JEE Main 2022 (29 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Functions: Let $f (x)$ and $g (x)$ be two real polynomials of degree 2 and 1 respectively. If $f (g (x)) = 8 x^{2} - 2 x$ , and $g (f (x)) = 4 x^{2} + 6 x + 1$ , then the value of $f (2) + g (2)$ is ______.

Q: Wait, why can't we just find $f(x)$ first by taking the inverse of $g(x)$?

That is a brilliant thought, but remember that $f$ and $g$ are both unknown. We don't know the inverse of $g$ until we know $a$ and $b$. By assuming the linear form $g(x) = ax + b$, we create a system of equations that allows us to solve for the constants $a$ and $b$ directly. It is much more robust than trying to invert an unknown function.

Q: Why do we assume $g(x) = ax + b$?

The problem explicitly states that $g(x)$ is a polynomial of degree 1. By definition, any polynomial of degree 1 is a linear function, which must take the form $ax + b$. This is not an assumption; it is a direct translation of the problem statement into mathematical language.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the problem as a locked box. You are given two composite functions,

f (g (x))

and

g (f (x))

. Since

g (x)

is a linear polynomial of degree 1, define it as

g (x) = a x + b

. The key is to treat

a

and

b

as unknown constants that will reveal themselves once you compare the coefficients of the resulting polynomials.

Hint 2

Hint 2: The Arsenal. You need the method of undetermined coefficients. Once you have the general form of

f (x)

derived from

f (g (x))

, you will have a quadratic expression in terms of

a

and

b

. Use the second equation,

g (f (x)) = 4 x^{2} + 6 x + 1

, to create a system of equations by equating the coefficients of

x^{2}

x

, and the constant term.

Hint 3

Hint 3: The Execution. Start by substituting

t = a x + b

into

f (g (x)) = 8 x^{2} - 2 x

. This allows you to write

f (t) = 8 (\frac{t - b}{a})^{2} - 2 (\frac{t - b}{a})

. Once you expand this, you have a clear expression for

f (x)

. Now, plug this

f (x)

into

g (f (x)) = a (f (x)) + b

and set it equal to

4 x^{2} + 6 x + 1

to solve for

a

and

b

Visualized Solution

Define $g (x)$ as a Linear Polynomial

Let $g (x) = a x + b$ be the linear polynomial (degree 1).
Here, $a$ and $b$ are real constants to be determined.

Substitute $g (x)$ into $f (g (x))$

Given: $f (g (x)) = 8 x^{2} - 2 x$
Substitute $g (x) = a x + b$ :
$f (a x + b) = 8 x^{2} - 2 x$

Change Variables to find $f (t)$

Let $t = a x + b \Rightarrow x = \frac{t - b}{a}$
Substitute $x$ in terms of $t$ into the equation for $f$ :
$f (t) = 8 (\frac{t - b}{a})^{2} - 2 (\frac{t - b}{a})$

Expand and Simplify $f (t)$

Expand the squared term:
$f (t) = \frac{8}{a ^{2}} (t^{2} - 2 b t + b^{2}) - \frac{2}{a} (t - b)$
Distribute the constants:
$f (t) = \frac{8}{a ^{2}} t^{2} - \frac{16 b}{a ^{2}} t + \frac{8 b ^{2}}{a ^{2}} - \frac{2}{a} t + \frac{2 b}{a}$

Express $f (x)$ in Standard Form

Replace $t$ with $x$ and group terms:
$f (x) = \frac{8}{a ^{2}} x^{2} - (\frac{16 b}{a ^{2}} + \frac{2}{a}) x + (\frac{8 b ^{2}}{a ^{2}} + \frac{2 b}{a})$

Set up the Equation for $g (f (x))$

Given: $g (f (x)) = 4 x^{2} + 6 x + 1$
Since $g (x) = a x + b$ , then $g (f (x)) = a f (x) + b$ .
Substitute the expression for $f (x)$ :
$g (f (x)) = a [\frac{8}{a ^{2}} x^{2} - (\frac{16 b}{a ^{2}} + \frac{2}{a}) x + (\frac{8 b ^{2}}{a ^{2}} + \frac{2 b}{a})] + b$

Simplify the $g (f (x))$ Expression

Distribute $a$ and simplify:
$g (f (x)) = \frac{8}{a} x^{2} - (\frac{16 b}{a} + 2) x + (\frac{8 b ^{2}}{a} + 2 b + b)$
$g (f (x)) = \frac{8}{a} x^{2} - (\frac{16 b}{a} + 2) x + (\frac{8 b ^{2}}{a} + 3 b)$

Compare $x^{2}$ Coefficients to find $a$

Compare coefficients of $x^{2}$ with $4 x^{2} + 6 x + 1$ :
$\frac{8}{a} = 4 \Rightarrow a = 2$

