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Maths-

General

Easy

Question

Minimum value of $y equals 256 s i n to the power of 2 end exponent invisible function application x plus 324 c o s invisible function application e c to the power of 2 end exponent x for all x element of R$ is

432
504
580
753

The correct answer is: 580

$y equals 256 s i n to the power of 2 end exponent invisible function application x plus fraction numerator 324 over denominator sin to the power of 2 end exponent invisible function application x end fraction$ , When sin²x = 1, $y equals 256 plus 324 equals 580$
$y equals open parentheses 16 s i n invisible function application x plus fraction numerator 18 over denominator sin invisible function application x end fraction close parentheses to the power of 2 end exponent minus 2 cross times 16 cross times 18$ , min value of $f left parenthesis x right parenthesis equals 16 s i n invisible function application x plus fraction numerator 18 over denominator sin invisible function application x end fraction$
$f to the power of ´ end exponent left parenthesis x right parenthesis equals 16 c o s invisible function application x minus fraction numerator 18 c o s invisible function application x over denominator sin to the power of 2 end exponent invisible function application x end fraction equals 0$ , $c o s invisible function application x equals 0 text or end text s i n to the power of 2 end exponent invisible function application x equals fraction numerator 18 over denominator 16 end fraction$ (not possible)
minimum value = 580.

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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?

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

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

parallel

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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

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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

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