If α, β ≠ 0 and f(n) = αn+ βn and = K(1-α)2(1-β)2(

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

If A = open square brackets table row cell 5 straight a end cell cell negative straight b end cell row 3 2 end table close square brackets and A adj A = AAT, then 5a +b is equal to

  • -1

  • 5

  • 4

  • 4

396 Views

2.

The system of linear equations x+λy−z=0; λx−y−z=0; x+y−λz=0 has a non-trivial solution for

  • infinitely many values of λ.

  • exactly one value of λ.

  • exactly two values of λ.

  • exactly two values of λ.

427 Views

3.

A = open square brackets table row 1 2 2 row 2 1 cell negative 2 end cell row straight a 2 straight b end table close square brackets is a matrix satisfying the equation AAT = 9I, Where I is 3 x 3 identity matrix, then the ordered pair (a,b) is equal to

  • (2,-1)

  • (-2,1)

  • (2,1)

  • (2,1)

459 Views

4.

The set of all values of λ for which the system of linear equations 

2x1-2x2+x3 = λx1
2x1- 3x2 + 2x3 = λx2
-x1 + 2x2 = λx3
a non- trivial solution.

  • is an empty set

  • is a singleton set

  • contains two elements

  • contains two elements

250 Views

Advertisement
Advertisement

5.

If α, β ≠ 0 and f(n) = αn+ βn and 

open vertical bar table row 3 cell 1 plus straight f left parenthesis 1 right parenthesis space space space space end cell cell 1 plus space straight f left parenthesis 2 right parenthesis end cell row cell 1 plus straight f left parenthesis 1 right parenthesis space space space space space end cell cell 1 plus straight f left parenthesis 2 right parenthesis space space space space space space end cell cell 1 plus straight f left parenthesis 3 right parenthesis end cell row cell 1 plus straight f left parenthesis 2 right parenthesis space space space end cell cell 1 plus straight f left parenthesis 3 right parenthesis space space space space space space space end cell cell 1 plus space straight f left parenthesis 4 right parenthesis end cell end table close vertical bar
= K(1-α)2(1-β)2(α- β)2, then K is equal to 

  • αβ 

  • 1/αβ 

  • 1

  • 1


C.

1

f(n) = αn + βn
f(1) = α + β
f(2) = α2 + β2 
f(3) =α3 + β3
f(4) = α4 + β4


Let Δ = open vertical bar table row 3 cell 1 plus straight f left parenthesis 1 right parenthesis space space space space end cell cell 1 plus space straight f left parenthesis 2 right parenthesis end cell row cell 1 plus straight f left parenthesis 1 right parenthesis space space space space space end cell cell 1 plus straight f left parenthesis 2 right parenthesis space space space space space space end cell cell 1 plus straight f left parenthesis 3 right parenthesis end cell row cell 1 plus straight f left parenthesis 2 right parenthesis space space space end cell cell 1 plus straight f left parenthesis 3 right parenthesis space space space space space space space end cell cell 1 plus space straight f left parenthesis 4 right parenthesis end cell end table close vertical bar

open vertical bar table row 3 cell 1 plus straight alpha space plus space straight beta space space space space end cell cell 1 plus space straight alpha squared space plus space straight beta squared space end cell row cell 1 plus straight alpha space plus space straight beta space space space space space end cell cell 1 plus straight alpha squared space plus space straight beta squared space space space space space space space end cell cell 1 plus straight alpha cubed space plus space straight beta cubed space end cell row cell 1 plus straight alpha squared space plus space straight beta squared space space end cell cell 1 plus straight alpha cubed space plus space straight beta cubed space space space space space space space space end cell cell 1 plus straight alpha to the power of 4 space plus space straight beta to the power of 4 space end cell end table close vertical bar
equals space open vertical bar table row cell 1.1 plus 1.1 plus 1.1 end cell cell 1.1 plus 1. straight alpha space plus 1 space. straight beta space space space space end cell cell 1 plus space 1. straight alpha squared space plus space 1. straight beta squared space end cell row cell 1.1 plus 1. straight alpha space plus 1. space straight beta space space space space space end cell cell 1.1 plus straight alpha. straight alpha space plus space straight beta. straight beta space space space space space space space end cell cell 1 plus straight alpha. straight alpha squared space plus space. ββ squared end cell row cell 1.1 plus 1. straight alpha squared space plus space 1. space straight beta squared space space end cell cell 1 plus straight alpha squared. straight alpha space plus space straight beta squared. straight beta space space space space space space space space end cell cell 1 plus straight alpha squared straight alpha squared space plus space straight beta squared. straight beta squared space end cell end table close vertical bar

equals space open vertical bar table row 1 1 1 row 1 straight alpha straight beta row 1 cell straight alpha squared end cell cell straight beta squared end cell end table close vertical bar open vertical bar table row 1 1 1 row 1 straight alpha straight beta row 1 cell straight alpha squared end cell cell straight beta squared end cell end table close vertical bar space equals space open vertical bar table row 1 1 1 row 1 straight alpha straight beta row 1 cell straight alpha squared end cell cell straight beta squared end cell end table close vertical bar squared
On expanding, we get
Δ = (1- α2 + (1-β2 )(α-β)2
Δ =  K(1-α)2(1-β)2(α- β)2
K=1
180 Views

Advertisement
6.

If A is a 3x3 non- singular matrix such that AAT = ATA, then BBT is equal to

  • l +B
  • l
  • B-1

  • B-1

284 Views

7.

Let P and Q be 3 × 3 matrices with P ≠ Q. If P3= Qand P2Q = Q2P, then determinant of(P2+ Q2) is equal to

  • -2

  • 1

  • 0

  • 0

309 Views

8.

The number of values of k for which the linear equations
4x + ky + 2z = 0
kx + 4y + z = 0
2x + 2y + z = 0
posses a non-zero solution is:

  • 3

  • 2

  • 1

  • 1

171 Views

Advertisement
9.

Consider the system of linear equation
x1 + 2x2 + x3 = 3
2x1 + 3x2 + x3 = 3
3x1 + 5x2 + 2x3 = 1
The system has

  • infinite number of solutions

  • exactly 3 solutions

  • a unique solution

  • a unique solution

146 Views

10.

Let A be a 2 × 2 matrix with non-zero entries and let A2 = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A.
Statement-1: Tr(A) = 0.
Statement-2: |A| = 1.

  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is true, Statement-2 is false.

152 Views

Advertisement