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Maths-

General

Easy

Question

Set of values of x satisfying the inequality $fraction numerator left parenthesis x minus 3 right parenthesis squared left parenthesis 2 x plus 5 right parenthesis squared left parenthesis x minus 7 right parenthesis over denominator open parentheses x squared plus x plus 1 close parentheses left parenthesis 3 x plus 6 right parenthesis squared end fraction less or equal than 0$ is $left square bracket a comma b right parenthesis union left parenthesis b comma c right square bracket$ then 2a + b + c is equal to

0
2
5
7

The correct answer is: 0

Related Questions to study

Number of integral values of x for which the inequality log₁₀ $open parentheses fraction numerator 2 x minus 2007 over denominator x plus 1 end fraction close parentheses$ $less or equal than$ 0 holds true, is

Number of integral values of x for which the inequality log₁₀ $open parentheses fraction numerator 2 x minus 2007 over denominator x plus 1 end fraction close parentheses$ $less or equal than$ 0 holds true, is

$x to the power of log subscript 5 invisible function application x end exponent greater than 5$ implies

$x to the power of log subscript 5 invisible function application x end exponent greater than 5$ implies

log4 (2x2 + x + 1) – log² (2x – 1) $less or equal than$ – tan $fraction numerator 7 pi over denominator 4 end fraction$

log4 (2x2 + x + 1) – log² (2x – 1) $less or equal than$ – tan $fraction numerator 7 pi over denominator 4 end fraction$

parallel

The solution set of the inequation log1/3 (x² + x + 1) + 1 > 0 is

The solution set of the inequation log1/3 (x² + x + 1) + 1 > 0 is

If $open parentheses a to the power of log subscript b end subscript invisible function application x end exponent close parentheses to the power of 2 end exponent$ –5 $x to the power of log subscript b end subscript invisible function application a end exponent$ + 6 = 0 where a > 0, b > 0 & ab $not equal to$ 1. Then the value of x is equal to

If $open parentheses a to the power of log subscript b end subscript invisible function application x end exponent close parentheses to the power of 2 end exponent$ –5 $x to the power of log subscript b end subscript invisible function application a end exponent$ + 6 = 0 where a > 0, b > 0 & ab $not equal to$ 1. Then the value of x is equal to

No. of ordered pair satisfying simultaneously the system of equation $2 to the power of square root of x end exponent$ . $2 to the power of square root of y end exponent$ = 256 & log10 $square root of x y end root$ – log10 1.5 = 1 is.

No. of ordered pair satisfying simultaneously the system of equation $2 to the power of square root of x end exponent$ . $2 to the power of square root of y end exponent$ = 256 & log10 $square root of x y end root$ – log10 1.5 = 1 is.

parallel

If $x to the power of left square bracket log subscript 3 end subscript invisible function application x to the power of 2 end exponent plus left parenthesis log subscript 3 end subscript invisible function application x right parenthesis to the power of 2 end exponent minus 10 right square bracket end exponent$ = 1/x², then x =

If $x to the power of left square bracket log subscript 3 end subscript invisible function application x to the power of 2 end exponent plus left parenthesis log subscript 3 end subscript invisible function application x right parenthesis to the power of 2 end exponent minus 10 right square bracket end exponent$ = 1/x², then x =

If x_n > x_n–1>...> x₂ > x₁ > 1 then the value of $log subscript straight x subscript 1 end subscript invisible function application log subscript straight x subscript 2 end subscript invisible function application log subscript straight x subscript 3 end subscript invisible function application horizontal ellipsis log subscript straight x subscript straight n end subscript invisible function application x subscript n$ $blank to the power of x subscript n minus 1 end subscript superscript up right diagonal ellipsis to the power of x subscript 1 end exponent end superscript end exponent$ is equal to-

If x_n > x_n–1>...> x₂ > x₁ > 1 then the value of $log subscript straight x subscript 1 end subscript invisible function application log subscript straight x subscript 2 end subscript invisible function application log subscript straight x subscript 3 end subscript invisible function application horizontal ellipsis log subscript straight x subscript straight n end subscript invisible function application x subscript n$ $blank to the power of x subscript n minus 1 end subscript superscript up right diagonal ellipsis to the power of x subscript 1 end exponent end superscript end exponent$ is equal to-

The expression logp where $p greater or equal than 2 comma p element of N semicolon n element of N$ when simplified is.

