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Chemistry-

General

Easy

Question

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

If both assertion (A) and reason (R) are correct and reason (R) is correct explanation of assertion (A).
If both assertion (A) and reason (R) are. correct but reason (R) is not correct explanation of assertion (A).
If assertion (A) is correct but reason(R) is incorrect,
If assertion (A)is incorrect but reason (R)is correct.

The correct answer is: If both assertion (A) and reason (R) are correct and reason (R) is correct explanation of assertion (A).

Related Questions to study

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

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?

parallel

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

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

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

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

parallel

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 $open parentheses c o t invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o t invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses c o t invisible function application alpha subscript n end subscript close parentheses equals 1$

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

parallel

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

The range of $c o s invisible function application x plus 4 square root of 2 s i n invisible function application open parentheses x minus fraction numerator pi over denominator 4 end fraction close parentheses plus 6$ is

The range of $c o s invisible function application x plus 4 square root of 2 s i n invisible function application open parentheses x minus fraction numerator pi over denominator 4 end fraction close parentheses plus 6$ is

parallel

Period of $s i n invisible function application left parenthesis s i n invisible function application x right parenthesis plus s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is

Period of $s i n invisible function application left parenthesis s i n invisible function application x right parenthesis plus s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is

Period of $t a n invisible function application open parentheses x plus 8 x plus 27 x plus midline horizontal ellipsis plus n to the power of 3 end exponent x close parentheses$ is

Period of $t a n invisible function application open parentheses x plus 8 x plus 27 x plus midline horizontal ellipsis plus n to the power of 3 end exponent x close parentheses$ is

$A equals s i n to the power of 8 end exponent invisible function application theta plus c o s to the power of 14 end exponent invisible function application theta$ then for all values of $theta$

$A equals s i n to the power of 8 end exponent invisible function application theta plus c o s to the power of 14 end exponent invisible function application theta$ then for all values of $theta$

parallel

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