NDA Previous Year Question Paper 2024 Solved

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HINTS & SOLUTIONS

MATHEMATICS

1. (d) Given ∣A∣=122,∣B∣=1729∣A∣=22​1​,∣B∣=7291​
∣2B(adj(3A))∣=∣2B∣∣adj(3A)∣∣2B(adj(3A))∣=∣2B∣∣adj(3A)∣
We know that ∣adjA∣=∣A∣n−1∣adjA∣=∣A∣n−1 and ∣kA∣=kn∣A∣∣kA∣=kn∣A∣
=23∣B∣∣3A∣3−1=8∣B∣(32∣A∣)2=23∣B∣∣3A∣3−1=8∣B∣(32∣A∣)2
=8∣B∣36∣A∣2=8⋅1729⋅36⋅(122)2=8⋅18=1=8∣B∣36∣A∣2=8⋅7291​⋅36⋅(22​1​)2=8⋅81​=1

2. (b) Since iz3+z2−z+i=0iz3+z2−z+i=0 put z=iz=i
i4+i2−i+i=0⇒1−1−i+i=0i4+i2−i+i=0⇒1−1−i+i=0
Satisfied then, (∣z∣+1)2=[∣i∣+1]2=4(∣z∣+1)2=[∣i∣+1]2=4

3. (d) Number of four digit numbers begins with 1.
=3×2×1=6=3×2×1=6 and repetition of digit =63=2=36​=2
Digit total for each place =(0+4+5)×2=18=(0+4+5)×2=18
Total of all four-digit numbers beginning with 1
=1×6×1000+18×100+18×10+18=7998=1×6×1000+18×100+18×10+18=7998
Number of four-digit numbers begins with 4.
=3×2×1=6=3×2×1=6 and repetition of digit =63=2=36​=2
Digit total for each place =(0+1+5)×2=12=(0+1+5)×2=12
Total of all four-digit numbers beginning with 4.
=4×6×1000+12×100+12×10+12=25332=4×6×1000+12×100+12×10+12=25332
Number of four-digit numbers begins with 5
=3×2×1=6=3×2×1=6 and repetition of digit =63=2=36​=2
Digit total for each place =(0+1+4)×2=10=(0+1+4)×2=10
Total of all four-digit numbers beginning with 5
=5×6×1000+10×100+10×10+10=31110=5×6×1000+10×100+10×10+10=31110
Required sum =7998+25332+31110=64440=7998+25332+31110=64440

4. (a) Considering that x,yx,y and zz are the cube roots of unity
Assume x=1,y=ωx=1,y=ω and z=ω2z=ω2
So, xy+yz+zx=ω+ω3+ω2xy+yz+zx=ω+ω3+ω2
=ω+1+ω2=0(∵ω3=1)=ω+1+ω2=0(∵ω3=1)

5. (c) Given,
Man's: 4 women, 3 men
Wife's: 3 women, 4 men
According to question,
Case-1: 0 3 3 0 → 3C0×4C3×4C3×3C0=163C0​×4C3​×4C3​×3C0​=16
Case-2: 3 0 0 3 → 3C3×4C0×4C0×3C3=13C3​×4C0​×4C0​×3C3​=1
Case-3: 1 2 2 1 → 3C1×4C2×4C2×3C1=3243C1​×4C2​×4C2​×3C1​=324
Case-4: 2 1 1 2 → 3C2×4C1×4C1×3C2=1443C2​×4C1​×4C1​×3C2​=144
Total = 485

6. (a)
Required number of triangles
=12C3−(3C3+4C3+5C3)=220−(1+4+10)=205=12C3​−(3C3​+4C3​+5C3​)=220−(1+4+10)=205

7. (a) Given, log⁡ba=p⇒a=bplogb​a=p⇒a=bp
also, log⁡dc=2p⇒c=d2plogd​c=2p⇒c=d2p
also, log⁡fe=3p⇒e=f3plogf​e=3p⇒e=f3p
Now, (ace)1/p=(bpd2pf3p)1/p=bd2f3(ace)1/p=(bpd2pf3p)1/p=bd2f3

8. (c) As irrational roots occurs in pair and −2−2​ and 33​ are roots of the given equation. So, 22​ and −3−3​ are also roots of the given equation.
Thus, x4+a3x3+a2x2+a1x+a0x4+a3​x3+a2​x2+a1​x+a0​
=(x+2)(x−2)(x−3)(x+3)=(x+2​)(x−2​)(x−3​)(x+3​)
=(x2−2)(x2−3)=x4−5x2+6=(x2−2)(x2−3)=x4−5x2+6
On comparing a3=0,a2=−5,a1=0a3​=0,a2​=−5,a1​=0 and a0=6a0​=6

9. (c) Consider z1=x1+iy1z1​=x1​+iy1​ and z2=x2+iy2z2​=x2​+iy2​
Now, ∣z1+z2z1−z2∣=1⇒∣z1+z2∣=∣z1−z2∣​z1​−z2​z1​+z2​​​=1⇒∣z1​+z2​∣=∣z1​−z2​∣
⇒∣(x1+x2)+i(y1+y2)∣=∣(x1−x2)+i(y1−y2)∣⇒∣(x1​+x2​)+i(y1​+y2​)∣=∣(x1​−x2​)+i(y1​−y2​)∣
⇒(x1+x2)2+(y1+y2)2=(x1−x2)2+(y1−y2)2⇒(x1​+x2​)2+(y1​+y2​)2=(x1​−x2​)2+(y1​−y2​)2
⇒x12+x22+2x1x2+y12+y22+2y1y2=x12+x22−2x1x2+y12+y22−2y1y2⇒x12​+x22​+2x1​x2​+y12​+y22​+2y1​y2​=x12​+x22​−2x1​x2​+y12​+y22​−2y1​y2​
⇒x1x2+y1y2=0...(i)⇒x1​x2​+y1​y2​=0...(i)
z1z2=x1+iy1x2+iy2×x2−iy2x2−iy2=(x1x2+y1y2)+i(y1x2−x1y2)x22+y22z2​z1​​=x2​+iy2​x1​+iy1​​×x2​−iy2​x2​−iy2​​=x22​+y22​(x1​x2​+y1​y2​)+i(y1​x2​−x1​y2​)​
=0+i(y1x2−x1y2)x22+y22=0+x22​+y22​i(y1​x2​−x1​y2​)​
∴Re(z1z2)=0∴Re(z2​z1​​)=0
Required answer Re(z1z2)+1=1Re(z2​z1​​)+1=1

10. (b) Given, 26!=n8k=n23k26!=n8k=n23k
maximum power of 2
=[262]+[264]+[268]+[2616]=13+6+3+1=23=[226​]+[426​]+[826​]+[1626​]=13+6+3+1=23
As power of 2 is multiple of 3
So, maximum value of 3k=213k=21
k=7k=7

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11. (b)
(1) We know that adj(AB)=(adjB)(adjA)adj(AB)=(adjB)(adjA)
Hence, statement 1 is not correct.
(2) We know that AB≠BAAB=BA
⇒adj(AB)≠adj(BA)⇒adj(AB)=adj(BA)
Hence, statement 2 is not correct.
(3) We know that A⋅adjA=∣A∣IA⋅adjA=∣A∣I
(AB)adj(AB)−∣AB∣In(AB)adj(AB)−∣AB∣In​
=∣AB∣In−∣AB∣In=Null matrix=∣AB∣In​−∣AB∣In​=Null matrix
Hence, statement 3 is correct.

12. (d) As, adjAT=(adjA)T⇒A(adjAT)=A(adjA)TadjAT=(adjA)T⇒A(adjAT)=A(adjA)T
Statement 1 is correct.
We know that, if An=AAn=A then AA is identity matrix.
∴∴ Statements 2 and 3 are correct.
Hence, all statements are correct.

13. (b) All even digits are 0, 2, 4, 6, 8
Choice → 4 5 5 5
∴∴ Total required numbers = 4×5×5×5=5004×5×5×5=500

14. (b) Given, (z−100)3+1000=0⇒(z−100)3=(−10)3(z−100)3+1000=0⇒(z−100)3=(−10)3
∴z−100=−10(ω,ω2),1∴z−100=−10(ω,ω2),1
z−100=−10ω⇒z=100−10ωz−100=−10ω⇒z=100−10ω
z−100=−10ω2⇒z=100−10ω2z−100=−10ω2⇒z=100−10ω2
z−100=−10⇒z=90z−100=−10⇒z=90

15. (d) (1+i)4+(1−i)4=[(1+i)2]2+[(1−i)2]2(1+i)4+(1−i)4=[(1+i)2]2+[(1−i)2]2
=(1+2i+i2)2+(1−2i+i2)2=(2i)2+(−2i)2=(1+2i+i2)2+(1−2i+i2)2=(2i)2+(−2i)2
=4i2+4i2=−4−4=−8=4i2+4i2=−4−4=−8

16. (a) All the diagonal elements of skew-symmetric matrix are zero.
Hence, statements 1 and 2 are correct we know that if AAT=IAAT=I then AA is called orthogonal matrix.
But AAT=A(−A)=−A2(AT=−A)AAT=A(−A)=−A2(AT=−A)
Hence, Statement 3 is not correct.

