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Maths-

General

Easy

Question

The number of values of x in the interval the number of values of x in the interval $blank left square bracket 05 pi right square bracket$ 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

0
5
6
10

The correct answer is: 6

Hint: $sin invisible function application x element of open square brackets negative 11 close square brackets$
$3 s i n to the power of 2 end exponent invisible function application x minus 6 s i n invisible function application x minus s i n invisible function application x plus 2 equals 0 left parenthesis 3 s i n invisible function application x minus 1 right parenthesis left parenthesis s i n invisible function application x minus 2 right parenthesis equals 0$
sin x = 2 not possible
$s i n invisible function application x equals fraction numerator 1 over denominator 3 end fraction comma 6 text solutions end text$

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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:

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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

parallel

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Among the following compounds which can be dehydrated very easily?

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

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

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
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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) $stack lim with x rightwards arrow infinity below invisible function application open curly brackets square root of x to the power of 2 end exponent plus 6 x plus 7 end root minus x close curly brackets$

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

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parallel

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parallel

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parallel

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