Maths Q18 – JEE-MAIN 2025

Home Maths Continuity and DifferentiabilityQuestion #157

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 :

Clue 1 of 3: Understanding Polynomial Properties

The degree of the polynomial and the conditions for extreme values and limits provide crucial information about its form and coefficients.

🧠 Full Solution Path

Step 1: Determine the form of

f (x)

using the limit condition✦ Active

Given that $f (x)$ is a polynomial of degree four and $lim_{x \to 0} \frac{f (x)}{x^{2}} = 5$ . For this limit to be finite and non-zero, $f (0)$ must be $0$ and $f^{'} (0)$ must be $0$ . Applying L'Hopital's rule twice, or by considering the Taylor expansion around $x = 0$ , we find that $f^{″} (0) = 10$ . Let $f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$ . From $f (0) = 0$ , we get $e = 0$ . From $f^{'} (0) = 0$ , we get $d = 0$ . From $f^{″} (0) = 10$ , we get $2 c = 10$ , so $c = 5$ . Thus, the polynomial takes the form $f (x) = a x^{4} + b x^{3} + 5 x^{2}$ .

Step 2: Determine the coefficients

a

and

b

using extreme value conditions○ Expand

The first derivative of $f (x)$ is $f^{'} (x) = 4 a x^{3} + 3 b x^{2} + 10 x$ . Since $f (x)$ has extreme values at $x = 4$ and $x = 5$ , we know that $f^{'} (4) = 0$ and $f^{'} (5) = 0$ . From Step 1, we also know $f^{'} (0) = 0$ . Therefore, $0, 4, 5$ are the roots of $f^{'} (x)$ . We can express $f^{'} (x)$ in factored form as $f^{'} (x) = K x (x - 4) (x - 5)$ for some constant $K$ . Expanding this, $f^{'} (x) = K x (x^{2} - 9 x + 20) = K (x^{3} - 9 x^{2} + 20 x)$ . Comparing the coefficient of $x$ in $f^{'} (x) = 4 a x^{3} + 3 b x^{2} + 10 x$ with $K (x^{3} - 9 x^{2} + 20 x)$ , we have $10 = 20 K$ , which implies $K = \frac{1}{2}$ . Substituting $K$ back, $f^{'} (x) = \frac{1}{2} x^{3} - \frac{9}{2} x^{2} + 10 x$ . Comparing the coefficients of $x^{3}$ and $x^{2}$ with $4 a x^{3} + 3 b x^{2} + 10 x$ : $4 a = \frac{1}{2} ⟹ a = \frac{1}{8}$ . And $3 b = - \frac{9}{2} ⟹ b = - \frac{3}{2}$ . So, $f (x) = \frac{1}{8} x^{4} - \frac{3}{2} x^{3} + 5 x^{2}$ .

Step 3: Calculate

f (2)

○ Expand

Now, substitute $x = 2$ into the expression for $f (x)$ :

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

Calculate the terms:

f (2) = \frac{1}{8} (16) - \frac{3}{2} (8) + 5 (4)

f (2) = 2 - 12 + 20

f (2) = 10

Thus, $f (2)$ is $10$ .

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Maths Question 18 – JEE-MAIN 2025

Clue 1 of 3: Understanding Polynomial Properties

🧠 Full Solution Path