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Chemistry-

General

Easy

Question

(A) Di-tert. butyl ether cannot be prepared by Williamson's ether synthesis.
(R) Tert. butyl bromide on n:eatment with sodium tert. butoxide preferentially undergoes elimination to form isobutylene and tert. butyl alcohol.

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

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

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In the following reaction,

‘A' is:

In the following reaction,

‘A' is:

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The number of solutions the equation $x to the power of 3 end exponent plus 2 x to the power of 2 end exponent plus 5 x plus 2 c o s invisible function application x equals 0 text in end text left square bracket 0 , 2 pi right square bracket$ is

The number of solutions the equation $x to the power of 3 end exponent plus 2 x to the power of 2 end exponent plus 5 x plus 2 c o s invisible function application x equals 0 text in end text left square bracket 0 , 2 pi right square bracket$ is

Which of the following is dihydric alcohol?

Which of the following is dihydric alcohol?

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The most general values of $theta$ which satisfy $s i n invisible function application theta equals negative fraction numerator 1 over denominator 2 end fraction comma t a n invisible function application theta equals fraction numerator 1 over denominator square root of 3 end fraction$ are

The most general values of $theta$ which satisfy $s i n invisible function application theta equals negative fraction numerator 1 over denominator 2 end fraction comma t a n invisible function application theta equals fraction numerator 1 over denominator square root of 3 end fraction$ are

The number of pairs $left parenthesis x comma y right parenthesis$ satisfying $4 x to the power of 2 end exponent minus 4 x plus 2 equals s i n to the power of 2 end exponent invisible function application y text and end text blank x to the power of 2 end exponent plus y to the power of 2 end exponent less or equal than 3$ are

The number of pairs $left parenthesis x comma y right parenthesis$ satisfying $4 x to the power of 2 end exponent minus 4 x plus 2 equals s i n to the power of 2 end exponent invisible function application y text and end text blank x to the power of 2 end exponent plus y to the power of 2 end exponent less or equal than 3$ are

The number of values of x in the interval the number of values of x in the interval $blank left square bracket 0 , 5 pi right square bracket$ satisfying the equation $text end text 3 s i n to the power of 2 end exponent invisible function application x minus 7 s i n invisible function application x plus 2 equals 0 text end text$ is

The number of values of x in the interval the number of values of x in the interval $blank left square bracket 0 , 5 pi right square bracket$ satisfying the equation $text end text 3 s i n to the power of 2 end exponent invisible function application x minus 7 s i n invisible function application x plus 2 equals 0 text end text$ is

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The number of solutions of the equation $vertical line s i n invisible function application x vertical line equals vertical line x minus 1 vertical line$ in the interval $text end text left square bracket 0 comma pi right square bracket text is end text$

The number of solutions of the equation $vertical line s i n invisible function application x vertical line equals vertical line x minus 1 vertical line$ in the interval $text end text left square bracket 0 comma pi right square bracket text is end text$

Reaction of esters with Grignard reagents give rise to:

Reaction of esters with Grignard reagents give rise to:

Chemistry-General

The set of values of x for which the inequality $blank s i n to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses plus c o s to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses less or equal than fraction numerator 1 over denominator 2 end fraction text end text$ holds is

The set of values of x for which the inequality $blank s i n to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses plus c o s to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses less or equal than fraction numerator 1 over denominator 2 end fraction text end text$ holds is

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The equation |sinx| + |cosx| = 3/2 has

The equation |sinx| + |cosx| = 3/2 has

Among the following compounds which can be dehydrated very easily?

Among the following compounds which can be dehydrated very easily?

Chemistry-General

$I f blank stack lim with x rightwards arrow infinity below invisible function application open parentheses fraction numerator x plus h over denominator x minus h end fraction close parentheses to the power of x end exponent equals 4 comma blank t h e n blank h equals$

$I f blank stack lim with x rightwards arrow infinity below invisible function application open parentheses fraction numerator x plus h over denominator x minus h end fraction close parentheses to the power of x end exponent equals 4 comma blank t h e n blank h equals$

parallel

Arrange the following limits in descending order
i) $stack lim with x rightwards arrow 3 to the power of minus end exponent below invisible function application fraction numerator 3 minus x over denominator open vertical bar x minus 3 close vertical bar end fraction$
ii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator left parenthesis 1 plus x right parenthesis to the power of 1 divided by 8 end exponent minus left parenthesis 1 minus x right parenthesis to the power of 1 divided by 8 end exponent over denominator x end fraction$
iii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator t a n x minus s i n x over denominator x to the power of 3 end exponent end fraction$
iv) $stack lim with x rightwards arrow infinity below invisible function application open curly brackets square root of x to the power of 2 end exponent plus 6 x plus 7 end root minus x close curly brackets$

Arrange the following limits in descending order
i) $stack lim with x rightwards arrow 3 to the power of minus end exponent below invisible function application fraction numerator 3 minus x over denominator open vertical bar x minus 3 close vertical bar end fraction$
ii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator left parenthesis 1 plus x right parenthesis to the power of 1 divided by 8 end exponent minus left parenthesis 1 minus x right parenthesis to the power of 1 divided by 8 end exponent over denominator x end fraction$
iii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator t a n x minus s i n x over denominator x to the power of 3 end exponent end fraction$
iv) $stack lim with x rightwards arrow infinity below invisible function application open curly brackets square root of x to the power of 2 end exponent plus 6 x plus 7 end root minus x close curly brackets$

$I f blank open square brackets x close square brackets d e n o t e s blank t h e blank g r e a t e s t blank i n t e g e r blank l e s s blank t h a n blank o r blank e q u a l blank t o blank x blank t h e n blank$ $stack lim with n rightwards arrow infinity below invisible function application fraction numerator 1 over denominator n to the power of 4 end exponent end fraction open curly brackets open square brackets 1 to the power of 3 end exponent x close square brackets plus open square brackets 2 to the power of 3 end exponent x close square brackets plus open square brackets 3 to the power of 3 end exponent x close square brackets plus horizontal ellipsis plus left square bracket n to the power of 3 end exponent x right square bracket close curly brackets equals$

$I f blank open square brackets x close square brackets d e n o t e s blank t h e blank g r e a t e s t blank i n t e g e r blank l e s s blank t h a n blank o r blank e q u a l blank t o blank x blank t h e n blank$ $stack lim with n rightwards arrow infinity below invisible function application fraction numerator 1 over denominator n to the power of 4 end exponent end fraction open curly brackets open square brackets 1 to the power of 3 end exponent x close square brackets plus open square brackets 2 to the power of 3 end exponent x close square brackets plus open square brackets 3 to the power of 3 end exponent x close square brackets plus horizontal ellipsis plus left square bracket n to the power of 3 end exponent x right square bracket close curly brackets equals$

$stack lim with x rightwards arrow 0 below invisible function application open parentheses c o s e c x minus fraction numerator 1 over denominator x end fraction close parentheses equals$

$stack lim with x rightwards arrow 0 below invisible function application open parentheses c o s e c x minus fraction numerator 1 over denominator x end fraction close parentheses equals$

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