JEE Main 2019 (8 April Shift 1)

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Animated Solution for Mathematics - Functions: If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ is equal to :

Q: Wait, why do we reject the negative root here? Or is there a domain issue?

Great observation! The constraint $|x| < 1$ is crucial. It ensures that the argument of the logarithm, $\frac{1-x}{1+x}$, is always positive. When we substitute $\frac{2x}{1+x^2}$, we are essentially working within the domain where the function is well-defined. In JEE problems, always check the constraints; they are there to keep the math valid!

Q: Why can't we just use a value like $x=0$ to check the options?

Using a test value like $x=0$ is a brilliant time-saving strategy in a competitive exam! If $x=0$, $f(0) = \log_e(1) = 0$. The expression becomes $f(0) = 0$. All options might yield zero, so it might not distinguish them. However, for a rigorous understanding, algebraic manipulation is the path to mastery.

Q: How did you know to use $(1-x)^2$ and $(1+x)^2$?

Pattern recognition is the heart of JEE math. When you see $1+x^2-2x$, your brain should immediately flag it as $(1-x)^2$. It is a classic algebraic identity. The more problems you solve, the more these patterns will jump out at you!

Select Answer:

Visualized Solution

The Given Function $f (x)$

Given function: $f (x) = lo g_{e} (\frac{1 - x}{1 + x})$
Constraint: $∣ x ∣ < 1$

The Objective

We need to evaluate: $f (\frac{2 x}{1 + x ^{2}})$
Strategy: Replace every instance of $x$ in the original function with $\frac{2 x}{1 + x ^{2}}$ .

Substituting the Expression

$f (\frac{2 x}{1 + x ^{2}}) = lo g_{e} (\frac{1 - ( \frac{2 x}{1 + x ^{2}} )}{1 + ( \frac{2 x}{1 + x ^{2}} )})$

Simplifying the Numerator

Numerator: $1 - \frac{2 x}{1 + x ^{2}}$
Taking LCM: $\frac{1 ( 1 + x ^{2} ) - 2 x}{1 + x ^{2}}$
Result: $\frac{1 + x ^{2} - 2 x}{1 + x ^{2}}$

Simplifying the Denominator

Denominator: $1 + \frac{2 x}{1 + x ^{2}}$
Taking LCM: $\frac{1 ( 1 + x ^{2} ) + 2 x}{1 + x ^{2}}$
Result: $\frac{1 + x ^{2} + 2 x}{1 + x ^{2}}$

Canceling Common Terms

The expression is now: $lo g_{e} (\frac{\frac{1 + x ^{2} - 2 x}{1 + x ^{2}}}{\frac{1 + x ^{2} + 2 x}{1 + x ^{2}}})$
The $(1 + x^{2})$ terms in the denominators cancel out.
Simplified to: $lo g_{e} (\frac{1 + x ^{2} - 2 x}{1 + x ^{2} + 2 x})$

Recognizing Perfect Squares

Recall the identities:
$(a - b)^{2} = a^{2} - 2 ab + b^{2}$
$(a + b)^{2} = a^{2} + 2 ab + b^{2}$

Condensing the Expression

Numerator: $1 - 2 x + x^{2} = (1 - x)^{2}$
Denominator: $1 + 2 x + x^{2} = (1 + x)^{2}$
The expression becomes: $lo g_{e} (\frac{( 1 - x ) ^{2}}{( 1 + x ) ^{2}})$

Grouping the Powers

Using the property $\frac{a ^{2}}{b ^{2}} = (\frac{a}{b})^{2}$
We can write: $lo g_{e} ((\frac{1 - x}{1 + x})^{2})$

The Logarithm Power Rule

Recall the logarithm property: $lo g_{e} (m^{n}) = n \cdot lo g_{e} (m)$
The exponent $2$ can be brought to the front.

Bringing the Power Down

Applying the rule: $lo g_{e} ((\frac{1 - x}{1 + x})^{2}) = 2 \cdot lo g_{e} (\frac{1 - x}{1 + x})$

Final Substitution

Notice that $lo g_{e} (\frac{1 - x}{1 + x})$ is exactly our original function $f (x)$ .
Therefore, $2 \cdot lo g_{e} (\frac{1 - x}{1 + x}) = 2 f (x)$
Final Answer: $2 f (x)$

00:00 / 00:00

The Sigma Insight: Composite Functions

Analyzing the Setup

We are given the function

f (x) = lo g_{e} (\frac{1 - x}{1 + x})

with the constraint

∣ x ∣ < 1

. Our objective is to evaluate the composite function

f (\frac{2 x}{1 + x ^{2}})

The first step is to commit to the substitution. We replace every instance of

x

in the original function with the expression

\frac{2 x}{1 + x ^{2}}

This yields the following setup:

f (\frac{2 x}{1 + x ^{2}}) = lo g_{e} (\frac{1 - ( \frac{2 x}{1 + x ^{2}} )}{1 + ( \frac{2 x}{1 + x ^{2}} )})

The Algebraic Dance

To simplify the fraction inside the logarithm, we address the numerator and denominator separately. For the numerator, we have:

1 - \frac{2 x}{1 + x ^{2}} = \frac{1 + x ^{2} - 2 x}{1 + x ^{2}}

Similarly, for the denominator, we have:

