Let f(x) = (x-a)^2+(x-b)^2+(x-c)^2 then f(x) has a minimum value at x = Option 1: a+b+c Option 2: √abc Option 3: abc Option 4: (a+b+c)/3

Answer: (a+b+c)/3. Solution: f'(x)=2(x-a)+2(x-b)+2(x-c)=0 gives 3x=a+b+c, so x=(a+b+c)/3. Since f''(x)=6>0, this critical point is a minimum.

Calculus

Application of Derivatives

Easy

common

Let $f(x) = (x-a)^2+(x-b)^2+(x-c)^2$ then $f(x)$ has a minimum value at $x =$

$a+b+c$

$\sqrt{abc}$

$abc$

$\frac{a+b+c}{3}$

✅ Correct Option: 4

$f'(x)=2(x-a)+2(x-b)+2(x-c)=0$ gives $3x=a+b+c$ , so $x=\frac{a+b+c}{3}$ . Since $f''(x)=6>0$ , this critical point is a minimum.

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For the function, $f(x) = \frac{-3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 - 350$ , which of the following statements are correct?

(A) $x = -3$ and $x = -5$ are the only critical points of the given function.

(B) $x = -3$ is a point of local minimum.

(D) $x = -5$ is a point of local maximum.

Choose the correct answer from the options given below:

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2026: 19 May Shift 1

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