If $$f(x) = \left\{ {\matrix{ {{{\sin (a + 2)x + \sin x} \over x};} & {x < 0} \cr {b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;} & {x = 0} \cr {{{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{ {1 \over 3}}}} \over {{x^{{4 \over 3}}}}};} & {x > 0} \cr } } \right.$$
is continuous at x = 0, then a + 2b is equal to :

f(0^-) = $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x + \sin x} \over x}$$ <br><br>= $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x} \over {\left( {a + 2} \right)x}} \times \left( {a + 2} \right)$$ + $\mathop {\lim }\limits_{x \to {0^ - }} {{\sin x} \over x}$ <br><br>= $\left( {a + 2} \right)$ + 1 <br><br>= $\left( {a + 3} \right)$ <br><br>f(0⁺) = $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{{1 \over 3}}}} \over {{x^{{4 \over 3}}}}}$$ <br><br>= $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {1 + 3x} \right)}^{{1 \over 3}}} - 1} \over {{x^{{1 \over 3}}}}}$$ <br><br>= $\mathop {\lim }\limits_{x \to {0^ + }} {{1 + x - 1} \over x}$ <br><br>= 1 <br><br>And f(0) = b <br><br>As f(x) is continuous at x = 0, then <br><br>f(0^-) = f(0) = f(0⁺) <br><br>$\Rightarrow$ $a + 3$ = b = 1 <br><br>$\therefore$ $a$ = -2 and b = 1 <br><br>$\therefore$ $a$ + 2b = -2 + 2 = 0