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Maths-

General

Easy

Question

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

0
2
4
None of these

The correct answer is: 2

$text Hint: end text 0 less or equal than s i n to the power of 2 end exponent invisible function application y less or equal than 1$
$table row cell rightwards double arrow sin to the power of 2 end exponent invisible function application y equals 4 x to the power of 2 end exponent minus 4 x plus 2 end cell row cell text end text text L end text text e end text text t end text text end text f left parenthesis x right parenthesis equals 4 x to the power of 2 end exponent minus 4 x plus 2 equals left parenthesis 2 x minus 1 right parenthesis to the power of 2 end exponent plus 1 end cell end table$
$f left parenthesis x right parenthesis text has minimum value end text$
$rightwards double arrow text at end text x equals fraction numerator 1 over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application y equals 1$
$f left parenthesis x right parenthesis text has end text D less than 0$
$rightwards double arrow f left parenthesis x right parenthesis greater or equal than 0 left parenthesis 2 x minus 1 right parenthesis to the power of 2 end exponent plus 1 equals s i n to the power of 2 end exponent invisible function application y$
$rightwards double arrow x equals fraction numerator 1 over denominator 2 end fraction comma c o s invisible function application y equals 0$
$text Therefore, end text$
$y equals left parenthesis 2 n plus 1 right parenthesis fraction numerator pi over denominator 2 end fraction text minimum value at end text x equals fraction numerator 1 over denominator 2 end fraction$
$x to the power of 2 end exponent plus y to the power of 2 end exponent less or equal than 3 rightwards double arrow fraction numerator 1 over denominator 4 end fraction plus fraction numerator left parenthesis 2 n plus 1 right parenthesis to the power of 2 end exponent pi to the power of 2 end exponent over denominator 4 end fraction less or equal than 3 blank$
$1 plus left parenthesis 2 n plus 1 right parenthesis to the power of 2 end exponent pi to the power of 2 end exponent less or equal than 12$
$left parenthesis 2 n plus 1 right parenthesis to the power of 2 end exponent pi to the power of 2 end exponent less or equal than 11$
Only satisfies when n = 0, −1
$text That is, only two solutions satisfy end text open parentheses table row cell fraction numerator 1 over denominator 2 end fraction comma blank fraction numerator pi over denominator 2 end fraction end cell row cell negative blank open parentheses fraction numerator 1 over denominator 2 end fraction comma fraction numerator negative pi over denominator 2 end fraction close parentheses end cell end table close curly brackets$

Related Questions to study

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

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

parallel

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

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

parallel

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

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

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

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

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

parallel

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

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

If f (x) = $fraction numerator x over denominator sin invisible function application x end fraction text and g end text left parenthesis x right parenthesis text = end text fraction numerator x over denominator tan invisible function application x end fraction$ where 0 < x ≤ 1 then in the interval

If f (x) = $fraction numerator x over denominator sin invisible function application x end fraction text and g end text left parenthesis x right parenthesis text = end text fraction numerator x over denominator tan invisible function application x end fraction$ where 0 < x ≤ 1 then in the interval

The function 2|x – 2| + |x – 3| is an increasing function when x belongs to the interval

The function 2|x – 2| + |x – 3| is an increasing function when x belongs to the interval

parallel

The function 2x³ – 9x² + 12x – 3 is increasing when x belongs to the interval

The function 2x³ – 9x² + 12x – 3 is increasing when x belongs to the interval

There is an error of 0.05cm in measuring the circumference 20cm of a circle. Then the percentage error in the area is

There is an error of 0.05cm in measuring the circumference 20cm of a circle. Then the percentage error in the area is

In the curve y = a log(x² – a²) the sum of the lengths and the sub tangent at any point varies as

In the curve y = a log(x² – a²) the sum of the lengths and the sub tangent at any point varies as

parallel

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