MATHEMATICS

SECTION-A

1. The positive integer n, for which the solutions of the equation

x(x+2)+(x+2)(x+4)+…+(x+2n−2)(x+2n)=8n3x(x+2)+(x+2)(x+4)+…+(x+2n−2)(x+2n)=38n?

are two consecutive even integers, is :-
(1) 3
(2) 6
(3) 12
(4) 9

Ans. (1)

Sol. x(x+2)+(x+2)(x+4)+…+(x+2n−2)(x+2n)=8n3x(x+2)+(x+2)(x+4)+…+(x+2n−2)(x+2n)=38n?
⇒∑r=1n(x+2r−2)(x+2r)=8n3⇒∑r=1n?(x+2r−2)(x+2r)=38n?
nx2+2x∑r=1n(2r−1)+4∑r=1nr(r−1)=8n3nx2+2x∑r=1n?(2r−1)+4∑r=1n?r(r−1)=38n?
nx2+2xn2+4n(n2−1)3−8n3=0nx2+2xn2+34n(n2−1)?−38n?=0
x2+2nx+4(n2−1)3−83=0x2+2nx+34(n2−1)?−38?=0
??α−β?=2⇒D?a?=2⇒D=4??α−β?=2⇒?a?D??=2⇒D=4
⇒4n2−4(4(n2−1)3−83)=4⇒4n2−4(43(n2−1)?−38?)=4
⇒n2−4n23=−3⇒n2−34n2?=−3
⇒n2=9⇒n2=9
⇒n=3⇒n=3

JEE Previous Year Question Paper

2. Let f:R→Rf:R→R be a twice differentiable function such that f′(x)>0f′(x)>0 for all x∈Rx∈R and f′(a−1)=0f′(a−1)=0, where a is real number. Let g(x)=f(tan?2x−2tan?x+a)g(x)=f(tan2x−2tanx+a), 0<x<π20<x<2π?. Consider the following two statements :
(I) gg is increasing in (0,π4)(0,4π?)
(II) gg is decreasing in (π4,π2)(4π?,2π?)
Then,
(1) Neither (I) nor (II) is True
(2) Only (II) is True
(3) Only (I) is True
(4) Both (I) and (II) are True

Ans. (1)

Sol. g(x)=f((tan?x−1)2+a−1)g(x)=f((tanx−1)2+a−1)
g′(x)=f′((tan?x−1)2+a−1).2(tan?x−1)sec?2xg′(x)=f′((tanx−1)2+a−1).2(tanx−1)sec2x
?f′(a−1)=0?f′(a−1)=0 and f′(x)>0f′(x)>0
∴f′((tan?x−1)2+a−1)>0∴f′((tanx−1)2+a−1)>0
g′(x)>0g′(x)>0 if (tan?x−1)>0(tanx−1)>0
gg is increasing in x∈(π4,π2)x∈(4π?,2π?)
g′(x)<0g′(x)<0 if tan?x−1<0tanx−1<0
gg is decreasing in x∈(0,π4)x∈(0,4π?)

3. Let f(x)=x2+x2f′(1)+2xf′′(2)+f′′(3)f(x)=x2+x2f′(1)+2xf′′(2)+f′′(3), x∈Rx∈R. Then the value of f′(5)f′(5) is :
(1) 625562?
(2) 65755657?
(3) 2552?
(4) 11755117?

Ans. (4)

Sol. f′(x)=3x2+2xf′(1)+2f′(2)f′(x)=3x2+2xf′(1)+2f′(2)
f′′(x)=6x+2f′(1)f′′(x)=6x+2f′(1)
f′′(2)=12+2f′(1)f′′(2)=12+2f′(1)
∴f′(x)=3x2+2xf′(1)+2(12+2f′(1))∴f′(x)=3x2+2xf′(1)+2(12+2f′(1))
f′(x)=3x2+2xf′(1)+24f′(x)=3x2+2xf′(1)+24
Putting, x = 1
f′(1)=3+6f′(1)+24f′(1)=3+6f′(1)+24
−5f′(1)=27⇒f′(1)=−275−5f′(1)=27⇒f′(1)=5−27?
∴f′′(2)=12+2(−275)=12−545=65∴f′′(2)=12+2(5−27?)=12−554?=56?
∴f′(x)=3x2−545x+125∴f′(x)=3x2−554?x+512?
∴f′(5)=75−54+125=1175∴f′(5)=75−54+512?=5117?

JEE Previous Year Question Paper

4. In the line αx+4y=7αx+4y=7?, where α∈Rα∈R, touches the ellipse 3x2+4y2=13x2+4y2=1 at the point P in the first quadrant, then one of the focal distances of P is :
(1) 13−12113?1?−211?1?
(2) 13+1253?1?+25?1?
(3) 13−1253?1?−25?1?
(4) 13+1273?1?+27?1?

Ans. (4)

Sol. αx+4y−7=0αx+4y−7?=0 touches 3x2+4y2=13x2+4y2=1
∴c2=a2m2+b2∴c2=a2m2+b2
716=13×α216+14⇒α=3,−3167?=31?×16α2?+41?⇒α=3,−3
Tangent is 3x+4y−7=03x+4y−7?=0
Let the point of contact is P(x1,y1)P(x1?,y1?)
∴∴ Tangent is 3xx1+4yy1=13xx1?+4yy1?=1
∴3x13=4y14=17∴33x1??=44y1??=7?1?
∴P(17,17)∴P(7?1?,7?1?)
e=1−34=12e=1−43??=21?
PS = e(PM)
=e(ae−17)=e(ea?−7?1?)
=12(23−17)=13−127=21?(3?2?−7?1?)=3?1?−27?1?
PS' = e(PM') = 12(ae+17)=12(23+17)21?(ea?+7?1?)=21?(3?2?+7?1?)
=13+127=3?1?+27?1?

JEE Previous Year Question Paper

5. Let y2=12xy2=12x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that ∠OPA=90?∠OPA=90?. Then the locus of the centroid of such triangles OPA is :
(1) y2−6x+4=0y2−6x+4=0
(2) y2−9x+6=0y2−9x+6=0
(3) y2−2x+8=0y2−2x+8=0
(4) y2−4x+8=0y2−4x+8=0

Ans. (3)

Sol. P(3t2,6t)P(3t2,6t)
mAP=−t2mAP?=2−t?
Equation of AP is
y−6t=−t2(x−3t2)y−6t=2−t?(x−3t2)
Put y=0⇒x=12+3t2y=0⇒x=12+3t2
⇒A(12+3t2,0)⇒A(12+3t2,0)
Let centroid of ΔOPAΔOPA be G(h, k)
⇒3h=0+3t2+12+3t2⇒3h=0+3t2+12+3t2
3k=0+6t+03k=0+6t+0
⇒t=k2,h=2t2+4⇒t=2k?,h=2t2+4
⇒h=2k24+4⇒h=24k2?+4
⇒⇒ Locus of (h, k) is
y2=2x−8y2=2x−8

6. Let one end of a focal chord of the parabola y2=16xy2=16x be (16, 16). If P(α,β)P(α,β) divides this focal chord internally in the ratio 5:25:2, then the minimum value of α+βα+β is equal to:
(1) 22
(2) 7
(3) 5
(4) 16

Ans. (2)

Sol. y2=16xy2=16x
parameter of point A is t=2t=2
⇒⇒ Parameter of point B is t=−12t=−21?
⇒⇒ Coordinates of B is (1,−4)(1,−4)
Case 1:
α=5+327=377α=75+32?=737?
β=−20+327=127β=7−20+32?=712?
⇒α+β=7⇒α+β=7
Case 2:
α=2+807,β=−8+807α=72+80?,β=7−8+80?
α+β=22α+β=22
So minimum value of α+β=7α+β=7

