First slide

Continuity

Question

The values of a, b and c which make the function $f (x)$ $= {\begin{matrix} \frac{\sin (a + 1) x + \sin x}{x}, x < 0 \\ c, x = 0 \\ \frac{\sqrt{x + b x^{2}} - \sqrt{x}}{b x^{3 / 2}}, x > 0 \end{matrix}$ continuous x = 0, are

Easy

A

$a = - \frac{3}{2}, c = \frac{1}{2}, b = 0$
B

$a = \frac{3}{2}, c = \frac{1}{2}, b \neq 0$
C

$a = - \frac{3}{2}, c = \frac{1}{2}, b \neq 0$
D

none of the above

Solution

In the definition of the function, $b \neq 0$ , then f (x) will be undefined in $x > 0$ .
               $∵ f (x) is continuous at x = 0$

               $∴ L H L = R H L = f (0)$

$\Rightarrow \lim_{\begin{matrix} x \to 0 \\ x < 0 \end{matrix}} \frac{\sin (a + 1) x + \sin x}{x} = \lim_{\begin{matrix} x \to 0 \\ x > 0 \end{matrix}} \frac{\sqrt{x + b x^{2}} - \sqrt{x}}{b x^{3 / 2}}$

$\Rightarrow \lim_{x \to 0} (\frac{\sin (a + 1) x}{x} + \frac{\sin x}{x}) = \lim_{x \to 0} \frac{\sqrt{1 + b x} - 1}{b x} = c$

$\Rightarrow (a + 1) + 1$ $= \lim_{x \to 0} \frac{(1 + b x) - 1}{b x (\sqrt{1 + b x} + 1)} = c$

$\Rightarrow a + 2$ $= \lim_{x \to 0} \frac{1}{(\sqrt{1 + b x} + 1)} = c$

$\Rightarrow a + 2 = \frac{1}{2} = c$ ,        $∴ a = - \frac{3}{2}, c = \frac{1}{2}, b \neq 0$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App