The expression logp where $p greater or equal than 2 comma p element of N semicolon n element of N$ when simplified is.

parallel

Let N= $open parentheses open parentheses square root of 7 close parentheses to the power of fraction numerator 2 over denominator log subscript 25 end subscript invisible function application 7 end fraction end exponent minus 12 5 to the power of log subscript 25 end subscript invisible function application 6 end exponent close parentheses$ Then log₂N has the value –

Let N= $open parentheses open parentheses square root of 7 close parentheses to the power of fraction numerator 2 over denominator log subscript 25 end subscript invisible function application 7 end fraction end exponent minus 12 5 to the power of log subscript 25 end subscript invisible function application 6 end exponent close parentheses$ Then log₂N has the value –

If a² + 4b² = 12ab, then log (a + 2b) =

If a² + 4b² = 12ab, then log (a + 2b) =

Given that log_px = α and log_qx = β, then value of log_p_/q x equals-

Given that log_px = α and log_qx = β, then value of log_p_/q x equals-

parallel

If f(x) is the primitive of $fraction numerator sin invisible function application root index 3 of x end root log invisible function application left parenthesis 1 plus 3 x right parenthesis over denominator left parenthesis tan to the power of – 1 end exponent invisible function application square root of x right parenthesis to the power of 2 end exponent left parenthesis e to the power of root index 3 of x end root end exponent – 1 right parenthesis end fraction$ (x $not equal to$ 0), then $stack l i m with x rightwards arrow 0 below$ f ' (x) is -

If f(x) is the primitive of $fraction numerator sin invisible function application root index 3 of x end root log invisible function application left parenthesis 1 plus 3 x right parenthesis over denominator left parenthesis tan to the power of – 1 end exponent invisible function application square root of x right parenthesis to the power of 2 end exponent left parenthesis e to the power of root index 3 of x end root end exponent – 1 right parenthesis end fraction$ (x $not equal to$ 0), then $stack l i m with x rightwards arrow 0 below$ f ' (x) is -

If f(x) is the primitive of $fraction numerator sin invisible function application root index 3 of x end root log invisible function application left parenthesis 1 plus 3 x right parenthesis over denominator left parenthesis tan to the power of negative 1 end exponent invisible function application square root of x right parenthesis to the power of 2 end exponent left parenthesis e to the power of root index 3 of x end root end exponent minus 1 right parenthesis end fraction$ (x $not equal to$ 0), then $stack l i m with x rightwards arrow 0 below f ´ left parenthesis x right parenthesis$ is-

If f(x) is the primitive of $fraction numerator sin invisible function application root index 3 of x end root log invisible function application left parenthesis 1 plus 3 x right parenthesis over denominator left parenthesis tan to the power of negative 1 end exponent invisible function application square root of x right parenthesis to the power of 2 end exponent left parenthesis e to the power of root index 3 of x end root end exponent minus 1 right parenthesis end fraction$ (x $not equal to$ 0), then $stack l i m with x rightwards arrow 0 below f ´ left parenthesis x right parenthesis$ is-

If $f left parenthesis x right parenthesis equals open vertical bar table row cell s i n invisible function application x end cell cell s i n invisible function application a end cell cell s i n invisible function application b end cell row cell c o s invisible function application x end cell cell c o s invisible function application a end cell cell c o s invisible function application b end cell row cell t a n invisible function application x end cell cell t a n invisible function application a end cell cell t a n invisible function application b end cell end table close vertical bar$ where $0 less than a less than b less than fraction numerator pi over denominator 2 end fraction$ hen the equation f(x) = 0 has, in the interval (a, b)

If $f left parenthesis x right parenthesis equals open vertical bar table row cell s i n invisible function application x end cell cell s i n invisible function application a end cell cell s i n invisible function application b end cell row cell c o s invisible function application x end cell cell c o s invisible function application a end cell cell c o s invisible function application b end cell row cell t a n invisible function application x end cell cell t a n invisible function application a end cell cell t a n invisible function application b end cell end table close vertical bar$ where $0 less than a less than b less than fraction numerator pi over denominator 2 end fraction$ hen the equation f(x) = 0 has, in the interval (a, b)

parallel

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