JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5$ , then $f (2)$ is equal to :

Q: Wait, why do we assume the constant term and the $x$ term are zero? Couldn't they be something else?

Great question! If the constant term $e$ were non-zero, then $\frac{e}{x^2}$ would approach infinity as $x \to 0$. Similarly, if the coefficient of $x$ were non-zero, $\frac{dx}{x^2} = \frac{d}{x}$ would also blow up. For the limit to be a finite value like $5$, those terms must vanish. The limit essentially 'filters' out the lower-degree terms.

Select Answer:

Visualized Solution

Analyzing the Limit at $x = 0$

Given limit: $lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5$
For the limit to be finite, $f (0)$ must be $0$ .
Since the denominator is $x^{2}$ , the terms $x^{0}$ and $x^{1}$ in the polynomial must have zero coefficients.

Determining the Polynomial Form

General form of degree 4 polynomial: $f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$
From limit: $e = 0$ , $d = 0$ , and $c = 5$
Resulting polynomial: $f (x) = a x^{4} + b x^{3} + 5 x^{2}$

Finding the Derivative $f^{'} (x)$

Differentiating $f (x)$ with respect to $x$ :
$\frac{d}{d x} (a x^{4} + b x^{3} + 5 x^{2})$
$f^{'} (x) = 4 a x^{3} + 3 b x^{2} + 10 x$

Applying Extreme Value at $x = 4$

Extreme values occur where $f^{'} (x) = 0$ .
At $x = 4$ , $f^{'} (4) = 0$
$4 a (4)^{3} + 3 b (4)^{2} + 10 (4) = 0$
$256 a + 48 b + 40 = 0$

Simplifying the First Equation

Divide the equation by $8$ :
$\frac{256 a}{8} + \frac{48 b}{8} + \frac{40}{8} = 0$
$32 a + 6 b + 5 = 0$
Equation 1: $32 a + 6 b = - 5$

Applying Extreme Value at $x = 5$

At $x = 5$ , $f^{'} (5) = 0$
$4 a (5)^{3} + 3 b (5)^{2} + 10 (5) = 0$
$500 a + 75 b + 50 = 0$

Simplifying the Second Equation

Divide the equation by $25$ :
$\frac{500 a}{25} + \frac{75 b}{25} + \frac{50}{25} = 0$
$20 a + 3 b + 2 = 0$
Equation 2: $20 a + 3 b = - 2$

Solving for Coefficient $a$

Multiply Equation 2 by $2$ : $40 a + 6 b = - 4$
Subtract Equation 1: $(40 a + 6 b) - (32 a + 6 b) = - 4 - (- 5)$
$8 a = 1 ⟹ a = \frac{1}{8}$

Solving for Coefficient $b$

Substitute $a = \frac{1}{8}$ into Equation 2:
$20 (\frac{1}{8}) + 3 b = - 2$
$\frac{5}{2} + 3 b = - 2 ⟹ 3 b = - 2 - \frac{5}{2} = - \frac{9}{2}$
$b = - \frac{3}{2}$

The Final Polynomial Expression

Substitute $a$ and $b$ back into $f (x)$ :
$f (x) = \frac{1}{8} x^{4} - \frac{3}{2} x^{3} + 5 x^{2}$

Calculating the Value of $f (2)$

Substitute $x = 2$ :
$f (2) = \frac{1}{8} (2)^{4} - \frac{3}{2} (2)^{3} + 5 (2)^{2}$
$f (2) = \frac{16}{8} - \frac{3}{2} (8) + 5 (4)$
$f (2) = 2 - 12 + 20 = 10$

Summary and Key Takeaways

Final Answer: $f (2) = 10$
Key Takeaway 1: $lim_{x \to 0} \frac{f ( x )}{x ^{n}} = k$ implies $f (x)$ starts with term $k x^{n}$ .
Key Takeaway 2: Extreme values at $x = c$ mean $f^{'} (c) = 0$ for polynomials.

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We are given the limit

lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5

. For a polynomial

f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e

, the expression becomes:

\frac{f ( x )}{x ^{2}} = a x^{2} + b x + c + \frac{d}{x} + \frac{e}{x ^{2}}

For this limit to equal a finite value of

5

x \to 0

, the coefficients of the singular terms must be zero. Thus, we must have

e = 0

and

d = 0

Evaluating the limit as

x \to 0

, the remaining terms

a x^{2}

and

b x

vanish, leaving us with

c = 5

. Consequently, the polynomial simplifies to:

f (x) = a x^{4} + b x^{3} + 5 x^{2}

The Geometry of Extremes

The function has extreme values at

x = 4

and

x = 5

. In calculus, this implies that the derivative

f^{'} (x)

must be zero at these points. We calculate the derivative as:

f^{'} (x) = 4 a x^{3} + 3 b x^{2} + 10 x

Setting

f^{'} (4) = 0

and

f^{'} (5) = 0

allows us to solve for the unknown coefficients

a

and

b

. Substituting

x = 4

4 a (64) + 3 b (16) + 10 (4) = 256 a + 48 b + 40 = 0

Dividing this equation by

8

yields:

32 a + 6 b = - 5

Substituting

x = 5

4 a (125) + 3 b (25) + 10 (5) = 500 a + 75 b + 50 = 0

Dividing this equation by

25

yields:

20 a + 3 b = - 2

The Algebraic Resolution

We now solve the system of linear equations:

32 a + 6 b = - 5

20 a + 3 b = - 2

Multiply the second equation by

2

to obtain

40 a + 6 b = - 4

. Subtracting the first equation from this result gives:

8 a = 1 \Rightarrow a = \frac{1}{8}

Substituting

a = \frac{1}{8}

into

20 a + 3 b = - 2

20 (\frac{1}{8}) + 3 b = - 2 \Rightarrow \frac{5}{2} + 3 b = - 2 \Rightarrow 3 b = - \frac{9}{2} \Rightarrow b = - \frac{3}{2}

The fully determined polynomial is:

f (x) = \frac{1}{8} x^{4} - \frac{3}{2} x^{3} + 5 x^{2}

Final Calculation

To find

f (2)

, we substitute

x = 2

into our derived polynomial:

f (2) = \frac{1}{8} (2)^{4} - \frac{3}{2} (2)^{3} + 5 (2)^{2}

f (2) = \frac{16}{8} - \frac{3}{2} (8) + 5 (4)

f (2) = 2 - 12 + 20 = 10

The final result is 10.

Common Traps & FAQs

Wait, why do we assume the constant term and the

x

term are zero? Couldn't they be something else?

Great question! If the constant term

e

were non-zero, then

\frac{e}{x ^{2}}

would approach infinity as

x \to 0

. Similarly, if the coefficient of

x

were non-zero,

\frac{d x}{x ^{2}} = \frac{d}{x}

would also blow up. For the limit to be a finite value like

5

, those terms must vanish. The limit essentially 'filters' out the lower-degree terms.

Why do we set

f^{'} (x) = 0

x = 4

and

x = 5

? Is it possible the extreme values are at the boundaries?

In the context of JEE problems involving polynomials, 'extreme values' almost always refers to local maxima or minima occurring at critical points where the derivative vanishes. Since polynomials are differentiable everywhere, the critical points are precisely where

f^{'} (x) = 0

Is there a faster way to solve this without finding

a

and

b

While you might be tempted to look for shortcuts, finding the coefficients is the most robust path. By determining

a

and

b

, you define the function completely. Once you have the function, calculating

f (2)

becomes a simple arithmetic task, ensuring you don't fall into traps by trying to guess the function's behavior.

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5$ , then $f (2)$ is equal to :

Select Answer:

Visualized Solution

Analyzing the Limit at $x = 0$

Determining the Polynomial Form

Finding the Derivative $f^{'} (x)$

Applying Extreme Value at $x = 4$

Simplifying the First Equation

Applying Extreme Value at $x = 5$

Simplifying the Second Equation

Solving for Coefficient $a$

Solving for Coefficient $b$

The Final Polynomial Expression

Calculating the Value of $f (2)$

Summary and Key Takeaways

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Geometry of Extremes

The Algebraic Resolution

Final Calculation

Similar Questions

Let $f (x)$ be a polynomial of degree four having extreme values at $x = 1$ and $x = 2$ . If $lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3$ , then $f (2)$ is equal to

Let $p (x)$ be a polynomial of degree 4 having extremum at $x = 1, 2$ and $lim_{x \to 0} (1 + \frac{p ( x )}{x ^{2}}) = 2$ . Then the value of $p (2)$ is

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Let $f (x)$ be a polynomial of degree $6$ in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x = - 1$ and $x = 1$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$ , then $5 \cdot f (2)$ is equal to

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $x = 2$ be a local minima of the function $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

The local maximum value of the function $f (x) = (\frac{2}{x})^{x^{2}}, x > 0$ , is

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Select Answer:

Visualized Solution

Analyzing the Limit at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Determining the Polynomial Form

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

Applying Extreme Value at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4

Simplifying the First Equation

Applying Extreme Value at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=5

Simplifying the Second Equation

Solving for Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solving for Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Final Polynomial Expression

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

Summary and Key Takeaways

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Geometry of Extremes

The Algebraic Resolution

Final Calculation

Similar Questions

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Analyzing the Limit at $x = 0$

Finding the Derivative $f^{'} (x)$

Applying Extreme Value at $x = 4$

Applying Extreme Value at $x = 5$

Solving for Coefficient $a$

Solving for Coefficient $b$

Calculating the Value of $f (2)$

Let $f (x)$ be a polynomial of degree four having extreme values at $x = 1$ and $x = 2$ . If $lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3$ , then $f (2)$ is equal to

Let $p (x)$ be a polynomial of degree 4 having extremum at $x = 1, 2$ and $lim_{x \to 0} (1 + \frac{p ( x )}{x ^{2}}) = 2$ . Then the value of $p (2)$ is

Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If $lim_{x \to 0} (\frac{f ( x )}{x ^{2}} + 1) = 3$ then f(-1) is equal to :-

Let $f (x)$ be a polynomial of degree $6$ in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x = - 1$ and $x = 1$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$ , then $5 \cdot f (2)$ is equal to

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $x = 2$ be a local minima of the function $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

The local maximum value of the function $f (x) = (\frac{2}{x})^{x^{2}}, x > 0$ , is

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at