JEE Previous Year Question Paper

7. Let the line L pass through the point (−3,5,2)(−3,5,2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2,r,1)(−2,r,1) is 143314??, then the sum of all possible values of r is :
(1) 12
(2) 16
(3) 6
(4) 10

Ans. (4)

Sol. Equation line is : x+31=y−51=z−21=λ1x+3?=1y−5?=1z−2?=λ
General point R on line is R(λ−3,λ+5,λ+2)R(λ−3,λ+5,λ+2)
P(−2,r,1)P(−2,r,1)
PR‾⋅d‾=0PR⋅d=0
⇒(λ−1)1+(λ+5−r)1+(λ+1)1=0⇒(λ−1)1+(λ+5−r)1+(λ+1)1=0
⇒3λ−r+5=0⇒3λ−r+5=0
⇒λ=r−53⇒λ=3r−5?
∴R=(r−143,r+103,r+13)∴R=(3r−14?,3r+10?,3r+1?)
PR=143⇒(PR)2=143PR=314??⇒(PR)2=314?
⇒(r−83)2+(10−2r3)2+(r−23)2=143⇒(3r−8?)2+(310−2r?)2+(3r−2?)2=314?
⇒6r2−60r+126=0⇒6r2−60r+126=0
⇒r2−10r+21=0⇒r2−10r+21=0
⇒r=3,7⇒r=3,7
sum of possible value of r is =10=10

8. Let the line L1L1? be parallel to the vector −3i^+2j^+4k^−3i^+2j^?+4k^ and pass through the point (2,6,7) and the line L2L2? be parallel to the vector 2i^+j^+3k^2i^+j^?+3k^ and pass through the point (4,3,5). If the line L3L3? is parallel to the vector −3i^+5j^+16k^−3i^+5j^?+16k^ and intersects the lines L1L1? and L2L2? at the points C and D, respectively, then ?CD→?2?CD?2 is equal to :
(1) 171
(2) 290
(3) 312
(4) 89

Ans. (2)

Sol. L1:x−2−3=y−62=z−74L1?:−3x−2?=2y−6?=4z−7?
Point C on L1:(−3λ1+2,2λ1+6,4λ1+7)L1?:(−3λ1?+2,2λ1?+6,4λ1?+7)
L2:x−42=y−31=z−53L2?:2x−4?=1y−3?=3z−5?
Point D on L2:(2λ2+4,λ2+3,3λ2+5)L2?:(2λ2?+4,λ2?+3,3λ2?+5)
L3:2λ2+3λ1+2−3=λ2−2λ1−35=3λ2−4λ1−216L3?:−32λ2?+3λ1?+2?=5λ2?−2λ1?−3?=163λ2?−4λ1?−2?
λ1=−3,λ2=2λ1?=−3,λ2?=2
C (11, 0, - 5)
D (8, 5, 11)
?CD→?2=32+52+162=290?CD?2=32+52+162=290

JEE Previous Year Question Paper

9. Let αα and ββ be the roots of equation x2+2ax+(3a+10)=0x2+2ax+(3a+10)=0 such that α<1<βα<1<β. Then the set of all possible values of a is :
(1) (−∞,−115)∪(5,∞)(−∞,5−11?)∪(5,∞)
(2) (−∞,−2)∪(5,∞)(−∞,−2)∪(5,∞)
(3) (−∞,−3)(−∞,−3)
(4) (−∞,−115)(−∞,5−11?)

Ans. (4)

Sol. ?α<1<β?α<1<β
f(1)<0f(1)<0
⇒1+2a+(3a+10)<0⇒1+2a+(3a+10)<0
⇒5a+11<0⇒5a+11<0
a<−115a<5−11?
∴a∈(−∞,−115)∴a∈(−∞,5−11?)

10. A random variable X takes values 0, 1, 2, 3 with probabilities 2a+130,8a−130,4a+130302a+1?,308a−1?,304a+1?, b respectively, where a, b ∈ R. Let μ and σ respectively be the mean and standard deviation of X such that σ2+μ2=2σ2+μ2=2. Then abba? is equal to :
(1) 30
(2) 3
(3) 60
(4) 12

Ans. (3)

Sol.

x	0	1	2	3
p(x)	2a+130302a+1?	8a−130308a−1?	4a+130304a+1?	b

σ2=∑x12p(x1)−μ2σ2=∑x12?p(x1?)−μ2
σ2+μ2=∑x12p(x1)σ2+μ2=∑x12?p(x1?)
=0+1(8a−130)+4(4a+130)+9b=0+1(308a−1?)+4(304a+1?)+9b
⇒24a+270b+330=2⇒3024a+270b+3?=2
24a+270b=5724a+270b=57
8a+90b=19…(1)8a+90b=19…(1)
Also ∑p(i)=1∑p(i)=1
2a+130+8a−130+4a+130+b=1302a+1?+308a−1?+304a+1?+b=1
14a+30b=29…(2)14a+30b=29…(2)
Solving (1) & (2)
a=2,b=130,ab=60a=2,b=301?,ba?=60

JEE Previous Year Question Paper

11. If the area of the region {(x,y):1−2x≤y≤4−x2,x≥0,y≥0}{(x,y):1−2x≤y≤4−x2,x≥0,y≥0} is αββα?, α,β,∈Nα,β,∈N, α,β=1α,β=1 then the value of (α+β)(α+β) is :
(1) 73
(2) 85
(3) 91
(4) 67

Ans. (1)

Sol. Required area =23×8−12×12×1=32?×8−21?×21?×1
=163−14=6112=αβ=316?−41?=1261?=βα?
⇒α+β=73⇒α+β=73

12. Let a1,a22,a322,…,a1029a1?,2a2??,22a3??,…,29a10?? be a G.P. of common ratio 122?1?. If a1+a3+…+a10=62a1?+a3?+…+a10?=62 then aiai? is equal to :
(1) 2(2−1)2(2?−1)
(2) 2−22−2?
(3) 2−12?−1
(4) 2(2−2)2(2−2?)

Ans. (1)

Sol. a22a1=a32a2=a42a3=…=a102a9=122a1?a2??=2a2?a3??=2a3?a4??=…=2a9?a10??=2?1?
∴a1,a2,a3,a10∴a1?,a2?,a3?,a10? are in G.P. with common ratio 22?
∑i=110ai=a1((2)10−1)2−1=62∑i=110?ai?=2?−1a1?((2?)10−1)?=62
⇒a1=2(2−1)⇒a1?=2(2?−1)

JEE Previous Year Question Paper

13. Let A={x:?x?2−10≤6}A={x:?x?2−10≤6} and B={x:?x−2?>1}B={x:?x−2?>1} Then
(1) A∪B=(−∞,1]∪(2,∞)A∪B=(−∞,1]∪(2,∞)
(2) A−B=[2,3)A−B=[2,3)
(3) A∩B=[−4,−2]∪[3,4]A∩B=[−4,−2]∪[3,4]
(4) B−A=(−∞,−4)∪(−2,1)∪(4,∞)B−A=(−∞,−4)∪(−2,1)∪(4,∞)

Ans. (4)

Sol. ?x2−10?≤6?x2−10?≤6
−6≤x2−10≤6−6≤x2−10≤6
4≤x2≤164≤x2≤16
A=[−4,−2]∪[2,4]A=[−4,−2]∪[2,4]
?x−2?>1?x−2?>1
B=(−∞,1)∪(3,∞)B=(−∞,1)∪(3,∞)
A∪B=(−∞,1)∪[2,∞)A∪B=(−∞,1)∪[2,∞)
A∩B=[−4,−2]∪(3,4]A∩B=[−4,−2]∪(3,4]
A−B=[2,3]A−B=[2,3]
B−A=(−∞,−4)∪(−2,1)∪(4,∞)B−A=(−∞,−4)∪(−2,1)∪(4,∞)

