JEE MAIN -Previous Year Solved Question paper 2024 

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INAL JEE–MAIN EXAMINATION – APRIL, 2024 (Held On Tuesday 09th April, 2024)

1. lim?x→0e−(1+2x)12xxlimx→0?xe−(1+2x)2x1?? is equal to : (1) e (2) −2ee−2? (3) 0 (4) e−e2e−e2 Ans. (1) Sol. lim?x→0e−e12x(ln?(1+2x))xlimx→0?xe−e2x1?(ln(1+2x))? =lim?x→0(−e)(eln?(1+2x)2x−1)x=limx→0?(−e)x(e2xln(1+2x)?−1)? =lim?x→0(−e)ln?(1+2x)−2x2x2=limx→0?(−e)2x2ln(1+2x)−2x? =(−e)×(−1)42×2=e=(−e)×(−1)2×24?=e 2. Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point (113,113,193)(311?,311?,319?) from the line L along the line 3x−112=3y−111=3z−19223x−11?=13y−11?=23z−19? is equal to : (1) 3 (2) 5 (3) 4 (4) 6 Ans. (1) Sol. x−12−1=y−23−2=z−35−32−1x−1?=3−2y−2?=5−3z−3? ⇒x−11=y−21=z−32=λ⇒1x−1?=1y−2?=2z−3?=λ

MATHEMATICS

TEST PAPER WITH SOLUTION

ABAB?

B(1+λ,2+λ,3+2λ)B(1+λ,2+λ,3+2λ) D.R.ofAB=<3λ−83,3λ−53,6λ−103>D.R.ofAB=<33λ−8?,33λ−5?,36λ−10?> B(53,83,133)3λ−83λ−5=21⇒3λ−8=6λ−10B(35?,38?,313?)3λ−53λ−8?=12?⇒3λ−8=6λ−10 3λ=23λ=2 λ=23λ=32? AB=36+9+363=93=3AB=336+9+36??=39?=3

1. Let ∫0x1−(y′(t))2dt=∫0xy(t)dt,0≤x≤3,y≥0,∫0x?1−(y′(t))2?dt=∫0x?y(t)dt,0≤x≤3,y≥0, y(0)=0y(0)=0 . Then at x=2x=2 y′′+y+1y′′+y+1 is equal to : (1) 1 (2) 2 (3) 22? (4) 1/2

(1) 1
(2) 2
(3) 22? (4) 1/2

Ans. (1)

Sol. 1−(y′(x))2=y(x)1−(y′(x))2?=y(x)

1−(dydx)2=y21−(dxdy?)2=y2 (dydx)2=1−y2(dxdy?)2=1−y2

1. Let zz be a complex number such that the real part of z−2iz+2iz+2iz−2i? is zero. Then, the maximum value of ?z−(6+8i)??z−(6+8i)? is equal to : (1) 12 (2) ∞∞ (3) 10 (4) 8 Ans. (1) z−2iz+2i=z‾+2iz‾−2i=0z+2iz−2i?=z−2iz+2i?=0 zz‾−2iz‾−2iz+4(−1)+zz‾+2zi+2z‾i+4(−1)=0⇒2?z?2=8⇒?z?=2?z−(6+8i)?maximum=10+2=12?zz−2iz−2iz+4(−1)+zz+2zi+2zi+4(−1)=0⇒2?z?2=8⇒?z?=2?z−(6+8i)?maximum?=10+2=12? 5. The area (in square units) of the region enclosed by the ellipse x2+3y2=18x2+3y2=18 in the first quadrant below the line y=xy=x is : (1) 3π+343?π+43? (2) 3π3?π (3) 3π−343?π−43? (4) 3π+13?π+1 Ans. (2) Sol. x218+y26=118x2?+6y2?=1

dy1−y2=dxORdy1−y2=−dx1−y2?dy?=dxOR1−y2?dy?=−dx ⇒sin?−1y=x+c,sin?−1y=−x+c⇒sin−1y=x+c,sin−1y=−x+c x=0,y=0⇒c=0x=0,y=0⇒c=0 sin?−1y=x,asy≥0sin−1y=x,asy≥0 sin?x=ysinx=y ⇒dydx=cos?x⇒dxdy?=cosx d2ydx2=−sin?xdx2d2y?=−sinx ⇒−sin?x+sin?x+1=1⇒−sinx+sinx+1=1

1. Let zz be a complex number such that the real part of z−2iz+2iz+2iz−2i? is zero. Then, the maximum value of ?z−(6+8i)??z−(6+8i)? is equal to : (1) 12 (2) ∞∞ (3) 10 (4) 8 Ans. (1) z−2iz+2i=z‾+2iz‾−2i=0z+2iz−2i?=z−2iz+2i?=0 zz‾−2iz‾−2iz+4(−1)zz−2iz−2iz+4(−1) +zz‾+2zi+2z‾i+4(−1)=0+zz+2zi+2zi+4(−1)=0 ⇒2?z?2=8⇒?z?=2⇒2?z?2=8⇒?z?=2 ?z−(6+8i)?maximum=10+2=12?z−(6+8i)?maximum?=10+2=12

Ans. (1)

Sol.z−2iz+2i+z‾+2iz‾−2i=0Sol.z+2iz−2i?+z−2iz+2i?=0 zz‾−2iz‾−2iz+4(−1)zz−2iz−2iz+4(−1) +zz‾+2zi+2z‾i+4(−1)=0+zz+2zi+2zi+4(−1)=0 ⇒2?z?2=8⇒?z?=2⇒2?z?2=8⇒?z?=2 ?z−(6+8i)?maximum=10+2=12?z−(6+8i)?maximum?=10+2=12

1. The area (in square units) of the region enclosed by the ellipse x2+3y2=18x2+3y2=18 in the first quadrant below the line y=xy=x is :

(1)3π+34(2)3π(1)3?π+43?(2)3?π (3)3π−34(4)3π+1(3)3?π−43?(4)3?π+1

Ans. (2)

Sol.x218+y26=1Sol.18x2?+6y2?=1

x218+3x218=1⇒4x2=18⇒x2=9218x2?+183x2?=1⇒4x2=18⇒x2=29? 18−x2333dx33?318−x2??dx

x218+3x218=1⇒4x2=18⇒x2=9218x2?+183x2?=1⇒4x2=18⇒x2=29? 18−x2333dx33?318−x2??dx =133(x18−x2+18xsin−1x23)x3223=33?1?(x18−x2?+18xsin−132?x?)32?x32?? =133(9xπ2−322x332−9xπ6)=33?1?(9x2π?−22?3?x2?33??−9x6π?)

Required Area

=12×92+(18π6−934)13=21?×29?+(618π?−493??)3?1? =3π=3?π

1. Let the foci of a hyperbola H coincide with the foci of the ellipse E:(x−1)2100+(y−1)275=1E:100(x−1)2?+75(y−1)2?=1 and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is αα and the length of its conjugate axis is ββ , then 3α2+2β23α2+2β2 is equal to : (1) 242 (2) 225 (3) 237 (4) 205

(1) 242

(2) 225

(3) 237

(4) 205

Ans. (2)

5

Fe12(1,1)Fe12?(1,1)

e1=1−75100=510=12e1?=1−10075??=105?=21? e2=2e2?=2 F1(6,1),F2(−4,1)F1?(6,1),F2?(−4,1) 2ae2=10⇒a=52⇒2a=52ae2?=10⇒a=25?⇒2a=5 ⇒α=5⇒α=5 4=1+b2a2⇒b2=3a24=1+a2b2?⇒b2=3a2 b=3×52b=3?×25? β=53β=53? 3α2+2β2=3×25+2×25×33α2+2β2=3×25+2×25×3 =225=225

1. Two vertices of a triangle ABC are A(3, -1) and B (-2, 3), and its orthocentre is P(1, 1). If the coordinates of the point C are (α,β)(α,β) and the centre of the circle circumscribing the triangle PAB is (h, k), then the value of (α+β)+2(h+k)(α+β)+2(h+k) equals :

(1) 51
(2) 81

(3) 5
(4) 15

Ans. (3)

Sol.

C(α,β)P(1,1)A(3,−1)DB(−2,3)C(α,β)P(1,1)A(3,−1)DB(−2,3)?MAB=4−5⇒MDP=54MAB?=−54?⇒MDP?=45?