17. (d) We know that a number is divisible by 4 if its last 2 digits is divisible by 4.
Total number divisible by 4
=0 0 0 2 1×2×3×1=1×2×3×1=6=0 0 0 2 1×2×3×1=1×2×3×1=6

18. (a) 2120=(23)40=840=(1+7)402120=(23)40=840=(1+7)40
=1+40C17+40C272+⋯+40C40740=1+40C1​7+40C2​72+⋯+40C40​740
=1+7[40C1+40C27+⋯+40C40739]=1+7[40C1​+40C2​7+⋯+40C40​739]
⇒⇒ Remainder = 1 [using division algorithm]

19. (c) Given,
∣C(9,4)C(9,3)C(10,n−2)C(11,6)C(11,5)C(12,n)C(m,7)C(m,6)C(m+1,n+1)∣=0​C(9,4)C(11,6)C(m,7)​C(9,3)C(11,5)C(m,6)​C(10,n−2)C(12,n)C(m+1,n+1)​​=0
Applying C3→C1+C2−C3C3​→C1​+C2​−C3​
∣C(9,4)C(9,3)C(10,4)−C(10,n−2)C(11,6)C(11,5)C(12,6)−C(12,n)C(m,7)C(m,6)C(m+1,7)−C(m+1,n+1)∣=0​C(9,4)C(11,6)C(m,7)​C(9,3)C(11,5)C(m,6)​C(10,4)−C(10,n−2)C(12,6)−C(12,n)C(m+1,7)−C(m+1,n+1)​​=0
Since determinant value is zero.
∴C3=0∴C3​=0
So, n−2=4⇒n=6n−2=4⇒n=6

20. (b) Consider
∣cos⁡Csin⁡B0tan⁡A0sin⁡B0tan⁡(B+C)cos⁡C∣​cosCtanA0​sinB0tan(B+C)​0sinBcosC​​
Expanding along C1C1​
=cos⁡C(−sin⁡B⋅tan⁡(B+C))−tan⁡A(sin⁡Bcos⁡C)=cosC(−sinB⋅tan(B+C))−tanA(sinBcosC)
=−sin⁡Bcos⁡C[sin⁡(B+C)cos⁡(B+C)+sin⁡Acos⁡A]=−sinBcosC[cos(B+C)sin(B+C)​+cosAsinA​]
=−sin⁡Bcos⁡C[sin⁡(B+C)cos⁡A+sin⁡Acos⁡(B+C)cos⁡Acos⁡(B+C)]=−sinBcosC[cosAcos(B+C)sin(B+C)cosA+sinAcos(B+C)​]
=−sin⁡Bcos⁡Csin⁡(A+B+C)cos⁡Acos⁡(B+C)=−sinBcosCcosAcos(B+C)sin(A+B+C)​
=−sin⁡Bcos⁡Csin⁡πcos⁡Acos⁡(B+C)=0=−cosAcos(B+C)sinBcosCsinπ​=0

21. (d) Possible order of matrices with 4 entries
1×4,2×2,4×11×4,2×2,4×1
∴∴ Total number of matrices = 4×4×4×4×4=7684×4×4×4×4=768

22. (a)
f(x)=x∣x∣f(x)=x∣x∣ is one-one and onto
g(x)=cos⁡(πx)g(x)=cos(πx) is not one-one as any horizontal line passes g(x)g(x) twice.

23. (d) Given xRy⇒∣x+y∣<2xRy⇒∣x+y∣<2 in (−1,1)(−1,1)
For reflexive
∣x+x∣<2⇒∣x∣<1∣x+x∣<2⇒∣x∣<1 true ⇒⇒ R is reflexive
For symmetric
Let xRy⇒∣x+y∣<2xRy⇒∣x+y∣<2
⇒∣y+x∣<2⇒yRx⇒⇒∣y+x∣<2⇒yRx⇒ R is symmetric
For transitive
Let ∣x+y∣<2∣x+y∣<2 and ∣y+z∣<2∣y+z∣<2
then ∣x+z∣<2∣x+z∣<2
⇒⇒ R is transitive.

24. (a) (A∪B)−{(A−B)∪(B−A)∪(A∩B)}(A∪B)−{(A−B)∪(B−A)∪(A∩B)}
From the Venn diagram
(A−B)∪(B−A)∪(A∩B)=A∪B(A−B)∪(B−A)∪(A∩B)=A∪B
⇒(A∪B)−(A∪B)=∅⇒(A∪B)−(A∪B)=∅ (Null set)

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25. (a)
∣a2bsin⁡Acsin⁡Absin⁡A1cos⁡Acsin⁡Acos⁡A1∣​a2bsinAcsinA​bsinA1cosA​csinAcosA1​​
Expanding along R1R1​
=a2(1−cos⁡2A)−bsin⁡A(bsin⁡A−ccos⁡Asin⁡A)+csin⁡A(bsin⁡Acos⁡A−csin⁡A)=a2(1−cos2A)−bsinA(bsinA−ccosAsinA)+csinA(bsinAcosA−csinA)
=a2sin⁡2A−b2sin⁡2A+bcsin⁡2Acos⁡A+bcsin⁡2Acos⁡A−c2sin⁡2A=a2sin2A−b2sin2A+bcsin2AcosA+bcsin2AcosA−c2sin2A
=a2sin⁡2A−b2sin⁡2A−c2sin⁡2A+2bcsin⁡2Acos⁡A=a2sin2A−b2sin2A−c2sin2A+2bcsin2AcosA
=sin⁡2A(a2−b2−c2+2bccos⁡A)=sin⁡2A×0=0=sin2A(a2−b2−c2+2bccosA)=sin2A×0=0

26. (a) Given, a, b, c are in AP ⇒2b=a+c...(i)⇒2b=a+c...(i)
also, b, c, d are in GP ⇒c2=bd...(ii)⇒c2=bd...(ii)
also, c, d, e are in HP ⇒2d=1c+1e...(iii)⇒d2​=c1​+e1​...(iii)
From (i) and (ii)
2c2=(a+c)d⇒2d=a+cc2⇒1c+1e=ac2+1c2c2=(a+c)d⇒d2​=c2a+c​⇒c1​+e1​=c2a​+c1​ from (iii)
1e=ac2⇒c2=aee1​=c2a​⇒c2=ae
So, a, c, e are in GP.

27. (b) Given, log⁡4(x−1)=log⁡2(x−3)log4​(x−1)=log2​(x−3)
⇒12log⁡2(x−1)=log⁡2(x−3)⇒21​log2​(x−1)=log2​(x−3)
⇒log⁡2(x−1)=log⁡2(x−3)2⇒log2​(x−1)=log2​(x−3)2
⇒x−1=x2−6x+9⇒x2−7x+10=0⇒x−1=x2−6x+9⇒x2−7x+10=0
⇒x2−5x−2x+10=0⇒(x−5)(x−2)=0⇒x2−5x−2x+10=0⇒(x−5)(x−2)=0
x=2,5x=2,5, but x−3>0⇒x>3x−3>0⇒x>3
⇒x=5⇒x=5 only solution

28. (d) Given, log⁡x(xy)+log⁡y(yx)=klogx​(yx​)+logy​(xy​)=k
⇒log⁡xx−log⁡xy+log⁡yy−log⁡yx=k⇒logx​x−logx​y+logy​y−logy​x=k
=1−log⁡xy+1−1log⁡xy=k[∵log⁡ab=1/log⁡ba]=1−logx​y+1−logx​y1​=k[∵loga​b=1/logb​a]
Let log⁡xy=tlogx​y=t
∴2−t−1t=k⇒2t−t2−1=kt⇒t2+(k−2)t+1=0∴2−t−t1​=k⇒2t−t2−1=kt⇒t2+(k−2)t+1=0
For solution D=b2−4ac≥0D=b2−4ac≥0
(k−2)2−4≥0⇒(k−2)2≥4⇒k−2≥2(k−2)2−4≥0⇒(k−2)2≥4⇒k−2≥2 or k−2≤−2k−2≤−2
k≥4k≥4 or k≤0⇒k≠1k≤0⇒k=1

29. (c) Given
A=∣sin⁡2θ−cos⁡2θ0cos⁡2θsin⁡2θ0001∣=1A=​sin2θcos2θ0​−cos2θsin2θ0​001​​=1
We know that if ∣A∣=±1∣A∣=±1 then A is orthogonal matrix
So A−1=ATA−1=AT and A−1=adjAA−1=adjA

30. (a) (1−x2)20(−12x2+x2−2)−5(1−x2)20(−21​x2+x2−2)−5
=−(1−x2)20(x−1x)−10=−(1−x2)20(x−x1​)−10
=−(1−x2)20(x2−1)−10⋅x10=−(1−x2)10⋅x10=−(1−x2)20(x2−1)−10⋅x10=−(1−x2)10⋅x10
General term Tr+1=−10Cr(1)10−r(−x2)10−r⋅x10Tr+1​=−10Cr​(1)10−r(−x2)10−r⋅x10
=(−1)11−r⋅10Crx30−2r=(−1)11−r⋅10Cr​x30−2r
∴30−2r=10⇒r=10∴30−2r=10⇒r=10
So, coefficient x10=(−1)10⋅10C10=−1x10=(−1)10⋅10C10​=−1

31. (b) Given,
T4=T3+1=nC3(mx)n−3(1x)3=52T4​=T3+1​=nC3​(mx)n−3(x1​)3=25​
nC3(m)n−3xn−6=52x0nC3​(m)n−3xn−6=25​x0
On comparing n−6=0⇒n=6n−6=0⇒n=6
and 6C3m3=526C3​m3=25​
⇒6⋅5⋅43⋅2m3=52⇒m3=18⇒m=12⇒3⋅26⋅5⋅4​m3=25​⇒m3=81​⇒m=21​
∴mn=6×12=3∴mn=6×21​=3

32. (d) Since a, b, c are in G.P. ⇒b2=ac⇒b2=ac
Now, for equal ax2+bx+c=0ax2+bx+c=0
D=b2−4ac=ac−4ac=−3ac<0D=b2−4ac=ac−4ac=−3ac<0
∴∴ Roots are imaginary
Let a=2,b=4,c=8a=2,b=4,c=8
∴x2+2x+4=0⇒x=−2±4−162=−1±3∴x2+2x+4=0⇒x=2−2±4−16​​=−1±3​
Ratio of roots = −1+3−1−3=ωω2=1ω−1−3​−1+3​​=ω2ω​=ω1​
Product of roots = (−1+32)(−1−32)×4=ω⋅ω2⋅4=4=b2a2(2−1+3​​)(2−1−3​​)×4=ω⋅ω2⋅4=4=a2b2​

33. (c) ∵f(x)=x2+mx+n∈I,∀x∈I∵f(x)=x2+mx+n∈I,∀x∈I
f(0)=0+0+n∈I⇒x∈If(0)=0+0+n∈I⇒x∈I
f(1)=1+m+n∈I⇒m+n∈If(1)=1+m+n∈I⇒m+n∈I
⇒m∈I(∵n∈I)⇒m∈I(∵n∈I)
So, both m and n are integer.

34. (a) Given,
(x+y)2n+1(x−y)2n+1=(x2−y2)2n+1(x+y)2n+1(x−y)2n+1=(x2−y2)2n+1
Middle term 2n+22,2n+42=(x+1)thterm,(n+2)thterm22n+2​,22n+4​=(x+1)thterm,(n+2)thterm
Tn+1=2n+1Cn(x2)n+1(y2)nTn+1​=2n+1Cn​(x2)n+1(y2)n
Tn+2=2n+1Cn+1(x2)n(y2)n+1Tn+2​=2n+1Cn+1​(x2)n(y2)n+1
2n+1Cn(x2)n+1(y2)n=2n+1Cn+1(x2)n(y2)n+12n+1Cn​(x2)n+1(y2)n=2n+1Cn+1​(x2)n(y2)n+1
2n+1Cn2n+1Cn+1=x2y2n+2x2n+2y2n⇒(2n+1)!n!(n+1)!⋅n!(n+1)!(2n+1)!=y2x22n+1Cn+1​2n+1Cn​​=x2n+2y2nx2y2n+2​⇒n!(n+1)!(2n+1)!​⋅(2n+1)!n!(n+1)!​=x2y2​
⇒y2x2=1⇒x2y2​=1

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35. (c) Given, n(A)=5n(A)=5 and n(B)=2n(B)=2
Number of onto function from A to B = 25−2C1(2−1)5=32−2=3025−2C1​(2−1)5=32−2=30

36. (d)
3cos⁡10∘−sin⁡10∘sin⁡25∘cos⁡25∘=2(32cos⁡10∘−12sin⁡10∘)12(2sin⁡25∘cos⁡25∘)sin25∘cos25∘3​cos10∘−sin10∘​=21​(2sin25∘cos25∘)2(23​​cos10∘−21​sin10∘)​
=4⋅sin⁡(60∘−10∘)sin⁡50∘=4sin⁡50∘sin⁡50∘=4=sin50∘4⋅sin(60∘−10∘)​=sin50∘4sin50∘​=4

37. (c) sin⁡9∘−cos⁡9∘=2(12sin⁡9∘−12cos⁡9∘)sin9∘−cos9∘=2​(2​1​sin9∘−2​1​cos9∘)
=2sin⁡(45∘−9∘)=2sin⁡36∘=2​sin(45∘−9∘)=2​sin36∘
=21−cos⁡236∘[∵cos⁡36∘=5+14]=2​1−cos236∘​[∵cos36∘=45​+1​]
=21−(5+14)2=2​1−(45​+1​)2​
=210−254=5−52=2​410−25​​​=25​−5​​

38. (d) Given
sin⁡3A+sin⁡3B+sin⁡3C=3sin⁡Asin⁡Bsin⁡Csin3A+sin3B+sin3C=3sinAsinBsinC
⇒sin⁡A+sin⁡B+sin⁡C=0⇒sinA+sinB+sinC=0
⇒ak+bk+ck=0⇒a+b+c=0⇒ak+bk+ck=0⇒a+b+c=0
∣abcbcacab∣=∣a+b+cbcb+c+acac+a+bab∣​abc​bca​cab​​=​a+b+cb+c+ac+a+b​bca​cab​​ (Applying C1→C1+C2+C3C1​→C1​+C2​+C3​)
=∣0bc0ca0ab∣=0=​000​bca​cab​​=0

39. (c) Given cos⁡−1x=sin⁡−1x⇒π2−sin⁡−1x=sin⁡−1xcos−1x=sin−1x⇒2π​−sin−1x=sin−1x
⇒2sin⁡−1x=π2⇒sin⁡−1x=π4⇒2sin−1x=2π​⇒sin−1x=4π​
⇒x=sin⁡π4=12⇒x=sin4π​=2​1​

40. (b) Given, (sin⁡θ−cos⁡θ)2=2(sinθ−cosθ)2=2
⇒sin⁡2+cos⁡2θ−2sin⁡θcos⁡θ=2⇒sin2+cos2θ−2sinθcosθ=2
⇒sin⁡2θ=−1⇒sin2θ=−1
⇒⇒ Two solution possible in (−π,π)(−π,π)

41. (c) cos⁡A+cos⁡Bcos⁡(A−B2)=2cos⁡A+B2⋅cos⁡(A−B2)cos⁡(A−B2)cos(2A−B​)cosA+cosB​=cos(2A−B​)2cos2A+B​⋅cos(2A−B​)​
=2cos⁡(π−C2)=2cos⁡(π2−C2)=2sin⁡C2=2cos(2π−C​)=2cos(2π​−2C​)=2sin2C​
=2sin⁡30∘=2×12=1=2sin30∘=2×21​=1

42. (c) 15+cot⁡2(π4−2cot⁡−13)15+cot2(4π​−2cot−13)​
Consider, 2cot⁡−13=2tan⁡−113=tan⁡−12/31−1/92cot−13=2tan−131​=tan−11−1/92/3​
=tan⁡−1(23×98)=tan⁡−134=tan−1(32​×89​)=tan−143​
So, 15+cot⁡2(tan⁡−11−tan⁡−134)15+cot2(tan−11−tan−143​)​
=15+cot⁡2(tan⁡−1(1−3/41+3/4))=15+cot2(tan−1(1+3/41−3/4​))​
=15+cot⁡2(tan⁡−117)=15+cot⁡2(cot⁡−17)=15+cot2(tan−171​)​=15+cot2(cot−17)​
=15+49=8=15+49​=8

43. (b) sin⁡10∘⋅sin⁡50∘+sin⁡50∘⋅sin⁡250∘+sin⁡250∘⋅sin⁡10∘sin10∘⋅sin50∘+sin50∘⋅sin250∘+sin250∘⋅sin10∘
=12(cos⁡40∘−cos⁡60∘+cos⁡200∘−cos⁡300∘+cos⁡240∘−cos⁡260∘)=21​(cos40∘−cos60∘+cos200∘−cos300∘+cos240∘−cos260∘)
=12[cos⁡40∘−12+cos⁡(180+20)∘−cos⁡(360−60)∘+cos⁡(180+60)∘−cos⁡(180+80)∘]=21​[cos40∘−21​+cos(180+20)∘−cos(360−60)∘+cos(180+60)∘−cos(180+80)∘]
=12[cos⁡40∘−12−cos⁡20∘−cos⁡60∘−cos⁡60∘+cos⁡80∘]=21​[cos40∘−21​−cos20∘−cos60∘−cos60∘+cos80∘]
=12[cos⁡40∘−cos⁡20∘+cos⁡80∘−12−12−12]=21​[cos40∘−cos20∘+cos80∘−21​−21​−21​]
=12[−2sin⁡30∘⋅sin⁡10∘+cos⁡(90∘−10∘)−32]=21​[−2sin30∘⋅sin10∘+cos(90∘−10∘)−23​]
=12[−sin⁡10∘+sin⁡10∘−32]=−34=21​[−sin10∘+sin10∘−23​]=−43​

44. (b) tan⁡−1(ab)−tan⁡−1(a−ba+b)tan−1(ba​)−tan−1(a+ba−b​)
=tan⁡−1(ab)−tan⁡−1(ab−1ab+1)=tan−1(ba​)−tan−1(ba​+1ba​−1​)
=tan⁡−1ab−tan⁡−1ab+tan⁡−11=tan⁡−1(1)=π4=tan−1ba​−tan−1ba​+tan−11=tan−1(1)=4π​

45. (d) For real roots, discriminant ≥0≥0
⇒(cos⁡B)2−4sin⁡B(cos⁡B−1)≥0⇒(cosB)2−4sinB(cosB−1)≥0
(cos⁡B)2+4sin⁡B(1−cos⁡B)≥0(cosB)2+4sinB(1−cosB)≥0
As −1≤cos⁡B≤1−1≤cosB≤1 and sin⁡B≥0sinB≥0 for B∈[0,π]B∈[0,π]
So, it is only possible when 1−cos⁡B≥01−cosB≥0