14. For the matrices A=[3−41−1]A=[31?−4−1?] and B=[−2949−1318]B=[−29−13?4918?] if (A15+B)[xy]=[00](A15+B)[xy?]=[00?], then among the following which one is true?
(1) x=5,y=7x=5,y=7
(2) x=18,y=11x=18,y=11
(3) x=11,y=2x=11,y=2
(4) x=16,y=3x=16,y=3

Ans. (3)

Sol. Here An=[2n+1−4nn−2n+1]An=[2n+1n?−4n−2n+1?]
⇒A15=[31−6015−29]⇒A15=[3115?−60−29?]
⇒A15+B=[2−112−11]⇒A15+B=[22?−11−11?]
Now (A15+B)[xy]=[00](A15+B)[xy?]=[00?]
⇒[2−112−11][xy]=[00]⇒[22?−11−11?][xy?]=[00?]
⇒2x−11y=0⇒2x−11y=0

JEE Previous Year Question Paper

15. For a triangle ABC, let p?=BC?,q?=CA?p?=BC,q?=CA and r?=BA?r=BA. If ?p??=23,?q??=2?p??=23?,?q??=2 and cos?θ=13cosθ=3?1?, where θ is the angle between p?p? and q?q?, then

?p?×(q?−3r?)?2+3?r??2?p?×(q?−3r)?2+3?r?2

is equal to:
(1) 340
(2) 220
(3) 410
(4) 200

Ans. (4)

Sol. p?+q?=r?p?+q?=r
cos?(π−θ)=?p??2+?q??2−?r??22?p???q??cos(π−θ)=2?p???q???p??2+?q??2−?r?2?
−13=12+4−?r??22⋅23⋅23?−1?=2⋅23?⋅212+4−?r?2?
?r??2=24?r?2=24
∴?p?×(q?−3r?)?2+3?r??2∴?p?×(q?−3r)?2+3?r?2
=?p?×(q?−3p?−3q?)?2+72=?p?×(q?−3p?−3q?)?2+72
=?p?×(−3p?−2q?)?2+72=?p?×(−3p?−2q?)?2+72
=?−2p?×q??2+72=?−2p?×q??2+72
=4?p??2?q??2×sin?2θ+72=4?p??2?q??2×sin2θ+72
=4⋅12⋅4⋅23+72=4⋅12⋅4⋅32?+72
=200=200

16. Let y=y(x)y=y(x) be the solution of the differential equation sec?xdydx−2y=2+3sin?xsecxdxdy?−2y=2+3sinx, x∈(−π2,π2)x∈(−2π?,2π?), y(0)=−74y(0)=−47?. Then y(π6)y(6π?) is equal to:
(1) −52−25?
(2) −54−45?
(3) −33−7−33?−7
(4) −32−7−32?−7

Ans. (1)

Sol. dydx−2ycos?x=2cos?x+3sin?x⋅cos?xdxdy?−2ycosx=2cosx+3sinx⋅cosx
I.F. = e−2sin?xe−2sinx
e−2sin?x⋅y=∫e−2sin?x(3sin?xcos?x+2cos?x)dxe−2sinx⋅y=∫e−2sinx(3sinxcosx+2cosx)dx
y⋅e−2sin?x=e−2sin?x(−32sin?x−74)+Cy⋅e−2sinx=e−2sinx(−23?sinx−47?)+C
⇒y=−32sin?x−74+C⋅e2sin?x⇒y=−23?sinx−47?+C⋅e2sinx
?y(0)=−74⇒C=0?y(0)=−47?⇒C=0
y(π6)=−32⋅12−74=−52y(6π?)=2−3?⋅21?−47?=2−5?

JEE Previous Year Question Paper

17. Let A = {2, 3, 5, 7, 9}. Let R be the relation on A defined by x R y if and only if 2x ≤ 3y. Let ? be the number of elements in R, and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then ? + m is equal to:
(1) 23
(2) 25
(3) 21
(4) 27

Ans. (2)

Sol. A = {2,3,5,7,9}
y≥2x3y≥32x?
x=2, y=2,3,5,7,9
x=3, y=2,3,5,7,9
x=5, y=5,7,9
x=7, y=5,7,9
x=9, y=7,9
→?=18→?=18
to make it symmetric elements to be added are {(5,2),(7,2),(9,2),(5,3),(7,3),(9,3),(9,5)}
m = 7
∴?+m=25∴?+m=25

18. If the system of equations
3x+y+4z=33x+y+4z=3
2x+αy−z=−32x+αy−z=−3
x+2y+z=4x+2y+z=4
has no solution, then the value of α is equal to:
(1) 19
(2) 4
(3) 13
(4) 23

Ans. (1)

Sol. for no solution Δ = 0
?3142α−1121?=0?321?1α2?4−11??=0
⇒3(α+2)+1(−1−2)+4(4−α)=0⇒3(α+2)+1(−1−2)+4(4−α)=0
⇒19−α=0⇒α=19⇒19−α=0⇒α=19
& for α = 19
Δx=?314−319−1421?=3(21)+1(−1)+4(−82)≠0Δx?=?3−34?1192?4−11??=3(21)+1(−1)+4(−82)?=0
∴∴ no solution for α = 19

19. Let z be the complex number satisfying ?z−5?≤3?z−5?≤3 and having maximum positive principal argument. Then 34?5z−125iz+16?234?5iz+165z−12??2 is equal to:
(1) 16
(2) 12
(3) 26
(4) 20

Ans. (4)

Sol. z=(4cos?θ,4sin?θ)z=(4cosθ,4sinθ)
≡(4⋅45,4⋅35)=(165,125)=165+12i5≡(4⋅54?,4⋅53?)=(516?,512?)=516?+512i?
Now, 34?5z−125iz+16?2=34?(16+12i)−12(16i−12)+16?234?5iz+165z−12??2=34?(16i−12)+16(16+12i)−12??2
=34?4+12i16i+4?2=34?16i+44+12i??2
=34(16+144256+16)=34(160272)=20=34(256+1616+144?)=34(272160?)=20

JEE Previous Year Question Paper

20. The largest n∈N, for which 7n7n divides 101!, is :
(1) 16
(2) 18
(3) 15
(4) 19

Ans. (1)

Sol. Exponent of 7 in 101!
=[1017]+[10172]+[10173]+…=[7101?]+[72101?]+[73101?]+…
=14+2=16=14+2=16

SECTION-B

21. Let [⋅][⋅] denote the greatest integer function and f(x)=lim?n→∞1n3∑k=1n[k23x]f(x)=limn→∞?n31?∑k=1n?[3xk2?]. Then 12∑j=1∞f(j)12∑j=1∞?f(j) is equal to ______.

Ans. (2)

Sol. ∑k=1n(k23x−1)<∑k=1n[k23x]≤∑k=1nk23x∑k=1n?(3xk2?−1)<∑k=1n?[3xk2?]≤∑k=1n?3xk2?
n(n+1)(2n+1)6⋅3x−n<∑k=1n[k23x]≤n(n+1)(2n+1)6⋅3x6⋅3xn(n+1)(2n+1)?−n<∑k=1n?[3xk2?]≤6⋅3xn(n+1)(2n+1)?
lim?n→∞n(n+1)(2n+1)6n3⋅3x−1n2<lim?n→∞1n3∑k=1n[k23x]≤lim?n→∞n(n+1)(2n+1)6⋅3x⋅n3limn→∞?6n3⋅3xn(n+1)(2n+1)?−n21?<limn→∞?n31?∑k=1n?[3xk2?]≤limn→∞?6⋅3x⋅n3n(n+1)(2n+1)?
13x+1<lim?n→∞1n3∑k=1n[k23x]≤13x+13x+11?<limn→∞?n31?∑k=1n?[3xk2?]≤3x+11?
⇒f(x)=13x+1⇒f(x)=3x+11?
⇒12∑j=1∞f(j)=12∑j=1∞13j+1=12[19+127+…]⇒12∑j=1∞?f(j)=12∑j=1∞?3j+11?=12[91?+271?+…]
=12⋅191−13=2=12⋅1−31?91??=2

JEE Previous Year Question Paper

22. If ∫014cot?−1(1−2x+4x2)dx=atan?−1(2)−blog?(5)∫01?4cot−1(1−2x+4x2)dx=atan−1(2)−blog(5), where a, b∈N, then (2a + b) is equal to ______.