Compare $x$ Coefficients to find $b$

Compare coefficients of $x$ and substitute $a = 2$ :
$- (\frac{16 b}{2} + 2) = 6$
$- (8 b + 2) = 6 \Rightarrow 8 b + 2 = - 6$
$8 b = - 8 \Rightarrow b = - 1$

Determine $g (x)$ and find $g (2)$

Substitute $a = 2, b = - 1$ into $g (x)$ :
$g (x) = 2 x - 1$
Calculate $g (2)$ :
$g (2) = 2 (2) - 1 = 3$

Determine $f (x)$ and find $f (2)$

Substitute $a = 2, b = - 1$ into the expression for $f (x)$ :
$f (x) = \frac{8}{2 ^{2}} x^{2} - (\frac{16 ( - 1 )}{2 ^{2}} + \frac{2}{2}) x + (\frac{8 ( - 1 ) ^{2}}{2 ^{2}} + \frac{2 ( - 1 )}{2})$
$f (x) = 2 x^{2} - (- 4 + 1) x + (2 - 1) = 2 x^{2} + 3 x + 1$
Calculate $f (2)$ :
$f (2) = 2 (2^{2}) + 3 (2) + 1 = 8 + 6 + 1 = 15$

Final Answer: $f (2) + g (2)$

Final calculation:
$f (2) + g (2) = 15 + 3 = 18$
Final Answer: 18

00:00 / 00:00

The Sigma Insight: Composite Functions

Analyzing the Setup

We are given two functions,

f (x)

and

g (x)

, where

g (x)

is a linear function defined as

g (x) = a x + b

. We are provided with the composite function:

f (g (x)) = 8 x^{2} - 2 x

Since

g (x)

is linear,

f (x)

must be a quadratic function to result in an

x^{2}

term. This observation is the foundation of our solution.

The Substitution Method

To isolate

f (t)

, we perform the substitution

t = a x + b

. Solving for

x

, we obtain:

x = \frac{t - b}{a}

Substituting this into the expression for

f (g (x))

, we get:

f (t) = 8 (\frac{t - b}{a})^{2} - 2 (\frac{t - b}{a})

Expanding this expression carefully, we arrive at the general form for

f (t)

f (t) = \frac{8}{a ^{2}} t^{2} - (\frac{16 b}{a ^{2}} + \frac{2}{a}) t + (\frac{8 b ^{2}}{a ^{2}} + \frac{2 b}{a})

Comparing Coefficients

We are also given the second composite function:

g (f (x)) = 4 x^{2} + 6 x + 1

Since

g (x) = a x + b

, we can write this as

a \cdot f (x) + b = 4 x^{2} + 6 x + 1

. Substituting our derived form of

f (x)

into this equation allows us to compare the coefficients of

x^{2}

and

x

Equating the coefficient of

x^{2}

\frac{8}{a} = 4 \Rightarrow a = 2

Equating the linear coefficient and solving for

b

using

a = 2

b = - 1

Final Calculation

With the constants determined, the functions are revealed as:

g (x) = 2 x - 1

f (x) = 2 x^{2} + 3 x + 1

To find the final result, we evaluate

f (2) + g (2)

f (2) = 2 (2)^{2} + 3 (2) + 1 = 8 + 6 + 1 = 15

g (2) = 2 (2) - 1 = 3

The final answer is $15 + 3 = 18$ .

Common Traps & FAQs

Wait, why can't we just find

f (x)

first by taking the inverse of

g (x)

That is a brilliant thought, but remember that

f

and

g

are both unknown. We don't know the inverse of

g

until we know

a

and

b

. By assuming the linear form

g (x) = a x + b

, we create a system of equations that allows us to solve for the constants

a

and

b

directly. It is much more robust than trying to invert an unknown function.

Is there a faster way than substituting

t = a x + b

You could potentially compare coefficients directly without the substitution, but it is prone to algebraic errors. The substitution

t = a x + b

is a standard, safe, and elegant technique in JEE Advanced problems to isolate the function

f (t)

. It keeps your algebra clean and organized, which is half the battle in competitive exams.

Why do we assume

g (x) = a x + b

The problem explicitly states that

g (x)

is a polynomial of degree 1. By definition, any polynomial of degree 1 is a linear function, which must take the form

a x + b

. This is not an assumption; it is a direct translation of the problem statement into mathematical language.

Animated Solution for Mathematics - Functions: Let $f (x)$ and $g (x)$ be two real polynomials of degree 2 and 1 respectively. If $f (g (x)) = 8 x^{2} - 2 x$ , and $g (f (x)) = 4 x^{2} + 6 x + 1$ , then the value of $f (2) + g (2)$ is ______.

