Q 11 :    

Let A=[α16β], α>0, such that det(A) = 0 and 2×2. If I denotes  identity matrix, then the matrix (I+A)8 is:          [2025]

  • [25764514127]

     

  • [102551120141024]

     

  • [4161]

     

  • [7662551530509]

     

(4)

We have, A=[α16β], |A|=0

 αβ+6=0  αβ=6

Also, α+β=1          [Given]

 α=3 and β=2          ( α>0)

Now, A=[3162]  A2=[3162][3162]=[3162]

 A2=A

So, A=A2=A3=A4=A5=A6=A7

  (I+A)8=I+C18A7+C28A6+...+C88A8

                   =I+A(C18+C28+...+C88)=I+A(281)

                   =[1001]+[7652551530510]=[7662551530509].



Q 12 :    

For some, a, b, let f(x)=|a+sin xx1ba1+sin xxba1b+sin xx|, x0, limx0f(x)=λ+μa+νb. Then (λ+μ+ν)2 is equal to :          [2025]

  • 25

     

  • 9

     

  • 16

     

  • 36

     

(3)

f(x)=|a+sin xx1ba1+sin xxba1b+sin xx|, x0

Since, limx0f(x)=λ+μa+νb

 λ+μa+νb=limx0|a+sin xx1ba1+sin xxba1b+sin xx|

=|a+11ba1+1ba1b+1|=|a+11ba2ba1b+1|

=(a+1)[2(b+1)b]1[a(b+1)ab]+b×(a2a)

=(a+1)(b+2)aab=a+b+2

On comparing, we get

λ=2, μ=1 and ν=1

Hence, (λ+μ+ν)2=(2+1+1)2=16.



Q 13 :    

Let A=[aij]=[log5128log45log58log425]. If Aij is the cofactor of aij,Cij=k=12aik Ajk, 1i, j2, and C=[Cij], then 8|C| is equal to :          [2025]

  • 242

     

  • 288

     

  • 262

     

  • 222

     

1

From the given matrix A|A|=112

Here, Cij=k=12aikAjk, 1i, j2

C11=k=12a1kA1k=a11A11+a12A12=|A|=112

C12=k=12a1kA2k=0

C21=k=12a2kA1k=0

C22=k=12a2kA2k=|A|112

Now, C=[11/20011/2]  |C|=1214

Hence, 8|C| = 242.



Q 14 :    

Let M and m respectively be the maximum and the minimum values of f(x)=|1+sin2xcos2x4sin4xsin2x1+cos2x4sin4xsin2xcos2x4sin4x|, xR. Then M4m4 is equal to :          [2025]

  • 1295

     

  • 1040

     

  • 1215

     

  • 1280

     

(4)

f(x)=|1+sin2xcos2x4sin4xsin2x1+cos2x4sin4xsin2xcos2x4sin4x|, xR

=|2cos2x4sin4x21+cos2x4sin4x1cos2x4sin4x|  (Applying C1C1+C2)

=|2cos2x4sin4x0101cos2x1+4sin4x|  (Applying R2R2R1)

Expanding along R2, we get f(x) = 2(1 + 4 sin 4x) – 4 sin 4x

 f(x)=2+4sin4x

Maximum value of f(x), M = 6

Minimum vlaue of f(x), m = –2

  M4m4=1280.



Q 15 :    

Let I be the identity matrix of order 3×3 and for the matrix A=[λ23456712], |A|=1. Let B the inverse of the matrix (A adj (A2)). Then |(λB+I)| is equal to __________.          [2025]



38

Given, A=[λ23456712]

|A|=[λ23456712]=1

 λ(16)2(34)+3(39)=1

 16λ=48  λ=3

Given, B1=adj(A·adj(A2))

LetC=A·adj(A2)

AC=A2adj(A2)=|A|2·I=I

 C=A1          [ |A| = – 1]

Now, B1=adj(A1)  B=adj(A)

Now, λB+I=3B+I

Let P = 3B + I

 P = 3 adj(A) + I

 AP = A 3 adj(A)+ (A)

 AP = 3|A· I + A  AP = A – 3I

 |AP| = |A – 3I|

|A|·|P|=[023426711]

             = 0 – 2(–46) + 3(–18)

             = 92 – 54 = 39

 |P| = 38.
 



Q 16 :    

Let integers a, b[3,3] be such that a+b0. Then the number of all possible ordered pairs (a, b) for which |zaz+b|=1 and |z+1ωω2ωz+ω21ω21z+ω|=1, zC, where ω and ω2 are the roots of x2+x+1=0, is equal to __________.          [2025]



10

We have, a, bz, 3a, b3 and a+b0.

Also, |zaz+b|=1 and |z+1ωω2ωz+ω21ω21z+ω|=1

Applying C1C1+C2+C3

|1ωω21z+ω2111z+ω|=1          [ 1+ω+ω2=0]

On expanding, we get

z[(z2+(ω+ω)z+ω31)ωzω2ω2+ωω2zω]=1

 z(z2)=1  x3=1  z=1,ω,ω2

Case 1 : z=ω, then a+b0 and ab = –1

a = –3, b = –2; a = –2; b = –1;

a = –1, b = 0; a = 0, b = 1

a = 1, b = 2; a = 2, b = 3

Case 2 : z = 1; then ab = 2 and a+b0

a = –1, b = –3; a = 0, b = –2; a = 2, b = 0; a = 3, b = 1

Total pairs = 0.