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Application of Derivatives question

2025 · 4 Apr · Shift 2 · Q30

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Application of Derivatives
2025 · 4 Apr · Shift 2 · Q30

Application of Derivatives question

2025 · 4 Apr · Shift 2 · Q30

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1

Let

\mathrm{a}\gt 0

. If the function

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

\mathrm{a}+x_1+x_2

is equal to :

A
15
B
13
C
24
D
18

View written solutionFree

Correct answer: D

Find the critical points

Given $f(x)=6x^3-45ax^2+108a^2x+1,$ with $a>0$ .

For local maxima/minima, we solve $f'(x)=0.$

Now, $f'(x)=18x^2-90ax+108a^2.$ Factor out $18$ : $f'(x)=18(x^2-5ax+6a^2)=18(x-2a)(x-3a).$

Hence the critical points are

\quad x=3a.$$ 2. **Identify maximum and minimum points** Since $a>0$, we have $2a<3a$. For a cubic with positive leading coefficient, the smaller critical point is the local maximum and the larger is the local minimum. So, $$x_1=2a,\qquad x_2=3a.$$ 3. **Use the given condition** We are given $$x_1x_2=54.$$ Substitute: $$ (2a)(3a)=54 $$ $$ 6a^2=54 $$ $$ a^2=9. $$ Since $a>0$, $$ a=3. $$ 4. **Compute $a+x_1+x_2$** Now, $$x_1=2a=6,\qquad x_2=3a=9.$$ Therefore, $$a+x_1+x_2=3+6+9=18.$$ 5. **Check options** The value is $$18,$$ which corresponds to **Option D**.

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