Let be continuous at x = 0. Then is equal to : [2025]

Question

Let $f (x)$ $= {\begin{matrix} {(1 + a x)}^{1 / x}, & x < 0 \\ 1 + b, & x = 0 \\ \frac{{(x + 4)}^{1 / 2} - 2}{{(x + c)}^{1 / 3} - 2} & x > 0 \end{matrix} \begin{matrix} \end{matrix}$ be continuous at x = 0. Then $e^{a} b c$ is equal to : [2025]

Smart Achievers · Accepted Answer

(1)

L.H.L. = $f (0^{-}) = e^{\lim_{x \to 0} \frac{1}{x} \log (1 + a x)} = e^{\lim_{x \to 0} \frac{a}{1 + a x}} = e^{a}$

R.H.L. = $f (0^{+}) = \lim_{x \to 0} \frac{{(x + 4)}^{1 / 2} - 2}{{(x + c)}^{1 / 3} - 2} = \frac{0}{c^{1 / 3} - 2}$

Since, f(x) is continuous at x = 0

$∴$ Right hand limit exists

$\Rightarrow c^{1 / 3} - 2 = 0 \Rightarrow c^{1 / 3} = 2 \Rightarrow c = 8$ ... (i)

Now, $f (0^{+}) = \frac{2 - 2}{2 - 2}$ $(\frac{0}{0} form)$

$= \frac{\frac{1}{2 \sqrt{0 + 4}}}{\frac{1}{3} {(0 + c)}^{- 2 / 3}}$ [Using L'Hospital's Rule]

$= \frac{3}{4} c^{2 / 3} = \frac{3}{4} {(8)}^{2 / 3} = 3$ [From (i)]

Now, $f (0) = f (0^{-}) = f (0^{+})$

$\Rightarrow 1 + b = e^{a} = 3 \Rightarrow b = 2 and e^{a} = 3$

$∴ e^{a} b c = 3 \times 2 \times 8 = 48$ .

Solution

Q.
Let $f (x)$ $= {\begin{matrix} {(1 + a x)}^{1 / x}, & x < 0 \\ 1 + b, & x = 0 \\ \frac{{(x + 4)}^{1 / 2} - 2}{{(x + c)}^{1 / 3} - 2} & x > 0 \end{matrix} \begin{matrix} \end{matrix}$ be continuous at x = 0. Then $e^{a} b c$ is equal to : [2025]

1 48

2 72

3 36

4 64

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