For each t \in R , let [t] be the greatest integer less than or equal to t Then \mathop \lim \limits_x \to 1^ + ( 1 - | x | + \sin | 1 - x | )\sin ( \pi \over 2[ 1 - x ] ) \over | 1 - x |.[ 1 - x ]

Limits Continuity and Differentiability PYQ — JEE Main 2019 · 10 Jan · Shift 1 · Q30 with Solution

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Correct answer: C

Since $x \to 1^+$ , we have $x>1$ and very close to $1$ .

So, $|x|=x, \qquad |1-x|=x-1.$ Also, $1-x<0 \quad \text{and} \quad -1<1-x<0$ for $x$ sufficiently close to $1$ from the right.

Hence, $[1-x]=-1$ because the greatest integer less than or equal to a number in $(-1,0)$ is $-1$ .

We get $\sin\left(\frac{\pi}{2}[1-x]\right)=\sin\left(\frac{\pi}{2}(-1)\right)=\sin\left(-\frac{\pi}{2}\right)=-1.$

Also, $|1-x|\,[1-x]=(x-1)(-1)=-(x-1).$

The numerator is $\left(1-|x|+\sin|1-x|\right)\sin\left(\frac{\pi}{2}[1-x]\right).$

Since $|x|=x$ and $|1-x|=x-1$ , $1-|x|+\sin|1-x| = 1-x+\sin(x-1).$

Multiplying by $-1$ , $\text{numerator}=-(1-x+\sin(x-1)).$

Thus the whole expression becomes $\frac{-(1-x+\sin(x-1))}{-(x-1)} = \frac{1-x+\sin(x-1)}{x-1}.$

Now note that $1-x=-(x-1),$ so

= -1 + \frac{\sin(x-1)}{x-1}.$$ --- 4. **Take the limit** Let $h=x-1$. Then as $x\to 1^+$, $h\to 0^+$. So the limit becomes $$\lim_{h\to 0^+}\left(-1+\frac{\sin h}{h}\right).$$ Using the standard limit $$\lim_{h\to 0}\frac{\sin h}{h}=1,$$ we get $$-1+1=0.$$ --- 5. **Evaluate the options** - A: equals $-1$ ❌ - B: equals $1$ ❌ - C: equals $0$ ✅ - D: does not exist ❌ Therefore, the correct option is $$\boxed{\text{C}}.$$