Equation of PC is y−1=54(x−1)y−1=45?(x−1) .....(1)

MAP=2−2=−1⇒MBC=+1MAP?=−22?=−1⇒MBC?=+1

Equation of BC is y−3=(x+2)y−3=(x+2) .....(2)

On solving (1) and (2)

x+4=54(x−1)⇒4x+16=5x−5⇒α=21x+4=45?(x−1)⇒4x+16=5x−5⇒α=21⇒β=y=x+5=26⇒β=y=x+5=26α+β=47α+β=47

Equation of ⊥⊥ bisector of AP

y−0=(x−2)y−0=(x−2) .....(3)

Equation of ⊥⊥ bisector of AB

y−1=54(x−12)……………………………(4)y−1=45?(x−21?)……………………………(4)

On solving (3) & (4)

(x−3)4=5x−52(x−3)4=5x−25?x=−192=hx=2−19?=hy=−232=ky=2−23?=k⇒2(h+k)=−42⇒2(h+k)=−42

1. If the variance of the frequency distribution is 160, then the value of c ∈ N is

<table>xc2c3c4c5c6cf211111</table>

(1) 5
(2) 8
(3) 7
(4) 6

Ans. (3)

Sol.

<table>xC2C3C4C5C6Cf211111</table>x‾=(2+2+3+4+5+6)C7=22C7x=7(2+2+3+4+5+6)C?=722C?

Var (x) = c2(2+22+32+42+52+62)77c2(2+22+32+42+52+62)?
−(22c7)2−(722c?)2
= 92c27−c2×48449792c2?−c2×49484?
= (644−484)c249=160c24949(644−484)c2?=49160c2?
160=160×c249⇒c=7160=49160×c2?⇒c=7

1. Let the range of the function
   f(x)=12+sin?3x+cos?3x,x∈IRf(x)=2+sin3x+cos3x1?,x∈IR be [a, b].
   If αα and ββ are respectively the A.M. and the G.M. of a and b, then αββα? is equal to :
   (1) 22? (2) 2
   (3) ππ? (4) ππ

Ans. (1)

Sol. f(x)=12+sin?3x+cos?3xf(x)=2+sin3x+cos3x1?
[12+2,12−2][2+2?1?,2−2?1?]
αβ=a+b2ab=12(ab+ba)βα?=2ab?a+b?=21?(ba??+ab??)
= 12(2−22+2+2+22−2)21?(2+2?2−2???+2−2?2+2???)
= (2−2)+(2+2)2×2=22×2?(2−2?)+(2+2?)?=2?

1. Between the following two statements :
   Statement-I : Let a?=i^+2j^−3k^a=i^+2j^?−3k^ and b?=2i^+j^−k^b=2i^+j^?−k^. Then the vector r?r satisfying a?×r?=a?×b?a×r=a×b and a?⋅r?=0a⋅r=0 is of magnitude 1010?.
   Statement-II : In a triangle ABC, cos?2A+cos?2B+cos?2C≥−32cos2A+cos2B+cos2C≥−23?.

(1) Both Statement-I and Statement-II are incorrect
(2) Statement-I is incorrect but Statement-II is correct
(3) Both Statement-I and Statement-II are correct
(4) Statement-I is correct but Statement-II is incorrect

Ans. (2)

Sol. a?=i^+2j^−3k^a=i^+2j^?−3k^
a?=2i^+j^−k^a=2i^+j^?−k^
a?×r?=a?×b?;a?⋅r?=0a×r=a×b;a⋅r=0
⇒a?×(r?−b?)=0⇒a×(r−b)=0
⇒a?=λ(r?−b?)⇒a=λ(r−b)
a?⋅a?=λ(a?⋅r?−a?⋅b?)a⋅a=λ(a⋅r−a⋅b)
14=−7λ⇒λ=−214=−7λ⇒λ=−2
−a?2=r?−b?⇒r?=b?−a?2−2a?=r−b⇒r=b−2a?
= 2b?−a?2=3i^+k^222b−a?=23i^+k^?
Statement (I) is incorrect

cos?2A+cos?2B+cos?2C≥−32cos2A+cos2B+cos2C≥−23?
2A+2B+2C=2π2A+2B+2C=2π
cos?2A+cos?2B+cos?2Ccos2A+cos2B+cos2C
= −1−4cos?Acos?Bcos?C−1−4cosAcosBcosC
≥−1−4×12×12×12≥−1−4×21?×21?×21?
= −32−23?
Statement (II) is correct.

1. lim?x→π2(∫x3(π/2)3(sin?(2t1/3)+cos?(t1/3))dt(x−π2)2)limx→2π??(∫x3(π/2)3?(sin(2t1/3)+cos(t1/3))dt(x−2π?)2?) is equal to :

(1)9π28(2)11π210(3)3π22(4)5π29(4)(1)89π2?(3)23π2??(2)1011π2?(4)95π2??(4)

Ans. (1)

limx→π20−{sin?(2x)+cos?(x)}.3x22(x−π2)limx→2π??2(x−2π?)0−{sin(2x)+cos(x)}.3x2? =lim?x→π2−{2sin?xcos?x+cos?x}3x22(x−π2)=limx→2π??2(x−2π?)−{2sinxcosx+cosx}3x2?

 λ =lim?x→π2{2sin?xsin?(π2−x)sin?(π2−x)}2(x−π2)+sin?(π2−x)2(π2−x)}3x2=(1(1)+12)3(π2)2=9π28(1)? λ =limx→2π??2(x−2π?){2sinxsin(2π?−x)sin(2π?−x)}?+2(2π?−x)sin(2π?−x)?}3x2=(1(1)+21?)3(2π?)2=89π2??(1)

1. The sum of the coefficient of x2/3x2/3 and x−2/5x−2/5 in the binomial expansion of (x2/3+12x−2/5)9(x2/3+21?x−2/5)9 is :

(1) 21/4
(2) 69/16
(3) 63/16
(4) 19/4

Ans. (1)

Trr+1=9Cr(x2/3)9−r(x−2/52)rTrr+1?=9Cr?(x2/3)9−r(2x−2/5?)r =9Cr(12)r(r)(6−2r3−2r33)=9Cr?(21?)r(r)(36−32r?−32r??)

for coefficient of x2/3x2/3 put 6−2r3−2r5=236−32r?−52r?=32? ⇒r=5⇒r=5 Coefficient of x2/3x2/3 is =9C5(15)5=9C5?(51?)5 For coefficient of x−2/5x−2/5 put 6−2r3−2r5=−256−32r?−52r?=−52? ⇒r=6⇒r=6 Coefficient of x−2/5x−2/5 is 9C6(12)69C6?(21?)6 Sum=9C5(12)5+9C6(12)6=214Sum=9C5?(21?)5+9C6?(21?)6=421?

1. Let B=[1315]B=[11?35?] and A be a 2×22×2 matrix such that AB−1=A−1AB−1=A−1 . If BCB−1=ABCB−1=A and C4+αC2+βI=OC4+αC2+βI=O then 2β−α2β−α is equal to : (1) 16 (2) 2 (3) 8 (4) 10

Ans. (4)

Sol. BCB−1=ABCB−1=A ⇒(BCB−1)(BCB−1)=A.A⇒(BCB−1)(BCB−1)=A.A ⇒BCICB−1=A2⇒BCICB−1=A2 ⇒BC2B−1=A2⇒BC2B−1=A2 ⇒B−1(BC2B−1)B=B−1(A.A)B⇒B−1(BC2B−1)B=B−1(A.A)B From equation (1) C2=A−1.A.BC2=A−1.A.B C2=BC2=B Also AB−1=A−1AB−1=A−1 ⇒AB−1.A=A−1A=I⇒AB−1.A=A−1A=I ⇒A−1(AB−1A)=A−1I⇒A−1(AB−1A)=A−1I B−1A=A−1B−1A=A−1 Now characteristics equation of C2C2 is ?C2−λI?=0?C2?−λI?=0 ?B−λI?=0?B−λI?=0

1. If log?ey=3sin?−1xloge?y=3sin−1x , then (1−x)2y′′−xy′(1−x)2y′′−xy′ at x=12x=21? is equal to : (1) 9eπ/69eπ/6 (2) 3eπ/63eπ/6 (3) 3eπ/23eπ/2 (4) 9eπ/29eπ/2 Ans. (4) Sol. ln?(y)=3sin?−1xln(y)=3sin−1x 1y⋅y′=3(11−x2)y1?⋅y′=3(1−x2?1?) ⇒y′=3y1−x2atx=12⇒y′=1−x2?3y?atx=21? ⇒y′=33eπ632=23eπ2⇒y′=323?3e6π???=23?e2π? ⇒y′′=3(1−x2y′−y121−x2(−2x)(1−x2))⇒y′′=3((1−x2)1−x2?y′−y21−x2?1?(−2x)?) ⇒(1−x2)y′′=3(3y+xy1−x2)⇒(1−x2)y′′=3(3y+1−x2?xy?) ↓atx=12,y=e3sin?−1(12)=e(π6)=eπ2↓atx=21?,y=e3sin−1(21?)=e(6π?)=e2π?

