JEE Advanced 1982

LEVELBoard

Animated Solution for Mathematics - Functions: $y = 1 0^{x}$ is the reflection of $y = lo g_{10} x$ in the line whose equation is .........

Q: Does the base of the logarithm change the line of reflection?

Great question! No, it does not. Whether it is $\log_{10} x$, $\ln x$, or $\log_2 x$, the inverse of $y = a^x$ is always $y = \log_a x$. The line of reflection remains $y = x$ because the fundamental operation of swapping coordinates is independent of the base.

Visualized Solution

Introduction to the Functions

We are given two functions: $y = 1 0^{x}$ and $y = lo g_{10} x$ .
Let's first plot the exponential function $y = 1 0^{x}$ on the coordinate plane.

Plotting the Logarithmic Function

Now, let's plot the second function: $y = lo g_{10} x$ .
Observe the shape of this curve compared to the exponential one.

The Concept of Inverse Functions

Two functions $f (x)$ and $g (x)$ are inverses if one undoes the operation of the other.
To find the inverse of a function, we swap the roles of $x$ and $y$ .

Finding the Inverse of $y = 1 0^{x}$

Let's start with our first function: $y = 1 0^{x}$ .
Step 1 for finding the inverse: Interchange $x$ and $y$ .
This gives us: $x = 1 0^{y}$ .

Converting to Logarithmic Form

We have $x = 1 0^{y}$ .
To isolate $y$ , we convert this exponential equation into its logarithmic form.
By definition, $a^{b} = c ⟹ b = lo g_{a} c$ .

The Inverse is the Logarithm

Applying the definition to $x = 1 0^{y}$ , we get:
$y = lo g_{10} x$ .
This is exactly our second given function!

Geometric Meaning of Swapping $x$ and $y$

Let's pick a point on $y = 1 0^{x}$ , for example, $(0, 1)$ .
If we swap the coordinates, we get $(1, 0)$ .
Notice that $(1, 0)$ lies perfectly on the curve $y = lo g_{10} x$ .

The Line of Reflection

The transformation $(x, y) \to (y, x)$ represents a reflection.
The mirror for this reflection is the line where $x$ and $y$ are equal.

Final Conclusion

The line acting as the mirror is $y = x$ .
Thus, $y = 1 0^{x}$ is the reflection of $y = lo g_{10} x$ in the line $y = x$ .
Final Answer: $y = x$

00:00 / 00:00

The Sigma Insight: Inverse of a Function

Solution Diagram

Analyzing the Mathematical Landscape

Welcome, fellow traveler of the mathematical landscape! Today, we embark on a journey to uncover the hidden symmetry between two of the most fundamental functions in mathematics: the exponential and the logarithmic.

Imagine standing before a vast coordinate plane. On one side, we have the exponential function,

y = 1 0^{x}

. It starts near the

x

-axis and rockets upwards, growing with breathtaking speed.

On the other side, we have its counterpart,

y = lo g_{10} x

. It hugs the

y

-axis and climbs slowly, a mirror image of the exponential's rapid ascent.

The Concept of Inverse Functions

The secret to this relationship lies in the concept of inverse functions. An inverse function acts as a mathematical "undo" button.

f (x)

takes an input and gives an output,

f^{- 1} (x)

takes that output and returns the original input. Algebraically, we find this by swapping the variables

x

and

y

When we take

y = 1 0^{x}

and swap the variables, we obtain:

x = 1 0^{y}

By the definition of logarithms, this equation is equivalent to:

y = lo g_{10} x

Geometric Symmetry

Geometrically, swapping

x

and

y

is equivalent to a reflection across the line

y = x

. This line acts as the perfect mirror, bisecting the first and third quadrants.

Every point

(a, b)

on the exponential curve has a corresponding point

(b, a)

on the logarithmic curve. They are perfectly balanced across the line

y = x

This is not merely a coincidence; it is the elegant architecture of mathematics.

Conclusion for JEE Mastery

As you continue your JEE preparation, remember that functions are not just static equations; they are dynamic relationships.

When you see

y = 1 0^{x}

and

y = lo g_{10} x

, do not just see symbols. See the exponential rocket and the logarithmic shadow, forever dancing around the mirror of

y = x

Keep exploring, keep questioning, and let the beauty of these symmetries guide your path to mastery.

Common Traps & FAQs

Why do we swap

x

and

y

to find the inverse? Is this just a rule, or is there a deeper meaning?