Enter Numerical Value:

Visualized Solution

Define $g (x)$ as a Linear Polynomial

Substitute $g (x)$ into $f (g (x))$

Change Variables to find $f (t)$

Expand and Simplify $f (t)$

Express $f (x)$ in Standard Form

Set up the Equation for $g (f (x))$

Simplify the $g (f (x))$ Expression

Compare $x^{2}$ Coefficients to find $a$

Compare $x$ Coefficients to find $b$

Determine $g (x)$ and find $g (2)$

Determine $f (x)$ and find $f (2)$

Final Answer: $f (2) + g (2)$

The Sigma Insight: Composite Functions

Analyzing the Setup

The Substitution Method

Comparing Coefficients

Final Calculation

Similar Questions

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:

Let $f : R \to R$ be a function defined by $f (x) = [2 (1 - \frac{x ^{25}}{2}) (2 + x^{25})]^{\frac{1}{50}}$ . If the function $g (x) = f (f (f (x))) + f (f (x))$ , then the greatest integer less than or equal to $g (1)$ is ______.

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 $f, g : (1, \infty) \to R$ be defined as $f (x) = \frac{2 x + 3}{5 x + 2}$ and $g (x) = \frac{2 - 3 x}{1 - x}$ . If the range of the function $f \circ g : [2, 4] \to R$ is $[α, β]$ , then $\frac{1}{β - α}$ 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

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:

For $x \in R$ , two real valued functions $f (x)$ and $g (x)$ are such that, $g (x) = x + 1$ and $f (g (x)) = x + 3 - x$ . Then $f (0)$ is equal to

If $f (x) = \frac{4 x + 3}{6 x - 4}, x \neq = \frac{2}{3}$ and $(f \circ f) (x) = g (x)$ , where $g : R - {\frac{2}{3}} \to R - {\frac{2}{3}}$ , then $(g \circ g \circ g) (4)$ is equal to

If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ 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

Enter Numerical Value:

Visualized Solution

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) as a Linear Polynomial

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

Change Variables to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Expand and Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) in Standard Form

Set up the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f(x))

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

Compare .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2 Coefficients to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Compare .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x Coefficients to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Determine .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) and find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(2)

Determine .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) and find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)+g(2)

The Sigma Insight: Composite Functions

Analyzing the Setup

The Substitution Method

Comparing Coefficients

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If g(x)=3x2+2x−3,f(0)=−3 and 4g(f(x))=3x2−32x+72, then f(g(2)) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If g(x)=x2+x−1 and (gof)(x)=4x2−10x+5, then f(45​) is equal to.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f,g:(1,∞)→R be defined as f(x)=5x+22x+3​ and g(x)=1−x2−3x​. If the range of the function f∘g:[2,4]→R is [α,β], then β−α1​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={2+2x,1−3x​,​−1≤x<00≤x≤3​;g(x)={−x,x,​−3≤x≤00<x≤1​, then range of (f∘g(x)) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x+1x−1​,x∈R−{0,−1,1}. If fn+1(x)=f(fn(x)) for all n∈N, then f6(6)+f7(7) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈R, two real valued functions f(x) and g(x) are such that, g(x)=x​+1 and f(g(x))=x+3−x​. Then f(0) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=6x−44x+3​,x=32​ and (f∘f)(x)=g(x), where g:R−{32​}→R−{32​}, then (g∘g∘g)(4) is equal to

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=xloge​(1234)−(tan1∘)(tan1∘)x+loge​(123)​, x>0, then the least value of f(f(x))+f(f(x4​)) is

Define $g (x)$ as a Linear Polynomial

Substitute $g (x)$ into $f (g (x))$

Change Variables to find $f (t)$

Expand and Simplify $f (t)$

Express $f (x)$ in Standard Form

Set up the Equation for $g (f (x))$

Simplify the $g (f (x))$ Expression

Compare $x^{2}$ Coefficients to find $a$

Compare $x$ Coefficients to find $b$

Determine $g (x)$ and find $g (2)$

Determine $f (x)$ and find $f (2)$

Final Answer: $f (2) + g (2)$

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

Let $f, g : (1, \infty) \to R$ be defined as $f (x) = \frac{2 x + 3}{5 x + 2}$ and $g (x) = \frac{2 - 3 x}{1 - x}$ . If the range of the function $f \circ g : [2, 4] \to R$ is $[α, β]$ , then $\frac{1}{β - α}$ 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

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:

For $x \in R$ , two real valued functions $f (x)$ and $g (x)$ are such that, $g (x) = x + 1$ and $f (g (x)) = x + 3 - x$ . Then $f (0)$ is equal to

If $f (x) = \frac{4 x + 3}{6 x - 4}, x \neq = \frac{2}{3}$ and $(f \circ f) (x) = g (x)$ , where $g : R - {\frac{2}{3}} \to R - {\frac{2}{3}}$ , then $(g \circ g \circ g) (4)$ is equal to

If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ 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