If a2 + b2 + c2 = -2 and  then f(x) is a polynomial of degree

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 Multiple Choice QuestionsMultiple Choice Questions

211.

Let A and B be two symmetric matrices of order 3.
Statement-1: A(BA) and (AB)A are symmetric matrices.
Statement-2: AB is symmetric matrix if matrix multiplication of A with B is commutative.

  • 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 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. 

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212.

The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is 

  • less than 4

  • 5

  • 6

  • 6

178 Views

213.

Let f : R → R be defined by
straight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell straight k minus 2 straight x comma if space straight x space less or equal than space minus 1 end cell row cell 2 straight x space plus 3 comma space if space straight x greater than negative 1 end cell end table close
If f has a local minimum at x = - 1 then a possible value of k is

  • 1

  • 0

  • -1/2

  • -1/2

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214.

If S is the set of distinct values of 'b' for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
has no solution, then S is

  • a singleton

  • an empty set

  • an infinite set

  • an infinite set

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215.

Let  ω be a complex number such that 2ω +1 = z where z = √-3. if

open vertical bar table row 1 1 1 row 1 cell negative straight omega squared end cell cell straight omega squared end cell row 1 cell straight omega squared end cell cell straight omega to the power of 7 end cell end table close vertical bar space equals space 3 straight k
then k is equal to 

  • 1

  • -z

  • z

  • z

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216.

Let a, b, c be such that 0 (a +c) ≠ . If open vertical bar table row cell space space space straight a end cell cell straight a plus 1 end cell cell space space straight a minus 1 end cell row cell negative straight b end cell cell straight b plus 1 end cell cell space space straight b minus 1 end cell row cell space space space straight c end cell cell space straight c minus 1 end cell cell space space straight c plus 1 end cell end table close vertical bar plus open vertical bar table row cell straight a plus 1 end cell cell straight b plus 1 end cell blank row cell straight a minus 1 end cell cell straight b minus 1 end cell cell straight c plus 1 end cell row cell left parenthesis negative 1 right parenthesis to the power of straight n plus 2 end exponent straight a end cell cell left parenthesis negative 1 right parenthesis to the power of straight n plus 1 end exponent straight b end cell cell left parenthesis negative 1 right parenthesis to the power of straight n space straight c end cell end table close vertical bar space equals space 0,then the value of 'n' is 

  • 0

  • any even integer

  • any odd integer

  • any odd integer

167 Views

217.

Let A be a 2 × 2 matrix
Statement 1 : adj (adj A) = A
Statement 2 : |adj A| = |A|

  • 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.

178 Views

218.

If A and B are square matrices of size n × n such that A2 − B2 = (A − B) (A + B), then which of the following will be always true?

  • A = B

  • AB = BA

  • either of A or B is a zero matrix

  • either of A or B is a zero matrix

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219.

If A2 – A + I = 0, then the inverse of A is

  • A + I

  • A

  • A – I

  • A – I

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220.

If a2 + b2 + c2 = -2 and straight f left parenthesis straight x right parenthesis space equals space open vertical bar table row cell 1 plus straight a squared straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row cell left parenthesis 1 plus straight a squared right parenthesis straight x end cell cell 1 plus straight b squared straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row cell left parenthesis 1 plus straight a squared right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell 1 plus straight c squared straight x end cell end table close vertical bar then f(x) is a polynomial of degree

  • 1

  • 0

  • 2

  • 2


C.

2

straight f left parenthesis straight x right parenthesis space equals space open vertical bar table row cell 1 space plus left parenthesis straight a squared plus straight b squared plus straight c squared plus 2 right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis straight x end cell row cell 1 plus left parenthesis straight a squared plus straight b squared space plus straight c squared space plus 2 right parenthesis straight x end cell cell 1 plus straight b squared straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis straight x end cell row cell 1 plus space left parenthesis straight a squared plus straight b squared space plus straight c squared plus 2 right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell 1 plus straight c squared straight x end cell end table close vertical bar
Applying space straight C subscript 1 space rightwards arrow with space on top space straight C subscript 1 space plus straight C subscript 2 plus space straight C subscript 3
space equals space open vertical bar table row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row 1 cell 1 plus straight b squared space straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell 1 plus straight c squared space straight x end cell end table close vertical bar space therefore space straight a squared space plus straight b squared space plus straight c squared space plus 2 space equals space 0
straight f left parenthesis straight x right parenthesis space equals space space open vertical bar table row 0 cell straight x minus 1 end cell 0 row 0 cell 1 minus straight x end cell cell straight x minus 1 end cell row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell 1 plus straight c squared space straight x end cell end table close vertical bar space space semicolon space Applying space straight R subscript 1 space rightwards arrow space straight R subscript 1 minus straight R subscript 2
straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight x minus 1 right parenthesis squared
Hence space degree space equals space 2
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