It is not just a rule! Think of a function as a machine that takes an input

x

and produces an output

y

. The inverse function is the machine running in reverse: it takes the output

y

and returns the original input

x

. By swapping

x

and

y

, we are mathematically defining this 'reverse' process.

Does this reflection property hold for all inverse functions?

Absolutely! This is the beauty of coordinate geometry. For any function

f (x)

and its inverse

f^{- 1} (x)

, the graph of

f^{- 1} (x)

is always the reflection of

f (x)

across the line

y = x

. It is a universal truth in mathematics.

Does the base of the logarithm change the line of reflection?

Great question! No, it does not. Whether it is

lo g_{10} x

ln x

, or

lo g_{2} x

, the inverse of

y = a^{x}

is always

y = lo g_{a} x

. The line of reflection remains

y = x

because the fundamental operation of swapping coordinates is independent of the base.

Animated Solution for Mathematics - Functions: $y = 1 0^{x}$ is the reflection of $y = lo g_{10} x$ in the line whose equation is .........

Visualized Solution

Introduction to the Functions

Plotting the Logarithmic Function

The Concept of Inverse Functions

Finding the Inverse of $y = 1 0^{x}$

Converting to Logarithmic Form

The Inverse is the Logarithm

Geometric Meaning of Swapping $x$ and $y$

The Line of Reflection

Final Conclusion

The Sigma Insight: Inverse of a Function

Analyzing the Mathematical Landscape

The Concept of Inverse Functions

Geometric Symmetry

Conclusion for JEE Mastery

Similar Questions

Suppose $f (x) = (x + 1)^{2}$ for $x \geq - 1$ . If $g (x)$ is the function whose graph is the reflection of the graph of $f (x)$ with respect to the line $y = x$ , then $g (x)$ equals

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + 1/ x$ , then $f^{- 1} (x)$ equals

If $f (x) = 3 x - 5$ , then $f^{- 1} (x)$

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x}$ , where $b$ is a constant such that $0 < b < 1$ . Then

If $y = f (x) = \frac{x + 2}{x - 1}$ then

Let $f$ be a real-valued function defined on the interval (-1, 1) such that $e^{- x} f (x) = 2 + \int_{0}^{x} t^{4} + 1 d t$ , for all $x \in (- 1, 1)$ , and let $f^{- 1}$ be the inverse function of $f$ . Then $(f^{- 1})^{'} (2)$ is equal to

Let $f$ be a one-one function with domain ${x, y, z}$ and range ${1, 2, 3}$ . It is given that exactly one of the following statements is true and the remaining two are false $f (x) = 1, f (y) \neq = 1, f (z) \neq = 2$ determine $f^{- 1} (1)$ .

Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=10x is the reflection of y=log10​x in the line whose equation is .........

Visualized Solution

Introduction to the Functions

Plotting the Logarithmic Function

The Concept of Inverse Functions

Finding the Inverse of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=10x

Converting to Logarithmic Form

The Inverse is the Logarithm

Geometric Meaning of Swapping .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

The Line of Reflection

Final Conclusion

The Sigma Insight: Inverse of a Function

Analyzing the Mathematical Landscape

The Concept of Inverse Functions

Geometric Symmetry

Conclusion for JEE Mastery

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose f(x)=(x+1)2 for x≥−1. If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to the line y=x, then g(x) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f:[1,∞)→[1,∞) is defined by f(x)=2x(x−1), then f−1(x) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The inverse function of f(x)=82x+8−2x82x−8−2x​, x∈(−1,1), is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The inverse function of f(x)=82x+8−2x82x−8−2x​, x∈(−1,1), is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:[1,∞)→[2,∞) is given by f(x)=x+1/x, then f−1(x) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=3x−5, then f−1(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,1)→R be defined by f(x)=1−bxb−x​, where b is a constant such that 0<b<1. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=f(x)=x−1x+2​ then

Animated Solution for Mathematics - Functions: $y = 1 0^{x}$ is the reflection of $y = lo g_{10} x$ in the line whose equation is .........

Finding the Inverse of $y = 1 0^{x}$

Geometric Meaning of Swapping $x$ and $y$

Suppose $f (x) = (x + 1)^{2}$ for $x \geq - 1$ . If $g (x)$ is the function whose graph is the reflection of the graph of $f (x)$ with respect to the line $y = x$ , then $g (x)$ equals

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + 1/ x$ , then $f^{- 1} (x)$ equals

If $f (x) = 3 x - 5$ , then $f^{- 1} (x)$

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x}$ , where $b$ is a constant such that $0 < b < 1$ . Then

If $y = f (x) = \frac{x + 2}{x - 1}$ then