Q52 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
If $f(x) = \begin{cases} \dfrac{(8 - 2x)^{\frac{1}{3}} - 2}{3 - (243 + 5x)^{\frac{1}{5}}}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$, then $k =$
- A$\dfrac{5}{2}$
- B$-\dfrac{5}{2}$
- Correct answerC$\dfrac{27}{2}$
- D$\dfrac{-27}{2}$
Solution
Since the function $f(x)$ is continuous at $x = 0$, we have $k = \lim_{x \to 0} f(x)$.
$k = \lim_{x \to 0} \dfrac{(8 - 2x)^{1/3} - 2}{3 - (243 + 5x)^{1/5}}$
This is a $\dfrac{0}{0}$ form. Applying L'Hospital's Rule, we differentiate the numerator and the denominator with respect to $x$:
$k = \lim_{x \to 0} \dfrac{\dfrac{1}{3}(8 - 2x)^{-2/3}(-2)}{-\dfrac{1}{5}(243 + 5x)^{-4/5}(5)}$
Substituting $x = 0$ into the expression:
$k = \dfrac{\dfrac{1}{3}(8)^{-2/3}(-2)}{-(243)^{-4/5}}$
$k = \dfrac{-\dfrac{2}{3} \times \dfrac{1}{4}}{-\dfrac{1}{81}}$
$k = \dfrac{-\dfrac{1}{6}}{-\dfrac{1}{81}}$
$k = \dfrac{81}{6} = \dfrac{27}{2}$
$k = \lim_{x \to 0} \dfrac{(8 - 2x)^{1/3} - 2}{3 - (243 + 5x)^{1/5}}$
This is a $\dfrac{0}{0}$ form. Applying L'Hospital's Rule, we differentiate the numerator and the denominator with respect to $x$:
$k = \lim_{x \to 0} \dfrac{\dfrac{1}{3}(8 - 2x)^{-2/3}(-2)}{-\dfrac{1}{5}(243 + 5x)^{-4/5}(5)}$
Substituting $x = 0$ into the expression:
$k = \dfrac{\dfrac{1}{3}(8)^{-2/3}(-2)}{-(243)^{-4/5}}$
$k = \dfrac{-\dfrac{2}{3} \times \dfrac{1}{4}}{-\dfrac{1}{81}}$
$k = \dfrac{-\dfrac{1}{6}}{-\dfrac{1}{81}}$
$k = \dfrac{81}{6} = \dfrac{27}{2}$
Hence, the correct answer is option: 'C'.