JEE Main - 2018 Paper - 2 (B.Arch.) Code R Code

33 Pages·2017·2.31 MB·English

Hkkx
Part-I / -I
xf.kr
Mathematics /
1. If f(x) + 2f(1(cid:150)x) = x2 + 1, xR, then the range of f is :
 1  1   1 1 
, (2)  , (3)  , (4) 
(1)  3  3   3 3 1,

;fn f(x) + 2f(1(cid:150)x) = x2 + 1, lHkh xR, ds fy, rks f dk ifjlj gS: 3 
 1  1   1 1 
, (2)  , (3)  , (4) 
(1)  3  3   3 3 1,

3 
Ans. (2)
Sol. f(x) + 2f(1 (cid:150) x) = x2 (cid:150) 1 (cid:133)(i)
x  1 (cid:150) x;
f(1 (cid:150) x) + 2f(x) = (1 (cid:150) x)2 + 1 (cid:133)(ii)
Solving (i) & (ii)
f(x) + 2f(1 (cid:150) x) = x2 + 1
2f(x) + f(1 (cid:150) x) = (1 (cid:150) x)2 + 1) (cid:215) 2
(cid:150) (cid:150) (cid:150)
(cid:150)3f(x) = (x2 + 1) (cid:150) 2[(1 (cid:150) x)2 + 1]
= (x2 + 1) (cid:150) 2(1 + x2 (cid:150) 2x + 1)
(cid:150)3f(x) = (cid:150) x2 + 4x (cid:150) 3
f(x) = (cid:150)x2 + 4x (cid:150) 3
f(x)= 1/3(x2 (cid:150) 4x + 3)  f(x) [(cid:150) 1/3, )

2. Let A {zC: |z| = 25) and B = {z C : |z + 5 + 12i| = 4}. Then the minimum value of |z (cid:150)for zA
and B, is :
(1) 6 (2) 7 (3) 8 (4) 9
ekuk rFkk gS] rks rFkk ds fy, dk
A {zC: |z| = 25) B = {z C : |z + 5 + 12i| = 4}. zA B |z (cid:150)
U;qure eku gS%

(1) 6 (2) 7 (3) 8 (4) 9
Ans. (3)

Sol.

((cid:150)5 (cid:150)12i)
min.
 |Z (cid:150) w| = 8
min.
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3. If the product of the roots of the equation x2 (cid:150) 5kx + 2e2loge|k| (cid:150)1 = 0 is 49, then the sum of the
squares of the roots of the equation is :
(1) 525 (2) 527 (3) 576 (4) 627
;fn lehdj.k x2 (cid:150) 5kx + 2e2loge|k| (cid:150)1 = 0 ds ewyksa dk xq.kuQy 49 gS] rks lehdj.k ds ewyksa ds oxksZ dk
;ksx gS%

(1) 525 (2) 527 (3) 576 (4) 627
Ans. (2)
Sol. x2 (cid:150) 5kx + 2e2logek (cid:150)1 = 0
x2 (cid:150) 5kx + (2k2 (cid:150) 1) = 0  (, )
 +  = 5k
 = 2k2 (cid:150)1 = 49
 k = –5
 2 + 2 = ( + )2 (cid:150) 2
= (25)2 (cid:150) 98
= 527

2 52 152
 
4. If A = 4 106 358 , then the determinant of the matrix adj (2A) is equal to :
 
6 162 620
 
2 52 152
;fn   gS] rks vkO;wg dk lkjf.kd cjkcj gS%
A = 4 106 358 , adj (2A) (Determinant)
 
6 162 620
 
(1) 64 (2) 256 (3) 2048 (4) 4096
Ans. (4)
Sol. adj(2A)  2An1 = 2n(n-1). An1
A 8
 adj(2A) 26.82 = 212 = 4096
5. Let S be the set of all real values of  for which the system of linear equations
x + y + z = 5
2x + 2y (cid:150) z = 1
3y + z = 9
has infinitely many solutions. Then S;
(1) equals R (2) is a singleton.
(3) contains exactly two elements (4) is an empty set

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;fn ds lHkh okLrfod ekuksa dk leqPp; gS ftuds fy, jSf[kd lehdj.k fudk;
 S
x + y + z = 5
2x + 2y (cid:150) z = 1
3y + z = 9
ds vuar gy gSa] rks
S :
ds cjkcj gSA ,d leqPp; gSA
(1) R (2) (singleton)
esa ek=k nks vo;o gSA ,d fjDr leqPp; gSA
(3) (4)
Ans. (4)
Sol. x + y + z = 5
 2x + 2y (cid:150) z = 1
3y + z = 9
These is no such value of '' for which  = x =y =z = 0.
Hence, S is an empty set.