Ans. (9)

Sol. Let I=∫01cot?−1(1−2x+4x2)dxI=∫01?cot−1(1−2x+4x2)dx
I=∫01(cot?−1(2x−1)−cot?−1(2x))dx…(1)I=∫01?(cot−1(2x−1)−cot−1(2x))dx…(1)
Applying king
I=∫01(−cot?−1(2x−1)+cot?−1(2x−2))dx…(2)I=∫01?(−cot−1(2x−1)+cot−1(2x−2))dx…(2)
From (1) & (2)
2I=∫01(cot?−1(2x−2)−cot?−1(2x))dx2I=∫01?(cot−1(2x−2)−cot−1(2x))dx
=∫01cot?−1(2x−2)dx−∫01cot?−1(2x)dx=∫01?cot−1(2x−2)dx−∫01?cot−1(2x)dx
Applying King
=∫01cot?−1(−2x)dx−∫01cot?−1(2x)dx=∫01?cot−1(−2x)dx−∫01?cot−1(2x)dx
=∫01(π−cot?−1(2x))dx−∫01cot?−1(2x)dx=∫01?(π−cot−1(2x))dx−∫01?cot−1(2x)dx
=∫01(π−2cot?−1(2x))dx=∫01?(π−2cot−1(2x))dx
=π−2∫01(cot?−1(2x))⋅1dx=π−2∫01?(cot−1(2x))⋅1dx
By parts
=π−2([xcot?−1(2x)]01+∫012x1+4x2dx)=π−2([xcot−1(2x)]01?+∫01?1+4x22x?dx)
Let 1+4x2=t1+4x2=t, 8xdx=dt8xdx=dt
=π−2[cot?−12+14∫15dtt]=π−2[cot−12+41?∫15?tdt?]
=π−2cot?−12−12ln?5=π−2cot−12−21?ln5
2I=2tan?−12−12ln?52I=2tan−12−21?ln5
⇒4I=4tan?−12−ln?5⇒4I=4tan−12−ln5
∴2a+b=8+1=9∴2a+b=8+1=9

23. Let the maximum value of (sin?−1x)2+(cos?−1x)2(sin−1x)2+(cos−1x)2 for x∈[−32,12]x∈[−23??,2?1?] be mnπ2nm?π2, where gcd(m, n) = 1. Then m + n is equal to ______.

Ans. (65)

Sol. (sin?−1x)2+(cos?−1x)2(sin−1x)2+(cos−1x)2
=(sin?−1x+cos?−1x)2−2sin?−1xcos?−1x=(sin−1x+cos−1x)2−2sin−1xcos−1x
=π24−2(sin?−1x)(π2−sin?−1x)=4π2?−2(sin−1x)(2π?−sin−1x)
=2(sin?−1x−π4)2+π28=2(sin−1x−4π?)2+8π2? where sin?−1x∈[−π3,π4]sin−1x∈[−3π?,4π?]
Then max value occurs at sin?−1x=−π3sin−1x=−3π?
Which is 2(π3+π4)2+π28=29π2362(3π?+4π?)2+8π2?=3629π2?
⇒m=29⇒m=29 and n = 36
∴m+n=65∴m+n=65

JEE Previous Year Question Paper

24. If (115C0+115C1)(115C1+115C2)…(115C12+115C13)=α1314C0⋅14C1…14C12(15C0?1?+15C1?1?)(15C1?1?+15C2?1?)…(15C12?1?+15C13?1?)=14C0?⋅14C1?…14C12?α13?, then 30α is equal to ______.

Ans. (32)

Sol. ∏r=012(115Cr+115Cr+1)=∏r=0121615Cr⋅15Cr+1∏r=012?(15Cr?1?+15Cr+1?1?)=∏r=012?15Cr?⋅15Cr+1?16?
=∏r=01216(r+1)⋅15r+1⋅14Cr=∏r=0121614Cr=∏r=012?(r+1)⋅r+115?⋅14Cr?16?=∏r=012?14Cr?16?
=(1615)1314C0⋅14C1…14C12⇒α=1615=14C0?⋅14C1?…14C12?(1516?)13?⇒α=1516?
⇒30α=32⇒30α=32

25. If P is a point on the circle x2+y2=4x2+y2=4, Q is a point on the straight line 5x+y+2=05x+y+2=0 and x−y+1=0x−y+1=0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such point P is

Ans. (2)

Sol. Mid point of PQ lies on x−y+1=0x−y+1=0
2cos?θ+α2−2sin?θ−5α−22+1=022cosθ+α?−22sinθ−5α−2?+1=0
2cos?θ+α−2sin?θ+5α+2+2=02cosθ+α−2sinθ+5α+2+2=0
cos?θ−sin?θ+3α+2=0…(1)cosθ−sinθ+3α+2=0…(1)
Slope of PQ is −1−1
2sin?θ+5α+22cos?θ−α=−12cosθ−α2sinθ+5α+2?=−1
2sin?θ+5α+2=−2cos?θ+α2sinθ+5α+2=−2cosθ+α
sin?θ+cos?θ+2α+1=0…(2)sinθ+cosθ+2α+1=0…(2)
eliminate α from (1) and (2)
⇒cos?θ+5sin?θ=1,θ∈[0,2π]⇒cosθ+5sinθ=1,θ∈[0,2π]
⇒5×2sin?θ2cos?θ2=2sin?2θ2⇒5×2sin2θ?cos2θ?=2sin22θ?
∴sin?θ2=0⇒cos?θ=1∴sin2θ?=0⇒cosθ=1
or sin?θ2=5⇒cos?θ=−1213sin2θ?=5⇒cosθ=−1312?
Sum of all possible values of abscissa of point P is
=2×1+2(−1213)=213=2×1+2(13−12?)=132?
13 times sum of all possible values of abscissa of point P is 2.

JEE Previous Year Question Paper

PHYSICS

SECTION-A

26. Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is :
(Take k=14π?0k=4π?0?1? and assume kQ2>Gm2kQ2>Gm2 where G is the Gravitational constant)
(1) kQ24mR(1−Gm2kQ2)4mRkQ2?(1−kQ2Gm2?)?
(2) kQ24mR(1+Gm2kQ2)4mRkQ2?(1+kQ2Gm2?)?
(3) kQ22mR(1−Gm2kQ2)2mRkQ2?(1−kQ2Gm2?)?
(4) kQ22mR(1−Gm22kQ2)2mRkQ2?(1−2kQ2Gm2?)?

Ans. (1)

Sol. Using energy conservation
(2)(12mu2)−Gm24r+KQ24r=−Gm22r+KQ22r(2)(21?mu2)−4rGm2?+4rKQ2?=−2rGm2?+2rKQ2?
u=14mr(KQ2−Gm2)u=4mr1?(KQ2−Gm2)?

