MCQ
જો $f(x) = \left\{ {\begin{array}{*{20}{c}}{\frac{{{x^2} - 9}}{{x - 3}}\,,}&{{\rm{if \,\,}}x \ne 3}\\{2x + k\,,}&{{\rm{otherwise}}}\end{array}} \right.$, એ $x = 3$ આગળ સતત હોય તો $k = $
- A$3$
- ✓$0$
- C$-6$
- D$1/6$
and $f(3) = 2(3) + k = 6 + k$
$\because f $ is continuous at $x = 3$;
$\therefore $ $6 + k = 6 \Rightarrow k = 0$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.