If f(x)=sin(a+2)x+sinxx; x 0 is continuous at x=0, then a+2b is equal to

limx→0sin(a+2)x+sinxx=b=(x+3x2)13−x13x43⇒(a+2)+1=b=limx→0(1+3x)13−1x⇒a+3=b=limx→0((1+3x)13−1(1+3x)13−1).3⇒a+3=b=13.3=1⇒a=−2, b=1So, a+2b=0

If f(x)=sin(a+2)x+sinxx; x 0 is continuous at x=0, then a+2b is equal to

limx→0sin(a+2)x+sinxx=b=(x+3x2)13−x13x43⇒(a+2)+1=b=limx→0(1+3x)13−1x⇒a+3=b=limx→0((1+3x)13−1(1+3x)13−1).3⇒a+3=b=13.3=1⇒a=−2, b=1So, a+2b=0

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$I f f (x) = \{\begin{cases} \frac{\sin (a + 2) x + \sin x}{x}; x < 0 \\ b; x = 0 \\ \frac{{(x + 3 x^{2})}^{\frac{1}{3}} - x^{\frac{1}{3}}}{x^{\frac{4}{3}}}; x > 0 \end{cases}$ is continuous at x=0, then $a + 2 b$ is equal to

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Detailed Solution

$\begin{array}{l} \lim_{x \to 0} \frac{\sin (a + 2) x + \sin x}{x} = b = \frac{{(x + 3 x^{2})}^{\frac{1}{3}} - x^{\frac{1}{3}}}{x^{\frac{4}{3}}} \\ \Rightarrow (a + 2) + 1 = b = \lim_{x \to 0} \frac{{(1 + 3 x)}^{\frac{1}{3}} - 1}{x} \\ \Rightarrow a + 3 = b = \lim_{x \to 0} (\frac{{(1 + 3 x)}^{\frac{1}{3}} - 1}{{(1 + 3 x)}^{\frac{1}{3}} - 1}) .3 \\ \Rightarrow a + 3 = b = \frac{1}{3} .3 = 1 \\ \Rightarrow a = - 2, b = 1 \\ S o, a + 2 b = 0 \end{array}$