Q80 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
The value of f(0) so that the function $f(x) = \dfrac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}}$, $x \neq 0$ is continuous at $x = 0$, is
- A$-\dfrac{1}{8}$
- Correct answerB$\dfrac{1}{8}$
- C$\dfrac{1}{64}$
- D$8$
Solution
For $f(x)$ to be continuous at $x = 0$, we must have $f(0) = \lim_{x \to 0} f(x)$.
$\lim_{x \to 0} \dfrac{(256 - 8x)^{\dfrac{1}{4}} - 4}{16 - 4(64 + 3x)^{\dfrac{1}{3}}}$
This is a $\dfrac{0}{0}$ indeterminate form. Applying L'Hospital's rule:
$\lim_{x \to 0} \dfrac{\dfrac{1}{4}(256 - 8x)^{-\dfrac{3}{4}} \times (-8)}{-4 \times \dfrac{1}{3}(64 + 3x)^{-\dfrac{2}{3}} \times 3}$
$= \lim_{x \to 0} \dfrac{-2(256 - 8x)^{-\dfrac{3}{4}}}{-4(64 + 3x)^{-\dfrac{2}{3}}}$
Substituting $x = 0$:
$= \dfrac{-2(256)^{-\dfrac{3}{4}}}{-4(64)^{-\dfrac{2}{3}}}$
$= \dfrac{-2(4^4)^{-\dfrac{3}{4}}}{-4(4^3)^{-\dfrac{2}{3}}}$
$= \dfrac{-2(4^{-3})}{-4(4^{-2})}$
$= \dfrac{-2 \times \dfrac{1}{64}}{-4 \times \dfrac{1}{16}}$
$= \dfrac{-\dfrac{1}{32}}{-\dfrac{1}{4}} = \dfrac{1}{8}$
Answer: $\dfrac{1}{8}$
$\lim_{x \to 0} \dfrac{(256 - 8x)^{\dfrac{1}{4}} - 4}{16 - 4(64 + 3x)^{\dfrac{1}{3}}}$
This is a $\dfrac{0}{0}$ indeterminate form. Applying L'Hospital's rule:
$\lim_{x \to 0} \dfrac{\dfrac{1}{4}(256 - 8x)^{-\dfrac{3}{4}} \times (-8)}{-4 \times \dfrac{1}{3}(64 + 3x)^{-\dfrac{2}{3}} \times 3}$
$= \lim_{x \to 0} \dfrac{-2(256 - 8x)^{-\dfrac{3}{4}}}{-4(64 + 3x)^{-\dfrac{2}{3}}}$
Substituting $x = 0$:
$= \dfrac{-2(256)^{-\dfrac{3}{4}}}{-4(64)^{-\dfrac{2}{3}}}$
$= \dfrac{-2(4^4)^{-\dfrac{3}{4}}}{-4(4^3)^{-\dfrac{2}{3}}}$
$= \dfrac{-2(4^{-3})}{-4(4^{-2})}$
$= \dfrac{-2 \times \dfrac{1}{64}}{-4 \times \dfrac{1}{16}}$
$= \dfrac{-\dfrac{1}{32}}{-\dfrac{1}{4}} = \dfrac{1}{8}$
Answer: $\dfrac{1}{8}$
Hence, the correct answer is option: 'B'.