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 Multiple Choice QuestionsMultiple Choice Questions

1.

If D = open vertical bar table row 1 1 1 row 1 cell 1 plus straight x end cell 1 row 1 1 cell 1 plus straight y end cell end table close vertical barfor x ≠ 0, y ≠ 0 then D is

  • divisible by neither x nor y

  • divisible by both x and y

  • divisible by x but not y

  • divisible by y but not x


B.

divisible by both x and y

straight D space equals space open vertical bar table row 1 1 1 row 1 cell 1 plus straight x end cell 1 row 1 1 cell 1 plus straight y end cell end table close vertical bar
C2 → C2 – C1 & C3 → C3 – C1
open vertical bar table row 1 0 0 row 1 straight x 0 row 1 0 straight y end table close vertical bar space equals space xy

Hence D is divisible by both x and y.
138 Views

2.

The system of equations

x + y + z = 0

2x + 3y + z = 0

and x + 2y = 0

has

  • a unique solution; x = 0, y = 0, z = 0

  • infinite solutions

  • no solution

  • finite number of non-zero solutions


B.

infinite solutions

The given system of equations are

    x + y + z = 0,

2x + 3y + z = 0,

and  x + 2y = 0

Here, 111231120 = 10 - 2 - 10 - 1 + 14 - 3                      = - 2 + 1 + 1 = 0  This system has infinite solutions.


3.

Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr (A), the sum of diagonal entries of A. Assume that A2= I.
Statement −1: If A ≠ I and A ≠ − I, then det A = − 1.
Statement −2: If A ≠ I and A ≠ − I, then tr (A) ≠ 0.

  • Statement −1 is false, Statement −2 is true

  • 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. 


D.

Statement − 1 is true, Statement − 2 is false. 

Let space straight A space equals space open square brackets table row straight a straight b row straight c straight d end table close square brackets space so space that space straight A squared space equals space open square brackets table row cell straight a squared plus bc end cell cell ab space plus space bd end cell row cell ac space plus dc end cell cell bc plus straight d squared end cell end table close square brackets space equals space open square brackets table row 1 0 row 0 1 end table close square brackets
rightwards double arrow space straight a squared space plus bc space equals space 1 space equals space bc plus straight d squared
and space left parenthesis straight a plus straight d right parenthesis straight c space equals space 0 space left parenthesis straight a plus straight d right parenthesis straight b.
Since space space straight A space not equal to space straight I comma space straight A space not equal to space 1 comma space straight a space equals space minus straight d space and space hence space det space straight A space equals space open vertical bar table row cell square root of 1 minus bc end root end cell straight b row straight c cell negative square root of 1 minus bc end root end cell end table close vertical bar
space equals space minus 1 plus bc minus bc space equals space minus 1
Statement space 1 space is space true
But space straight A space equals space 0 space space and space hence space statement space 2 space is space false.
175 Views

4.

Let A =  open square brackets table row 5 cell 5 straight alpha end cell straight alpha row 0 straight alpha cell 5 straight alpha end cell row 0 0 5 end table close square brackets  If |A2| = 25, then |α| equals

  • 52

  • 1

  • 1/5

  • 5


C.

1/5

straight A squared space equals space open square brackets table row 5 cell 5 straight alpha end cell straight alpha row 0 straight alpha cell 5 straight alpha end cell row 0 0 5 end table close square brackets open square brackets table row 5 cell 5 straight alpha end cell straight alpha row 0 straight alpha cell 5 straight alpha end cell row 0 0 5 end table close square brackets
straight A squared space equals space open square brackets table row 25 cell 25 straight alpha plus 5 straight alpha squared end cell cell 5 straight alpha end cell row 0 cell straight alpha squared end cell cell 5 straight alpha squared plus 25 straight alpha end cell row 0 0 25 end table close square brackets
625 space straight alpha squared space equals space 25
rightwards double arrow space vertical line straight alpha vertical line space equals space 1 fifth
143 Views

5.

