Maths-

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Statement - I For any real value of $theta not equal to left parenthesis 2 n plus 1 right parenthesis pi$ or $left parenthesis 2 straight n plus 1 right parenthesis pi divided by 2 comma straight n element of straight I$ the value of the expression $straight y equals fraction numerator cos squared begin display style space end style theta minus 1 over denominator cos squared begin display style space end style theta plus cos space theta end fraction text is end text straight y less or equal than$ 0 or y≥2 (either less than or equal to zero or greater than or equal to two)
Statement - II $sec space theta element of left parenthesis negative straight infinity comma negative 1 right square bracket union left square bracket 1 comma straight infinity right parenthesis$ for all real values of $theta$ .

Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I.
Statement-I is true, Statement-II is false
Statement-I is false, Statement-II is true

hint Hint:

In this question, given two statement. It is like assertion and reason. Statement1 is assertion and statement 2 is reason, Find the statement 1 is correct or not and the statement 2 correct or not if correct then is its correct explanation.

The correct answer is: Statement-I is false, Statement-II is true

Here we have to find the which statement is correct and if its correct explanation or not.
Firstly,
Statement-1: For any real value θ ≠ (2n + 1) π or (2n + 1) $straight pi over 2$ , nϵ I, the value of expression
y =( cos^2 θ -1)/ ( cos^2 θ + cos θ) is y ≤ 0 or y ≥ 2 ( either less than or equal to zero or greater than to two.
We have,
$y equals c o s squared theta minus 1 divided by divided by c o s squared theta plus c o s theta$
$equals negative s i n squared theta divided by c o s theta left parenthesis c o s theta plus 1 right parenthesis left square bracket s i n squared theta plus c o s squared theta equals 1 right square bracket$

$equals negative s i n squared theta divided by 2 c o s theta. C o s squared theta divided by 2 equals left parenthesis 4 s i n squared theta divided by 2. c o s squared theta divided by 2 right parenthesis divided by left parenthesis 2 c o s theta. C o s squared theta divided by 2 right parenthesis equals negative left parenthesis 2 s i n squared theta divided by 2 right parenthesis divided by c o s theta equals left parenthesis 1 minus c o s theta right parenthesis divided by c o s equals negative left square bracket s e c theta minus 1 right square bracket equals 1 minus s e c theta.$
Or
$y equals 1 minus s e c theta$
$s e c theta equals 1 minus y$
As the range of secx is ≤−1 and ≥1
Or
1−y<−1 and 1−y>1
Or
y>2 and y<0.
So statement I is False,
Statement-II – secθ∈(−∞,−1]∪[1,∞) for all real values of θ.
So we know that for all , range of sec x ≤−1 and ≥1,
So secθ∈(−∞,−1]∪[1,∞) for all real values of θ.
Therefore, statement-II is correct.
The correct answer is Statement – I is false, Statement – II is true.

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason

If A = $open square brackets table row cell cos invisible function application alpha end cell cell sin invisible function application alpha end cell row cell negative sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close square brackets$ , then A² equals-

Maths-General

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The root of the equation [x 1 2] $open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell 0 1 1 end cell row cell 1 0 1 end cell row cell 1 1 0 end cell end table close square brackets$ $open square brackets table row x row cell negative 1 end cell row 1 end table close square brackets$ = 0 is-

Maths-General

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If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

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If A and B are matrices of order m × n and n × n respectively, then which of the following are defined-

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The multiplicative inverse of A = $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ is

maths-General

General

chemistry-

(A) and (B) are:

chemistry-General

General

maths-

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

maths-General

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maths-

Inverse of the matrix $open square brackets table row 3 cell – 2 end cell cell – 1 end cell row cell – 4 end cell 1 cell – 1 end cell row 2 0 1 end table close square brackets$ is

maths-General

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maths-

If A = $open square brackets table row 2 cell – 1 end cell row cell – 1 end cell 2 end table close square brackets$ and A² – 4A – n I = 0, then n is equal to

maths-General

General

maths-

The value of x for which the matrix product $open square brackets table attributes columnalign center center center columnspacing 1em end attributes row 2 0 7 row 0 1 0 row 1 cell negative 2 end cell 1 end table close square brackets$ $open square brackets table row cell – x end cell cell 14 x end cell cell 7 x end cell row 0 1 0 row x cell – 4 x end cell cell – 2 x end cell end table close square brackets$ equal an identity matrix is :

maths-General

General

maths-

If A = $open square brackets table row 1 2 row 3 5 end table close square brackets$ , then A^–1 is equal to :

maths-General

General

maths-

If A is a singular matrix, then adj A is :

maths-General

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Statement - I For any real value of or the value of the expression 0 or y≥2 (either less than or equal to zero or greater than or equal to two)Statement - II for all real values of .

Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I.Statement-I is true, Statement-II is falseStatement-I is false, Statement-II is true

In this question, given two statement. It is like assertion and reason. Statement1 is assertion and statement 2 is reason, Find the statement 1 is correct or not and the statement 2 correct or not if correct then is its correct explanation.

The correct answer is: Statement-I is false, Statement-II is true

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Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I.
Statement-I is true, Statement-II is false
Statement-I is false, Statement-II is true

If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

If A = $open square brackets table row 3 1 row 7 5 end table close square brackets$ and A² + kI = 8A, then k equals

For any square matrix A = [aij], aij = 0, when i $not equal to$ j, then A is-

For any square matrix A = [aij], aij = 0, when i $not equal to$ j, then A is-

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

Inverse of the matrix $open square brackets table row 3 cell – 2 end cell cell – 1 end cell row cell – 4 end cell 1 cell – 1 end cell row 2 0 1 end table close square brackets$ is

Inverse of the matrix $open square brackets table row 3 cell – 2 end cell cell – 1 end cell row cell – 4 end cell 1 cell – 1 end cell row 2 0 1 end table close square brackets$ is

If A = $open square brackets table row 2 cell – 1 end cell row cell – 1 end cell 2 end table close square brackets$ and A² – 4A – n I = 0, then n is equal to

If A = $open square brackets table row 2 cell – 1 end cell row cell – 1 end cell 2 end table close square brackets$ and A² – 4A – n I = 0, then n is equal to

If A = $open square brackets table row 1 2 row 3 5 end table close square brackets$ , then A^–1 is equal to :

If A = $open square brackets table row 1 2 row 3 5 end table close square brackets$ , then A^–1 is equal to :