If $fraction numerator a over denominator b plus c end fraction comma fraction numerator b over denominator c plus a end fraction comma fraction numerator c over denominator a plus b end fraction$ are in AP, then

The entries in a two-by-two determinant $open vertical bar table row cell a blank b end cell row cell c blank d end cell end table close vertical bar$ are integers that are chosen randomly and independently, and, for each entry, the probability that the entry is odd is p. If the probability that the value of the determinant is even is 1/ 2, then the value of p, is

Statement-I : $Cot to the power of negative 1 end exponent invisible function application open parentheses x close parentheses minus tan to the power of negative 1 end exponent invisible function application open parentheses x close parentheses greater than 0 text for all end text x less than 1$
Statement-II : Graph of $c o t to the power of negative 1 end exponent invisible function application left parenthesis x right parenthesis$ is always above the graph of $t a n to the power of negative 1 end exponent invisible function application left parenthesis x right parenthesis text for all end text x less than 1 text . end text$

If $c o s to the power of negative 1 end exponent invisible function application x minus c o s to the power of negative 1 end exponent invisible function application fraction numerator y over denominator 2 end fraction equals alpha comma$ then $4 x to the power of 2 end exponent minus 4 x y c o s invisible function application alpha plus y to the power of 2 end exponent text end text$ is equal to-

The solution of the equation $2 c o s to the power of negative 1 end exponent invisible function application x equals s i n to the power of negative 1 end exponent invisible function application open parentheses 2 x square root of 1 minus x to the power of 2 end exponent end root close parentheses$

The solution of the inequality $open parentheses tan to the power of negative 1 end exponent invisible function application x close parentheses to the power of 2 end exponent minus 3 t a n to the power of negative 1 end exponent invisible function application x plus 2 greater or equal than 0 text is - end text$

The value of tan $invisible function application open square brackets s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses plus t a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 over denominator 3 end fraction close parentheses close square brackets$ is

The image of the interval [- 1, 3] under the mapping specified by the function $f left parenthesis x right parenthesis equals 4 x to the power of 3 end exponent minus 12 x text end text$ is :

Hence the correct option in [-8,72]

The image of the interval [- 1, 3] under the mapping specified by the function $f left parenthesis x right parenthesis equals 4 x to the power of 3 end exponent minus 12 x text end text$ is :

Hence the correct option in [-8,72]

maths-

Let f(x) $equals fraction numerator X over denominator 1 plus x end fraction$ defined from $left parenthesis 0 comma infinity right parenthesis rightwards arrow left square bracket 0 comma infinity right parenthesis$ , then by f(x) is -

maths-General

If f : R $rightwards arrow$ R is a function defined by f(x) = [x] $c o s invisible function application pi$ $open parentheses fraction numerator 2 x minus 1 over denominator 2 end fraction close parentheses$ , where [x] denotes the greatest integer function, then f is :

The correct answer is choice 2

If f : R $rightwards arrow$ R is a function defined by f(x) = [x] $c o s invisible function application pi$ $open parentheses fraction numerator 2 x minus 1 over denominator 2 end fraction close parentheses$ , where [x] denotes the greatest integer function, then f is :

The correct answer is choice 2

Let ƒ : (–1, 1) $rightwards arrow$ B, be a function defined by ƒ(x) $equals t a n to the power of negative 1 end exponent invisible function application fraction numerator 2 x over denominator 1 minus x to the power of 2 end exponent end fraction comma$ then ƒ is both one-one and onto when B is the interval-

Hence, the range of the given function is $open parentheses negative straight pi over 2 comma straight pi over 2 close parentheses$ .

Let ƒ : (–1, 1) $rightwards arrow$ B, be a function defined by ƒ(x) $equals t a n to the power of negative 1 end exponent invisible function application fraction numerator 2 x over denominator 1 minus x to the power of 2 end exponent end fraction comma$ then ƒ is both one-one and onto when B is the interval-

Hence, the range of the given function is $open parentheses negative straight pi over 2 comma straight pi over 2 close parentheses$ .

The range of the function $f left parenthesis x right parenthesis equals fraction numerator 2 plus x over denominator 2 minus x end fraction comma x not equal to 2 text end text$ is-

Hence, range of the given function will be $R - {- 1}$

The range of the function $f left parenthesis x right parenthesis equals fraction numerator 2 plus x over denominator 2 minus x end fraction comma x not equal to 2 text end text$ is-

Hence, range of the given function will be $R - {- 1}$

A function whose graph is symmetrical about the origin is given by -

Hence, the function f(x+y)=f(x)+f(y) is symmetric about the origin.

A function whose graph is symmetrical about the origin is given by -

Hence, the function f(x+y)=f(x)+f(y) is symmetric about the origin.

The minimum value of $f left parenthesis x right parenthesis equals vertical line 3 minus x vertical line plus vertical line 2 plus x vertical line plus vertical line 5 minus x vertical line$ is

$f colon left square bracket negative 1 , 1 right square bracket rightwards arrow left square bracket negative 1 , 2 right square bracket comma f left parenthesis x right parenthesis equals x plus vertical line x vertical line comma$ is -

Hence, the given function is many one and onto.

$f colon left square bracket negative 1 , 1 right square bracket rightwards arrow left square bracket negative 1 , 2 right square bracket comma f left parenthesis x right parenthesis equals x plus vertical line x vertical line comma$ is -