JEE Advanced · Mathematics · 24. Differentiation

Let $f (x) = x sin π x$ , $x > 0$ . Then for all natural numbers $n, f^{'} (x)$ vanishes at

A A unique point in the interval $(n, n + \frac{1}{2})$
B A unique point in the interval $(n + \frac{1}{2}, n + 1)$
C A unique point in the interval (n, n + 1)
D Two points in the interval (n, n + 1)

Verified Solution

Answer & Solution

Correct Answer

(C) A unique point in the interval (n, n + 1)

Step-by-step Solution

Detailed explanation

f (x) = x sin π x, x > 0

f' (x)

vanishes,

f^{'} (x) = sin π x + π x cos π x = 0

tan π x = - π x

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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(P) \(\|z\|^2\) is equal to	(1) 12
(Q) \(\|z-\bar{z}\|^2\) is equal to	(2) 4
(R) \(\|z\|^2+\|z+\bar{z}\|^2\) is equal to	(3) 8
(S) \(\|z+1\|^2\) is equal to	(4) 10
	(5) 7

Let f⁡ x = x sin π x , x > 0 . Then for all natural numbers n , f⁡ ′ x vanishes at

Step-by-step Solution

Let $f (x) = x sin π x$ , $x > 0$ . Then for all natural numbers $n, f^{'} (x)$ vanishes at