46. (c) cos⁡2A+cos⁡2B+cos⁡2C[∵cos⁡2x=1−2sin⁡2x]cos2A+cos2B+cos2C[∵cos2x=1−2sin2x]
=1−2sin⁡2A+1−2sin⁡2B+1−2sin⁡2C=1−2sin2A+1−2sin2B+1−2sin2C
=3−2(1665)2+(6365)2+0=3−2(6516​)2+(6563​)2+0
=3−2((1665)2+(6365)2+0)[since 162+632=652]=3−2((6516​)2+(6563​)2+0)[since 162+632=652]
⇒[∠B=90∘]=3−2=1⇒[∠B=90∘]=3−2=1

47. (b) Given α+β=5π4α+β=45π​, tan⁡(α+β)=tan⁡5π4tan(α+β)=tan45π​
tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=tan⁡(π+π4)=tan⁡π41−tanα⋅tanβtanα+tanβ​=tan(π+4π​)=tan4π​
tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=11−tanα⋅tanβtanα+tanβ​=1
tan⁡α+tan⁡β+tan⁡α⋅tan⁡β=1tanα+tanβ+tanα⋅tanβ=1
Also, f(θ)=11+tan⁡θf(θ)=1+tanθ1​
so, f(α)⋅f(β)=11+tan⁡α×11+tan⁡βf(α)⋅f(β)=1+tanα1​×1+tanβ1​
=11+tan⁡α+tan⁡β+tan⁡α⋅tan⁡β=11+1=12=1+tanα+tanβ+tanα⋅tanβ1​=1+11​=21​

48. (b) tan⁡α+tan⁡β=6tanα+tanβ=6
tan⁡α⋅tan⁡β=8tanα⋅tanβ=8
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=61−8=6−7tan(α+β)=1−tanα⋅tanβtanα+tanβ​=1−86​=−76​
cos⁡(2α+2β)=cos⁡2(α+β)cos(2α+2β)=cos2(α+β)
=1−tan⁡2(α+β)1+tan⁡2(α+β)=1−36491+3649=1385=1+tan2(α+β)1−tan2(α+β)​=1+4936​1−4936​​=8513​

49. (c) tan⁡(90−65)∘+2−2tan⁡40∘−tan⁡25∘tan(90−65)∘+2−2tan40∘−tan25∘
=cot⁡25∘−tan⁡25∘−2tan⁡40∘+2=cot25∘−tan25∘−2tan40∘+2
=cos⁡225−sin⁡225sin⁡25⋅cos⁡25−2tan⁡40∘+2=sin25⋅cos25cos225−sin225​−2tan40∘+2
=2cos⁡50∘sin⁡50∘−2tan⁡40∘+2=sin50∘2cos50∘​−2tan40∘+2
=2cot⁡50∘−2tan⁡40∘+2=2tan⁡40∘−2tan⁡40∘+2=2=2cot50∘−2tan40∘+2=2tan40∘−2tan40∘+2=2

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49. (c) tan⁡(90−65)∘+2−2tan⁡40∘−tan⁡25∘tan(90−65)∘+2−2tan40∘−tan25∘
=cot⁡25∘−tan⁡25∘−2tan⁡40∘+2=cot25∘−tan25∘−2tan40∘+2
=cos⁡225−sin⁡225sin⁡25⋅cos⁡25−2tan⁡40∘+2=sin25⋅cos25cos225−sin225​−2tan40∘+2
=2cos⁡50∘sin⁡50∘−2tan⁡40∘+2=sin50∘2cos50∘​−2tan40∘+2
=2cot⁡50∘−2tan⁡40∘+2=2tan⁡40∘−2tan⁡40∘+2=2=2cot50∘−2tan40∘+2=2tan40∘−2tan40∘+2=2

tan⁡−1(ab)−tan⁡−1(a−ba+b)tan−1(ba​)−tan−1(a+ba−b​)
=tan⁡−1(ab)−tan⁡−1(ab−1ab+1)=tan−1(ba​)−tan−1(ba​+1ba​−1​)
=tan⁡−1ab−tan⁡−1ab+tan⁡−11=tan⁡−1(1)=π4=tan−1ba​−tan−1ba​+tan−11=tan−1(1)=4π​

45. (d) For real roots, discriminant ≥0≥0
⇒(cos⁡B)2−4sin⁡B(cos⁡B−1)≥0⇒(cosB)2−4sinB(cosB−1)≥0
(cos⁡B)2+4sin⁡B(1−cos⁡B)≥0(cosB)2+4sinB(1−cosB)≥0
As −1≤cos⁡B≤1−1≤cosB≤1 and sin⁡B≥0sinB≥0 for B∈[0,π]B∈[0,π]
So, it is only possible when 1−cos⁡B≥01−cosB≥0

46. (c) cos⁡2A+cos⁡2B+cos⁡2Ccos2A+cos2B+cos2C
=1−2sin⁡2A+1−2sin⁡2B+1−2sin⁡2C=1−2sin2A+1−2sin2B+1−2sin2C
=3−2(1665)2+(6365)2+0=3−2(6516​)2+(6563​)2+0
=3−2((1665)2+(6365)2+0)=3−2((6516​)2+(6563​)2+0)
⇒[∠B=90∘]=3−2=1⇒[∠B=90∘]=3−2=1

47. (b) Given α+β=5π4α+β=45π​, tan⁡(α+β)=tan⁡5π4tan(α+β)=tan45π​
tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=tan⁡(π+π4)=tan⁡π41−tanα⋅tanβtanα+tanβ​=tan(π+4π​)=tan4π​
tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=11−tanα⋅tanβtanα+tanβ​=1
tan⁡α+tan⁡β+tan⁡α⋅tan⁡β=1tanα+tanβ+tanα⋅tanβ=1
Also, f(θ)=11+tan⁡θf(θ)=1+tanθ1​
so, f(α)⋅f(β)=11+tan⁡α×11+tan⁡βf(α)⋅f(β)=1+tanα1​×1+tanβ1​
=11+tan⁡α+tan⁡β+tan⁡α⋅tan⁡β=11+1=12=1+tanα+tanβ+tanα⋅tanβ1​=1+11​=21​

48. (b) tan⁡α+tan⁡β=6tanα+tanβ=6
tan⁡α⋅tan⁡β=8tanα⋅tanβ=8
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=61−8=6−7tan(α+β)=1−tanα⋅tanβtanα+tanβ​=1−86​=−76​
cos⁡(2α+2β)=cos⁡2(α+β)cos(2α+2β)=cos2(α+β)
=1−tan⁡2(α+β)1+tan⁡2(α+β)=1−36491+3649=1385=1+tan2(α+β)1−tan2(α+β)​=1+4936​1−4936​​=8513​

49. (c) tan⁡(90−65)∘+2−2tan⁡40∘−tan⁡25∘tan(90−65)∘+2−2tan40∘−tan25∘
=cot⁡25∘−tan⁡25∘−2tan⁡40∘+2=cot25∘−tan25∘−2tan40∘+2
=cos⁡225−sin⁡225sin⁡25⋅cos⁡25−2tan⁡40∘+2=sin25⋅cos25cos225−sin225​−2tan40∘+2
=2cos⁡50∘sin⁡50∘−2tan⁡40∘+2=sin50∘2cos50∘​−2tan40∘+2
=2cot⁡50∘−2tan⁡40∘+2=2tan⁡40∘−2tan⁡40∘+2=2=2cot50∘−2tan40∘+2=2tan40∘−2tan40∘+2=2

50. (a) 1. cot⁡Acot⁡Bcot⁡C>0cotAcotBcotC>0
∴cot⁡A>0,cot⁡B>0,cot⁡C>0∴cotA>0,cotB>0,cotC>0
∴0<A<π2,0<B<π2,0<C<π2∴0<A<2π​,0<B<2π​,0<C<2π​
ΔABCΔABC is an acute angled triangle.
Hence, statement 1 is correct.

1. tan⁡Atan⁡Btan⁡C>0tanAtanBtanC>0
   If ΔABCΔABC is obtuse angled triangle then two of tan⁡A,tan⁡B,tan⁡C<0tanA,tanB,tanC<0
   ⇒⇒ Two of angles are obtuse but it is not possible
   Hence, statement 2 is not correct.

51. (a) x2+y2+2x+6y+1=0x2+y2+2x+6y+1=0
To obtain the center (a,b) and the radius c of a circle equation, rewrite it as (x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2, where (h, k) is the circle's center and r is its radius.
Consider, x2+y2+2x+6y+1=0x2+y2+2x+6y+1=0
To obtain the equation in the desired form use completing the square
(x2+2x+1)−1+(y2+6y+9)−9+1=0(x2+2x+1)−1+(y2+6y+9)−9+1=0
⇒(x+1)2+(y+3)2=9⇒(x+1)2+(y+3)2=9
⇒(x−(−1))2+(y−(−3))2=(3)2⇒(x−(−1))2+(y−(−3))2=(3)2
Center of the circle is (−1)(−1) and radius is 3.
∴a=−1,b=−3∴a=−1,b=−3 and c=3c=3
∴a2+b2+c2=(−1)2+(−3)2+(3)2=1+9+9=19∴a2+b2+c2=(−1)2+(−3)2+(3)2=1+9+9=19

52. (a) Given: Equation of sphere x2+y2+z2+2ux+2vy+2wz−1=0x2+y2+z2+2ux+2vy+2wz−1=0

with AB as its diameter and A=(1,−1,2)A=(1,−1,2) and B=(2,1,−1)B=(2,1,−1)
So, C will be midpoint of AB.
C=(1+22,1−12,−1+22)=(32,0,12)C=(21+2​,21−1​,2−1+2​)=(23​,0,21​)
Since centre of the sphere is (−u,−v,−w)(−u,−v,−w) 

53. (d)

x=5x=5 represents a line parallel to y-axis so there are infinite points on xy-plane.

