Maths Q17 – JEE-MAIN 2025

Home Maths Continuity and DifferentiabilityQuestion #154

Let

a > 0

. If the function

f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1

attains its local maximum and minimum values at the points

x_{1}

and

x_{2}

respectively such that

x_{1} x_{2} = 54

, then

a + x_{1} + x_{2}

is equal to :

Clue 1 of 3: Critical Points

Local maximum and minimum values of a function occur at critical points, where the first derivative is zero or undefined.

🧠 Full Solution Path

Step 1: Find the first derivative and critical points✦ Active

The function is $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ . To find the local maximum and minimum values, we first find the derivative $f^{'} (x)$ and set it to zero.

f^{'} (x) = \frac{d}{d x} (6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1) = 18 x^{2} - 90 a x + 108 a^{2}

The critical points $x_{1}$ and $x_{2}$ are the roots of $f^{'} (x) = 0$ . Dividing the equation by 18, we get:

x^{2} - 5 a x + 6 a^{2} = 0

Step 2: Use Vieta's formulas and the given condition to find 'a'○ Expand

For the quadratic equation $x^{2} - 5 a x + 6 a^{2} = 0$ , the sum and product of the roots $x_{1}$ and $x_{2}$ are given by Vieta's formulas:

x_{1} + x_{2} = - (- 5 a) / 1 = 5 a

x_{1} x_{2} = 6 a^{2} / 1 = 6 a^{2}

We are given that $x_{1} x_{2} = 54$ . Equating this with the product from Vieta's formulas:

6 a^{2} = 54 ⟹ a^{2} = 9

Since $a > 0$ , we take the positive root:

a = 3

💡 Teacher's Secret Hint

Remember to consider the condition $a > 0$ when solving for $a$ .

Step 3: Calculate the required expression○ Expand

Now substitute the value of $a = 3$ into the sum of roots expression:

x_{1} + x_{2} = 5 a = 5 (3) = 15

The question asks for the value of $a + x_{1} + x_{2}$ . Substitute the values of $a$ and $(x_{1} + x_{2})$ :

a + x_{1} + x_{2} = 3 + 15 = 18

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Maths Question 17 – JEE-MAIN 2025

Clue 1 of 3: Critical Points

🧠 Full Solution Path