(1−x2)y′′?atx=12=3(3eπ2+13)(1−x2)y′′?atx=21???=3(3e2π?+3?1?) =3eπ2(3+13)=3e2π?(3+3?1?) (1−x2)y′′−xy′?atx=12(1−x2)y′′−xy′?atx=21??? =3eπ2(3+13)−12(23eπ2)=9eπ2=3e2π?(3+3?1?)−21?(23?e2π?)=9e2π?

1. The integral ∫1/43/4cos?(2cot?−11−x1+x)dx∫1/43/4?cos(2cot−11+x1−x??)dx is equal to: (1) −1/2−1/2 (2) 1/41/4 (3) 1/21/2 (4) −1/4−1/4 Ans. (4)

Ans. (4)

I=∫1/43/4cos?(2cot?−1(1−x1+x)dx)I=∫1/43/4?cos(2cot−1(1+x1−x??)dx) ∫1/43/4cos?(2(tan?−11+x1+x))dx∫1/43/4?cos(2(tan−11+x1+x??))dx ∫1/43/41−tan?2(tan?−11+x1−x)dx∫1/43/4?1−tan2(tan−11−x1+x??)dx ∫1/43/41+tan?2(tan?−11+x1−x)dx∫1/43/4?1+tan2(tan−11−x1+x??)dx =∫1/43/41−(1+x1−x)(1+x1−x)dx=∫1/43/4−2x2dx=∫1/43/4?(1−x1+x?)1−(1−x1+x?)?dx=∫1/43/4?2−2x?dx =∫1/43/4(−x)dx=−(x22)1/43/4=∫1/43/4?(−x)dx=−(2x2?)1/43/4? =−12[916−116]=−21?[169?−161?] =−14=−41?

1. Let a, ar, ar², .......be an infinite G.P. If ∑n=0∞arn=57∑n=0∞?arn=57 and ∑n=0∞a3r3n=9747∑n=0∞?a3r3n=9747, then a + 18r is equal to : (1) 27 (2) 46 (3) 38 (4) 31 Ans. (4) Sol. ∑n=0∞arn=57∑n=0∞?arn=57 a + ar + ar² + ∞ = 57 a1−r=571−ra?=57 ............ (I) ∑n=0∞a3r3n=9747∑n=0∞?a3r3n=9747 a³ + a³·r³ + a³·r? + ..........∞ = 9746 a31−r3=97461−r3a3?=9746 ............ (II) (I)3(II)⇒a3(1−r)3a31−r3=5739717=19(II)(I)3?⇒1−r3a3?(1−r)3a3??=9717573?=19 On solving, r = 2332? and r = 3223? (rejected) a = 19 ∴ a + 18r = 19 + 18 × 2332? = 31

2. If an unbiased dice is rolled thrice, then the probability of getting a greater number in the i?? roll than the number obtained in the (i-1)?? roll, i = 2, 3, is equal to : (1) 3/54 (2) 2/54 (3) 5/54 (4) 1/54 Ans. (3)

Sol. Favourable cases = ?C? Total out comes = 6³ Probability of getting greater number than previous one = 6C3r3=20216=554r36C3??=21620?=545?

1. The value of the integral ∫−12log?e(x+x2+1)dx∫−12?loge?(x+x2+1?)dx is : (1) 5−2+log?e(9+451+2)5?−2?+loge?(1+2?9+45??) (2) 2−5+log?e(9+451+2)2?−5?+loge?(1+2?9+45??) (3) 5−2+log?e(7+451+2)5?−2?+loge?(1+2?7+45??) (4) 2−5+log?e(7+451+2)2?−5?+loge?(1+2?7+45??) Ans. (2) Sol. I = ∫−121.log?e(x+x2+1)dx∫−12?1.loge?(x+x2+1?)dx = xlog?e(x+x2+1)−∫−12(1+xx2+1x+x2+1)dxxloge?(x+x2+1?)−∫−12?(x+x2+1?1+x2+1?x??)dx = xlog?e(x+x2+1)−∫−12xx2+1dxxloge?(x+x2+1?)−∫−12?x2+1?x?dx = xlog?e(x+x2+1)−x2+1?−12xloge?(x+x2+1?)−x2+1??−12? = (2log?e(2+5)−5)−(−log?e(−1+2)−2)(2loge?(2+5?)−5?)−(−loge?(−1+2?)−2?) = log?e(2+5)2−5+log?e(2−1)+2loge?(2+5?)2−5?+loge?(2?−1)+2? = log?e(2+5)2−5+log?e(2−1)+2loge?(2+5?)2−5?+loge?(2?−1)+2? = 2−5+log?e((2+5)22+1)2?−5?+loge?(2?+1(2+5?)2?) = 2−5+log?e(9+452+1)2?−5?+loge?(2?+19+45??)

2. Let α, β; α > β, be the roots of the equation x² - √2x - √3 = 0. Let P? = α? - β?, n ∈ N. Then (11√3 - 10√2) P?? + (11√2 + 10) P?? - 11P?? is equal to : (1) 10√2P? (2) 10√3P? (3) 11√2P? (4) 11√3P? Ans. (2) Sol. x² - √2x - √3 = 0 (α, β) α??² - √2α??¹ - √3α? = 0 and β??² - √2β??¹ - √3β? = 0 Subtracting (α??² - β??²) - √2(α??¹ - β??¹) - √3(α? - β?) = 0 ⇒ P??? - √2P??? - √3P? = 0 Put n = 10 P?? - √2P?? - √3P?? = 0 n = 9 P?? - √2P?? - √3P? = 0 11(√3P?? + √2P?? - P??) - 10(√2P?? - P??) = 0 - 10(-√3P?) = 10√3P?

3. Let a?=2i^+αj^+k^a=2i^+αj^?+k^, b?=−i^+k^b=−i^+k^, c?=βj^−k^c=βj^?−k^, where α and β are integers and αβ = -6. Let the values of the ordered pair (α, β) for which the area of the parallelogram of diagonals a?+b?a+b and b?+c?b+c is 212221??, be (α?, β?) and (α?, β?). Then α?² + β?² - α?β? is equal to (1) 17 (2) 24 (3) 21 (4) 19 Ans. (4) Sol. Area of parallelogram = 12?d?1×d?2?21??d1?×d2?? A = 12?(a?+b?)×(b?+c?)?=21221??(a+b)×(b+c)?=221?? so, a?+b?=i^+αj^+2k^a+b=i^+αj^?+2k^ b?+c?=−i^+βj^b+c=−i^+βj^? (a?+b?)×(b?+c?)=?i^j^k^1α2−1β0?(a+b)×(b+c)=?i^1−1?j^?αβ?k^20?? = i^(−2β)−j^(2)+k^(β+α)i^(−2β)−j^?(2)+k^(β+α) ?(a?+b?)×(b?+c?)?=4β2+4+(α+β)2=21?(a+b)×(b+c)?=4β2+4+(α+β)2?=21? 4β² + 4 + α² + β² + 2αβ = 21 α² + 5β² - 12 = 17 α² + 5β² = 29 and αβ = -6 and given α, β are integers so, α = -3, β = 2 or α = 3, β = -2 (α?, β?) = (-3, 2) (α?, β?) = (3, -2) α?² + β?² - α?β? = 9 + 4 + 6 = 19

4. Consider the circle C : x² + y² = 4 and the parabola P : y² = 8x. If the set of all values of α for which three chords of the circle C on three distinct lines passing through the point (α, 0) are bisected by the parabola P is the interval (p, q), then (2q - p)² is equal to Ans. (80) Sol.

(x1,y1)(2t2,4t)(α, 0)(x1?,y1?)(2t2,4t)(α, 0)?