6. In order to get through in an examination of nine papers a candidate has to pass in more papers
than the number of papers in which he fails. The number of ways in which he can fail, in this
examination is :
iz’u i=kksa dh ,d ijh{kk ikl djus ds fy,] ,d ijh{kkFkhZ dks mu iz’uksa i=kksa dh la[;k ftlesa og ikl ugha
9
gS] mlls vf/kd iz’u i=kksa esa ikl gksuk vko’;d gSA bl ijh{kk esa vuq(cid:217)kh.kZ gksus ds rjhdksa dh la[;k gS%
(1) 128 (2) 255 (3) 256 (4) 9 (8)!
Ans. (3)
Sol. Number of ways in which he can fail
= 9C + 9C + 9C + 9C + 9C
5 6 7 8 9
= 126 + 84 + 36 + 9 + 1
= 256
125
7. Let T denote the rth term in the binomial expansion of (a+1)50. If T + T = T then the sum
r 25 27 26
52
of all the values of a is :
(a+1)50 ds f}in izlkj esa ekuk r oka in T gSA ;fn T + T = 125 T gS] rks a ds lHkh ekuksa dk ;ksx gS%
r 25 27 26
52
1 3 5
(1) (2) (3) 2 (4)
2 2 2
Ans. (4)
Sol. T = 50C .a51(cid:150)r
r r-1
125
T +T = .T
25 27   26
 52 
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 50C24.a26 + 50C26.a24 = 12550C25.a25
 52 
 a2 + 1 = 152255500CC2254.a
5a
 a2 + 1 =
2
5
 2a2 (cid:150) 5a +2 = 0  (a , a )  a + a =
1 2 1 2
2
8. In an ordered set of four numbers, the first 3 are in A.P and the last 3 are in G.P. whose common
ratio is 7/4. If the product of the first and fourth of these number is 49, then the product of the
second and third of these is :
pkj la[;kvksa ds ,d dfFkr leqPp; esa izFke lekarj Js.kh esa gS rFkk vafre xq.kks(cid:217)kj Js.kh esa gS ftldk
3 3
lkoZvuqikr gSA ;fn blesa ls izFke rFkk pkSFkh la[;kvksa dk xq.kuQy gS] rks nwljh rFkk rhljh la[;kvksa
7/4 49
dk xq.kuQy gS%
(1) 60 (2) 112 (3) 128 (4) 144
Ans. (2)
16 28 49
Sol. Let a, , ,  four numbers
a a a

A.P.
32 28
 a  a2 + 28 = 32  a2 = 4
a a
 product of 2nd and 3rd = 1628 = 1628 = 112
 a  a  4

9. If e(sin2xsin4xsin6x.....adinf.)loge2 0 x  satisfies the equation y2(cid:150)5y+4 = 0, then
 2
sinx
is equal to :
cosxsinx
;fne(sin2xsin4xsin6x.....adinf.)loge2 0 x lehdj.k y2(cid:150)5y+4 = 0, dks larq"V djrk gS] rks
 2
sinx cjkcj gS%
cosxsinx
   
(1)  2 2 (2)  21 (3) 2 (cid:150)1 (4) 2 + 2
Ans. (1)

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Sol. esin2xsin4xsin6x......loge2
 sin2x 
= e1(cid:150)sin2x.log 2 = etan2x.loge2 = 2tan2x
e
y2 (cid:150) 5y + 4 = 0
(y (cid:150) 1) (y (cid:150) 4) = 0
y = 1, 4
 2tan2x = 20, 22
tan2x = 0, 2
tan x = 2 [ x  (0, /2)]
sinx tanx 2
 
   (cid:150) 2 2
cosx(cid:150)sinx 1(cid:150)tanx 1(cid:150) 2

1
10. Let f(x) = x for all x()R where for each tR, [t] denotes the greatest integer less than or
 
x
equal to t. Then:
ekuk lHkh ds fy, 1 tgk¡ izR;sd ds fy, lcls cM+k iw.kkZd n’kkZrk gS tks ds
x()R f(x) = x tR [t] t
 
x
cjkcj vFkok ls NksVk gS%
t
(1) lim f(x)0 (2) lim f(x)1 (3) lim f(x)1 (4) lim f(x)1
x0 x1 x1 x2
3 2
Ans. (2)
1 1 1 
Sol. f(x) = x x = xx x = 1xx
1
= lim f(x)1, lim f(x)1, lim f(x) , lim f(x)0
x0 x1/3 x1 2 x2
2
72x 9x 8x 1

