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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Continuity and Differentiability

If f(x)=\ array cc a|x|+x^ 2 -2( |x|)( |x|) x &, x ≠ 0 b &, x=0...

If f(x)=\left\{a∣x∣+x2−2(sin⁡∣x∣)(cos⁡∣x∣)x,x≠0b,x=0\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=0x=0, then a+ba+b is equal to

Options

  1. A4
  2. B1
  3. C2
  4. D0

Correct answer

C. 2

Step-by-step solution

Note 2sin⁡∣x∣cos⁡∣x∣=sin⁡(2∣x∣)2\sin|x|\cos|x| = \sin(2|x|).
For x>0x > 0: f(x)=a+x−sin⁡2xx→x→0+a−2f(x) = a + x - \frac{\sin 2x}{x} \xrightarrow{x \to 0^+} a - 2.
For x<0x < 0: f(x)=−a+x+sin⁡2xx→x→0−−a+2f(x) = -a + x + \frac{\sin 2x}{x} \xrightarrow{x \to 0^-} -a + 2.
For continuity: a−2=2−a⇒a=2a - 2 = 2 - a \Rightarrow a = 2, and b=a−2=0b = a - 2 = 0.
a+b=2a + b = 2.
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