If the function f:ℝ→ℝ defined by f(x)=sin⁡(a+1)x+sin⁡xx ,x 0is continuous on ℝ then a + b =

Let. f (x) is continuous at x=0⇒Ltx→0- f(x)=Ltx→0+ f(x)=f(0)⇒Ltx→0−sin⁡(a+1)x+sin⁡xx=Ltx→0+x2+x−xx3/2=b⇒Ltx→0− ⁡sin⁡(a+1)xx+sin⁡xx=Ltx→0+⁡x(x+1−1)xx=b⇒(a+1)+1=,Ltx→0+⁡x+1−1x=b⇒a+2=Ltx→0+⁡(x+1)−1x(x+1+1)=b⇒a+2=Ltx→0+1x+1+1=b⇒a+2=12=b⇒a=−32,b=12, ∴a+b=−1

If the function f:ℝ→ℝ defined by f(x)=sin⁡(a+1)x+sin⁡xx ,x 0is continuous on ℝ then a + b =

Let. f (x) is continuous at x=0⇒Ltx→0- f(x)=Ltx→0+ f(x)=f(0)⇒Ltx→0−sin⁡(a+1)x+sin⁡xx=Ltx→0+x2+x−xx3/2=b⇒Ltx→0− ⁡sin⁡(a+1)xx+sin⁡xx=Ltx→0+⁡x(x+1−1)xx=b⇒(a+1)+1=,Ltx→0+⁡x+1−1x=b⇒a+2=Ltx→0+⁡(x+1)−1x(x+1+1)=b⇒a+2=Ltx→0+1x+1+1=b⇒a+2=12=b⇒a=−32,b=12, ∴a+b=−1

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If the function $f : ℝ \to ℝ$ defined by $f (x) = \{\begin{cases} \frac{\sin (a + 1) x + \sin x}{x}, x < 0 \\ b, x = 0 \\ \frac{\sqrt{x + x^{2}} - \sqrt{x}}{x^{3 / 2}}, x > 0 \end{cases}$ is continuous on $ℝ$ then a + b =

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Detailed Solution

Let. f (x) is continuous at $x = 0 \Rightarrow \underset{x \to 0 -}{L t} f (x) = \underset{x \to 0 +}{L t} f (x) = f (0)$

$\Rightarrow \underset{x \to 0 -}{L t} \frac{\sin (a + 1) x + \sin x}{x} = \underset{x \to 0 +}{L t} \frac{\sqrt{x^{2} + x} - \sqrt{x}}{x^{3 / 2}} = b$

$\Rightarrow \underset{x \to 0 -}{L t} (\frac{\sin (a + 1) x}{x} + \frac{\sin x}{x}) = \underset{x \to 0 +}{L t} \frac{\sqrt{x} (\sqrt{x + 1} - 1)}{x \sqrt{x}} = b$

$\Rightarrow (a + 1) + 1 =, \underset{x \to 0 +}{L t} \frac{\sqrt{x + 1} - 1}{x} = b \Rightarrow a + 2 = \underset{x \to 0 +}{L t} \frac{(x + 1) - 1}{x (\sqrt{x + 1} + 1)} = b$

$\Rightarrow a + 2 = \underset{x \to 0 +}{L t} \frac{1}{\sqrt{x + 1} + 1} = b \Rightarrow a + 2 = \frac{1}{2} = b \Rightarrow a = - \frac{3}{2}, b = \frac{1}{2}, ∴ a + b = - 1$