27. The charge stored by the capacitor C in the given circuit in the steady state is ______ μC.
(1) 12.5
(2) 10
(3) 7.5
(4) 5

Ans. (2)

Sol. i=2.5/5=0.5Ai=2.5/5=0.5A
Vc=4×0.5Vc?=4×0.5
Vc=2VVc?=2V
charge Q=CVcQ=CVc?
=5×2=5×2
=10μC=10μC

JEE Previous Year Question Paper

28. The total length of potentiometer wire AB is 50cm50cm in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is ______ cm.
(1) 15
(2) 30
(3) 25
(4) 20

Ans. (2)

Sol. 6RAP=4RPBRAP?6?=RPB?4?
?AP+?PB=50?AP?+?PB?=50
RAPRPB=?AP?PB=32RPB?RAP??=?PB??AP??=23?
?AP=35×50=30cm?AP?=53?×50=30cm

29. A capacitor C is first charged fully with potential difference of VoVo? and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In t s 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is ______ s.
(1) πLC33πLC??
(2) πLC66πLC??
(3) πLC22πLC??
(4) πLC2π2LC??

Ans. (2)

Sol. Ue=75%UeUe?=75%Ue?
QF2=34Qi2QF2?=43?Qi2?
Qicos?ωt=32Qi⇒t=T12Qi?cosωt=23??Qi?⇒t=12T?
t=π6LCt=6π?LC?

30. The r.m.s speed of oxygen molecules at 47?C47?C is equal to that of the hydrogen molecules kept at ______ °C. (Mass of oxygen molecule/mass of hydrogen molecule = 32/2)
(1) -235
(2) -100
(3) -253
(4) -20

Ans. (3)

Sol. Vrms=3RTMVrms?=M3RT??
VrmsO2=VrmsH2VrmsO2??=VrmsH2??
TO2=273+47=320KTO2??=273+47=320K
3RTO2MO2=3RTH2MH2MO2??3RTO2????=MH2??3RTH2????
TO2MO2=TH2MH2MO2??TO2???=MH2??TH2???
32032=TH2232320?=2TH2???
TH2=20KTH2??=20K
TH2=−253?CTH2??=−253?C

JEE Previous Year Question Paper

31. Two cars A and B each of mass 103kg103kg are moving on parallel tracks separated by a distance of 10m10m in same direction with speeds 72km/h72km/h and 36km/h36km/h. The magnitude of angular momentum of car A with respect to car B is ______ J.s.
(1) 3.6×1053.6×105
(2) 105105
(3) 3×1053×105
(4) 2×1052×105

Ans. (2)

Sol. L=m.VrelrLL=m.Vrel?rL?
=1000×(36×518)×10=1000×(36×185?)×10
=105kgm2/s=105kgm2/s

32. The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is (assume no slipping of string on pulley.)
(1) [(M−m)g[136]M+m][[613?]M+m(M−m)g?]
(2) (M−m)gM+mM+m(M−m)g?
(3) [(M−m)g[83]M+m][[38?]M+m(M−m)g?]
(4) (M−m)g2M+m2M+m(M−m)g?

Ans. (3)

Sol. Mg−T2=Ma…(1)Mg−T2?=Ma…(1)
T1−mg=ma…(2)T1?−mg=ma…(2)
(T2−T1)r=Iar…(3)(T2?−T1?)r=Ira?…(3)
(1)+(2)+(3)(1)+(2)+(3)
(M−m)g=(M+m+Ir2)a(M−m)g=(M+m+r2I?)a
Here I=Mr2+M×(2r)212×2I=Mr2+12M×(2r)2?×2
=(1+23)Mr2=(1+32?)Mr2
=53Mr2=35?Mr2
(M−m)g=[M+m+5M3]a(M−m)g=[M+m+35M?]a
a=(M−m)g[M+m+5M3]a=[M+m+35M?](M−m)g?

33. The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz. [take π = 22/7]
(1) 14
(2) 88
(3) 28
(4) 176

Ans. (1)

Sol. ω=176ω=176 rad/sec
fk=ω2π=1762×22×7fk?=2πω?=2×22176?×7
=17644×7=44176?×7
=4×7=28Hz=4×7=28Hz
So frequency of oscillator
=fk2=14Hz=2fk??=14Hz

JEE Previous Year Question Paper

34. Given below are two statements:
Statement I : In a Young's double slit experiment, the angular separation of fringes will increase as the screen is moved away from the plane of the slits
Statement II : In a Young's double slit experiment, the angular separation of fringes will increase when monochromatic source is replaced by another monochromatic source of higher wavelength
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is false but Statement II is true
(4) Statement I is true but Statement II is false

Ans. (3)

Sol. Angular fringe width θ=λdθ=dλ?

35. A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when :
(1) R=2rR=2r
(2) R=r2R=2r?
(3) R=2rR=2?r
(4) R=rR=r

Ans. (4)

Sol. For maximum power drawn across load
Rload=RinternalRload?=Rinternal?
R=rR=r

36. Keeping the significant figures in view, the sum of the physical quantities 52.01m, 153.2 m and 0.123 m is :
(1) 205 m
(2) 205.333 m
(3) 205.33 m
(4) 205.3 m

Ans. (4)

Sol. L=52.01+153.2+0.123L=52.01+153.2+0.123
=205.333=205.333
=205.3=205.3

37. A spherical body of radius r and density σ falls freely through a viscous liquid having density ρ and viscosity η and attains a terminal velocity ν?. Estimated maximum error in the quantity η is : (Ignore errors associated with σ, ρ and g, gravitational acceleration)
(1) 2Δrr−Δν0ν02rΔr?−ν0?Δν0??
(2) 2Δrr+Δν0ν02rΔr?+ν0?Δν0??
(3) 2[Δrr+Δν0ν0]2[rΔr?+ν0?Δν0??]
(4) Δrr+Δν0ν0rΔr?+ν0?Δν0??

Ans. (2)

Sol. ν0=2r2g9η(ρB−ρL)ν0?=9η2r2g?(ρB?−ρL?)
η=2r2g9v0(ρB−ρL)η=9v0?2r2g?(ρB?−ρL?)
Δηη=2Δrr+Δv0v0ηΔη?=r2Δr?+v0?Δv0??

JEE Previous Year Question Paper

38. Surface tension of two liquids (having same densities), T1T1? and T2T2?, are measured using capillary rise method utilizing two tubes with inner radii of r1r1? and r2r2? where r1>r2r1?>r2?. The measured liquid heights in these tubes are h1h1? and h2h2? respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights h1h1? and h2h2? and surface tensions T1T1? and T2T2? satisfy the relation :
(1) h1<h2h1?<h2? and T1=T2T1?=T2?
(2) h1=h2h1?=h2? and T1=T2T1?=T2?
(3) h1>h2h1?>h2? and T1=T2T1?=T2?
(4) h1>h2h1?>h2? and T1<T2T1?<T2?

Ans. (1)

Sol. h=2Tρgrh=ρgr2T?
h∝1rh∝r1?
If r1>r2⇒h2>h1r1?>r2?⇒h2?>h1?