11. If f(x) =  2 1cosx is a continuous function in the interval [0, 2), then k is equal to :

k 2loge2loge3,x 0
72x 9x 8x 1
;fn  varjky esa larr Qyu gS] rks cjkcj gS%
f(x) =  2 1cosx [0, 2), k

k 2loge2loge3,x 0
(1) 4 (2) 18 (3) 24 (4) 36
Ans. (3)
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 9x 18x 1
 ;x 0
Sol. f(x) = 2 1cosx

k 2.loge2.loge3; x 0
9x 18x 1
9x 18x 1  x  x   
=lim = lim . 2 1cosx
x0 2 1cosx x0 1cosx
 x2 
  k 2 n2. n3= 2(n9) (n8)  2 2 k = 24

12. If y = y(x) is an implicit function of x given by ycosx + xcosy =  ; then y"(0) is equal to :
;fn esa ,d vLi"V Qy gS tks }kjk izn(cid:217)k GS] rks cjkcj gS%
y = y(x) x (implicit) ycosx + xcosy =  y"(0)
(1)  (2) (cid:150) (3) 0 (4) 2

Ans. (1)
Sol. y.cosx + x.cosy =  ; y (0) = 
 y'. cosx + y ((cid:150)sinx) + x.((cid:150)siny).y1 + cosy = 0
put, x = 0 y' + cosy = 0
y' = 1
(cosx (cid:150) x.siny)y' = sinx.y (cid:150) cosy
 (cosx (cid:150) x.siny)y'' + ((cid:150)sinx (cid:150) x.cosy.y' (cid:150) siny) y' = cosx.y + sinx.y' + siny.y'
put, x = 0;
y''(0) = 

13. For each x  R , let f(x) = |x (cid:150) 1|, |x (cid:150) 1|, g(x) = cosx and (x) = f(g(2sinx) (cid:150) g(f(x)). Then  is :
(1) differnentaible at each point of R (2) not differentiable at 0
  
(3) not differentiable at 1 (4) differentiable only in (cid:150) , 
 2 2
lHkh ds fy, ekuk rFkk gS] rks
x  R f(x) = |x (cid:150) 1|, |x (cid:150) 1|, g(x) = cosx (x) = f(g(2sinx) (cid:150) g(f(x)) 
ds izR;sd fcUnq ij vodyuh; gSA ij vodyuh; ugha gSA
(1) R (2) 0
ij vodyuh; ugha gSA dsoy    esa vodyuh; gSA
(3) 1 (4) (cid:150) , 
 2 2
Ans. (1)
Sol. f(x)=|(cid:215)(cid:150)1|
g(x) = cosx
(x) = f (g(2sinx)) (cid:150)g (f(x)) = cos(sinx)(cid:150)1(cid:150)cos|x(cid:150)1|
1(cid:150)cos(2sinx)(cid:150)cos |x(cid:150)1|
differentiable at each point of R.
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14. If f(x) = |x2 (cid:150) 16| for all xR, then the total number of points of R at which f : RR attains local
extreme values is :
;fn lHkh x  R ds fy, f(x) = |x2 (cid:150) 16| gS] rks R ds mu fcanqvksa dh la[;k tgk¡ f : R  R LFkkuh; ije eku
ysrk gS] gS

(1) 1 (2) 2 (3) 3 (4) 4
Ans. (1)
Sol. f(x) = |x2 (cid:150) 16|

extreme value
(cid:150)4 4

15. Let
ex e(cid:150)x
I =  dx, dx, then J (cid:150) i equals :
e4x e2x 1 e(cid:150)4x e(cid:150)2x 1
ekuk

ex e(cid:150)x gS] rks cjkcj gS %
I =  dx, dx J (cid:150) i
e4x e2x 1 e(cid:150)4x e(cid:150)2x 1