Let straight A space equals space open parentheses table row 1 2 row 3 4 end table close parentheses space and space straight B space equals open parentheses table row straight a 0 row 0 straight b end table close parentheses a , b ∈ N. Then

  • there cannot exist any B such that AB = BA

  • there exist more than one but finite number of B’s such that AB = BA

  • there exists exactly one B such that AB = BA

  • there exist infinitely many B’s such that AB = BA


D.

there exist infinitely many B’s such that AB = BA

straight A space equals open square brackets table row 1 2 row 3 4 end table close square brackets space space space space straight B space equals open square brackets table row straight a 0 row 0 straight b end table close square brackets
AB space equals open square brackets table row straight a cell 2 straight b end cell row cell 3 straight a end cell cell 4 straight b end cell end table close square brackets
BA space equals space open square brackets table row straight a 0 row 0 straight b end table close square brackets open square brackets table row 1 2 row 3 4 end table close square brackets space equals space open square brackets table row straight a cell 2 straight a end cell row cell 3 straight b end cell cell 4 straight b end cell end table close square brackets
AB = BA only when a = b
146 Views

6.

Let A be a square matrix all of whose entries are integers. Then which one of the following is true?

  • If det A = ± 1, then A–1 exists but all its entries are not necessarily integers

  • If detA ≠ ± 1, then A–1 exists and all its entries are non-integers

  • If detA = ± 1, then A–1 exists and all its entries are integers

  • If detA = ± 1, then A–1 need not exist


C.

If detA = ± 1, then A–1 exists and all its entries are integers

Each entry of A is integer, so the cofactor of every entry is an integer and hence each entry in the adjoint of matrix A is integer. Now detA = ± 1 and A–1 =(1/ det(A)) (adj A)

⇒ all entries in A–1 are integers

195 Views

7.

The simultaneous equations Kx + 2y - z = 1, (K - I)y - 2z = 2 and (K + 2)z = 3 have only one solution when :

  • K = - 2

  • K = - 1

  • K = 0

  • K = 1


B.

K = - 1

The system of given equations are

  Kx + 2y - z = 1             ...(i)

 (K - 1)y - 2z = 2           ...(ii)

and (K + 2)z = 3           ...(iii)

This system of equations has a unique solution, if

K2- 10K - 1- 200K + 2  0 K + 2K20K - 1  0 K + 2KK - 1  0 K  - 2, 0, 1i.e., K = - 1, is a required answer.


8.

If x = - 5 is a root of 2x + 14822x2762x = 0, then the other roots are :

  • 3, 3.5

  • 1, 3.5

  • 1, 7

  • 2, 7


B.

1, 3.5

2x + 14822x2762x = 0 2x + 102x + 102x + 1022x2762x = 0                            R1  R1 + R2 + R3 2x + 1011122x2762x = 0 2x + 1010022x - 207- 12x - 7 = 0          C3  C3 - C1 and C2  C2 - C1 2x +102x - 22x - 7 = 0 x = - 5, 1, 72Hence, other roots are 1 and 72.


9.

If a1, a2, a3,…, an,… are in G.P., then the determinant

space increment space equals space open vertical bar table row cell log space straight a subscript straight n end cell cell log space straight a subscript straight n plus 1 end subscript end cell cell log space straight a subscript straight n plus 2 end subscript end cell row cell log space straight a subscript straight n plus 3 end subscript end cell cell log space straight a subscript straight n plus 4 end subscript end cell cell log space straight a subscript straight n plus 5 end subscript end cell row cell log space straight a subscript straight n plus 6 end subscript end cell cell log space straight a subscript straight n plus 7 end subscript end cell cell log space straight a subscript straight n plus 8 end subscript end cell end table close vertical bar space is space equal space to

  • 1

  • 0

  • 4

  • 2


B.

0

C1 – C2, C2 – C3 two rows becomes identical Answer: 0

480 Views

10.

Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to

  • 2

  • -1

  • 0

  • 1


D.

1

The system of equations x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 have non-trivial solution if

open vertical bar table row 1 cell negative straight c end cell cell negative straight b end cell row straight c cell negative 1 end cell straight a row straight b straight a cell negative 1 end cell end table close vertical bar space equals space 0 space
rightwards double arrow space 1 left parenthesis space 1 minus straight a squared right parenthesis space plus space straight c left parenthesis negative straight c minus ab right parenthesis minus straight b left parenthesis ca plus straight b right parenthesis space equals space 0 space
rightwards double arrow space straight a squared space plus straight b squared space plus space straight c squared space plus space 2 abc space equals space 1

391 Views