39. A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ______ and ______ respectively.
(1) 20 s and 100 m
(2) 40 s and 0 m
(3) 40 s and 200 m
(4) 40 s and 100 m

Ans. (3)

Sol. Minimum time : tmin=20010=20sectmin?=10200?=20sec
For round trip = 40 sec.
Displacement along river bank =40×5=200m=40×5=200m

40. Two known resistance of RΩRΩ and 2RΩ2RΩ and one unknown resistance XΩXΩ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is XΩXΩ, then the value of X is ______ Ω.
(1) (3−1)R(3?−1)R
(2) RR
(3) 2(3−1)R2(3?−1)R
(4) (3+1)R(3?+1)R

Ans. (1)

Sol. (2R+x)⋅(R)3R+x=x3R+x(2R+x)⋅(R)?=x
x2+2Rx−2R2=0x2+2Rx−2R2=0
x=(3−1)Rx=(3?−1)R

JEE Previous Year Question Paper

41. The energy of an electron in an orbit of the Bohr's atom is −0.04E0eV−0.04E0?eV where E0E0? is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then 2πLhh2πL? is
(1) 2
(2) 4
(3) 5
(4) 6

Ans. (3)

Sol. Angular momentum L=nh2πL=2πnh?
n=2πLhn=h2πL?
Energy E=−E0n2E=−n2E0??
⇒−E0n2=−0.04E0⇒−n2E0??=−0.04E0?
n2=25,n=5n2=25,n=5

42. An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :
(1) μ02πI(π+1)i^2πμ0??I(π+1)i^
(2) −μ02πI(π−1)i^−2πμ0??I(π−1)i^
(3) μ02πI(π−1)i^2πμ0??I(π−1)i^
(4) −μ02πI(π+1)i^−2πμ0??I(π+1)i^

Ans. (2)

Sol. B?0=B?AB+B?DE+B?BCDB0?=BAB?+BDE?+BBCD?
=μ0i4πri^+μ0i4πri^−μ0i2ri^=4πrμ0?i?i^+4πrμ0?i?i^−2rμ0?i?i^
=μ0i2πri^−μ0i2ri^=2πrμ0?i?i^−2rμ0?i?i^
=μ0i2πr(1−π)i^=2πrμ0?i?(1−π)i^
=−μ0i2πr(π−1)i^=−2πrμ0?i?(π−1)i^

JEE Previous Year Question Paper

43. A large drum having radius R is spinning around its axis with angular velocity ω as shown in figure. The minimum value of ω so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M is μ, is :
(1) μgRRμg??
(2) 2gμRμR2g??
(3) g2μR2μRg??
(4) gμRμRg??

Ans. (4)

Sol. N=mω2r,mg=μNN=mω2r,mg=μN
μ×mω2r=mgμ×mω2r=mg
ω=gμrω=μrg??

44. A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by x(t)=αt2+βt+γx(t)=αt2+βt+γ m, where α = 1 m/s², β = 1 m/s and γ = 1 m. The work done on the body during the time interval t = 2 s to t = 3 s, is ______ J.
(1) 49
(2) 42
(3) 24
(4) 12

Ans. (3)

Sol. x(t)=t2+t+1x(t)=t2+t+1
v(t)=2t+1v(t)=2t+1
a(t)=2a(t)=2
F=4NF=4N
Displacement =x(3)−x(2)=x(3)−x(2)
=13−7=6m=13−7=6m
W=F⋅S=4×6=24JW=F⋅S=4×6=24J

45. As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface. If refractive index of the material of prism is 22? the angle θ of prism is.
(1) 60?60?
(2) 75?75?
(3) 90?90?
(4) 45?45?

Ans. (1)

Sol. For grazing emergence
sin?r2=1μsinr2?=μ1?
By Snell's Law at incident surface
1×12=2sin?r11×2?1?=2?sinr1?
r1=30r1?=30
r1+r2=Ar1?+r2?=A
A=75A=75
75+45+θ=180?75+45+θ=180?
θ=60?θ=60?

JEE Previous Year Question Paper

SECTION-B

46. An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, σ = 10 mho/m. The ratio of maximum conducting current density to maximum displacement current density is ______. [Take 14π?0=9×109Nm2/C24π?0?1?=9×109Nm2/C2]

Ans. (1800)

Sol. jc=σEjc?=σE
E⇒Essin?(ωt−kx)E⇒Es?sin(ωt−kx)
jc=σEssin?(ωt−kx)jc?=σEs?sin(ωt−kx)
⇒(jc)max=σE0…(i)⇒(jc?)max?=σE0?…(i)
jd=?0dEdt=?0ωE0cos?(ωt−kx)jd?=?0?dtdE?=?0?ωE0?cos(ωt−kx)
(jd)max=?0ωE0…(ii)(jd?)max?=?0?ωE0?…(ii)
(jc)max(jd)max=σE0?0ωE0=σ?0ω(jd?)max?(jc?)max??=?0?ωE0?σE0??=?0?ωσ?
=10×4π×9×1092π×100×106=2π×100×10610×4π×9×109?
=1800=1800

47. The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is 20 cm/s. The terminal velocity of another ball of same material and having radius 3 mm in the same fluid will be ______ cm/s.

Ans. (5)

Sol. Terminal velocity ∝ (radius)²
(vT)1=(63)2(vT?)1?=(36?)2
(vT)2=(vT)14=5cm/sec(vT?)2?=4(vT?)1??=5cm/sec

48. A particle having electric charge 3×10−193×10−19 C and mass 6×10−276×10−27 kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α×10−12α×10−12 m. The value of α is ______. (Take Planck's constant = 6.6×10−346.6×10−34 J.s)

Ans. (10)

Sol. λ=h2mqVλ=2mqV?h?
λ=6.6×10−342×18×10−46×1.21λ=2×18×10−46?×1.216.6×10−34?
λ=10−11m=10×10−12mλ=10−11m=10×10−12m
α=10α=10

JEE Previous Year Question Paper

49. In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 μm is introduced in front of one of the slits then centre bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by α1010α?, where α is ______.

Ans. (14)

Sol. yshift=(μ−1)tDdyshift?=d(μ−1)tD?
20×10−3=(μ−1)×20×10−6×10.4×10−320×10−3=0.4×10−3(μ−1)×20×10−6×1?
(μ−1)=0.4(μ−1)=0.4
μ=1.4μ=1.4
α10=1.4,α=1410α?=1.4,α=14

CHEMISTRY

SECTION-A

51. Consider the following spectral lines for atomic hydrogen:
A. First line of Paschen series
B. Second line of Balmer series
C. Third line of Paschen series
D. Fourth line of Bracket series
The correct arrangement of the above lines in ascending order of energy is:
(1) D<C<A<BD<C<A<B
(2) A<B<C<DA<B<C<D
(3) C<D<B<AC<D<B<A
(4) D<A<C<BD<A<C<B

Ans. (4)

Sol. ΔE=13.6Z2(1n12−1n22)ΔE=13.6Z2(n12?1?−n22?1?)
Series n?, n?
(A) Paschen (1st line) 3, 4
(B) Balmer (2nd line) 2, 4
(C) Paschen (3rd line) 3, 6
(D) Bracket (4th line) 4, 8
So correct ascending order of energy of above lines is :
D<A<C<BD<A<C<B

52. Match List-I with List-II.
List-I (Pair of Compounds)
A. 2-Methylpropene and but-1-ene
B. Cis-but-2-ene and trans-but-2-ene
C. 2-Butanol and diethyl ether
D. But-1-ene and but-2-ene
List-II (Type of Isomers)
I. Stereoisomers
II. Position isomers
III. Chain isomers
IV. Functional group isomers
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-III, B-I, C-II, D-IV
(3) A-I, B-IV, C-III, D-II
(4) A-II, B-I, C-IV, D-III

Ans. (2)

Sol. (A) Chain isomer
(B) Stereoisomers
(C) Functional isomers
(D) Positional isomers

JEE Previous Year Question Paper

53. Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be:
(1) 1
(2) 6
(3) 5
(4) 3

Ans. (3)

Sol. Number of Br atom in major product (P) = 5

54. Aqueous HCl reacts with MnO2(s)MnO2?(s) to form MnCl2(aq)MnCl2?(aq), Cl2(g)Cl2?(g) and H2O(l)H2?O(l). What is the weight (in g) of Cl2Cl2? liberated when 8.7 g of MnO2(s)MnO2?(s) is reacted with excess aqueous HCl solution? (Given Molar mass in g mol?¹ Mn = 55, Cl = 35.5, O = 16, H = 1)
(1) 7.1
(2) 71
(3) 21.3
(4) 14.2

Ans. (1)

