JEE Main 2026 (28 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Functions: 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:

Q: Why did we get two possible values for $a$ ($\pm 1/2$) but only choose one?

Excellent observation! When we equated the coefficients of $x^2$, we got $12a^2 = 3$, which leads to $a^2 = 1/4$, so $a = \pm 1/2$. However, we must also satisfy the coefficient of the $x$ term. When we equated the coefficients of $x$, we found $-64a = -32$, which forces $a$ to be exactly $1/2$. The negative value is inconsistent with the $x$ term, so we must reject it.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look at the degree of the polynomials. If

g (x)

is a quadratic polynomial (degree 2) and the composite function

4 g (f (x))

is also a quadratic polynomial (degree 2), what does this imply about the degree of the inner function

f (x)

? Remember that the degree of a composite function

g (f (x))

is the product of the degrees of

g

and

f

Hint 2

Hint 2: The Arsenal. Since

f (x)

must be linear, let

f (x) = a x + b

. You are given

f (0) = - 3

. Use this immediately to find the value of

b

. This reduces your problem to finding only one unknown constant,

a

Hint 3

Hint 3: The Execution. Substitute

f (x) = a x - 3

into the expression for

g (x)

, which is

3 x^{2} + 2 x - 3

. Expand the resulting expression

g (f (x)) = 3 (a x - 3)^{2} + 2 (a x - 3) - 3

and multiply by

4

. Finally, compare the coefficients of

x^{2}

and

x

with the given expression

3 x^{2} - 32 x + 72

to solve for

a

Visualized Solution

Analyze the Degree of $f (x)$

Given $g (x) = 3 x^{2} + 2 x - 3$ (Degree 2)
Given $4 g (f (x)) = 3 x^{2} - 32 x + 72$ (Degree 2)
Since the degree of $g (f (x))$ is the same as the degree of $g (x)$ , the inner function $f (x)$ must be a linear function.
Let $f (x) = a x + b$

Use the Initial Condition $f (0) = - 3$

Given $f (0) = - 3$
Substitute $x = 0$ into $f (x) = a x + b$ :
$f (0) = a (0) + b = - 3$
$\Rightarrow b = - 3$

Define the Linear Form of $f (x)$

Using the value of $b$ , we can write:
$f (x) = a x - 3$

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

Substitute $f (x) = a x - 3$ into $g (x) = 3 x^{2} + 2 x - 3$ :
$g (f (x)) = 3 (a x - 3)^{2} + 2 (a x - 3) - 3$

Expand the Square Term

Expand $(a x - 3)^{2}$ using $(u - v)^{2} = u^{2} - 2 uv + v^{2}$ :
$g (f (x)) = 3 (a^{2} x^{2} - 6 a x + 9) + 2 (a x - 3) - 3$
$g (f (x)) = 3 a^{2} x^{2} - 18 a x + 27 + 2 a x - 6 - 3$

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

Combine like terms:
$g (f (x)) = 3 a^{2} x^{2} + (- 18 a + 2 a) x + (27 - 6 - 3)$
$g (f (x)) = 3 a^{2} x^{2} - 16 a x + 18$

Multiply by Four

Multiply the entire expression by $4$ :
$4 g (f (x)) = 4 (3 a^{2} x^{2} - 16 a x + 18)$
$4 g (f (x)) = 12 a^{2} x^{2} - 64 a x + 72$

Compare Coefficients of $x^{2}$

Compare $12 a^{2} x^{2} - 64 a x + 72$ with $3 x^{2} - 32 x + 72$ :
For $x^{2}$ : $12 a^{2} = 3$
$a^{2} = \frac{3}{12} = \frac{1}{4}$
$a = \pm \frac{1}{2}$

Verify with Coefficient of $x$

For $x$ : $- 64 a = - 32$
$a = \frac{- 32}{- 64} = \frac{1}{2}$
This matches $a = \pm \frac{1}{2}$ from the previous step.

Finalize the Function $f (x)$

Substitute $a = \frac{1}{2}$ and $b = - 3$ back into $f (x)$ :
$f (x) = \frac{1}{2} x - 3$

Calculate $g (2)$

To find $f (g (2))$ , first calculate $g (2)$ :
$g (2) = 3 (2)^{2} + 2 (2) - 3$
$g (2) = 3 (4) + 4 - 3$
$g (2) = 12 + 4 - 3 = 13$

Calculate $f (g (2))$

Now substitute $g (2) = 13$ into $f (x) = \frac{1}{2} x - 3$ :
$f (g (2)) = f (13) = \frac{1}{2} (13) - 3$
$f (13) = \frac{13}{2} - 3$
$f (13) = \frac{13 - 6}{2} = \frac{7}{2}$

Conclusion and Summary

Key Takeaway:
1. Analyze the degree of composite functions to determine the form of the unknown function.
2. Use given initial conditions to find constants quickly.
3. Compare coefficients of like terms to solve for unknown parameters.
Final Answer: $f (g (2)) = \frac{7}{2}$

00:00 / 00:00

The Sigma Insight: Composite Functions

Analyzing the Setup

We are given the function

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

and the composite expression

4 g (f (x)) = 3 x^{2} - 32 x + 72

. Our objective is to determine the value of

f (g (2))

The function

g (x)

is a quadratic polynomial of degree

2

. Since the composite function

4 g (f (x))

is also a quadratic polynomial, the inner function

f (x)

must be a linear function.