T = S?
xx? + yy? = x?² + y?²
a x? = x?² + y?³
a(2t²) = 4t? + 16t²
a = 2t² + 8
a−82=t22a−8?=t2
As o, 4t? + 16t² - 4 < 0
t² = -2 + √5
a = 4 + 2√5
∴ a ∈ (8, 4 + 2√5)
∴ (2q - p)² = 80

1. Let the set of all values of p, for which f(x) = (p² - 6p + 8)(sin²2x - cos²2x) + 2(2 - p)x + 7 does not have any critical point, be the interval (a, b). Then 16ab is equal to Ans. (252) Sol. f(x) = -(p² - 6p + 8)cos 4n + 2(2 - p)n + 7 f¹(x) = +4(p² - 6p + 8)sin 4x + (4 - 2p) ≠ 0 sin 4x ≠ 2p−44(p−4)(p−2)4(p−4)(p−2)2p−4? sin 4x ≠ 2(p−2)4(p−4)(p−2)4(p−4)(p−2)2(p−2)? p ≠ 2 sin 4x ≠ 12(p−4)2(p−4)1? ⇒ ?12(p−4)?>1?2(p−4)1??>1 on solving we get ∴ p ∈ (72,92)(27?,29?) Hence a = 7227?, b = 9229? ∴ 16ab = 252

2. For a differentiable function f : IR → IR, suppose f'(x) = 3f(x) + α, where α ∈ IR, f(0) = 1 and lim_{x→∞} f(x) = 7. Then 9f(-log?3) is equal to Ans. (61) Sol. dydx−3y=αdxdy?−3y=α IF = e^{∫ -3dx} = e^{-3x} ∴ y e^{-3x} = ∫ e^{-3x}·α dx y e^{-3x} = αe−3x−3+c−3αe−3x?+c (*e^{3x}) y = α−3+C⋅e3x−3α?+C⋅e3x on substituting x = 0, y = 1 x → -∞, y = 7 we get y = 7 - 6e^{3x} on substituting x = 0, y = 1 x → -∞, y = 7 we get y = 7 - 6e^{3x} 9f(-log?3) = 61

3. The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is ______. Ans. (70) Sol. N = a b c (i) All distinct digits a + b + c = 14 a ≥ 1 b, c ∈ {0 to 9} by hit & trial : 8 cases (6, 5, 3) (8, 6, 0) (9, 4, 1) (7, 6, 1) (8, 5, 1) (9, 3, 2) (7, 5, 2) (8, 4, 2) (7, 4, 3) (9, 5, 0) (ii) 2 same, 1 diff a = b ; c 2a + c = 14 by values : (3, 8) (4, 6) (5, 4) (6, 2) (7, 0) Total 3!2!×5−12!3!?×5−1 = 14 cases (iii) all same : 3a = 14 a = 143×314?× rejected 0 cases Hence, Total cases : 8 × 3! + 2 × (4) + 14 = 48 + 22 = 70

4. Let A = {(x, y) : 2x + 3y = 23, x, y ∈ N} and B = {x : (x, y) ∈ A}. Then the number of one-one functions from A to B is equal to ______. Ans. (24) Sol. 2x + 3y = 23 x = 1 y = 7 x = 4 y = 5 x = 7 y = 3 x = 10 y = 1 A B (1, 7) 1 (4, 5) 4 (7, 3) 7 (10, 1) 10 The number of one-one functions from A to B is equal to 4!

5. Let A, B and C be three points on the parabola y² = 6x and let the line segment AB meet the line L through C parallel to the x-axis at the point D. Let M and N respectively be the feet of the perpendiculars from A and B on L. Then (AM⋅BNCD)2(CDAM⋅BN?)2 is equal to ______. Ans. (36) Sol.

ABt1t2DNCML(at12,2at1)(at22,2at2)ABt1?t2?DNCML(at12?,2at1?)(at22?,2at2?)?

m_{AB} = m_{AD} ⇒ 2t1+t2=2a(t1−t3)at12−αt1?+t2?2?=at12?−α2a(t1?−t3?)? ⇒ at12−α=a[t12−t1t3+t1t2−t2t3]at12?−α=a[t12?−t1?t3?+t1?t2?−t2?t3?] ⇒ α = a(t_1t_3 + t_2t_3 - t_1t_2) AM = |2a(t_1 - t_3)|, BN = |2a(t_2 - t_3)|, CD = |at_3^2 - α|

CD = |at_3^2 - a(t_1t_3 + t_2t_3 - t_1t_2)| = a|t_3^2 - t_1t_3 - t_2t_3 + t_1t_2| = a|t_3(t_3 - t_1) - t_2(t_3 - t_1)| CD = a|(t_3 - t_2)(t_3 - t_1)| (AM⋅BNCD)2=[2a(t1−t3)⋅2a(t2−t3)a(t3−t2)(t3−t1)]2(CDAM⋅BN?)2=[a(t3?−t2?)(t3?−t1?)2a(t1?−t3?)⋅2a(t2?−t3?)?]2 16a² = 16 × 9449? = 36

1. The square of the distance of the image of the point (6, 1, 5) in the line x−13=y2=z−243x−1?=2y?=4z−2?, from the origin is ______. Ans. (62) Sol.

IMb?=3i^+2j^+4k^LA(6,1,5)IMb=3i^+2j^?+4k^LA(6,1,5)?

Let M(3λ + 1, 2λ, 4λ + 2)
AM?⋅b?=0AM⋅b=0
⇒ 9λ - 15 + 4λ - 2 + 16λ - 12 = 0
⇒ 29λ = 29
⇒ λ = 1
M (4, 2, 6), I = (2, 3, 7)
Required Distance = √(4 + 9 + 49) = √62
Ans. 62

1. If (1α+1+1α+2+...+1α+1012)−(12⋅1+14⋅3+16⋅5+...+12024⋅2023)=12024(α+11?+α+21?+...+α+10121?)−(2⋅11?+4⋅31?+6⋅51?+...+2024⋅20231?)=20241?, then α is equal to- Ans. (1011) Sol. (1α+1+1α+2+...+1α+2012)−{(11−12)+(13−14)+...+(12023−12024)}=12024(α+11?+α+21?+...+α+20121?)−{(11?−21?)+(31?−41?)+...+(20231?−20241?)}=20241? ⇒ (1α+1+1α+2+...+1α+2012)−{(11+12+13+14)+...+12023−12024−2(12+14+...+12022)}=12024(α+11?+α+21?+...+α+20121?)−{(11?+21?+31?+41?)+...+20231?−20241?−2(21?+41?+...+20221?)}=20241? ⇒ (1α+1+1α+2+...+1α+2012)−(11+12+...+12023)+12024+(11+12+...+11011)=12024(α+11?+α+21?+...+α+20121?)−(11?+21?+...+20231?)+20241?+(11?+21?+...+10111?)=20241? ⇒ 1α+1+1α+2+...+1α+2012=11012+11013+...+12023α+11?+α+21?+...+α+20121?=10121?+10131?+...+20231? ⇒ α = 1011

2. Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2sin?¹x + 3cos?¹x = 2π552π?, is ______. Ans. (0) Sol. 2sin?¹x + 3cos?¹x = 2π552π? ⇒ π + cos?¹x = 2π552π? ⇒ cos?¹x = −3π55−3π? Not possible Ans. 0

3. Consider the matrices : A = [2−53m][23?−5m?], B = [20m][20m?] and X = [xy][xy?]. Let the set of all m, for which the system of equations AX = B has a negative solution (i.e., x < 0 and y < 0), be the interval (a, b). Then 8∫{a}^{b}|A|dm is equal to Ans. (450) Sol. A = (2−53m)(23?−5m?), B = (20m)(20m?) X = (xy)(xy?) x - 5y = 20 3x + my = m ⇒ y = 2m−602m+152m+152m−60? y < 0 ⇒ m ∈ (−152,30)(2−15?,30) x = 25m2m+152m+1525m? x < 0 ⇒ m ∈ (−152,0)(2−15?,0) ⇒ m ∈ (−152,0)(2−15?,0) |A| = 2m + 15 Now, 8∫{-15/2}^{0}(2m + 15)dm = 8|m² + 15m|_{-15/2}^{0} ⇒ 8{-(\frac{225}{4} - \frac{225}{2})} = 8 × \frac{225}{4} = 450

4. A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2:1. After disintegration they will move :- (1) In opposite directions with speed in the ratio of 1:2 respectively (2) In opposite directions with speed in the ratio of 2:1 respectively (3) In the same direction with same speed. (4) In opposite directions with the same speed. Ans. (1) Sol. By conservation of momentum p? = p_f O = m?u? + m?u? u1u2=−[12]u2?u1??=−[21?] as m1m2=21m2?m1??=12? move in opposite direction with speed ratio 1 : 2

5. The following figure represents two biconvex lenses L? and L? having focal length 10 cm and 15 cm respectively. The distance between L? & L? is :

L1L2L1?L2??

