Q26 / 75 · JEE Main 2026 (23 January Shift 2) · Mathematics
If $f(x)=\left\{\begin{array}{cc}\frac{a|x|+x^{2}-2(\sin |x|)(\cos |x|)}{x} &, x \neq 0 \\ b &, x=0\end{array}\right.$
is continuous at $x=0$, then $a+b$ is equal to
is continuous at $x=0$, then $a+b$ is equal to
- A4
- B1
- Correct answerC2
- D0
Solution
Note $2\sin|x|\cos|x| = \sin(2|x|)$.
For $x > 0$: $f(x) = a + x - \frac{\sin 2x}{x} \xrightarrow{x \to 0^+} a - 2$.
For $x < 0$: $f(x) = -a + x + \frac{\sin 2x}{x} \xrightarrow{x \to 0^-} -a + 2$.
For continuity: $a - 2 = 2 - a \Rightarrow a = 2$, and $b = a - 2 = 0$.
$a + b = 2$.
For $x > 0$: $f(x) = a + x - \frac{\sin 2x}{x} \xrightarrow{x \to 0^+} a - 2$.
For $x < 0$: $f(x) = -a + x + \frac{\sin 2x}{x} \xrightarrow{x \to 0^-} -a + 2$.
For continuity: $a - 2 = 2 - a \Rightarrow a = 2$, and $b = a - 2 = 0$.
$a + b = 2$.
Hence, the correct answer is option: 'C'.