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Maths-

General

Easy

Question

If $f left parenthesis x right parenthesis equals fraction numerator 10 to the power of x minus 10 to the power of negative x end exponent over denominator 10 to the power of x plus 10 to the power of negative x end exponent end fraction text then end text f to the power of negative 1 end exponent left parenthesis x right parenthesis equals$

$\log_{10} (2 - x)$
$\frac{1}{2} \log_{10} \frac{1 + x}{1 - x}$
$\frac{1}{2} \log_{10} (2 x - 1)$
$\frac{1}{4} \log_{10} \frac{2 x}{2 - x}$

hint Hint:

In this question, we have to find the inverse of f(x) where f(x) = $fraction numerator 10 to the power of x minus 10 to the power of negative x end exponent over denominator 10 to the power of x plus 10 to the power of negative x end exponent end fraction$ . Firstly, we will consider f(x) to be y and then using the equation find the value of x. Further, after we get the value of x we will replace x with $f to the power of negative 1 end exponent left parenthesis x right parenthesis$ and y with x.

The correct answer is: $1 half log subscript 10 invisible function application fraction numerator 1 plus x over denominator 1 minus x end fraction$

$f left parenthesis x right parenthesis equals fraction numerator 10 to the power of x minus 10 to the power of negative x end exponent over denominator 10 to the power of x plus 10 to the power of negative x end exponent end fraction equals y rightwards double arrow fraction numerator 10 to the power of x left parenthesis 10 to the power of 2 x end exponent minus 1 right parenthesis over denominator 10 to the power of x left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis end fraction equals y rightwards double arrow fraction numerator left parenthesis 10 to the power of 2 x end exponent minus 1 right parenthesis over denominator left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis end fraction equals y rightwards double arrow 10 to the power of 2 x end exponent minus 1 space equals y left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis rightwards double arrow 10 to the power of 2 x end exponent minus 1 equals y.10 to the power of 2 x end exponent plus y rightwards double arrow 10 to the power of 2 x end exponent minus y.10 to the power of 2 x end exponent equals y plus 1 rightwards double arrow 10 to the power of 2 x end exponent left parenthesis 1 minus y right parenthesis equals 1 plus y rightwards double arrow 10 to the power of 2 x end exponent equals fraction numerator 1 plus y over denominator 1 minus y end fraction P u t t i n g space log space o n space b o t h space t h e space s i d e s comma rightwards double arrow log space 10 to the power of 2 x end exponent space equals log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses rightwards double arrow 2 x space log space 10 space equals log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses rightwards double arrow 2 x space equals fraction numerator log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses over denominator log space 10 end fraction rightwards double arrow x space equals 1 half log subscript 10 open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses rightwards double arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals space 1 half log subscript 10 open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses$

Related Questions to study

If $f left parenthesis x right parenthesis equals fraction numerator e to the power of x plus e to the power of negative x end exponent over denominator 2 end fraction$ ,then the inverse function of $f left parenthesis x right parenthesis text is end text$

If $f left parenthesis x right parenthesis equals fraction numerator e to the power of x plus e to the power of negative x end exponent over denominator 2 end fraction$ ,then the inverse function of $f left parenthesis x right parenthesis text is end text$

If $f left parenthesis x right parenthesis equals x squared plus x comma x equals 10 comma delta x equals 0.1$ observe the following
$text l) end text delta f equals 2.11$
$text II) end text d f equals 2.1$
$text III) Relative error in end text x text is end text 1$
The true statements are

If $f left parenthesis x right parenthesis equals x squared plus x comma x equals 10 comma delta x equals 0.1$ observe the following
$text l) end text delta f equals 2.11$
$text II) end text d f equals 2.1$
$text III) Relative error in end text x text is end text 1$
The true statements are

Statement I : $f colon A not stretchy rightwards arrow B text is one - one and end text g colon B not stretchy rightwards arrow C text is a one-one function, then gof end text colon A not stretchy rightwards arrow C$ is one - one
Statement II : $text If end text f colon A not stretchy rightwards arrow B comma g colon B not stretchy rightwards arrow A$ are two functions such that $g o f equals I subscript A text and end text fog equals I subscript B comma text then end text f equals$ $g to the power of negative 1 end exponent$
Staatement III : $f left parenthesis x right parenthesis equals sec squared invisible function application x minus tan squared invisible function application x g left parenthesis x right parenthesis equals cosec squared invisible function application x minus cot squared invisible function application x comma text then end text f equals g$
Which of the above statement/s is/are true.

Statement I : $f colon A not stretchy rightwards arrow B text is one - one and end text g colon B not stretchy rightwards arrow C text is a one-one function, then gof end text colon A not stretchy rightwards arrow C$ is one - one
Statement II : $text If end text f colon A not stretchy rightwards arrow B comma g colon B not stretchy rightwards arrow A$ are two functions such that $g o f equals I subscript A text and end text fog equals I subscript B comma text then end text f equals$ $g to the power of negative 1 end exponent$
Staatement III : $f left parenthesis x right parenthesis equals sec squared invisible function application x minus tan squared invisible function application x g left parenthesis x right parenthesis equals cosec squared invisible function application x minus cot squared invisible function application x comma text then end text f equals g$
Which of the above statement/s is/are true.

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