1 e4x (cid:150)e2x 1 1 e2x ex 1
(1) log + C (2) log + C
2 e e4x e2x 1 2 e e2x (cid:150)ex 1
1 e2x (cid:150)ex 1 1 e4x (cid:150)e2x 1
(3) log + C (4) log + C
2 e e2x ex 1 2 e e4x (cid:150)e2x 1
Ans. (3)
ex
Sol.   
e4x e2x 1
e(cid:150)x
  dx
e(cid:150)4x e(cid:150)2x 1
e3x
  dx
e4x ex 1
e3x (cid:150)ex
J(cid:150)   dx
e4x e2x 1
ex(e2x (cid:150)1)
J(cid:150)   dx
e4x e2x 1
ex = t
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t2 (cid:150)1
J(cid:150)   dt
t4 t2 1
1
1(cid:150)
t2
J(cid:150)   dt
1
t2 1
t2
 1 
1(cid:150) dt
 t2 
J(cid:150)  
2
 1
t  (cid:150)1
 t
1
t z
t
 1 
1(cid:150) dt dz
 t2
dz
J(cid:150)  
z2 (cid:150)1
1 z(cid:150)1
 In c
2 z1
1
t (cid:150)1
1 t
 In c
2 1
t 1
t
1 t2 (cid:150)t1
 In c
2 t2 t1
1 e2x (cid:150)ex 1
 In c
2 e2x ex 1

1x2
16. If x5 dx = m + n, then the ordered pair (m,n) is equal to
1(cid:150)x2
;fn x5 1x2 dx = m + n, gS] rks (cid:216)fer ;qXe (m,n) cjkcj gS %
1(cid:150)x2

1 1 1 2 1 1 1 1
(1) , (2) , (3) , (4) ,
       
3 8 8 3 4 3 8 3
Ans. (4)
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Sol. Use formula
 
41cos4 4cos6(cid:150)3cos2
=  d +  d
2 4
0 0
 1
= + = m + n
8 3
1 1
(m, n) = , option (4)
 
8 3
17. The area (in sq. units ) of the region bounded by the curve , 12y = 36 (cid:150) x2 and the tangents drawn
to it at the points , where the curve intersects the x-axis is
o(cid:216) 12y = 36 (cid:150) x2 rFkk ml ij mu fcUnqvksa] tgk¡ o(cid:216) x-v{k d izfrPNsn djrh gS] ij [khaph xbZ Li’kZjs[kk ds
chp f?kjs dk {ks=kQy …oxZ bdkb;ksa esa‰ gS %

(1) 12 (2) 18 (3) 27 (4) 6
Ans. (1)
Sol x2 = 36 (cid:150) 12 y
x2 = (cid:150) 12 (y (cid:150) 3)
(0, 6)
(0, 3)
(-6, 0) (6, 0)

Required Area =
1 636x2   1  216
= 2 266 12 dx = 2 1812366 3 
0
 1 
= 2 18 21672= 2 (18 (cid:150) 12) = 12
 12 

18. Let y = y(x) be the solution of the differential equation :
dy
xlogex + y = 3x log x, (x > 1) If y(e) = 0, then y(e2) is equal to
e
dx
ekuk vody lehdj.k
y = y(x)
xlogexdy + y = 3x log x, (x > 1) dk gy gSA ;fn y(e) = 0 gS] rks y(e2) cjkcj gS %
e
dx
1 3
(1) e2 (2) e2 (3) e2 (4) 3e2
2 2
Ans. (3)
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dy
Sol. xlnx + y = 3xlnx
dx
dy 1
+ y = 3
dx xlnx
1
 dx
IF = e xlnx
= lnx
y.lnx = 3.lnx dx  6
y.lnx = 3.x(lnx(cid:150)1) + 6
y(e) = 0
0 = 3e(0) + 1
C = 0
ylnx = 3x(lnx (cid:150) 1)
y(e2) = ?
2y = 3e2(2 (cid:150) 1)
3
y = e2
2
19. Let the straight lines , 5x (cid:150) 3y + 15 = 0 and 5x + 3y (cid:150) 15 form a triangle with the x-axis. Then the
radius of the circle circumscribing this triangle is
ekuk js[kk,sa rFkk rFkk v{k ds lkFk ,d f=kHkqt cukrs gS] rks f=kHkqt ds
5x (cid:150) 3y + 15 = 0 5x + 3y (cid:150) 15 = 0 x-
ifjo‘(cid:217)k dh f=kT;k gS %

8 17 12 16
(1) (2) (3) (4)
5 5 5 5
Ans. (4)
Sol. 5x (cid:150) 3y + 15 = 0
5x + 3y (cid:150) 15 = 0

A (0, 5)
B ((cid:150)3, 0) C ((cid:150)3, 0)

Circumcentre (0, 9/5)
16
Radius =
5
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