
1
3 (9 – 3h2) = 0
9 – 3h2 = 0
h2 = 3 , h = 3
V = 1
3() (6) 3 = 2 3
19. If cos–1 2
3x
+ cos–1 3
4x
= 2
, x >
3
4 then x =
(1) 145
11 (2)
145
12 (3)
146
10 (4)
146
11
Ans. (2)
Sol. cos–1 12 3cos
3x 4x 2
3x
4
cos–1 2 2
2 3 4 91 1
3x 4x 29x 16x
2
1
2x =
2 2
2
9x 4 16x 9
12x
6 = 2 29x 4 16x 9
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph.No. : +91-744-3012222, 6635555 | To Know more : sms RESO at 56677
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PAGE # 9
square both side
36 = 144x4 – 81x2 – 64x2 + 36
144x4 = 145x2
x4 = 2145x
144 x = ±
145, 0
12
x > 3
4 hence x =
145
12
20. If px + qx + r = 0 represent a family of straight lines such that 3p + 2q + 4r = 0 then
(1) All lines are parallel (2) All lines are inconsistence
(3) All lines are concurrent at 3 1
,4 2
(4) All lines are concurrent at 3, 2
Ans. (3)
Sol. Given relation is 3p + 2q + 4r = 0
3
4p +
q
2+ r = 0 ...........(ii)
comparing (1) and (2) we get x = 3
4, y =
1
2
hence we can say that these lines are concurrent at 3 1
,4 2
21. Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x + 3y + (a2 – 1) z = a + 1 then
(1) system has a unique solution for |a| = 3
(2) system is inconsistence for |a| = 3
(3) system is inconsistence for a = 4
(4) system is inconsistence for a = 3
Ans. (2)
Sol. D = 2
1 1 1
2 3 2
2 3 a 1
= 3( a2 –1) – 6 – 2(a2 – 1) + 4
= a2 – 1–2 = a2 – 3
If |a| 3 system has unique solution
if |a| = 3
x y z 1
2x 3y 2z 1
2x 3y 2z 3 1
Hence system is inconsistent for |a| = 3
22. The value of 3(cos – sin)4 + 6(sin + cos)2 + 4sin6 is where ,4 2
(1) 13 – 4 cos4 (2) 13 – 4 cos6
(3) 13 – 4 cos6 + 2sin4 cos2 (4) 13 – 4 cos4 + 2sin4 cos2
Ans. (2)
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
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PAGE # 10
Sol. 3(sin – cos)4 + 6(sin + cos)2 + 4sin6
= 3(1 – sin2)2 + 6(1 + sin2) + 4(1 – cos2)3
= 3(1 – 2sin2 + sin22) + 6(1 + sin2) + 4(1 – 3cos2 + 3cos4 – cos6)
= 13 + 3sin22 + 12cos2(cos2 – 1) – 4cos6
= 13 + 12sin2 cos2 – 12sin2 cos2 – 4cos6
= 13 – 4cos6
23. 3 circles of radii a, b, c (a < b < c) touch each other externally and have x-axis as a common tangent
then
(1) a, b, c are in A.P. (2) 1
b=
1
a+
1
c
(3) a , b , c are in A.P. (4) 1
c +
1
b =
1
a
Ans. (4)
Sol.