Sol. MnO2+4HCl→MnCl2+Cl2+2H2OMnO2?+4HCl→MnCl2?+Cl2?+2H2?O
8.787=0.1878.7?=0.1 mole
Wt. of Cl2Cl2? obtained =0.1×71=7.1g=0.1×71=7.1g

55. By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer) (Given, molar mass in g mol?¹: O = 16, Mg = 24, P = 31)
(1) 50
(2) 30
(3) 20
(4) 40

Ans. (1)

Sol. %% of P=nMg2P2O7×2×31W(unknowncompound)×100P=W(unknowncompound)?nMg2?P2?O7??×2×31?×100
=(1.79222×2×31)1×100=1(2221.79?×2×31)?×100
=49.99%≈50%=49.99%≈50%

JEE Previous Year Question Paper

56. Consider the following data:
ΔHfo(methane,g)=−XkJmol−1ΔHfo?(methane,g)=−XkJmol−1
Enthalpy of sublimation of graphite = Y kJ mol?¹
Dissociation enthalpy of H2H2? = Z kJ mol?¹
The bond enthalpy of C–H bond is given by:
(1) X+Y+2Z44X+Y+2Z?
(2) X+Y+4Z22X+Y+4Z?
(3) X+Y+ZX+Y+Z
(4) −X+Y+Z44−X+Y+Z?

Ans. (1)

Sol. C(s)+2H2(g)→CH4(g)C(s)+2H2?(g)→CH4?(g)
−x=(ΔHsubofcarbon)+2×(B.E.ofH−H)−4×(B.E.ofC−H)−x=(ΔHsub?ofcarbon)+2×(B.E.ofH−H)−4×(B.E.ofC−H)
−x=y+2z−4(B.E.ofC−H)−x=y+2z−4(B.E.ofC−H)
B.E.ofC−H=y+2z+x4B.E.ofC−H=4y+2z+x?

57. Match List-I with List-II.
List-I (Reagents Involving aldehydes)
A. H2H2? Pd-BaSO?
B. SnCl2SnCl2? HCl
C. CrO2Cl2CrO2?Cl2? CS?
D. CO2CO2? HCl Anhyd. AlCl3AlCl3?
List-II (Reaction Name)
I. Etard Reaction
II. Rosenmund Reduction
III. Gatterman-Koch Reaction
IV. Stephen Reaction
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-IV, B-III, C-I, D-II
(3) A-IV, B-I, C-II, D-III
(4) A-II, B-IV, C-I, D-III

Ans. (4)

Sol. NCERT Name reaction theory based

58. Decomposition of A is a first order reaction at T(K) and is given by A(g)→B(g)+C(g)A(g)→B(g)+C(g). In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min?¹) of the reaction? (log 2 = 0.3)
(1) 6.9×10−16.9×10−1
(2) 6.9×10−36.9×10−3
(3) 6.9×10−26.9×10−2
(4) 6.9×10−46.9×10−4

Ans. (2)

Sol. Ag→Bg+CgAg?→Bg?+Cg?
1 - -
1 - P P P
Ptotal=1+PPtotal?=1+P
1.5=1+P1.5=1+P
P=0.5P=0.5
K=1100ln?10.5K=1001?ln0.51?
=0.693100=1000.693?
=6.9×10−3min−1=6.9×10−3min−1

JEE Previous Year Question Paper

59. The correct order of reactivity of the following benzyl halides towards reaction with KCN is:
(1) a>b>c>da>b>c>d
(2) b>a>d>cb>a>d>c
(3) b>a>c>db>a>c>d
(4) a>b>d>da>b>d>d

Ans. (2)

Sol. This is SN1SN?1 reaction.
Rate of SN1SN?1 reaction ∝ stability of carbocation

60. Given below are two statements:
Statement-I: The correct order in terms of atomic/ionic radii is Al>Mg>Mg2+>Al3+Al>Mg>Mg2+>Al3+.
Statement-II: The correct order in terms of the magnitude of electron gain enthalpy is Cl>Br>S>OCl>Br>S>O.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are false
(2) Statement I is false but Statement II is true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true

Ans. (2)

Sol. Correct order of size is Mg>Al>Mg2+>Al3+Mg>Al>Mg2+>Al3+
Atomic size depends mainly upon ZeffectiveZeffective? and shell number.
Generally on moving down the group electron affinity decreases and on moving across the period electron affinity increase.
In the periodic table Cl has maximum electron affinity. Halogen has higher electron affinity than Chalcogen. Cl>Br>S>OCl>Br>S>O

61. The correct statements are:
A. Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis.
B. During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact.
C. Nucleotides are joined together by glycosidic linkage between C? and C? carbons of the pentose sugar
D. Quaternary structure of proteins represents overall folding of the polypeptide chain.
Choose the correct answer from the options given below:
(1) A, C and D Only
(2) A, B and D Only
(3) A and B Only
(4) B and C Only

Ans. (3)

JEE Previous Year Question Paper

62. The correct order of the rate of the reaction for the following reaction with respect to nucleophiles is:
CH3Br+NuΘ?CH3Nu+BrΘCH3?Br+NuΘ?CH3?Nu+BrΘ
(1) PhO−>OH−>CH3COO−>ClO4−PhO−>OH−>CH3?COO−>ClO4−?
(2) ClO4−>CH3COO−>OH−>PhO−ClO4−?>CH3?COO−>OH−>PhO−
(3) CH3COO−>PhO−>OH−>ClO4−CH3?COO−>PhO−>OH−>ClO4−?
(4) OH−>PhO−>CH3COO−>ClO4−OH−>PhO−>CH3?COO−>ClO4−?

Ans. (4)

Sol. Stability order of anion ClO4−>CH3COO−>PhO−>OH−ClO4−?>CH3?COO−>PhO−>OH−
and nucleophilicity order reverse of stability of anion
OH−>PhO−>CH3COO−>ClO4−OH−>PhO−>CH3?COO−>ClO4−?

63. Given below are two statements:
Statement I: Crystal Field Stabilization Energy (CFSE) of [Cr(H2O)6]2+[Cr(H2?O)6?]2+ is greater than that of [Mn(H2O)6]2+[Mn(H2?O)6?]2+
Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium Ferrocyanide.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true

Ans. (1)

Sol. [Mn(H2O)6]2+⇒CFSE[Mn(H2?O)6?]2+⇒CFSE value is zero because of d? configuration with WFL in coordination number 6
[Cr(H2O)6]2+⇒CFSE[Cr(H2?O)6?]2+⇒CFSE value is −0.6Δ0−0.6Δ0? because of d? configuration with WFL in coordination number 6.
For K3[Fe(CN)6]K3?[Fe(CN)6?]: μ=1(1+2)=3B.M.μ=1(1+2)?=3?B.M.
For Na4[Fe(CN)6]Na4?[Fe(CN)6?]: μ=0B.M.μ=0?B.M.

64. The correct increasing order of C–H(A), C–O(B), C=O(C) and C≡N(D) bonds in terms of covalent bond length is:
(1) A<B<C<DA<B<C<D
(2) A<D<C<BA<D<C<B
(3) D<C<B<AD<C<B<A
(4) D<C<A<BD<C<A<B

Ans. (2)

Sol. C–H(A) 107 pm
C≡N(D) 116 pm
C–O(B) 143 pm
C=O(C) 121 pm

JEE Previous Year Question Paper

65. Given below are four compounds:
(a) n-propyl chloride
(b) iso-propyl chloride
(c) sec-butyl chloride
(d) neo-pentyl chloride
Percentage of carbon in the one which exhibits optical isomerism is:
(1) 52
(2) 56
(3) 46
(4) 40

Ans. (1)

Sol. 2-Chlorobutane is optically active and chiral molecule
Molecular formula ⇒C4H9Cl⇒C4?H9?Cl
Molar mass =48+9+35.5=92.5=48+9+35.5=92.5
%C=4892.5×100=51.89%%C=92.548?×100=51.89%

66. Given below are some of the statements about Mn and Mn2O7Mn2?O7?. Identify the correct statements
A. Mn forms the oxide Mn2O7Mn2?O7? in which Mn is in its highest oxidation state.
B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds with Mn
C. Mn2O7Mn2?O7? is an ionic oxide.
D. The structure of Mn2O7Mn2?O7? consists of one bridged oxygen.
Choose the correct answer from the options given below:
(1) A, B, C and D
(2) A, B and D Only
(3) A, C and D Only
(4) A, B and C Only

Ans. (2)

Sol. Mn2O7Mn2?O7?: Mn in +7 oxidation state.