(1) 10 cm (2) 15 cm (3) 25 cm (4) 35 cm Ans. (3) Sol.

f1f2f1?f2??

D = f? + f? = 25 cm Paraxial parallel rays pass through focus and ray from focus of convex lens will become parallel

TEST PAPER WITH SOLUTION

1. The temperature of a gas is -78°C and the average translational kinetic energy of its molecules is K. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2K is : (1) -39°C (2) 117°C (3) 127°C (4) -78°C Ans. (2) K.E = nf1RT22nf1?RT? T? = -78°C → 273 + [-78°C] = 195 K K.E α T To double the K.E energy temp also become double T_f = 390 K T_f = 117°C

2. A hydrogen atom in ground state is given an energy of 10.2 eV. How many spectral lines will be emitted due to transition of electrons ? (1) 6 (2) 3 (3) 10 (4) 1 Ans. (4) Sol. Hydrogen will be in first excited state therefore it will emit one spectral line corresponding to transition b/w energy level 2 to 1

3. The magnetic field in a plane electromagnetic wave is B_y = (3.5 × 10??) sin (1.5 × 10³x + 0.5 × 10¹¹t) T. The corresponding electric field will be E_y = 1.17 sin (1.5 × 10³x + 0.5 × 10¹¹t) Vm?¹ E_z = 105 sin (1.5 × 10³x + 0.5 × 10¹¹t) Vm?¹ E_z = 1.17 sin (1.5 × 10³x + 0.5 × 10¹¹t) Vm?¹ E_y = 10.5 sin (1.5 × 10³x + 0.5 × 10¹¹t) Vm?¹ Ans. (2) Sol. E? = B?C E? = 3 × 10? × (35 × 10??) sin (1.5 × 10³x + 0.5 × 10¹¹t) E? = 105 sin (1.5 × 10³x + 0.5 × 10¹¹t) Vm?¹ Data inconsistent while calculating speed of wave. You can challenge for data.

4. A square loop of side 15 cm being moved towards right at a constant speed of 2 cm/s as shown in figure. The front edge enters the 50 cm wide magnetic field at t = 0. The value of induced emf in the loop at t = 10 s will be :

2cm/sB=1.0T15cm50cm2cm/sB=1.0T15cm50cm?

(1) 0.3 mV (2) 4.5 mV (3) zero (4) 3 mV Ans. (3) Sol. At t = 10 sec complete loop is in magnetic field therefore no change in flux

B?vB?vB?vB?v?

e = dφdtdtdφ? = 0 e = 0 for complete loop

1. Two cars are travelling towards each other at speed of 20 m s?¹ each. When the cars are 300 m apart, both the drivers apply brakes and the cars retard at the rate of 2 m s?². The distance between them when they come to rest is: (1) 200 m (2) 50 m (3) 100 m (4) 25 m Ans. (3)

AB20 m/s20 m/s300 mAB20 m/s20 m/s300 m?

?u¨BA?=40m/s?u¨BA??=40m/s
?a¨BA?=4m/s?a¨BA??=4m/s
Apply (v² = u² + 2as)_{relative}
O = (40)² + 2(-4)(S)
S = 200 m
Remaining distance = 300 - 200 = 100 m

1. The I - V characteristics of an electronic device shown in the figure. The device is :

IV(volt)5(μA)IV(volt)5(μA)?

(1) a solar cell (2) a transistor which can be used as an amplifier (3) a zener diode which can be used as voltage regulator (4) a diode which can be used as a rectifier Ans. (3) Sol. Theory Zener diode used as voltage regulator

1. The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is: (1) 1:9 (2) 1:3 (3) 1:81 (4) 1:27 Ans. (4)

P1P2r1r2P1?P2?r1?r2??

P? - P? = 4Tr1r1?4T? P? - P? = 4Tr2r2?4T? P? - P? = 3(P? - P?) 4Tr1=34Tr2r1?4T?=3r2?4T? r? = 3r? V1V2=43πr1343πr23=127V2?V1??=34?πr23?34?πr13??=271?

1. The de-Broglie wavelength associated with a particle of mass m and energy E is h/√(2mE). The dimensional formula for Planck's constant is : (1) [ML?¹T?²] (2) [ML?²T?¹] (3) [ML?²] (4) [M²L?²T?²] Ans. (2) Sol. λ = h2mE2mE?h? or E = hv [ML²T?²] = h[T?¹] h = [ML²T?¹]

2. A satellite of 10³ kg mass is revolving in circular orbit of radius 2R. If 104R6J6104R?J energy is supplied to the satellite, it would revolve in a new circular orbit of radius : (use g = 10 m/s², R = radius of earth) (1) 2.5 R (2) 3 R (3) 4 R (4) 6 R Ans. (4)

mMR2RmMR2R?

Total energy = −Gm2(2R)2(2R)−Gm? if energy = 104R66104R? is added then −Gm4R+104R6=−Gm2r4R−Gm?+6104R?=2r−Gm? where r is new radius of revolving and g = GMR2R2GM? −mgR4+104R6=−mgR22r4−mgR?+6104R?=2r−mgR2? (m = 10³ kg) 103×10×R4+104R6=103×10×R22r4103×10×R?+6104R?=2r103×10×R2? −14+16=−R2r−41?+61?=−2rR? r = 6R

1. The effective resistance between A and B, if resistance of each resistor is R, will be

RRABRRRRRRABRRRR?

(1) 23R32?R (2) 8R338R? (3) 5R335R? (4) 4R334R? Ans. (2) Sol. From symmetry we can remove two middle resistance. New circuit is

RRABRRRR⇒2RAB2R2R⇒AB2R/3⇒AB8R/3RRABRRRR?⇒2RAB2R2R?⇒AB2R/3?⇒AB8R/3?

1. Five charges +q, +5q, -2q, +3q and -4q are situated as shown in the figure. The electric flux due to this configuration through the surface S is :

+5q-4qSq-2q+3q+5q-4qSq-2q+3q?

(1) 5qε0ε0?5q? (2) 4qε0ε0?4q? Ans. (2) Sol. As per gauss theorem, φ = qinε0=q+(−2q)+5qε0=4qε0ε0?qin??=ε0?q+(−2q)+5q?=ε0?4q?

1. A proton and a deuteron q = +e, m = 2.0u having same kinetic energies enter a region of uniform magnetic field B, moving perpendicular to B. The ratio of the radius r_d of deuteron path to the radius r_p of the proton path is : (1) 1:1 (2) 1:√2 (3) √2:1 (4) 1:2 Ans. (3) Sol. In uniform magnetic field, R = mvqB=2m(K.E)qBqBmv?=qB2m(K.E)?? Since same K.E R ∝ mqqm?? ∴ RdeutronRproton=mdmp×qpqdRproton?Rdeutron??=mp?md???×qd?qp?? = √2 × 1 ∴ γ_d : γ_p = √2 : 1

2. UV light of 4.13 eV is incident on a photosensitive metal surface having work function 3.13 eV. The maximum kinetic energy of ejected photoelectrons will be : (1) 4.13 eV (2) 1 eV (3) 3.13 eV (4) 7.26 eV Ans. (2) Sol. E_{photon} = (work function) + K.E_{max} ∴ 4.13 = 3.13 + K.E_{max} ∴ K.E_{max} = 1 eV

3. The energy released in the fusion of 2 kg of hydrogen deep in the sun is E_H and the energy released in the fission of 2 kg of ²³?U is E_U. The ratio E_H/E_U is approximately : (Consider the fusion reaction as 4?¹H + 2e? → ??He + 2v + 6y + 26.7 MeV, energy released in the fission reaction of ²³?U is 200 MeV per fission nucleus and N_A = 6.023 × 10²³) (1) 9.13 (2) 15.04 (3) 7.62 (4) 25.6 Ans. (3) Sol. In each fusion reaction, 4?¹H nucleus are used. Energy released per Nuclei of ?¹H = 26.74426.7? MeV Energy released by 2 kg hydrogen (E_H) = 20001×NA×26.7412000?×NA?×426.7? MeV & Energy released by 2 kg Vranium (E_v) = 2000235×NA×2002352000?×NA?×200 MeV So, EHEv=235×26.74×200=7.84Ev?EH??=235×4×20026.7?=7.84 Approximately close to 7.62

4. A real gas within a closed chamber at 27°C undergoes the cyclic process as shown in figure. The gas obeys PV³ = RT equation for the path A to B. The net work done in the complete cycle is (assuming R = 8 J/mol/K):

P(N/m2)20A10CBV(m3)24P(N/m2)20A10CBV(m3)24?

