Let f:R→R be given by fx=x-1x-2x-5. Define Fx=∫0xftdt, x>0. Then which of the following options is/are correct?

(B) F has a local maximum at x=2

(D) F has two local maxima and one local minimum in 0, ∞

JEE Advanced · Mathematics · 25. AOD

Let $f : R \to R$ be given by $f (x) = (x - 1) (x - 2) (x - 5) .$ Define $F (x) = \int_{0}^{x} f (t) d t, x > 0 .$ Then which of the following options is/are correct?

A $F$ has a local minimum at $x = 1$
B $F$ has a local maximum at $x = 2$
C $F (x) \neq 0$ for all $x \in (0, 5)$
D $F$ has two local maxima and one local minimum in $(0, \infty)$

Verified Solution

Answer & Solution

Correct Answer

(A) $F$ has a local minimum at $x = 1$

Step-by-step Solution

Detailed explanation

F (x) = \int_{0}^{x} f (t) . d t

F^{'} (x) = f (x) = (x - 1) (x - 2) (x - 5)

\Rightarrow F (x)

has maxima at

x = 2

and minima at

x = 1

and

x = 5

now

F (2) = \int_{0}^{2} (x^{3} - 8 x^{2} + 17 x - 10) . d x

= {[\frac{x^{4}}{4} - \frac{8 x^{3}}{3} + \frac{17 x^{2}}{2} - 10 x]}_{0}^{2}

F (2) = \frac{- 10}{3}

which is a maxima in

x \in (0, 5)

F (0) = 0

Hence

F (x) < 0 \forall x \in (0, 5)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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