Mathematics · Application of Derivatives

JEE Main 2025 — 4 April, Evening Shift — Question 35

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 $x_{1} x_{2}=54$ , then $a+x_{1}+x_{2}$ is equal to

Answer: C

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

For maxima or minima $f^{\prime}(x)=0$

$f(x)=18 x^{2}-90 x+108 a^{2}=0 \underbrace{x_{1}}_{x_{2}}$

$x_{1} x_{2}=\frac{108 a^{2}}{18}=54$

$\Rightarrow a^{2}=9 \Rightarrow a=3$

Now, $a+x_{1}+x_{2}=3+\frac{90}{6}=3+15=18$

Answer key and solution verified before publishing.

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