67. For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?
(1) EcelloEcello? vs time graph
(2) EcelloEcello? constant vs time graph
(3) EcelloEcello? decreasing vs time graph
(4) EcelloEcello? increasing vs time graph

Ans. (2)

Sol. EcelloEcello? remain constant with time.

JEE Previous Year Question Paper

68. Given below are two statements:
Statement-I: Compound (X), shown below, dissolves in NaHCO? solution and has two chiral carbon atoms
Statement-II: Compound (Y), shown below, has two carbons with sp³ hybridization, one carbon with sp² and one carbon with sp hybridization
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is true but Statement II is false
(2) Statement I is false but Statement II is true
(3) Both Statement I and Statement II are true
(4) Both Statement I and Statement II are false

Ans. (3)

Sol. Two chiral centre and due to presence of –COOH compound dissolves in NaHCO?.

69. Given below are two statements:
Statement I: The correct order in terms of bond dissociation enthalpy is Cl2>Br2>F2>I2Cl2?>Br2?>F2?>I2?
Statement II: The correct trend in the covalent character of the metal halides is [SnCl4>SnCl2][SnCl4?>SnCl2?], [PbCl4>PbCl2][PbCl4?>PbCl2?] and [UF4>UF2][UF4?>UF2?]
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is true but Statement II is false
(2) Both Statement I and Statement II are true
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are false

Ans. (1)

Sol. Statement-I: Bond energy order is Cl2>Br2>F2>I2Cl2?>Br2?>F2?>I2?
Bond energy increases with increase in bond order.
Statement-II: Correct order of covalent character according to Fajan's rule, higher the charge on cation, greater is the covalent character.
PbCl2<PbCl4PbCl2?<PbCl4?, UF6>UF4UF6?>UF4?, SnCl4>SnCl2SnCl4?>SnCl2?

70. On heating a mixture of common salt and K2Cr2O7K2?Cr2?O7? in equal amount along with concentrated H2SO4H2?SO4? in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are:
(1) CrO2Cl2CrO2?Cl2? and +5
(2) CrO2Cl2CrO2?Cl2? and +6
(3) Cr2O2Cl2Cr2?O2?Cl2? and +6
(4) Cr2O2Cl2Cr2?O2?Cl2? and +3

Ans. (2)

Sol. 4NaCl+K2Cr2O7+6H2SO4?2KHSO4+2CrO2Cl2+4NaHSO4+3H2O4NaCl+K2?Cr2?O7?+6H2?SO4??2KHSO4?+2CrO2?Cl2?+4NaHSO4?+3H2?O
(Chromyl chloride)
In Chromyl chloride Cr is in +6 oxidation state.

JEE Previous Year Question Paper

SECTION-B

71. The first and second ionization constants of H2XH2?X are 2.5×10−82.5×10−8 and 1.0×10−131.0×10−13 respectively. The concentration of X2−X2− in 0.1 M H2XH2?X solution is ______ × 10?¹? M. (Nearest Integer)

Ans. (100)

Sol. H2X?H++HX−H2?X?H++HX−
0.1−x,x+y,x−y0.1−x,x+y,x−y
2.5×10−8=(x+y)(x−y)0.1−x2.5×10−8=0.1−x(x+y)(x−y)?
HX−?H++X2−HX−?H++X2−
x−y,x+y,yx−y,x+y,y
1×10−13=(x+y)(y)x−y1×10−13=x−y(x+y)(y)?
Approximate: Ka1>>Ka2⇒x>>yKa1?>>Ka2?⇒x>>y
x+y≈x,x−y≈xx+y≈x,x−y≈x
10−13=x⋅yx10−13=xx⋅y?
y=10−13y=10−13
[X2−]=10−13[X2−]=10−13
[X2−]=100×10−15[X2−]=100×10−15

72. The osmotic pressure of a living cell in 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is ______ gL?¹. (Nearest integer) Given: R = 0.08 L atm K?¹ mol?¹. Assume complete dissociation of NaCl (Given: Molar mass of Na and Cl are 23 and 35.5 g mol?¹ respectively.)

Ans. (15)

Sol. π=iCRTπ=iCRT
12=2×C×0.08×30012=2×C×0.08×300
12=2×C×2412=2×C×24
C=14C=41? mole/L
then strength of NaCl solution
=14×58.5g/L=41?×58.5g/L
=14.625g/L=14.625g/L
=15g/L=15g/L

JEE Previous Year Question Paper

73. A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol?¹) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is ______ × 10?². (Nearest integer) [Given: K_b of the solvent = 5.0 K kg mol?¹] Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Ans. (3)

Sol. ΔTb=i×Kb×mΔTb?=i×Kb?×m
0.5=i×m×50.5=i×m×5
i×m=0.55=0.1i×m=50.5?=0.1
i×a=1.51000i×a=10001.5? (where a = moles of solute)
Now, Po−PsPo=i×Xsolute=i×aa+150300PoPo?−Ps??=i×Xsolute?=i×a+300150?a?
=i×a1/2=1.5/10001/2=301000=3×10−2=3

=i×1/2a?=1/21.5/1000?=100030?=3×10−2=3

74. Identify the metal ions among Co2+Co2+, Ni2+Ni2+, Fe2+Fe2+, V3+V3+ and Ti2+Ti2+ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is ______.

Ans. (7)

Sol. V3+=(Ar)183d2V3+=(Ar)18?3d2
Ti2+=(Ar)183d2Ti2+=(Ar)18?3d2
Ni2+=(Ar)183d8Ni2+=(Ar)18?3d8
Fe2+=(Ar)183d6Fe2+=(Ar)18?3d6
Co2+=(Ar)183d7Co2+=(Ar)18?3d7
Only for Fe2+Fe2+ and Co2+Co2+, μμ is more than 3.0 B.M.
Fe2+3d6Fe2+3d6: n = 4, μ>3μ>3
Co2+3d7Co2+3d7: n = 3, μ>3μ>3
Number of unpaired electrons =4+3=7=4+3=7

75. MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K
MX(s)?M+(aq)+X−(aq):Ksp=10−10MX(s)?M+(aq)+X−(aq):Ksp?=10−10
If the standard reduction potential for M+(aq)+e−→M(s)M+(aq)+e−→M(s) is EM+/Mo=0.79VEM+/Mo?=0.79V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode EX−/M(X)(s)oEX−/M(X)(s)o? is ______ mV. (nearest integer)
[Given: 2.303RTF=0.059VF2.303RT?=0.059V]

Ans. (200)

Sol. EX−/M(X)(s)o=EM+/Mo+0.059nlog?KspEX−/M(X)(s)o?=EM+/Mo?+n0.059?logKsp?
=0.79+0.0591log?10−10=0.79+10.059?log10−10
=0.79−0.59=0.79−0.59
=0.20V=200mV=0.20V=200mV

JEE Previous Year Question Paper 2026- solved PDF Download link

JEE MAIN Previous Year Question Paper with Free PDF Download link 2025

JEE Main Previous Year Question paper 2024

JEE MAIN Previous Year Question paper 2023

JEE MAIN 2012 to 2026 PDF Download Link