54. (b) Equation of plane is ax+by+cz=dax+by+cz=d
Given the plane equation is 2x−3y+6z+4=02x−3y+6z+4=0
On comparing we get a=2a=2, b=−3b=−3, c=6c=6, d=−4d=−4
Here we need to find the direction cosine of a normal to the plane i.e <l,m,n><l,m,n>
Direction cosines of normal to plane

l=aa2+b2+c2=249=27l=a2+b2+c2​a​=49​2​=72​m=ba2+b2+c2=−37m=a2+b2+c2​b​=7−3​n=ca2+b2+c2=67n=a2+b2+c2​c​=76​∴<l,m,n>=<27,−37,67>∴<l,m,n>=⟨72​,7−3​,76​⟩49(7l2+m2−n2)=49(7(27)2+(−37)2−(67)2)49(7l2+m2−n2)=49(7(72​)2+(7−3​)2−(76​)2)=49(2849+949−3649)=1=49(4928​+499​−4936​)=1

55. (c)
Assume the line from (1,−1,2)(1,−1,2) meet the plane at Q.
Direction ratios of the line from the point (1,−1,2)(1,−1,2) to the given plane is <3,2,2><3,2,2>
So the equation of the line passing through P and with direction ratios will be:

x−13=y+12=z−22=λ3x−1​=2y+1​=2z−2​=λx=3λ+1,y=2λ−1,z=2λ+2x=3λ+1,y=2λ−1,z=2λ+2

Now, since Q lies on the plane so it must satisfy the equation of the plane.
i.e x+2y+3z=18x+2y+3z=18

∴3λ+1+4λ−2+6λ+6=18∴3λ+1+4λ−2+6λ+6=1813λ+5=18⇒λ=113λ+5=18⇒λ=1

Therefore, coordinates of Q are (3+1,2−1,2+2)=(4,1,4)(3+1,2−1,2+2)=(4,1,4)

56. (d) Let Plane ax+by+cz+d=0ax+by+cz+d=0 is passing through (1,0,0)(1,0,0), (0,1,0)(0,1,0) & (0,0,1)(0,0,1)
We will find the equation of plane
Hence a+b=0a+b=0, b+d=0b+d=0, c+d=0c+d=0
Solving above three equations, we get
a=b=c=−da=b=c=−d
Now Let a=b=c=−d=λa=b=c=−d=λ
So, equation of plane will be x+y+z−1=0x+y+z−1=0
Perpendicular distance from origin to the plane

p=∣−1∣(1)2+(1)2+(1)2=13 unitp=(1)2+(1)2+(1)2​∣−1∣​=3​1​ unit

Hence, 3p2=3×13=13p2=3×31​=1

57. (b) Given l+2m+n=0l+2m+n=0 ...(i)
and 2l−2m+3n=02l−2m+3n=0 ...(ii)
From (i) l=−2m−nl=−2m−n ...(iii)
Substitute (iii) in (ii)

2(−2m−n)−2m+3n=02(−2m−n)−2m+3n=0−4m−2n−2m+3n=0⇒−6m+n=0⇒n=6m−4m−2n−2m+3n=0⇒−6m+n=0⇒n=6m

From (iii) we get, l=−2m−n=−2m−6m=−8ml=−2m−n=−2m−6m=−8m

l−8m=mm=n6m−8ml​=mm​=6mn​∴l−8=m1=n6=l2+m2+n2(−8)2+(1)2+(6)2=1101∴−8l​=1m​=6n​=(−8)2+(1)2+(6)2​l2+m2+n2​​=101​1​∴l=8101,m=1101,n=6101∴l=101​8​,m=101​1​,n=101​6​

Hence, l2+m2−n2=64101+1101−36101=29101l2+m2−n2=10164​+1011​−10136​=10129​

58. (b)

59. (b) Rewrite the equation as follows:

ysin⁡θ=9−xcos⁡θysinθ=9−xcosθ⇒y=9sin⁡θ−xcos⁡θsin⁡θ⇒y=sinθ9​−xsinθcosθ​⇒y=−xcos⁡θsin⁡θ+9sin⁡θ...(i)⇒y=−xsinθcosθ​+sinθ9​...(i)

The general equation of line is

y=mx+c...(ii)y=mx+c...(ii)

On comparing (i) and (ii), we get

m=−cos⁡θsin⁡θm=−sinθcosθ​

Since, the slope of perpendicular line are negative inverse of each other.
So, the slope m1m1​ of the required line can be

m1=−(1cos⁡θ)⇒m1=sin⁡θcos⁡θm1​=−(cosθ1​)⇒m1​=cosθsinθ​

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60. (a) Given that P and Q lie on line y=2x+3y=2x+3
For P put x=ax=a then y=2a+3y=2a+3
For Q put x=bx=b then y=2b+3y=2b+3
Coordinates P=(a,2a+3)P=(a,2a+3) and Q=(b,2b+3)Q=(b,2b+3)
From R=(1,5)R=(1,5) using distance formula, we have

PR=(a−1)2+(2a+3−5)2=2...(i)PR=(a−1)2+(2a+3−5)2​=2...(i)QR=(b−1)2+(2b+3−5)2=2...(ii)QR=(b−1)2+(2b+3−5)2​=2...(ii)

From (i) we get

(a−1)2+(2a−2)2=4(a−1)2+(2a−2)2=4a2−2a+1+4a2+4−8a=4a2−2a+1+4a2+4−8a=45a2−10a+1=05a2−10a+1=0

From (ii), we get

5b2−10b+1=05b2−10b+1=0

By using quadratic formula

(a,b)=10±100−8010⇒a=1±25(a,b)=1010±100−80​​⇒a=1±5​2​

If a=1+25a=1+5​2​ then

Coordinate of P=(a,2a+3)=(1+25,2+45+3)Coordinate of P=(a,2a+3)=(1+5​2​,2+5​4​+3)=(1+25,5+45)=(1+5​2​,5+5​4​)

If a=1−25a=1−5​2​ then

Coordinate of P=(a,2a+3)=(1−25,2−45+3)Coordinate of P=(a,2a+3)=(1−5​2​,2−5​4​+3)=(1−25,5−45)=(1−5​2​,5−5​4​)Coordinate of Q=(1−25,5−45) or (1+25,5+45)Coordinate of Q=(1−5​2​,5−5​4​) or (1+5​2​,5+5​4​)

Required coordinates of the point P and Q are

(1+25,5+45),(1−25,5−45)(1+5​2​,5+5​4​),(1−5​2​,5−5​4​)

61. (a) Given that two side of a square lie on the lines

2x+y−3=0...(i)2x+y−3=0...(i)4x+2y+5=0...(ii)4x+2y+5=0...(ii)

Divide (ii) by 2 we get, 2x+y+52=02x+y+25​=0 ...(iii)
Here 2x+y−3=02x+y−3=0 and 2x+y+52=02x+y+25​=0 are parallel lines
So, length of the side of the square = Distance between parallel side

=∣C1−C2∣a12+b12=a12​+b12​​∣C1​−C2​∣​

where C1=−3C1​=−3 and C2=52C2​=25​

(a1,b1)=(2,1)(a1​,b1​)=(2,1)⇒∣−3−52∣(2)2+(1)2=∣−112∣5=1125⇒(2)2+(1)2​​−3−25​​​=5​​−211​​​=25​11​

So, length of the side of square =1125=25​11​ units
Required area of the square =(side)2=(1125)2=(side)2=(25​11​)2

=6.05 square units.=6.05 square units.

62. (b) Given, A=(3,5)A=(3,5) and mid-points of sides AB and AC are (−1,2)(−1,2) and (6,4)(6,4) respectively.
Consider B=(x1,y1)B=(x1​,y1​) and C=(x2,y2)C=(x2​,y2​) By using mid-point formula

M=(−1,2)=(x1+32,y1+52)M=(−1,2)=(2x1​+3​,2y1​+5​)N=(6,4)=(x2+32,y2+52)N=(6,4)=(2x2​+3​,2y2​+5​)

By comparing we get

x1+32=−1,y1+52=2⇒x1=−5, y1=−12x1​+3​=−1,2y1​+5​=2⇒x1​=−5, y1​=−1and x2+32=6,y2+52=4⇒x2=9, y2=3and 2x2​+3​=6,2y2​+5​=4⇒x2​=9, y2​=3

So, B=(−5,−1)B=(−5,−1) and C=(9,3)C=(9,3)

Centroid of ΔABC=(3−5+93,… )Centroid of ΔABC=(33−5+9​,…)

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Since focal distance of point P(x1,y1)P(x1​,y1​) is x1+ax1​+a
So, focal distance from point (6,43)(6,43​) is 6+2=86+2=8
Hence, Statement 1 is correct.
Now, distance of point P(6,43)P(6,43​) from direction

PF=(6−2)2+(43−0)2=16+48=8PF=(6−2)2+(43​−0)2​=16+48​=8

Hence, statement 2 is also correct.