(1) 225 J (2) 205 J (3) 20 J (4) -20 J Ans. (2) Sol. W_{AB} = ∫PdV (Assuming T to be constant) = RTdV = RT∫??V?³dV = 8 × 300 × (-1221?[142−122421?−221?]) = 225 J W_{BC} = P∫?²dV = 10(2 - 4) = -20 J W_{CA} = 0 ∴ W_{cycle} = 205 J Note : Data is inconsistent in process AB. So needs to be challenged.

1. A 1 kg mass is suspended from the ceiling by a rope of length 4m. A horizontal force 'F' is applied at the mid point of the rope so that the rope makes an angle of 45° with respect to the vertical axis as shown in figure. The magnitude of F is :

T1θ=45°FT21 KgT1?θ=45°FT2?1 Kg?

(1) 1022?10? N (2) 1 N (3) 110×210×2?1? N (4) 10 N Ans. (4) Sol. T? sin 45° = F T? cos 45° = T? = 1 × g ∴ tan 45° = FggF? ∴ F = 10 N

1. A spherical ball of radius 1 × 10?? m and density 10? kg/m³ falls freely under gravity through a distance h before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of h is approximately : (The coefficient of viscosity of water is 9.8 × 10?? N s/m²) (1) 2296 m (2) 2249 m (3) 2518 m (4) 2396 m Ans. (3)

2. V_T = 2g9R2[ρB−ρL]η92g?ηR2[ρB?−ρL?]? ⇒ V_T = 29×10×(10−4)29.8×10−6[105−103]92?×9.8×10−610×(10−4)2?[105−103] ⇒ V_T = 224.5 when ball fall from height (h) V = √(2gh) h = (V22g2gV2?) = 2518 m

3.  

ABEA|B|E0|0|00|1|X1|0|Y1|1|0ABEA|B|E0|0|00|1|X1|0|Y1|1|0?

In the truth table of the above circuit the value of X and Y are : (1) 1, 1 (2) 1, 0 (3) 0, 1 (4) 0, 0 Ans. (1) Sol. For x

0=A1=B01x=10=A1=B01x=1?

For y

1=A0=B01101=A0=B0110?

SECTION-B

1. A straight magnetic strip has a magnetic moment of 44 Am². If the strip is bent in a semicircular shape, its magnetic moment will be ........... Am² (Given π = 227722?) Ans. (28) Sol. Magnetic moment of straight wire = mx? = 44

?⇒R?=πR?⇒R?=πR?

Magnetic moment of arc = m × 2r = m × 2?ππ2?? = 44×2π=88ππ44×2?=π88? = 28

1. A particle of mass 0.50 kg executes simple harmonic motion under force F = -50(Nm?¹)x. The time period of oscillation is x3535x? s. The value of x is Ans. (22) Sol. m = 0.5 kg F = -50(x) ma = (-50x) 0.5a = -50x a = (-100x) W² = 100 ⇒ (w = 10) T = 2π10=(π5)=227×15=(2235)102π?=(5π?)=7×1522?=(3522?) π35=2235⇒[x=22]35π?=3522?⇒[x=22]

2. A capacitor of reactance 4√3 Ω and a resistor of resistance 4 Ω are connected in series with an ac source of peak value 8√2 V. The power dissipation in the circuit is W. Ans. (4)

XC=43ΩR=4ΩSol.XC?=43?ΩR=4ΩSol.?

Z = √(R² + X²_L) Z = √(4² + (4√3)²) = 8Ω V_{ms} = V2=822=(8V)2?V?=2?82??=(8V) I_{ms} = VmsZ=88=1AZVms??=88?=1A Power dissipated = I_{ms}² × R = 1 × 4 = (4W)

1. An electric field E = (2xi) NC?¹ exists in space. A cube of side 2 m is placed in the space as per figure given below. The electric flux through the cube is nm²/C.

Y2m0Z2mXY2m0Z2mX?

Ans. (16) Sol.

yE=4i^E=8i^(0,0)x=2x=4E=2xi^?=E.AyE=4i^E=8i^(0,0)x=2x=4E=2xi^?=E.A?

φ_{in} = -4 × 4 = -16 Nm²/c φ_{out} = 8 × 4 = 32 Nm²/c d_{net} = φ_{in} + φ_{out} = -16 + 32 = 16 Nm²/c

1. A circular disc reaches from top to bottom of an inclined plane of length l. When it slips down the plane, if takes t s. When it rolls down the plane then it takes (α22α?)^{1/2} t s, where α is ... Ans. (3) Sol. For slipping a = gsinθ ? = 12at2⇒t=2?gsinθ21?at2⇒t=gsinθ2??? For rolling a' = gsinθ1+k2R21+R2k2?gsinθ? [k = R22?R?] ⇒ a' = 2gsinθ332gsinθ? ? = 12a′(t′)221?a′(t′)2 ⇒ t' = 6?2gsinθ=α22?gsinθ2gsinθ6???=2α??gsinθ2??? ⇒ |α = 3|

2. To determine the resistance (R) of a wire, a circuit is designed below, The V-1 characteristic curve for this circuit is plotted for the voltmeter and the ammeter readings as shown in figure. The value of R is ... Ω.

V10 kΩRmAEV10 kΩRmAE?I(mA)4320468V(volt)I(mA)4320468V(volt)?

Ans. (2500) Req = 104R104+R104+R104R? E = 4V, I = 2mA I = EReq⇒2×10−3=4(104+R)104RReqE?⇒2×10−3=104R4(104+R)? ⇒ 20R = 40000 + 4R 16R = 40000 R = 2500Ω

1. The resultant of two vectors A and B is perpendicular to A and its magnitude is half that of B. The angle between vectors A and B is Ans. (150) Sol.

R=B/2θBAR=B/2θBA?

Bcosθ = B22B? ⇒ θ = 60° So, angle between A and B is 90° + 60° = 150°

1. Monochromatic light of wavelength 500 nm is used in Young's double slit experiment. An interference pattern is obtained on a screen When one of the slits is covered with a very thin glass plate (refractive index = 1.5), the central maximum is shifted to a position previously occupied by the 4th bright fringe. The thickness of the glass-plate is mm. Ans. (4) Sol. (μ - 1)t = nλ (1.5 - 1)t = 4 × 500 × 10?? m t = 4000 × 10?? m t = 4 μm

2. A force (3x² + 2x - 5) N displaces a body from x = 2 m to x = 4 m. Work done by this force is mm. Ans. (58) Sol. W = ∫{x?}^{x?}Fdx W = ∫{1/2}^{4/2}(3x² + 2x - 5)dx W = [x³ + x² - 5x]?? W = [60 - 2]J = 58J

3. At room temperature (27°C), the resistance of a heating element is 50Ω. The temperature coefficient of the material is 2.4 × 10?? °C?¹. The temperature of the element, when its resistance is 62Ω, is °C. Ans. (1027) Sol. R = R?(1 + αΔT) 62 = 50[1 + 2.4 × 10??ΔT] ΔT = 1000°C ⇒ T - 27° = 1000°C T = 1027°C

SECTION-A

1. The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 'A' × 10¹² hertz and that has a radiant intensity in that direction of 1B′B′1? watt per steradian. 'A' and 'B' are respectively (1) 540 and 16836831? (2) 540 and 683 (3) 450 and 16836831? (4) 450 and 683 Ans. (2) Sol. The candela is the luminous intensity of a source that emits monochromatic radiation of frequency radiation of frequency 540 × 10¹² Hz and has a radiant intensity in that direction of 16836831? w/sr. It is unit of Candela.

2. The correct stability order of the following resonance structures of CH?-CH=CH-CHO is

CH3−CH−CH=C−H↔CH3−CH−CH=C−HIII↔CH3−CH=CH−C−HIIICH3?−CH−CH=C−H↔CH3?−CH−CH=C−HIII↔CH3?−CH=CH−C−HIII?

(1) II > III > I (2) III > II > I (3) I > II > III (4) II > I > III Ans. (2) Sol. CH?-CH=CH-CH (III) Non Polar R.S. More No of covalent bond CH?-CH-CH=CH (II) Having -ve charge on more electronegative atom CH?-CH-CH=CH (I) Having -ve charge on less electronegative atom Stability order III > II > I

1. Total number of stereo isomers possible for the given structure:

BrBrBrBr?

