JEE Main 2021 (18 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Functions: Let $f : R - {3} \to R - {1}$ be defined by $f (x) = \frac{x - 2}{x - 3}$ . Let $g : R \to R$ be given as $g (x) = 2 x - 3$ . Then, the sum of all the values of $x$ for which $f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}$ is equal to

Q: Wait, why do we swap $x$ and $y$ to find the inverse?

That is a fantastic question! Think of a function $f$ as a machine that takes an input $x$ and gives an output $y$. The inverse function $f^{-1}$ is simply the machine running in reverse—it takes the output $y$ and tells you what the original input $x$ was. By swapping $x$ and $y$, we are essentially relabeling our variables to align with the standard convention where the input is $x$ and the output is $f^{-1}(x)$.

Q: What if I get a root that makes the denominator zero?

Always check your domain! In our case, $f^{-1}(x) = \frac{3x-2}{x-1}$, so $x$ cannot be $1$. If you had found $x=1$ as a solution, you would have to reject it because it would make the function undefined. Always verify your final answers against the domain of the original functions.

Select Answer:

Progressive Hints

Hint 1

The Spark: Think of the inverse function

f^{- 1} (x)

not just as an algebraic manipulation, but as a geometric reflection of the function

f (x)

across the line

y = x

. When you are asked to find

f^{- 1} (x)

, you are essentially asking: 'If the output was

y

, what was the input

x

that produced it?' This perspective shift is the key to unlocking the problem.

Hint 2

The Arsenal: Don't get intimidated by the rational expression. Your best friend here is the common denominator. Once you express both

f^{- 1} (x)

and

g^{- 1} (x)

as algebraic fractions, look for the least common multiple of their denominators to clear the fractions entirely. This will transform your complex rational equation into a much friendlier polynomial one.

Hint 3

The Execution: Start by setting up the equation:

\frac{3 x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2}

. Now, multiply the entire equation by

2 (x - 1)

to eliminate the denominators. This will lead you directly to a quadratic equation in the form

a x^{2} + b x + c = 0

Visualized Solution

Objective: Find the sum of $x$ values

Given functions: $f (x) = \frac{x - 2}{x - 3}$ and $g (x) = 2 x - 3$
Goal: Find all $x$ such that $f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}$
Calculate the sum of these $x$ values.

Finding $f^{- 1} (x)$ - Setup

Let $y = f (x) = \frac{x - 2}{x - 3}$
To find the inverse, we need to express $x$ in terms of $y$ .
Cross-multiply to remove the fraction: $y (x - 3) = x - 2$

Finding $f^{- 1} (x)$ - Execution

Expand: $y x - 3 y = x - 2$
Group $x$ terms: $y x - x = 3 y - 2$
Factor out $x$ : $x (y - 1) = 3 y - 2$
Isolate $x$ : $x = \frac{3 y - 2}{y - 1}$
Therefore, $f^{- 1} (x) = \frac{3 x - 2}{x - 1}$

Finding $g^{- 1} (x)$

Let $y = g (x) = 2 x - 3$
Add $3$ to both sides: $y + 3 = 2 x$
Divide by $2$ : $x = \frac{y + 3}{2}$
Therefore, $g^{- 1} (x) = \frac{x + 3}{2}$

Setting up the Equation

Substitute $f^{- 1} (x)$ and $g^{- 1} (x)$ into the given condition.
Given: $f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}$
Substitution: $\frac{3 x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2}$

Clearing Denominators

Equation: $\frac{3 x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2}$
Multiply the entire equation by the common denominator $2 (x - 1)$ .
Result: $2 (3 x - 2) + (x + 3) (x - 1) = 13 (x - 1)$

Expanding the Terms

Expand $2 (3 x - 2) ⟹ 6 x - 4$
Expand $(x + 3) (x - 1) ⟹ x^{2} + 2 x - 3$
Expand $13 (x - 1) ⟹ 13 x - 13$
Combine: $6 x - 4 + x^{2} + 2 x - 3 = 13 x - 13$

Simplifying to Quadratic Form

Left side: $x^{2} + (6 x + 2 x) + (- 4 - 3) = x^{2} + 8 x - 7$
Equation: $x^{2} + 8 x - 7 = 13 x - 13$
Move all terms to the left: $x^{2} + 8 x - 13 x - 7 + 13 = 0$
Final Quadratic: $x^{2} - 5 x + 6 = 0$

Solving the Quadratic Equation

Factorize: $x^{2} - 5 x + 6 = 0$
Find two numbers that multiply to $6$ and add to $- 5$ : $- 2$ and $- 3$ .
$(x - 2) (x - 3) = 0$
Possible values: $x = 2$ or $x = 3$