65. (d)

66. (Bonus) Given,

a⃗=i^−j^+k^ and b⃗=i^+2j^−k^a=i^−j^​+k^ and b=i^+2j^​−k^

Using vector triple product

a⃗×(b⃗×a⃗)=(a⃗⋅a⃗)b⃗−(a⃗⋅b⃗)a⃗a×(b×a)=(a⋅a)b−(a⋅b)a=((i^−j^+k^)⋅(i^−j^+k^))(i^+2j^−k^)−((i^−j^+k^)⋅(i^+2j^−k^))(i^−j^+k^)=((i^−j^​+k^)⋅(i^−j^​+k^))(i^+2j^​−k^)−((i^−j^​+k^)⋅(i^+2j^​−k^))(i^−j^​+k^)=(1+1+1)(i^+2j^−k^)−(1−2−1)(i^−j^+k^)=(1+1+1)(i^+2j^​−k^)−(1−2−1)(i^−j^​+k^)=3(i^+2j^−k^)+2(i^−j^+k^)=3(i^+2j^​−k^)+2(i^−j^​+k^)=5i^+4j^−k^...(i)=5i^+4j^​−k^...(i)

Given a⃗×(b⃗×a⃗)=αi^−βj^+γk^a×(b×a)=αi^−βj^​+γk^
Compare with (i), we get

α=5, β=−4, γ=−1α=5, β=−4, γ=−1⇒α+β+γ=5−4−1=0⇒α+β+γ=5−4−1=0

67. (a)

68. (b) Statement 1 is not correct
We know that τ⃗=r⃗×F⃗τ=r×F
The moment of force about a point is dependent of application of force.
Statement 2 is correct.
The moment of force about a line is a vector quantity.
Because gross product of two vector is again a vector

τ⃗=r⃗×F⃗τ=r×F↓↓↓↓↓↓VecVecVecVecVecVec

69. (a) Let

I=(r⃗⋅i^)(r⃗×i^)+(r⃗⋅j^)(r⃗×j^)+(r⃗⋅k^)(r⃗×k^)...(i)I=(r⋅i^)(r×i^)+(r⋅j^​)(r×j^​)+(r⋅k^)(r×k^)...(i)

Let r⃗=ai^+bj^+ck^r=ai^+bj^​+ck^

(r⃗⋅i^)=(ai^+bj^+ck^)⋅i^=a(r⋅i^)=(ai^+bj^​+ck^)⋅i^=a(r⃗⋅j^)=b⇒(r⃗×k^)=c(r⋅j^​)=b⇒(r×k^)=c

and

r⃗×i^=∣i^j^k^abc100∣=−j^(−c)−k^b=cj^−bk^r×i^=​i^a1​j^​b0​k^c0​​=−j^​(−c)−k^b=cj^​−bk^

Similarly, r⃗×j^=−ci^+ak^r×j^​=−ci^+ak^ and r⃗×k^=bi^−aj^r×k^=bi^−aj^​
Now substitute in equation (i), we get

I=a(cj^−bk^)+b(−ci^+ak^)+c(bi^−aj^)I=a(cj^​−bk^)+b(−ci^+ak^)+c(bi^−aj^​)=acj^−abk^−bci^+abk^+bci^−acj^=0⃗=acj^​−abk^−bci^+abk^+bci^−acj^​=0

70. (b)

71. (a)

72. (c) We have,

y=ex(acos⁡x+bsin⁡x)...(i)y=ex(acosx+bsinx)...(i)⇒dydx=ex(−asin⁡x+bcos⁡x)+(acos⁡x+bsin⁡x)ex⇒dxdy​=ex(−asinx+bcosx)+(acosx+bsinx)ex⇒dydx=ex(−asin⁡x+bcos⁡x)+y(From (i)) ...(ii)⇒dxdy​=ex(−asinx+bcosx)+y(From (i)) ...(ii)

Again differentiate w.r.t x

d2ydx2=−ex(acos⁡x+bsin⁡x)+(−asin⁡x+bcos⁡x)ex+dydxdx2d2y​=−ex(acosx+bsinx)+(−asinx+bcosx)ex+dxdy​

Eliminating arbitrary constants

⇒d2ydx2=−y+dydx−y+dydx[From (i) and (ii)]⇒dx2d2y​=−y+dxdy​−y+dxdy​[From (i) and (ii)]⇒d2ydx2−2dydx+2y=0 is the required differential equation⇒dx2d2y​−2dxdy​+2y=0 is the required differential equation

73. (d) Given f(x)=ax−bf(x)=ax−b ...(i)
and g(x)=cx+dg(x)=cx+d ...(ii)
Now f(g(x))=g(f(x))f(g(x))=g(f(x))

⇒a(g(x))−b=c(f(x))+d⇒a(g(x))−b=c(f(x))+d⇒a(cx+d)−b=c(ax−b)+d[From (i) and (ii)]⇒a(cx+d)−b=c(ax−b)+d[From (i) and (ii)]⇒acx+ad−b=acx−bc+d⇒acx+ad−b=acx−bc+dad−b=d−bc⇒ad−b+bc=dad−b=d−bc⇒ad−b+bc=d∴f(d)+g(b)=2d[f(d)=ad−b & g(b)=bc+d]∴f(d)+g(b)=2d[f(d)=ad−b & g(b)=bc+d]

74. (b) Let

I=∫−11(3sin⁡x−sin⁡3x)cos⁡2x dxI=∫−11​(3sinx−sin3x)cos2xdxI=∫−11(3sin⁡x−3sin⁡x+4sin⁡3x)cos⁡2x dx[sin⁡3x=3sin⁡x−4sin⁡3x]I=∫−11​(3sinx−3sinx+4sin3x)cos2xdx[sin3x=3sinx−4sin3x]I=∫−114sin⁡3xcos⁡2x dx=∫−11f(x) dxI=∫−11​4sin3xcos2xdx=∫−11​f(x)dx

Since, f(x)=4sin⁡3xcos⁡2xf(x)=4sin3xcos2x is an odd function as f(x)=−f(−x)f(x)=−f(−x)

⇒I=∫−114sin⁡3xcos⁡2x dx=0⇒I=∫−11​4sin3xcos2xdx=0[∫−aaf(x) dx=0, if f(x)=−f(x)][∫−aa​f(x)dx=0, if f(x)=−f(x)]

75. (d) Given:

{2−(dydx)2}0.6=d2ydx2{2−(dxdy​)2}0.6=dx2d2y​{2−(dydx)2}3/5=d2ydx2{2−(dxdy​)2}3/5=dx2d2y​

Raise power of 5 both sides

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76. (a) Given,

dydx=2exy3⇒dyy3=2exdxdxdy​=2exy3⇒y3dy​=2exdx

Integrating both sides wrt x

−12y2=2ex+C...(i)−2y21​=2ex+C...(i)

Using y(0)=12y(0)=21​, in equation (i)

−2=2+c⇒c=−4−2=2+c⇒c=−4

Now,

−12y2=2ex−4−2y21​=2ex−4⇒4y2(ex)−8y2=−1⇒4y2(2−ex)=1⇒4y2(ex)−8y2=−1⇒4y2(2−ex)=1

77. (d) Given,

P=∫abf(x)dx and q=∫ab∣f(x)∣dxP=∫ab​f(x)dx and q=∫ab​∣f(x)∣dx

As f(x)=e−xf(x)=e−x is always +ve ∀x∈R+ve ∀x∈R

⇒∣f(x)∣=∣e−x∣=e−x=f(x)⇒∣f(x)∣=∣e−x∣=e−x=f(x)⇒∫abf(x)dx=∫ab∣f(x)∣dx⇒∫ab​f(x)dx=∫ab​∣f(x)∣dx

Hence, p=qp=q

78. (a) Let

I=∫0π2a+sin⁡x2a+sin⁡x+cos⁡xdx...(i)I=∫02π​​2a+sinx+cosxa+sinx​dx...(i)⇒I=∫0π2a+cos⁡x2a+cos⁡x+sin⁡xdx...(ii)⇒I=∫02π​​2a+cosx+sinxa+cosx​dx...(ii)[∫0af(x)dx=∫0af(a−x)dx][∫0a​f(x)dx=∫0a​f(a−x)dx]

Adding equation (i) and (ii)

⇒2I=∫0π22a+sin⁡x+cos⁡x2a+sin⁡x+cos⁡xdx=∫0π2dx=[x]0π2⇒I=π4⇒2I=∫02π​​2a+sinx+cosx2a+sinx+cosx​dx=∫02π​​dx=[x]02π​​⇒I=4π​

79. (d) Given

f(x)=16x33−4bx2+xf(x)=316x3​−4bx2+xf′(x)=16x2−8bx+1f′(x)=16x2−8bx+1

Since f(x)f(x) is neither maximum nor minimum, then f′(x)≠0f′(x)=0

⇒f′(x)>0 or f′(x)<0 (but here a>0 for f′(x) so not possible)⇒f′(x)>0 or f′(x)<0 (but here a>0 for f′(x) so not possible)

for f′(x)>0f′(x)>0 Discriminant of f′(x)f′(x) is D<0D<0 and a>0a>0

D=64b2−64<0D=64b2−64<0b2−1<0 and b is non-negative is b≥0b2−1<0 and b is non-negative is b≥0b∈(−1,1) and b≥0⇒0≤b<1b∈(−1,1) and b≥0⇒0≤b<1

80. (a)

f(x)=1∣x∣−x for domain of f(x)=∣x∣−x>0f(x)=∣x∣−x​1​ for domain of f(x)=∣x∣−x>0

ie ∣x∣>x⇒x∈(−∞,0)∣x∣>x⇒x∈(−∞,0)

g(x)=1x−∣x∣, for domain of g(x), x−∣x∣>0g(x)=x−∣x∣​1​, for domain of g(x), x−∣x∣>0⇒∣x∣<x (Not possible : ∣x∣≥x)⇒∣x∣<x (Not possible : ∣x∣≥x)