(1) 8 (2) 2 (3) 4 (4) 3 Ans. (1)

BrBrBrBr?

There are three stereo center So No of stereoisomer = 2³ = 8

1. The correct increasing order for bond angles among BF?, PF? and ClF? is : (1) PF? < BF? < ClF? (2) BF? < PF? < ClF? (3) ClF? < PF? < BF? (4) BF? = PF? < ClF? Ans. (3)

2. Match List I with List II

FB120°FFClFF87.5°FPF97°FFB120°FFClFF87.5°FPF97°F?

Order of bond angle is ClF? < PF? < BF?

1. Match List I with List II

<table><td colspan="2">LIST-I (Test)<td colspan="2">LIST-II (Observation)A.Br2 water testI.Yellow orange or orange red precipitate formedB.Ceric ammonium nitrate testII.Reddish orange colour disappearsC.Ferric chloride testIII.Red colour appearsD.2, 4-DNP testIV.Blue, Green, Violet or Red colour appear</table>

Choose the correct answer from the options given below: (1) A-I, B-II, C-III, D-IV (2) A-II, B-III, C-IV, D-I (3) A-III, B-IV, C-I, D-II (4) A-IV, B-I, C-II, D-III Ans. (2) Sol. (A) Br? water test is test of unsaturation in which reddish orange colour of bromine water disappears. (B) Alcohols given Red colour with ceric ammonium nitrate. (C) Phenol gives Violet colour with natural ferric chloride. (D) Aldehyde & Ketone give Yellow/Orange/Red Colour compounds with 2, 4-DNP i.e., 2, 4-Dinitrophenyl hydrazine.

<table><td colspan="2">LIST-I (Cell)<td colspan="2">LIST-II (Use/Property/Reaction)A.Leclanche cellI.Converts energy of combustion into electrical energyB.Ni-Cd cellII.Does not involve any ion in solution and is used in hearing aidsC.Fuel cellIII.RechargeableD.Mercury cellIV.Reaction at anode Zn→Zn2++2e-</table>

Choose the correct answer from the options given below: (1) A-I, B-II, C-III, D-IV (2) A-III, B-I, C-IV, D-II (3) A-IV, B-III, C-I, D-II (4) A-II, B-III, C-IV, D-I Ans. (3) Sol. A-IV, B-III, C-I, D-II

1. Match List I with List II

<table><td colspan="2">LIST-I<td colspan="2">LIST-IIA.K2[Ni(CN)4]I.sp3B.[Ni(CO)4]II.sp3d2C.[Co(NH3)6]Cl3III.dsp2D.Na3[CoF6]IV.d2sp3</table>

Choose the correct answer from the options given below: (1) A-III, B-I, C-II, D-IV (2) A-III, B-II, C-IV, D-I (3) A-I, B-III, C-II, D-IV (4) A-III, B-I, C-IV, D-II Ans. (4)

1. The incorrect statement about Glucose is : (1) Glucose is soluble in water because of having aldehyde functional group (2) Glucose remains in multiple isomeric form in its aqueous solution (3) Glucose is an aldehyde (4) Glucose is one of the monomer unit in sucrose Ans. (1) Sol. Glucose is soluble in water due to presence of alcohol functional group and extensive hydrogen bonding. Glucose exist is open chain as well as cyclic forms in its aqueous solution. Glucose having 6C atoms so it is hexose and having aldehyde functional group so it is aldose. Thus, aldehyde. Glucose is monomer unit in sucrose with fructose.

2. In the above reaction product 'P' is Ans. (1) Sol. (A) K?[Ni(CN)?] Ni²? : [Ar]3d?4s?, (CN? is S.F.L) Pre hybridization state of Ni²?

3d4s4pdsp23d4s4pdsp2?

(B) [Ni(CO)4] Ni : [Ar] 3d?4s² CO is S.F.L, so pairing occur Pre hybridization state of Ni

3d4s4psp33d4s4psp3?

(C) [Co(NH?)?]Cl? Co³? : [Ar]3d?4s? With Co³? NH? act as S.F.L

3d4s4pd2sp33d4s4pd2sp3?

(d) Na?[CoF?] Co³? : [Ar]3d?(F? : W.F.L)

3d4s4p4dsp3d23d4s4p4dsp3d2?

1. The coordination environment of Ca²? ion in its complex with EDTA?? is : (1) tetrahedral (2) octahedral (3) square planar (4) trigonal prismatic Ans. (2) Sol. EDTA?? → Hexadentate ligand [Ca(EDTA)]²? So Coordination environment is octahedral

2.  

KCN(alc)Δ⇒Major Product ’P’KCN(alc)Δ⇒Major Product ’P’?CNOCH3CNOCH3??

(1)

CNOCH3CNOCH3??

(2)

OCH3CNOCH3?CN?

(3)

CNOCH3CNOCH3??

(4)

CNOCH3CNOCH3??

Ans. (1) Sol.

BrOCH3⇒CNBrOCH3?⇒CN?

Due to NGP effect of phenyl ring Nucleophilic substitution of Br will occurs.

1. Which of the following compound can give positive iodoform test when treated with aqueous KOH solution followed by potassium hypoxiodite.

O||(1) CH3CH2-C-CH2CH3Cl|(2) CH3CH2-C-CH3|Cl(3) CH3CH2CH2CHOO/(4) CH3CH2-CH-CH2O||(1) CH3?CH2?-C-CH2?CH3?Cl|(2) CH3?CH2?-C-CH3?|Cl(3) CH3?CH2?CH2?CHOO/(4) CH3?CH2?-CH-CH2??

Ans. (2) Sol.

Cl|CH3-CH2-C-CH3→aq. KOHCH3-CH2-C-CH3|ClOH-H2O↓O||CH3-CH2-C-CH3KOI↓CH3-CH2-COOK+CHI3↓Yellow pptCl|CH3?-CH2?-C-CH3?aq. KOH?CH3?-CH2?-C-CH3?|ClOH-H2?O↓O||CH3?-CH2?-C-CH3?KOI↓CH3?-CH2?-COOK+CHI3?↓Yellow ppt?

1. For a sparingly soluble salt AB? the equilibrium concentrations of A²? ions and B? ions are 1.2 × 10?? M and 0.24 × 10?³ M respectively. The solubility product of AB? is : (1) 0.069 × 10?¹² (2) 6.91 × 10?¹² (3) 0.276 × 10?¹² (4) 27.65 × 10?¹² Ans. (2) Sol. AB?(s) ? A²?(aq) + 2B?(aq) K_{sp} = [A?²][B?]² = 1.2 × 10?? × (2.4 × 10??)² = 6.91 × 10?¹² M³

2. Major product of the following reaction is

CNCO2CH3(i) CH3MgBr(excess)(ii) H3O+CN(1) HO-C-CH3|CH3O||(2) C-CH3HO-C-CH3|CH3O||(3) C-CH3CO2CH3CN(4) C-CH3OCNCO2?CH3?(i) CH3?MgBr(excess)(ii) H3?O+CN(1) HO-C-CH3?|CH3?O||(2) C-CH3?HO-C-CH3?|CH3?O||(3) C-CH3?CO2?CH3?CN(4) C-CH3?O?

Ans. (2) Sol.

C≡NCH3MgBr(excess)BrMgO-C-CH3|CH3H3O+↓O=C-CH3HO-C-CH3|CH3C≡NCH3?MgBr(excess)BrMgO-C-CH3?|CH3?H3?O+↓O=C-CH3?HO-C-CH3?|CH3??

1. Given below are two statements : Statement I : The higher oxidation states are more stable down the group among transition elements unlike p-block elements. Statement II : Copper can not liberate hydrogen from weak acids. 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) Both Statement I and Statement II are true (4) Statement I is true but Statement II is false Ans. (3) Sol. On moving down the group in transition elements, stability of higher oxidation state increases, due to increase in effective nuclear charge. ⇒ E°{Cu²?/Cu} = 0.34V ⇒ E°{H?/H?} = 0 SRP : Cu²? > H? Cu can't liberate hydrogen gas from weak acid.

2. The incorrect statement regarding ethyne is (1) The C-C bonds in ethyne is shorter than that in ethene (2) Both carbons are sp hybridised (3) Ethyne is linear (4) The carbon-carbon bonds in ethyne is weaker than that in ethene Ans. (4) Sol. The carbon-carbon bonds in ethyne is stronger than that in ethene. (H-C≡C-H) Ethyne is linear and carbon atoms are SP hybridised.