Final Step: Sum of Values

The valid values of $x$ are $2$ and $3$ .
Check domain: $x \neq = 1$ for $f^{- 1} (x)$ . Both $2$ and $3$ are valid.
Sum $= 2 + 3 = 5$
Final Answer: 5

00:00 / 00:00

The Sigma Insight: Inverse of a Function

Analyzing the Inverse of $f (x)$

To find the inverse of

f (x) = \frac{x - 2}{x - 3}

, we set

y = \frac{x - 2}{x - 3}

and solve for

x

Cross-multiplying yields

y (x - 3) = x - 2

, which expands to

y x - 3 y = x - 2

Rearranging to isolate

x

terms gives

y x - x = 3 y - 2

. Factoring out

x

, we obtain

x (y - 1) = 3 y - 2

, leading to:

f^{- 1} (x) = \frac{3 x - 2}{x - 1}

Analyzing the Inverse of $g (x)$

For the linear function

g (x) = 2 x - 3

, we set

y = 2 x - 3

Adding

3

to both sides gives

y + 3 = 2 x

. Dividing by

2

, we find the inverse:

g^{- 1} (x) = \frac{x + 3}{2}

Solving the Master Equation

We are given the condition

f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}

. Substituting our derived expressions, we have:

\frac{3 x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2}

To clear the denominators, we multiply the entire equation by

2 (x - 1)

2 (3 x - 2) + (x + 3) (x - 1) = 13 (x - 1)

Expanding the terms results in

6 x - 4 + x^{2} + 2 x - 3 = 13 x - 13

. Simplifying the left side, we get:

x^{2} + 8 x - 7 = 13 x - 13

Final Calculation

Moving all terms to one side yields the quadratic equation:

x^{2} - 5 x + 6 = 0

Factoring the quadratic, we obtain

(x - 2) (x - 3) = 0

, which provides the roots

x = 2

and

x = 3

We must verify the domain constraint for

f^{- 1} (x)

, which requires $x eq 1$. Since both roots satisfy this condition, the sum of the values is

2 + 3 = 5

The final answer is 5.

Common Traps & FAQs

Wait, why do we swap

x

and

y

to find the inverse?

That is a fantastic question! Think of a function

f

as a machine that takes an input

x

and gives an output

y

. The inverse function

f^{- 1}

is simply the machine running in reverse—it takes the output

y

and tells you what the original input

x

was. By swapping

x

and

y

, we are essentially relabeling our variables to align with the standard convention where the input is

x

and the output is

f^{- 1} (x)

Why can't we just solve

f (x) + g (x) = \frac{13}{2}

That is a very common trap! The problem specifically asks for the sum of values for

f^{- 1} (x) + g^{- 1} (x)

, not

f (x) + g (x)

. These are fundamentally different functions. Solving for the inverse first is non-negotiable because the inverse operation is not the same as the original operation.

What if I get a root that makes the denominator zero?

Always check your domain! In our case,

f^{- 1} (x) = \frac{3 x - 2}{x - 1}

, so

x

cannot be

1

. If you had found

x = 1

as a solution, you would have to reject it because it would make the function undefined. Always verify your final answers against the domain of the original functions.

Animated Solution for Mathematics - Functions: Let $f : R - {3} \to R - {1}$ be defined by $f (x) = \frac{x - 2}{x - 3}$ . Let $g : R \to R$ be given as $g (x) = 2 x - 3$ . Then, the sum of all the values of $x$ for which $f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}$ is equal to

Select Answer:

Visualized Solution

Objective: Find the sum of $x$ values

Finding $f^{- 1} (x)$ - Setup

Finding $f^{- 1} (x)$ - Execution

Finding $g^{- 1} (x)$

Setting up the Equation

Clearing Denominators

Expanding the Terms

Simplifying to Quadratic Form

Solving the Quadratic Equation

Final Step: Sum of Values

The Sigma Insight: Inverse of a Function

Analyzing the Inverse of $f (x)$

Analyzing the Inverse of $g (x)$

Solving the Master Equation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Objective: Find the sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x values

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f−1(x) - Setup

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f−1(x) - Execution

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g−1(x)

Setting up the Equation

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Simplifying to Quadratic Form

Solving the Quadratic Equation

Final Step: Sum of Values

The Sigma Insight: Inverse of a Function

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Analyzing the Inverse of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Solving the Master Equation

Final Calculation

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Finding $g^{- 1} (x)$

Analyzing the Inverse of $f (x)$

Analyzing the Inverse of $g (x)$

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