Sol (81 - 82)
Let

I=∫3cos⁡x+4sin⁡x2cos⁡x+5sin⁡xdxI=∫2cosx+5sinx3cosx+4sinx​dx

Let 3cos⁡x+4sin⁡x=A(2cos⁡x+5sin⁡x)+B(ddx(2cos⁡x+5sin⁡x))3cosx+4sinx=A(2cosx+5sinx)+B(dxd​(2cosx+5sinx))

3cos⁡x+4sin⁡x=A(2cos⁡x+5sin⁡x)+B(−2sin⁡x+5cos⁡x)3cosx+4sinx=A(2cosx+5sinx)+B(−2sinx+5cosx)

Comparing coefficient of sin and cos

3=2A+5B...(i)3=2A+5B...(i)4=5A−2B...(ii)4=5A−2B...(ii)

Solving (i) and (ii) we get

A=2629,B=729A=2926​,B=297​

Now

I=∫2629(2cos⁡x+5sin⁡x)+729(−2sin⁡x+5cos⁡x)2cos⁡x+5sin⁡xdxI=∫2cosx+5sinx2926​(2cosx+5sinx)+297​(−2sinx+5cosx)​dx⇒I=∫2629dx+729∫−2sin⁡x+5cos⁡x2cos⁡x+5sin⁡xdx⇒I=∫2926​dx+297​∫2cosx+5sinx−2sinx+5cosx​dx⇒I=2629x+729ln⁡∣2cos⁡x+5sin⁡x∣+c⇒I=2926​x+297​ln∣2cosx+5sinx∣+ca=26, β=7 (on comparing)a=26, β=7 (on comparing)

81. (d) α=26α=26

82. (a) β=7β=7

83. (d) Given

f(x)=xln⁡x (x>1)f(x)=lnxx​ (x>1)

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Differentiate w.r.t x

f′(x)=ln⁡x−1(ln⁡x)2f′(x)=(lnx)2lnx−1​

f′(x)f′(x) is +ve+ve when x∈(e,∞)x∈(e,∞) ie f(x)f(x) is increasing in interval (e,∞)⇒(e,∞)⇒ (A) is correct.
f′(x)f′(x) is −ve−ve, when x∈(1,e)x∈(1,e) ie f(x)f(x) is decreasing in the interval (1,e)⇒(1,e)⇒ (B) is correct.
Since ln⁡xlnx is an increasing function.
7>97>9?
ln⁡7>ln⁡9ln7>ln9?
9ln⁡7>7ln⁡9⇒9ln7>7ln9⇒ statement (C) is correct
Hence, statement (A), (B), and (C) are correct.

84. (d) Again differentiating w.r.t x

f′′(x)=(ln⁡x)2−2(ln⁡x)2+2ln⁡xx(ln⁡x)4f′′(x)=x(lnx)4(lnx)2−2(lnx)2+2lnx​f′′(e)=1−2+2e×1=1e (e>0)⇒statement (A) is correctf′′(e)=e×11−2+2​=e1​ (e>0)⇒statement (A) is correct

As f′′(e)>0f′′(e)>0 (+ve)
So, f(x)f(x) attains local minima at x=e⇒x=e⇒ statement (B) is correct
A local minimum value occurs at x=ex=e,
f(x)f(x) at x=ex=e, is f(e)=ef(e)=e

f(e)=eln⁡e=e⇒statement (C) is correctf(e)=lnee​=e⇒statement (C) is correct

Hence, statement (A), (B) and (C) are correct.

Sol (85 - 86)
Given:

g(x)=x−1x and f(g(x))=x3−1x3g(x)=x−x1​ and f(g(x))=x3−x31​f(x−1x)=x3−1x3⇒f(x−1x)=(x−1x)3+3(x−1x)f(x−x1​)=x3−x31​⇒f(x−x1​)=(x−x1​)3+3(x−x1​)

85. (a) f(x)=x3+3xf(x)=x3+3x ...(i)
So,

g[f(x)−3x]=f(x)−3x−1f(x)−3xg[f(x)−3x]=f(x)−3x−f(x)−3x1​g[f(x)−3x]=x3+3x−3x−1x3+3x−3x=x3−1x3g[f(x)−3x]=x3+3x−3x−x3+3x−3x1​=x3−x31​g[f(x)−3x]=x3−1x3g[f(x)−3x]=x3−x31​

86. (d) f′(x)=3x2+3f′(x)=3x2+3
f′′(x)=6xf′′(x)=6x

Sol (87 - 88)
Given f(x)=∣x∣+1f(x)=∣x∣+1 and g(x)=[x]−1g(x)=[x]−1,

h(x)=f(x)g(x)=∣x∣+1[x]−1h(x)=g(x)f(x)​=[x]−1∣x∣+1​

87. (a) (A) for x<0x<0, f(x)=−x+1f(x)=−x+1
so f′(x)=−1f′(x)=−1 then f(x)f(x) is differentiable ∀x<0∀x<0
⇒⇒ statement (A) is correct
(B) at x=.0001x=.0001
g(x)g(x) is continuous every where except for integer value as [x][x] is continuous ∀x∈R∀x∈R except integers.
So g(x)g(x) is continuous at x=.0001⇒x=.0001⇒ statement (B) is correct
(C) g(x)=[x]−1g(x)=[x]−1, [x]⇒[x]⇒ Always a integer

g′(x)=ddx[x]−0g′(x)=dxd​[x]−0

g′(x)=0−0=0⇒g′(x)=0−0=0⇒ statement (C) is not correct.
(A), and (B) only correct.

88. (a) Consider

lim⁡x→0−h(x)+lim⁡x→0+h(x)x→0−lim​h(x)+x→0+lim​h(x)=lim⁡x→0−∣x∣+1[x]−1+lim⁡x→0+∣x∣+1[x]−1=x→0−lim​[x]−1∣x∣+1​+x→0+lim​[x]−1∣x∣+1​=0+1−1−1+0+10−1{[0−h]=−1}=−1−10+1​+0−10+1​{[0−h]=−1}=−12−1=−32=−21​−1=−23​

89. (d) Given

ϕ(a)=∫aa+100π∣sin⁡x∣dxϕ(a)=∫aa+100π​∣sinx∣dx

Since period of ∣sin⁡x∣=π∣sinx∣=π and ∣sin⁡x∣=sin⁡x∣sinx∣=sinx in the interval 0 to ππ

ϕ(a)=∫0100π∣sin⁡x∣dx=100∫0π∣sin⁡x∣dx=100∫0πsin⁡xdxϕ(a)=∫0100π​∣sinx∣dx=100∫0π​∣sinx∣dx=100∫0π​sinxdxϕ(a)=100(−cos⁡x)0π=100×2=200ϕ(a)=100(−cosx)0π​=100×2=200

90. (a) ϕ(a)=200ϕ(a)=200
Differentiating w.r.t a
ϕ′(a)=0ϕ′(a)=0

Sol (91 - 92)
Given,

y=2f(x)+ax−b...(i)y=2f(x)+ax−b...(i)

Differentiating w.r.t. 'x' we get,

⇒dydx=2f′(x)+a(1)...(ii)⇒dxdy​=2f′(x)+a(1)...(ii)

Again differentiating w.r.t. x, we get,

⇒d2ydx2=2f′′(x)...(iii)⇒dx2d2y​=2f′′(x)...(iii)

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91. (c) Since f(x)f(x) has a local maximum at x=0x=0. Then

f′(0)=0 and f′′(0)<0...(iv)f′(0)=0 and f′′(0)<0...(iv)

So, option (c) is correct.

92. (b) Since yy has a relative maxima at x=0x=0 Then

(dydx)x=0=0(dxdy​)x=0​=0⇒2f′(0)+a=0{from (i)}⇒2f′(0)+a=0{from (i)}⇒2(0)+a=0{from (i)}⇒2(0)+a=0{from (i)}a=0a=0

also

(d2ydx2)x=0<0⇒2f′′(0)<0(dx2d2y​)x=0​<0⇒2f′′(0)<0⇒f′′(0)<0⇒f′′(0)<0

So, yy has a relative maxima for a=0a=0 and all value of bb

Sol (93 - 94): We have given

f(x)=∣x−1∣...(i)f(x)=∣x−1∣...(i)g(x)=[x]...(ii)g(x)=[x]...(ii)

and h(x)=f(x)⋅g(x)h(x)=f(x)⋅g(x)

⇒h(x)=∣x−1∣⋅[x]...(iii)⇒h(x)=∣x−1∣⋅[x]...(iii)

93. (a) Now

∫−10h(x)dx=∫−10f(x)⋅g(x)⋅dx=∫−10∣x−1∣⋅[x]dx∫−10​h(x)dx=∫−10​f(x)⋅g(x)⋅dx=∫−10​∣x−1∣⋅[x]dx=∫−10(1−x)⋅(−1)dx=∫−10(x−1)dx=∫−10​(1−x)⋅(−1)dx=∫−10​(x−1)dx=[x22−x]−10=[0−(12+1)]=−32=[2x2​−x]−10​=[0−(21​+1)]=−23​

94. (d) Now

∫02h(x)dx=∫01∣x−1∣[x]dx+∫12∣x−1∣[x]dx∫02​h(x)dx=∫01​∣x−1∣[x]dx+∫12​∣x−1∣[x]dx=∫01(1−x)⋅(0)dx+∫12(x−1)(1)dx=∫01​(1−x)⋅(0)dx+∫12​(x−1)(1)dx=∫12(x−1)dx=[x22−x]12=∫12​(x−1)dx=[2x2​−x]12​=(42−2)−(12−1)=12=(24​−2)−(21​−1)=21​

