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Maths-

General

Easy

Question

The no. of real solutions of (x,y) where $vertical line y vertical line equals sin invisible function application x comma y equals cos to the power of negative 1 end exponent invisible function application left parenthesis cos invisible function application x right parenthesis comma left parenthesis negative 2 pi 2 pi right parenthesis$ are

2
1
3
4

The correct answer is: 1

Related Questions to study

The molecules
are:

The molecules
are:

Chemistry-General

(A) The boiling point 6f cis-l,2-dichloroethene is higher than corresponding trans-isomer.
(R) The dipole moment of cis-l,2-dichloroethene is higher than trans-isomers.

(A) The boiling point 6f cis-l,2-dichloroethene is higher than corresponding trans-isomer.
(R) The dipole moment of cis-l,2-dichloroethene is higher than trans-isomers.

Chemistry-General

(A) Metamers can also be chain or position isomers :
(R) The term tautomerism was introduced to explain the reactivity of a substance . according to two possible structures.

(A) Metamers can also be chain or position isomers :
(R) The term tautomerism was introduced to explain the reactivity of a substance . according to two possible structures.

Chemistry-General

parallel

(A) Following amide exist in two Structural forms:
I)
II)
(R) Rotation about carbon nitrogen bond is restricted due to resonance.·

(A) Following amide exist in two Structural forms:
I)
II)
(R) Rotation about carbon nitrogen bond is restricted due to resonance.·

Chemistry-General

If $square root of 2$ cosA = cosB + cos³B, $square root of 2$ sinA = sinB - sin³B. Then |sin(A-B)| =

If $square root of 2$ cosA = cosB + cos³B, $square root of 2$ sinA = sinB - sin³B. Then |sin(A-B)| =

Which of the following is chiral?

Which of the following is chiral?

Chemistry-General

parallel

statement I: Minimum value of $left parenthesis 3 s i n invisible function application theta minus 4 c o s invisible function application theta right parenthesis left parenthesis 3 c o s invisible function application theta plus 4 s i n invisible function application theta right parenthesis$ is $negative fraction numerator 25 over denominator 2 end fraction$
statement II: Minimum value of $9 t a n to the power of 2 end exponent invisible function application x plus 16 c o t to the power of 2 end exponent invisible function application x$ is 24 Which of the above statements is correct?

statement I: Minimum value of $left parenthesis 3 s i n invisible function application theta minus 4 c o s invisible function application theta right parenthesis left parenthesis 3 c o s invisible function application theta plus 4 s i n invisible function application theta right parenthesis$ is $negative fraction numerator 25 over denominator 2 end fraction$
statement II: Minimum value of $9 t a n to the power of 2 end exponent invisible function application x plus 16 c o t to the power of 2 end exponent invisible function application x$ is 24 Which of the above statements is correct?

$open curly brackets open x element of R divided by l o g open square brackets open parentheses 1.6 close parentheses to the power of 1 minus x 2 end exponent minus open parentheses 0.625 close parentheses to the power of 6 open parentheses 1 plus x close parentheses end exponent close square brackets element of R close curly brackets equals close$

$open curly brackets open x element of R divided by l o g open square brackets open parentheses 1.6 close parentheses to the power of 1 minus x 2 end exponent minus open parentheses 0.625 close parentheses to the power of 6 open parentheses 1 plus x close parentheses end exponent close square brackets element of R close curly brackets equals close$

The extreme values of $4 c o s invisible function application open parentheses x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x to the power of 2 end exponent close parentheses$ over $R$ are

The extreme values of $4 c o s invisible function application open parentheses x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x to the power of 2 end exponent close parentheses$ over $R$ are

parallel

The period of $open parentheses t a n invisible function application theta minus fraction numerator 1 over denominator 3 end fraction t a n to the power of 3 end exponent invisible function application theta close parentheses open parentheses fraction numerator 1 over denominator 3 end fraction minus t a n to the power of 2 end exponent invisible function application theta close parentheses to the power of negative 1 end exponent$ , where $t a n to the power of 2 end exponent invisible function application theta not equal to fraction numerator 1 over denominator 3 end fraction$ is

The period of $open parentheses t a n invisible function application theta minus fraction numerator 1 over denominator 3 end fraction t a n to the power of 3 end exponent invisible function application theta close parentheses open parentheses fraction numerator 1 over denominator 3 end fraction minus t a n to the power of 2 end exponent invisible function application theta close parentheses to the power of negative 1 end exponent$ , where $t a n to the power of 2 end exponent invisible function application theta not equal to fraction numerator 1 over denominator 3 end fraction$ is

The minimum value of $27 t a n to the power of 2 end exponent invisible function application theta plus 3 c o t to the power of 2 end exponent invisible function application theta$ is

The minimum value of $27 t a n to the power of 2 end exponent invisible function application theta plus 3 c o t to the power of 2 end exponent invisible function application theta$ is

The maximum value of $open parentheses c o s invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o s invisible function application alpha subscript 2 end subscript close parentheses$ $open parentheses c o s invisible function application alpha subscript n end subscript close parentheses$ , under the restrictions $0 less or equal than alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma horizontal ellipsis. alpha subscript n end subscript less or equal than fraction numerator pi over denominator 2 end fraction$ and $open parentheses c o t invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o t invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses c o t invisible function application alpha subscript n end subscript close parentheses equals 1$

The maximum value of $open parentheses c o s invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o s invisible function application alpha subscript 2 end subscript close parentheses$ $open parentheses c o s invisible function application alpha subscript n end subscript close parentheses$ , under the restrictions $0 less or equal than alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma horizontal ellipsis. alpha subscript n end subscript less or equal than fraction numerator pi over denominator 2 end fraction$ and $open parentheses c o t invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o t invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses c o t invisible function application alpha subscript n end subscript close parentheses equals 1$

parallel

The maximum value of $s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is equal to

The maximum value of $s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is equal to

Range of $c o s to the power of 4 end exponent invisible function application x minus s i n to the power of 4 end exponent invisible function application x$ is

Range of $c o s to the power of 4 end exponent invisible function application x minus s i n to the power of 4 end exponent invisible function application x$ is

The maximum value of $16 c o s to the power of 5 end exponent invisible function application theta minus 20 c o s to the power of 3 end exponent invisible function application theta plus 5 c o s invisible function application theta$ is

The maximum value of $16 c o s to the power of 5 end exponent invisible function application theta minus 20 c o s to the power of 3 end exponent invisible function application theta plus 5 c o s invisible function application theta$ is

parallel

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