JEE Mathematics 2026 Question with Solution

Given:

Find: The value of $a+b$ such that $f(x)$ is continuous at $x=0$.

Using $2\sin\theta\cos\theta = \sin 2\theta$, for $x \neq 0$ we get

For the right hand limit, when $x \to 0^+$, we have $|x|=x$. Therefore,

For the left hand limit, when $x \to 0^-$, we have $|x|=-x$. Therefore,

Since $\sin(-2x)=-\sin(2x)$,

For continuity at $x=0$, the right hand limit and left hand limit must be equal.

So,

Now the common limit is

Since $f(0)=b$ and the function is continuous at $x=0$, we must have

Thus,

Therefore, the correct option is D.

Given: The function is continuous at $x=0$.

Find: $a+b$.

Near $x=0$, use $\sin(2|x|) \sim 2|x|$. Then for $x \neq 0$,

For $x>0$, this becomes

For $x<0$, this becomes

Equating both gives

Then the limit at $x=0$ is $0$, so by continuity $b=0$. Hence $a+b=2$, and the correct option is D.