Logo
Jeeify
MCQMediumJEE 2025Applications of Derivatives (Monotonicity, Extrema)

JEE Mathematics 2025 Question with Solution

MATHEMATICS SECTION-A

Let a>0a > 0. If the function f(x)=6x3−45ax2+108a2x+1f(x) = 6x^3 - 45ax^2 + 108a^2x + 1 attains its local maximum and minimum values at the points x1x_1 and x2x_2 respectively such that x1x2=54x_1x_2 = 54, then a+x1+x2a + x_1 + x_2 is equal to:

  • A

    1515

  • B

    1818

  • C

    2424

  • D

    1313

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: f(x)=6x3−45ax2+108a2x+1f(x) = 6x^3 - 45ax^2 + 108a^2x + 1 and the local maximum and minimum occur at x1x_1 and x2x_2 with x1x2=54x_1x_2 = 54.

Find: a+x1+x2a + x_1 + x_2.

For local maxima and minima, set the first derivative equal to zero.

f′(x)=18x2−90ax+108a2f'(x) = 18x^2 - 90ax + 108a^2

Critical points satisfy

18x2−90ax+108a2=018x^2 - 90ax + 108a^2 = 0

Dividing by 1818,

x2−5ax+6a2=0x^2 - 5ax + 6a^2 = 0

Solving,

x=5a±25a2−24a22=5a±a2x = \frac{5a \pm \sqrt{25a^2 - 24a^2}}{2} = \frac{5a \pm a}{2}

Hence the critical points are

x1=2a,x2=3ax_1 = 2a, \quad x_2 = 3a

Using the given condition,

x1x2=(2a)(3a)=6a2=54x_1x_2 = (2a)(3a) = 6a^2 = 54

So,

a2=9a^2 = 9

Since a>0a > 0,

a=3a = 3

Therefore,

x1=6,x2=9x_1 = 6, \quad x_2 = 9

Now,

a+x1+x2=3+6+9=18a + x_1 + x_2 = 3 + 6 + 9 = 18

Therefore, the correct option is B.

Using the derivative roots explicitly

Given: f(x)=6x3−45ax2+108a2x+1f(x) = 6x^3 - 45ax^2 + 108a^2x + 1.

Find: a+x1+x2a + x_1 + x_2 where x1x_1 and x2x_2 are the points of local maximum and minimum.

Step 1: Differentiate the function.

f′(x)=ddx(6x3−45ax2+108a2x+1)=18x2−90ax+108a2f'(x) = \frac{d}{dx}\left(6x^3 - 45ax^2 + 108a^2x + 1\right) = 18x^2 - 90ax + 108a^2

Step 2: Set the derivative equal to zero.

18x2−90ax+108a2=018x^2 - 90ax + 108a^2 = 0

Divide throughout by 1818:

x2−5ax+6a2=0x^2 - 5ax + 6a^2 = 0

Step 3: Solve the quadratic equation.

x=−(−5a)±(−5a)2−4(1)(6a2)2(1)x = \frac{-(-5a) \pm \sqrt{(-5a)^2 - 4(1)(6a^2)}}{2(1)}x=5a±25a2−24a22x = \frac{5a \pm \sqrt{25a^2 - 24a^2}}{2}x=5a±a2x = \frac{5a \pm a}{2}

So the two roots are

x1=3a,x2=2ax_1 = 3a, \quad x_2 = 2a

Step 4: Use the condition x1x2=54x_1x_2 = 54.

(3a)(2a)=54(3a)(2a) = 546a2=546a^2 = 54a2=9a^2 = 9

Since a>0a > 0,

a=3a = 3

Step 5: Compute the required sum.

x1=9,x2=6x_1 = 9, \quad x_2 = 6

Thus,

a+x1+x2=3+9+6=18a + x_1 + x_2 = 3 + 9 + 6 = 18

Therefore, the value of a+x1+x2a + x_1 + x_2 is 1818.

Common mistakes

  • Students may differentiate −45ax2-45ax^2 incorrectly by treating aa as a variable of differentiation. Here aa is a constant with respect to xx, so ddx(−45ax2)=−90ax\frac{d}{dx}(-45ax^2) = -90ax. Always differentiate only with respect to xx.

  • Students may use a=±3a = \pm 3 after getting a2=9a^2 = 9. This is incorrect because the question explicitly gives a>0a > 0. Therefore, only a=3a = 3 is valid.

  • Students may confuse which values are the critical points and directly add coefficients from the quadratic. The roots of x2−5ax+6a2=0x^2 - 5ax + 6a^2 = 0 must first be found as 2a2a and 3a3a before using the condition x1x2=54x_1x_2 = 54.

Practice more Applications of Derivatives (Monotonicity, Extrema) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Applications of Derivatives (Monotonicity, Extrema)

Related questions