JEE 2023 — Mathematics PYQ

Continuity ke liye ${\lim }_{x\to 0}f(x)=f(0)$ hona chahiye.

Hum jante hain ki $-1\le \sin \left(\frac{1}{x}\right)\le 1$.

Jab $x\to 0$, tab $x^2\to 0$.

Isliye, $0\times (\text{a value between -1 and 1})=0$.

Yahan ${\lim }_{x\to 0}f(x)=0$ aur $f(0)=0$ hai.

Isliye, $f(x)$ continuous hai.

2. Differentiability check karein ($x=0$ par):

First Principle use karte hain:

$f^{\prime }(0)={\lim }_{h\to 0}\frac{f(0+h)-f(0)}{h}$

$f^{\prime }(0)={\lim }_{h\to 0}\frac{h^2\sin \left(\frac{1}{h}\right)-0}{h}$

$f^{\prime }(0)={\lim }_{h\to 0}h\sin \left(\frac{1}{h}\right)$

Jaise upar dekha, $h\to 0$ hone par ye puri limit $0$ ho jayegi.

Isliye, $f(x)$ differentiable hai aur $f^{\prime }(0)=0$.

3. $f^{\prime }(x)$ ki continuity check karein:

Ab $x\ne 0$ ke liye derivative nikalte hain (Product Rule se):

$f^{\prime }(x)=\frac{d}{dx}[x^2\sin (1/x)]$

$f^{\prime }(x)=2x\sin \left(\frac{1}{x}\right)+x^2\cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$

$f^{\prime }(x)=2x\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)$

Ab check karte hain ki kya ${\lim }_{x\to 0}{f}^{\prime }(x)=f^{\prime }(0)$ hai?

Yahan $2x\sin (1/x)\to 0$ ho jayega, lekin ${\lim }_{x\to 0}\cos (1/x)$ exist nahi karti (ye -1 aur 1 ke beech oscillate karta rehta hai).

Kyuki limit exist nahi karti, isliye $f^{\prime }(x)$ continuous nahi hai.

Final Result:

$f$ continuous hai, lekin $f^{\prime }$ continuous nahi hai.

Sahi option (D) hai.