3. Match List I with List II

<table>List-I (Element)<td colspan="3">List-II (Electronic Configuration)A.NI.[Ar] 3d10 4s2 4p5B.SII.[Ne] 3s2 3p4C.BrIII.[He] 2s2 2p3DKrIV.[Ar] 3d10 4s2 4p6</table>

Choose the correct answer from the options given below : (1) A-IV, B-III, C-II, D-I (2) A-III, B-II, C-I, D-IV (3) A-I, B-IV, C-III, D-II (4) A-II, B-I, C-IV, D-III Ans. (2) Sol. (A) N : He[2s²2p³] (B) ??S : Ne[2s²3p?] (C) ??Br : Ar[3d¹?4s²4p?] (D) ??Kr : Ar[3d¹?4s²4p?]

1. Match List I with List II

<table>List-I<td colspan="2">List-IIA.Melting point [K]I. T1 &gt; In &gt; Ga &gt; Al &gt; BB.Ionic Radius [M+3/pm]II. B &gt; T1 &gt; Al &gt; Ga &gt; InC.ΔH1 [kJ mol-1]III. T1 &gt; In &gt; Al &gt; Ga &gt; BDAtomic Radius [pm]IV. B &gt; Al &gt; T1 &gt; In &gt; Ga</table>

Choose the correct answer from the options given below : (1) A-III, B-IV, C-I, D-II (2) A-II, B-III, C-IV, D-I (3) A-IV, B-I, C-II, D-III (4) A-I, B-II, C-III, D-IV Ans. (3)

Sol. Melting point : B > Al > Tl > In > Ga Ionic radius (M³?/pm) : Tl > In > Ga > Al > B (Δ_{IE}H) [kJ/mol] : B > Tl > Al ≈ Ga > In Atomic radius (in pm) : Tl > In > Al > Ga > B

1. Which of the following compounds will give silver mirror with ammoniacal silver nitrate? (A) Formic acid (B) Formaldehyde (C) Benzaldehyde (D) Acetone Choose the correct answer from the options given below : (1) C and D only (2) A, B and C only (3) A only (4) B and C only Ans. (2) Sol. Apart from aldehyde, Formic acid also gives silver mirror test with ammoniacal silver nitrate.

2. Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning : (1) Λ²_m C - K_a Λ²_m + K_a Λ_m Λ°_m = 0 (2) Λ_m - Λ°_m + AC^{1/2} = 0 (3) Λ_m - Λ°_m - AC^{1/2} = 0 (4) Λ²_m C + K_a Λ²_m - K_a Λ_m Λ°_m = 0 Ans. (1) Sol. HA(aq) ? H?(aq) + A?(aq) K_a = α2C1−α1−αα2C? α²C + K_aα - K_a = 0 (λmλ°mλ°m?λm??)² C + K_a λmλ°mλ°m?λm?? - K_a = 0 λ²_m C + K_aλ_mλ°_m - K_a(λ°_m)² = 0

3. The electronic configuration of Einsteinium is : (Given atomic number of Einsteinium = 99) (1) [Rn] 5f¹² 6d? 7s² (2) [Rn] 5f¹¹ 6d? 7s² (3) [Rn] 5f¹³ 6d? 7s² (4) [Rn] 5f¹? 6d? 7s² Ans. (2) Sol. Einsteinium (atomic No = 99) : [Rn] 5f¹¹ 6d? 7s²

SECTION-B

1. Number of oxygen atoms present in chemical formula of fuming sulphuric acid is ______. Ans. (7) Sol. Fuming sulphuric acid is a mixture of conc. H?SO? + SO? Or H?S?O? So, Number of Oxygen atoms = 7

2. A transition metal 'M' among Sc, Ti, V, Cr, Mn and Fe has the highest second ionisation enthalpy. The spin only magnetic moment value of M? ion is ______ BM (Near integer) (Given atomic number Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26) Ans. (6) Sol. Among given metals, Cr has maximum IE? because Second electron is removed from stable configuration 3d? Cr? : [Ar] 3d? 4s? ∴ No. of unpaired e? in Cr? is 5, n = 5 So, Magnetic moment = √(n(n + 2)) B.M = √(5(5 + 2)) = 5.92 BM ≈ 6

3. The vapour pressure of pure benzene and methyl benzene at 27°C is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapour phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _ × 10?² (nearest integer) Ans. (23) Sol. X_{methylbenzene} = 0.5 Y_{methylbenzene} = PmethylbenzenePtotalPtotal?Pmethylbenzene?? Y_{methylbenzene} = 0.5×240.5×80+0.5×240.5×80+0.5×240.5×24? = 1240+12=0.23=23×10−240+1212?=0.23=23×10−2

4. Consider the following test for a group-IV cation. M²? + H?S → A (Black precipitate) + byproduct A + aquaregia → B + NOCl + S + H?O B + KNO? + CH?COOH → C + byproduct The spin only magnetic moment value of the metal complex C is BM. (Nearest integer) Ans. (0) Sol. Co²? + H?S → CoS↓ (Black) (A) CoS + Aqua-regia → Co²?(aq) + NOCl + S + H?O (A) (B) Co²?(aq) + KNO? + CH?COOH ↓ K?[Co(NO?)?] + NO + S + H?O In K?[Co(NO?)?], Co³? : 3d? 4s? Co³? : d²sp³ Hybridisation Number of unpaired e? = 0 Magnetic moment = √(n(n + 2)) = 0 B.M

5. Consider the following first order gas phase reaction at constant temperature A(g) → 2B(g) + C(g) If the total pressure of the gases is found to be 200 torr after 23 sec. and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is _ × 10?² s?¹ (nearest integer) [Given : log??(2) = 0.301] Ans. (3) Sol. A(g) → 2B(g) + C(g) P?? = P? + 2x = 200 P_e = 3P? = 300 P? = 100 K = 1tln?Pe−P0Pe−Ptt1?lnPe?−Pt?Pe?−P0?? K = 2.323log?300−100300−200232.3?log300−200300−100? = 2.3×0.30123=0.0301=3.01×10−2sec−1232.3×0.301?=0.0301=3.01×10−2sec−1

6.  

Top1cmSolvent front10cmBA1cmBottomTop1cmSolvent front10cmBA1cmBottom?

In the given TLC, the distance of spot A & B are 5 cm & 7 cm, from the bottom of TLC plate, respectively. R_f value of B is x × 10?¹ times more than A. The value of x is Ans. (15)

1. Sol. Compounds which can not undergo Friedel Crafts reaction are Sol. R_f = Distance moved by substance from base line / Distance moved by solvent from base line

Top1cmSolvent front10cm6cmBA4cm1cmBottomBase lineTop1cmSolvent front10cm6cmBA4cm1cmBottomBase line?

(R_f)_A = 4884? (R_f)_B = 6886? (R_f)_B = 68×8486?×48? (R_f)_B = 1.5(R_f)_A x = 15

1. Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10?¹? m is × 10? ms?¹ (nearest integer) [Given : mass of electron = 9.1 × 10?³¹ kg, Plank's constant (h) = 6.626 × 10?³? Js] (Value of π = 3.14) Ans. (58) Sol. ΔV.Δx = h4π4πh? ΔV = 6.626×10−349.1×10−31×10−15×4×3.149.1×10−31×10−15×4×3.146.626×10−34? = 57.97 × 10?? m/sec

2. Number of compounds from the following which cannot undergo Friedel-Crafts reactions is : toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, m-nitroaniline, m-dinitrobenzene Ans. (4)

NO2NH2NH2NO2Nitrobenzene Aniline m-nitroaniline m-dinitrobenzeneNO2?NH2?NH2?NO2?Nitrobenzene Aniline m-nitroaniline m-dinitrobenzene?

1. Total number of electron present in (π) molecular orbitals of O?, O?? and O?? is Ans. (6) O?(16e) : (σ_{1s})²(σ{1s})²(σ{2s})²(σ{2s})² (σ{2p})²[(π_{2p})²=(π_{2p})²][(π{2p})¹=(π*{2p})¹] Number of e? present in (π) of O? = 2 Number of e? present in (π) of O?? = 1 Number of e? present in (π) of O?? = 3 So total e? in (π) = 2 + 1 + 3 = 6

2. When ΔH_{vap} = 30 kJ/mol and ΔS_{vap} = 75 Jmol?¹K?¹ then the temperature of vapour, at one atmosphere is K.

Ans. (400) Sol. At equilibrium ΔG_{PT} = 0 ΔH_{vap} = TΔS_{vap} 30 × 1000 = T × 75 T = 400 K

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