Sol. (95 - 96) We have given,

∫dxx+1−x−1=α(x+1)3/2+β(x−1)3/2+C...(i)∫x+1​−x−1​dx​=α(x+1)3/2+β(x−1)3/2+C...(i)

Let

I=∫dxx+1−x−1=∫(x+1+x−1)dx(x+1−x−1)(x+1+x−1)I=∫x+1​−x−1​dx​=∫(x+1​−x−1​)(x+1​+x−1​)(x+1​+x−1​)dx​⇒I=12∫x+1dx+12∫x−1dx⇒I=21​∫x+1​dx+21​∫x−1​dx⇒I=12⋅(x+1)3/2(3/2)+12⋅(x−1)3/2(3/2)+C⇒I=21​⋅(3/2)(x+1)3/2​+21​⋅(3/2)(x−1)3/2​+C⇒I=13(x+1)3/2+13(x−1)3/2+C⇒I=31​(x+1)3/2+31​(x−1)3/2+C⇒∫dxx+1−x−1=13(x+1)3/2+13(x−1)3/2+C...(ii)⇒∫x+1​−x−1​dx​=31​(x+1)3/2+31​(x−1)3/2+C...(ii)

95. (a) Comparing (i) and (ii) we get

α=13α=31​

96. (c) Comparing (i) & (ii) we get

β=13β=31​

Sol (97 - 98)
Equation of circle

x2+y2−2x=0x2+y2−2x=0⇒(x−1)2+y2=1⇒(x−1)2+y2=1⇒y2=1−(x−1)2⇒y2=1−(x−1)2

Area of minor segment (A2)=∫01(1−(x−1)2−x)dx(A2​)=∫01​(1−(x−1)2​−x)dx

⇒A2=[x−121−(x−1)2+12sin⁡−1(x−1)−x22]01⇒A2​=[2x−1​1−(x−1)2​+21​sin−1(x−1)−2x2​]01​⇒A2=[(0+12(0)−12)−(0+12sin⁡−1(−1)−0)]⇒A2​=[(0+21​(0)−21​)−(0+21​sin−1(−1)−0)]⇒A2=π−24⇒A2​=4π−2​

97. (d) Since Area of given circle =π(1)2=π=π(1)2=π

⇒A1+A2=π⇒A1​+A2​=π⇒A1=π−π−24=3π+24⇒A1​=π−4π−2​=43π+2​

98. (a)

2(A1+A2)A1−3A2=2(π)(3π+24)−3(π−2)4=πA1​−3A2​2(A1​+A2​)​=(43π+2​)−43(π−2)​2(π)​=π

PART – A : ENGLISH (Solutions)

1. (b) The meaning of the word ‘deplorable’ is morally bad or deserving disapproval. The antonym for deplorable is ‘commendable’ (deserving praise).

2. (a) The meaning of the word ‘counteract’ is to reduce the effect of something by acting against it. The antonym is ‘exacerbate’ (to make something worse).

3. (c) The meaning of the word ‘persevere’ is to continue trying to achieve something that is difficult. The antonym is ‘to give up’.

4. (b) The meaning of the word ‘hideous’ is very ugly or unpleasant. The antonym is ‘beautiful’.

5. (d) The meaning of the word ‘squander’ is to waste time or money. The antonym is ‘manage’.

6. (c) Continually – The mentioned word is appropriately placed as the sentence talks about climate which keeps changing continuously.

7. (a) The mentioned word follows the first statement (changes have already been stated previously).

8. (b) Since the activities listed are carried out by humankind, the appropriate word is ‘human’.

9. (a) The word ‘assimilate’ means to be taken in or absorbed. The sentence becomes: “emissions from industry and transport… which have led to the assimilation of…”

10. (c) Article ‘the’ is the correct word to fill in the blank since the sentence is a stated fact.

11. (a) The greenhouse effect allows greenhouse gases to trap the heat from the earth’s surface and raise air temperatures.

12. (d) The greenhouse gases act on the surface of the earth (the planet) in order to protect it.

13. (a) The preposition ‘for’ is appropriate as the sentence talks about assessment of climate change.

14. (d) “have initiated” correctly fits the sentence.

15. (b) The preposition ‘at’ completes the sentence (activity takes place at regular intervals).

16. (b) The complete sentence is “He paused for a few moments.”

17. (a) In relation to vehicles, they “break down” when something goes wrong.

18. (b) “except” is the correct word before the remaining two countries.

19. (a) In the case of electricity, the lights are always “turned off” or “switched off”.

20. (c) When you refer to an individual, you use ‘who’.

21. (c) When referring to months, the preposition ‘in’ is used.

22. (b) The determiner ‘that’ is used to modify or introduce a noun.

23. (b) The grammatically correct sentence is: “The thieves denied stealing the money.”

24. (b) The correct sentence is: “Did you drop in to see Sunita on your way home?”

25. (b) “Would you like to meet her?” is the grammatically correct sentence.

26. (c) The idiom “be bad news” refers to someone who is considered undesirable.

27. (a) The idiom “back to the drawing board” refers to starting planning again because the previous plan has failed.

28. (d) The idiom “be in the eye of the storm” refers to being in the midst of a controversy.

29. (d) The idiom “life in the fast lane” refers to an exciting and eventful lifestyle.

30. (b) The idiom “to pass the buck” refers to shifting the responsibility to someone else.

31. (a) Correct sentence: “The fundamental rights are basic rights and include basic freedoms guaranteed to the individual.”

32. (a) Correct sentence: “The Indian nation is the product of a historical process that has been in the making for many millennia.”

33. (a) Correct sentence: “India’s Independence represented for its people the start of an era…”

34. (a) Correct sentence: “All students have been instructed to be present before the principal…”

35. (a) Correct sentence: “With the captain of the team injured, the team has two bowlers and two batsmen respectively.”

36. (a) ‘Abode’ is a noun (residence, dwelling or habitat).

37. (a) ‘Beyond’ is used as an adverb.

38. (d) ‘Except’ acts as a conjunction.

39. (b) ‘Nicely’ is used as an adverb.

40. (a) ‘This’ has been used as a demonstrative determiner.

41. (c) Human belongs to abiotic community? (No – correction in context of biotic/abiotic).

42. (b) The nature of biotic community largely depends on climate.

43. (a) Algae is a living organism.

44. (b) Aquatic community refers to the biotic community living in water.

45. (a) Species are defined on the basis of both similarities as well as differences.

46. (a) ‘Deported’ is a synonym for ‘banished’.

47. (a) ‘Brilliant’ is a synonym for ‘dazzling’.

48. (b) ‘Rejected’ is a synonym for ‘rebuffed’.

49. (c) ‘Abandoned’ is a synonym for ‘disowned’.

50. (c) ‘Beautiful’ is a synonym for ‘exquisite’.

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PART – B : GENERAL KNOWLEDGE (Solutions)

51. (b) Equivalent circuit and net resistance calculations lead to the correct option.

52. (d) Potential difference and current through resistor R calculated using series-parallel combination.

53. (a) Charge Q = I × t = 15 × 10 × 60 = 9000 C.

54. (b) Lightning conductor is made of metal rod with sharp pointed edge to provide low resistance path.

55. (d) Four fundamental forces: Gravitational, Electromagnetic, Strong nuclear, Weak nuclear.

56. (b) Density and pressure relations for large sphere.

57. (a) Human eye has variable focal length and variable aperture size.

58. (c) Article 371A – special provisions for Nagaland.

59. (c) Article 191 – Disqualifications for membership.

60. Important international conventions dates:

• Universal Declaration of Human Rights – 10 December 1948
• Convention Relating to the Status of Refugees – 28 July 1951
• Convention on the Elimination of all Forms of Discrimination Against Women – 1979

61. (c) Main objectives of the Second Five Year Plan (1956-61): rapid industrialisation, increase in national income, expansion of employment.

62. Mahajanapadas period: 600–300 BCE.

63. (a) Earliest deciphered inscriptions – Ashoka’s edicts in Prakrit using Brahmi script (3rd century BCE).

64. (a) Abundance of ¹⁰B ≈ 19–20%, ¹¹B ≈ 80–81%.

65. (b) Portland cement: lime, silica, alumina (with gypsum).

66. (c) Litmus is derived from lichens.

67. (a) Oxidation number of iron in Fe₃O₄ is +8/3 (or mixture of +2 and +3).

68. (a)

• Graphite – soft and slippery
• Diamond – hardest natural substance
• Fullerene – light and strong
• Graphene – thinnest and strongest

69. (a) International Date Line does not cut across any country.

70. (d) Black soil (Regur) formed by weathering of Deccan lava.

71–150. Remaining solutions cover:

• Cloud classification
• Fronts and weather
• Soils (red, laterite, black)
• Ports (Mormugao, Rourkela Steel Plant)
• Rivers (Godavari, Brahmaputra, Chambal, Betwa)
• ITCZ and tropical climates
• Temperate climates and deciduous forests
• Chlorophyll (absorbs blue & red, reflects green)
• Prokaryotic chromosome number = 1
• Bacterial DNA is naked
• Lymphocytes produce antibodies
• Nuclear reactor components
• LIGO experiment
• Work-energy theorem applications
• Current affairs (Desert Cyclone exercise, 16th Finance Commission, Wetland City Accreditation – Indore, Bhopal, Udaipur, National Youth Day – 12 January, Simultaneous Elections Committee, All-girls Sainik School at Vrindavan, Free Movement Regime with Myanmar scrapped, book “Why Bharat Matters” by S. Jaishankar)

NDA Previous Year Question Paper 2024

NDA Previous Year Question paper 2024 Free pdf download with Solution