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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
Question
Let $f(x) = \left\{ \begin{array}{l}{x^2} + k,\;\;\;\;{\rm{when}}\;\;x \ge 0\\ - {x^2} - k,\;\;{\rm{when\,\, }}x < 0\end{array} \right.$. If the function $f(x)$ be continuous at $x = 0$, then $k =$
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Answer

Here $\mathop {\lim }\limits_{x \to 0 + } f(x) = k,\mathop {\lim }\limits_{x \to 0 - } f(x) = - k$ and $f(0) = k$
But $f(x)$ is continuous at $x = 0,$
therefore $k$ must be zero.

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