a c b
2 2(a b) – (a –b) + 2 2(a c) – (a – c) = 2 2(b c) – (b – c)
ab + ac = bc
1
c +
1
b =
1
a
24. If f(x) = 1
x, f2 (x) = 1 – x, f3 (x) =
1
1 xthen find J(x) such that f2 o J o f1 (x) = f3 (x)
(1) f1 (x) (2) 1
xf3 (x) (3) f3 (x) (4) f2(x)
Ans. (3)
Sol. f2 (Jo(f1(x))) = f3(x)
1 – J (f1(x)) = f3 (x)
1 – J 1
x
= 1
1 x
J 1
x
= 1 – 1
1 x= –
x
1 x
J(x) =
1
x1
1x
= 1
1 x J(x) = f3 (x)
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
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Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph.No. : +91-744-3012222, 6635555 | To Know more : sms RESO at 56677
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PAGE # 11
25. Find the equation of line through (–4, 1, 3) & parallel to the plane x + y + z = 3 while the line intersects
another line whose equation is x + y – z = x + 2y – 3z + 5
(1) x 4 y 1 z 3
3 2 1
(2)
x 4
1
=
y 1
2
=
z 3
3
(3) x 4 y 1 z 3
3 2 1
(4)
x 4
1
=
y 1
2
=
z 3
3
Ans. (3)
Sol. Family of plan containing the line of intersection of planes of x + y – z = 0 = x + 2y – 3z + 5 is
(x + y – z) + (x + 2y – 3z + 5) = 0
If this plane passes through (–4, 1, 3), we get (–4 + 1 – 3) + (–4 + 2 – 9 + 5) = 0 –6 + (–6)
= –1
Hence equation of this plane is (x + y – z) – (x + 2y – 3z + 5) = 0
now required line lies in this plane and is parallel to x + y + z = 5
Hence vector, v (say), along the line is v =
ˆ ˆ ˆi j k
ˆ ˆ ˆ1 1 1 3i 2 j k
0 1 2
so required line is x 4 y 1 z 3
3 2 1
26. Consider the curves y = x2 + 2 and y = 10– x2. Let be the angle between both the curves at point of
intersection, then find |tan|
(1) 8
15 (2)
5
17 (3)
3
17 (4)
8
17
Ans. (1)
Sol. y = x2 + 2 & y = 10 – x2 meet at (±2, 6)
dy
2xdx
for first curve and dy
2xdx
for second.
Hence slope are 4 and –4
so |tan| = 4 ( 4) 8
1 4( 4) 15
27. A plane parallel to y-axis passing through line of intersection of planes x + y + z = 1 & 2x + 3y – z – 4 =
0 which of the point lie on the plane
(1) (3, 2, 1) (2) (–3, 0, 1) (3) (–3, 1, 1) (4) (3, –1, 1)
Ans. (4)
Sol. Equation of required plane is
(x + y + z – 1) + (2x + 3y – z – 4) = 0 (1 + 2)x + (1 + 3)y +(1 – ) = 0
since given plane is parallel to y-axis 3 + 1 = 0 = 1
3
Hence equation of plane is x 4z 1
03 3 3 x + 4z + 1 = 0
| JEE MAIN-2019 | DATE : 09-01-2019 (SHIFT-1) | PAPER-1 | MEMORY BASED | MATHEMATICS
RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph.No. : +91-744-3012222, 6635555 | To Know more : sms RESO at 56677
Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
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PAGE # 12
28. Find common tangent of the two curve y2 = 4x and x2 + y2 – 6x = 0
(1) y = x
33 (2) y =
x3
3
(3) y =
x3
3 (4) y =
x3
3
Ans. (4)
Sol. Tangent to y2 = 4x is y = mx + 1
m
If it is tangent to given circle it's distance from (3, 0) is equal to 3
Hence 2
13m
m
1 m
= 3
|3m2 + 1| = 3m 21 m
squaring both sides, we get
9m4 + 6m2 + 1 = 9m2 + 9m4 m = 1
3
Hence common tangents are y = x
3+ 3 or y = –
x3
3
29. If the area bounded by the curve y = x2 – 1 tangent to it at (2, 3) and y-axis is
(1) 2
3 (2)
4
3 (3)
8
3 (4) 1
Ans. (3)
Sol.
3
(2, 3)
–1
–5
area =3 3
5 1
xdy xdy
= 3 3
5 1
y 5y 1 dy
4
=
32
3
3 / 2
1
5
y5y
22 (y 1)4 3
=
9 2515 25
2 2
4
= |16
3 | =
8
3
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