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 $t a n to the power of negative 1 end exponent invisible function application left parenthesis x right parenthesis text for all end text x less than 1 text . end text$

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 -

Hence, the given function is many one and onto.

If f(x) is a polynomial function satisfying the condition f(x). f(1/x) = f(x) + f(1/x) and f(2) = 9 then -

Fill in the blank with the appropriate transition.
The movie managed to fetch decent collections ______ all the negative reviews it received.

GeneralGeneral

If R be a relation '<' from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e. (a, b) $element of$ R iff a < b, then $R O R to the power of negative 1 end exponent text end text$ is $text-end text$

Values of $R O R to the power of negative 1 end exponent text end text$ are {(3, 3), (3, 5), (5, 3), (5, 5)}

If R be a relation '<' from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e. (a, b) $element of$ R iff a < b, then $R O R to the power of negative 1 end exponent text end text$ is $text-end text$