Question

If  \[f\left( x \right) = \begin{cases}\frac{\sin (a + 1) x + \sin x}{x} , & x < 0 \\ c , & x = 0 \\ \frac{\sqrt{x + b x^2} - \sqrt{x}}{bx\sqrt{x}} , & x > 0\end{cases}\]is continuous at x = 0, then 

विकल्प

• a =  \[- \frac{3}{2}\] , b = 0, c = \[\frac{1}{2}\]

•  a = \[- \frac{3}{2}\] , b = 1, c = \[- \frac{1}{2}\] 

• a =\[- \frac{3}{2}\], b ∈ R − {0}, c = \[\frac{1}{2}\] 

• none of these

Answer 1

a =\[- \frac{3}{2}\], b ∈ R − {0}, c = \[\frac{1}{2}\] 

The given function can be rewritten as

\[f\left( x \right) = \begin{cases}\frac{\sin \left( a + 1 \right) x + x \sin x}{x}, \text{ for}  x < 0 \\ c ,\text{  for }  x = 0 \\ \frac{\sqrt{x + b x^2} - \sqrt{x}}{b x^\frac{3}{2}} , \text{ for } x > 0\end{cases}\]

\[\Rightarrow f\left( x \right) = \begin{cases}\frac{\sin \left( a + 1 \right)x + \sin x}{x}, \text{ for }  x < 0 \\ c , \text{ for } x = 0 \\ \frac{\sqrt{1 + bx} - 1}{bx} , \text{ for }  x > 0\end{cases}\]

We have
(LHL at x = 0) =  \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right)\]

\[= \lim_{h \to 0} \left[ \frac{- \sin \left( a + 1 \right)h - \sin \left( - h \right)}{h} \right] = \lim_{h \to 0} \left[ \frac{- \sin \left( a + 1 \right)h}{h} - \frac{\sin h}{h} \right]\]

\[= - \left( a + 1 \right) \lim_{h \to 0} \left[ \frac{\sin \left( a + 1 \right)h}{\left( a + 1 \right)h} \right] - \lim_{h \to 0} \frac{\sin h}{h} = - a - 1\]

(RHL at x = 0) =  \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right)\]

\[\lim_{h \to 0} \left( \frac{\sqrt{1 + bh} - 1}{bh} \right) = \lim_{h \to 0} \left( \frac{bh}{bh\left( \sqrt{1 + bh} + 1 \right)} \right) = \lim_{h \to 0} \left( \frac{1}{\left( \sqrt{1 + bh} + 1 \right)} \right) = \frac{1}{2}\]

Also,  \[f\left( 0 \right) = c\] 

If  \[f\left( x \right)\]  is continuous at x = 0, then

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\] 

\[\Rightarrow - a - 1 = \frac{1}{2} = c\]

\[\Rightarrow - a - 1 = \frac{1}{2} \text{ and } c = \frac{1}{2}\]

\[\Rightarrow a = \frac{- 3}{2}\] \[, c = \frac{1}{2}\]

Now,

\[\frac{\sqrt{1 + bx} - 1}{bx}\]  exists only if  \[bx \neq 0 \Rightarrow b \neq 0\]

Thus,  \[b \in R - \left\{ 0 \right\}\].