1 + \frac{2 x}{1 + x ^{2}} = \frac{1 + x ^{2} + 2 x}{1 + x ^{2}}

When we substitute these back into the logarithm, the common denominator

(1 + x^{2})

cancels out perfectly. We are left with:

lo g_{e} (\frac{1 + x ^{2} - 2 x}{1 + x ^{2} + 2 x})

The Hidden Identity

Observe the terms

1 + x^{2} - 2 x

and

1 + x^{2} + 2 x

. These are perfect squares, specifically

(1 - x)^{2}

and

(1 + x)^{2}

Substituting these into our expression, we obtain:

lo g_{e} (\frac{( 1 - x ) ^{2}}{( 1 + x ) ^{2}})

Using the property

\frac{a ^{2}}{b ^{2}} = (\frac{a}{b})^{2}

, we can rewrite the expression as:

lo g_{e} ((\frac{1 - x}{1 + x})^{2})

Final Calculation

We now apply the logarithmic power rule,

lo g_{e} (m^{n}) = n \cdot lo g_{e} (m)

. Bringing the exponent

2

to the front, we get:

2 \cdot lo g_{e} (\frac{1 - x}{1 + x})

Since the term

lo g_{e} (\frac{1 - x}{1 + x})

is exactly our original function

f (x)

, the entire expression simplifies to:

$2 f (x)$

Common Traps & FAQs

Wait, why do we reject the negative root here? Or is there a domain issue?

Great observation! The constraint

∣ x ∣ < 1

is crucial. It ensures that the argument of the logarithm,

\frac{1 - x}{1 + x}

, is always positive. When we substitute

\frac{2 x}{1 + x ^{2}}

, we are essentially working within the domain where the function is well-defined. In JEE problems, always check the constraints; they are there to keep the math valid!

Why can't we just use a value like

x = 0

to check the options?

Using a test value like

x = 0

is a brilliant time-saving strategy in a competitive exam! If

x = 0

f (0) = lo g_{e} (1) = 0

. The expression becomes

f (0) = 0

. All options might yield zero, so it might not distinguish them. However, for a rigorous understanding, algebraic manipulation is the path to mastery.

How did you know to use

(1 - x)^{2}

and

(1 + x)^{2}

Pattern recognition is the heart of JEE math. When you see

1 + x^{2} - 2 x

, your brain should immediately flag it as

(1 - x)^{2}

. It is a classic algebraic identity. The more problems you solve, the more these patterns will jump out at you!

Animated Solution for Mathematics - Functions: If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ is equal to :

Select Answer:

Visualized Solution

The Given Function $f (x)$

The Objective

Substituting the Expression

Simplifying the Numerator

Simplifying the Denominator

Canceling Common Terms

Recognizing Perfect Squares

Condensing the Expression

Grouping the Powers

The Logarithm Power Rule

Bringing the Power Down

Final Substitution

The Sigma Insight: Composite Functions

Analyzing the Setup

The Algebraic Dance

The Hidden Identity

Final Calculation

Similar Questions

If $g (x) = x^{2} + x - 1$ and $(g o f) (x) = 4 x^{2} - 10 x + 5$ , then $f (\frac{5}{4})$ is equal to.

Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

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If $f (x) = \frac{( t a n 1 ^{\circ} ) ^{x} + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{\circ} )}$ , $x > 0$ , then the least value of $f (f (x)) + f (f (\frac{4}{x}))$ is

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If $f (x) = {2 + 2 x, 1 - \frac{x}{3}, - 1 \leq x < 0 0 \leq x \leq 3; g (x) = {- x, x, - 3 \leq x \leq 0 0 < x \leq 1$ , then range of $(f \circ g (x))$ is

Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=loge​(1+x1−x​),∣x∣<1, then f(1+x22x​) is equal to :

Select Answer:

Visualized Solution

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

The Objective

Substituting the Expression

Simplifying the Numerator

Simplifying the Denominator

Canceling Common Terms

Recognizing Perfect Squares

Condensing the Expression

Grouping the Powers

The Logarithm Power Rule

Bringing the Power Down

Final Substitution

The Sigma Insight: Composite Functions

Analyzing the Setup

The Algebraic Dance

The Hidden Identity

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Functions: If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ is equal to :

The Given Function $f (x)$

If $g (x) = x^{2} + x - 1$ and $(g o f) (x) = 4 x^{2} - 10 x + 5$ , then $f (\frac{5}{4})$ is equal to.

Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

Let $f (x) = \frac{x - 1}{x + 1}, x \in R - {0, - 1, 1}$ . If $f^{n + 1} (x) = f (f^{n} (x))$ for all $n \in N$ , then $f^{6} (6) + f^{7} (7)$ is equal to:

If $f (x) = \frac{( t a n 1 ^{\circ} ) ^{x} + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{\circ} )}$ , $x > 0$ , then the least value of $f (f (x)) + f (f (\frac{4}{x}))$ is

If $g (x) = 3 x^{2} + 2 x - 3, f (0) = - 3$ and $4 g (f (x)) = 3 x^{2} - 32 x + 72$ , then $f (g (2))$ is equal to:

If $f (x) = {2 + 2 x, 1 - \frac{x}{3}, - 1 \leq x < 0 0 \leq x \leq 3; g (x) = {- x, x, - 3 \leq x \leq 0 0 < x \leq 1$ , then range of $(f \circ g (x))$ is