We define the linear function as

f (x) = a x + b

. This assumption is the foundation of our solution.

The Detective Work

The problem provides the condition

f (0) = - 3

. Substituting

x = 0

into our linear form

f (x) = a x + b

, we obtain:

f (0) = a (0) + b = - 3

This immediately reveals that

b = - 3

. Consequently, our function simplifies to

f (x) = a x - 3

The Algebraic Battle

We now substitute

f (x)

into

g (x)

to find the expression for

g (f (x))

g (f (x)) = 3 (a x - 3)^{2} + 2 (a x - 3) - 3

Expanding the square term

(a x - 3)^{2} = a^{2} x^{2} - 6 a x + 9

, we substitute this back into the expression:

g (f (x)) = 3 (a^{2} x^{2} - 6 a x + 9) + 2 a x - 6 - 3

g (f (x)) = 3 a^{2} x^{2} - 18 a x + 27 + 2 a x - 9

g (f (x)) = 3 a^{2} x^{2} - 16 a x + 18

Multiplying the entire expression by

4

to match the given equation, we get:

4 g (f (x)) = 12 a^{2} x^{2} - 64 a x + 72

Coefficient Matching

We compare our derived expression

12 a^{2} x^{2} - 64 a x + 72

with the given

3 x^{2} - 32 x + 72

. Equating the coefficients of

x^{2}

yields:

12 a^{2} = 3 \Rightarrow a^{2} = \frac{1}{4} \Rightarrow a = \pm \frac{1}{2}

Next, we equate the coefficients of

x

to determine the correct sign:

- 64 a = - 32 \Rightarrow a = \frac{32}{64} = \frac{1}{2}

The negative root is discarded. Thus, our function is defined as

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

Final Calculation

To find

f (g (2))

, we first evaluate the inner function

g (2)

g (2) = 3 (2)^{2} + 2 (2) - 3 = 12 + 4 - 3 = 13

Finally, we substitute this result into

f (x)

f (13) = \frac{1}{2} (13) - 3 = \frac{13}{2} - \frac{6}{2} = \frac{7}{2}

The final answer is $\frac{7}{2}$ .

Common Traps & FAQs

Wait, why do we assume

f (x)

is linear? Could it be a constant or something else?

That is a fantastic question! Let us look at the degrees. If

f (x)

were a constant,

g (f (x))

would also be a constant, which contradicts the fact that

4 g (f (x))

is a quadratic polynomial. If

f (x)

were a quadratic,

g (f (x))

would be a quartic (degree 4) polynomial. Since

g (f (x))

is quadratic, the only possibility is that

f (x)

is a linear function of the form

a x + b

Why did we get two possible values for

a

(

\pm 1/2

) but only choose one?

Excellent observation! When we equated the coefficients of

x^{2}

, we got

12 a^{2} = 3

, which leads to

a^{2} = 1/4

, so

a = \pm 1/2

. However, we must also satisfy the coefficient of the

x

term. When we equated the coefficients of

x

, we found

- 64 a = - 32

, which forces

a

to be exactly

1/2

. The negative value is inconsistent with the

x

term, so we must reject it.

Is there a faster way to find

f (g (2))

without finding the full expression for

f (x)

While it is tempting to look for shortcuts, finding the explicit form of

f (x)

is the most robust method here. Because

f (x)

is a simple linear function, determining

a

and

b

is very fast. Once you have

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

, calculating

f (g (2))

becomes a straightforward substitution, which is less prone to errors than trying to manipulate the composite function directly.

Animated Solution for Mathematics - Functions: 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:

Select Answer:

Visualized Solution

Analyze the Degree of $f (x)$

Use the Initial Condition $f (0) = - 3$

Define the Linear Form of $f (x)$

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

Expand the Square Term

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

Multiply by Four

Compare Coefficients of $x^{2}$

Verify with Coefficient of $x$

Finalize the Function $f (x)$

Calculate $g (2)$

Calculate $f (g (2))$

Conclusion and Summary

The Sigma Insight: Composite Functions

Analyzing the Setup

The Detective Work

The Algebraic Battle

Coefficient Matching

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Functions: .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:

Select Answer:

Visualized Solution

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

Use the Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)=−3

Define the Linear Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

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

Expand the Square Term

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

Multiply by Four

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

Verify with Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

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

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

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

Conclusion and Summary

The Sigma Insight: Composite Functions

Analyzing the Setup

The Detective Work

The Algebraic Battle

Coefficient Matching

Final Calculation

Similar Questions

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.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

Animated Solution for Mathematics - Functions: 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:

Analyze the Degree of $f (x)$

Use the Initial Condition $f (0) = - 3$

Define the Linear Form of $f (x)$

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

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

Compare Coefficients of $x^{2}$

Verify with Coefficient of $x$

Finalize the Function $f (x)$

Calculate $g (2)$

Calculate $f (g (2))$

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

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

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 ______.

If $g (f (x)) = ∣ sin x ∣$ and $f (g (x)) = (sin x)^{2}$ , then

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

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

Let $f, g : (1, \infty) \to R$ 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