The inverse function of $f\left(x\right) = \frac{8^{2x}-8^{-2x}}{8^{2x} + 8^{-2x}}, x\epsilon \left(-1, 1\right),$ is __________.
- $\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1-x}{1+x}\right)$
- $\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1+x}{1-x}\right)$
- $\frac{1}{4} log_{e} \left(\frac{1+x}{1-x}\right)$
- $\frac{1}{4} log_{e} \left(\frac{1- x}{1+ x}\right)$
The Correct Option is B
Solution and Explanation
so, $8^{4x}+1=\frac{2}{1-y} \Rightarrow 8^{4x}=\frac{1+y}{1-y}$
$\Rightarrow x=\ell n\left(\frac{1+y}{1-y}\right)\times \frac{1}{4\ell n8}=f ^{-1}\left(y\right)$
Hence, $f ^{-1}\left(x\right)=\frac{1}{4}log_{g}\,e\ell n\left(\frac{1+x}{1-x}\right)$
Top Questions on Functions
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \] - In \( I(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx \), where \( m, n > 0 \), then \( I(9, 14) + I(10, 13) \) is:
- Let \( f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32} \). Then the value of \[ 8 \left( f\left( \frac{1}{15} \right) + f\left( \frac{2}{15} \right) + \dots + f\left( \frac{59}{15} \right) \right) \] is equal to:
- Let \( f: \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = \left( 2 + 3a \right)x^2 + \left( \frac{a+2}{a-1} \right)x + b, a \neq 1 \). If \[ f(x + y) = f(x) + f(y) + 1 - \frac{2}{7}xy, \] then the value of \( 28 \sum_{i=1}^5 f(i) \) is:
- The area of the region enclosed by the curves \( y = e^x \), \( y = |e^x - 1| \), and the y-axis is:
Questions Asked in JEE Main exam
- Consider ‘n’ is the number of lone pair of electrons present in the equatorial position of the most stable structure of \( \text{ClF}_3 \). The ions from the following with ‘n’ number of unpaired electrons are:
A. \( \text{V}^{3+} \)
B. \( \text{Ti}^{3+} \)
C. \( \text{Cu}^{2+} \)
D. \( \text{Ni}^{2+} \)
E. \( \text{Ti}^{2+} \)
Choose the correct answer from the options given below:- JEE Main - 2025
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Find the equivalent capacitance between A and B, where \( C = 16 \, \mu F \).
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If the equation of the parabola with vertex \( \left( \frac{3}{2}, 3 \right) \) and the directrix \( x + 2y = 0 \) is \[ ax^2 + b y^2 - cxy - 30x - 60y + 225 = 0, \text{ then } \alpha + \beta + \gamma \text{ is equal to:} \]
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions