Test Index

MHTCET 13 Apr 2026 Shift 1 Paper

© examsnet.com

Question : 121 of 150

Marks: +1, -0

If

f(x)=\left\{\begin{array}{lll}\frac{(8-2 x)^{\frac{1}{3}}-2}{3-(243+5 x)^{\frac{1}{5}}}, & \text { if } & x \neq 0 \\ k, & \text { if } & x=0\end{array}\right.

is continuous at

x=0

k=

$\frac{5}{2}$
$-\frac{5}{2}$
$\frac{27}{2}$
$\frac{-27}{2}$

Solution:

Concept:

For continuity at

x=0

, the value

k

must equal

\lim_{x \to 0} f(x)

.

Explanation:

Since

f(x)

is continuous at

x=0

, we need:

k=\lim_{x \to 0} \frac{(8-2x)^{1/3}-2}{3-(243+5x)^{1/5}}

Both numerator and denominator tend to

0

at

x=0

.

Use the approximation

(1+u)^n \approx 1+nu

for small

u

.

Write the numerator as:

(8-2x)^{1/3}=2\left(1-\frac{x}{4}\right)^{1/3}\approx 2\left(1-\frac{x}{12}\right)

So,

(8-2x)^{1/3}-2\approx 2\left(1-\frac{x}{12}\right)-2=-\frac{x}{6}

Similarly,

(243+5x)^{1/5}=3\left(1+\frac{5x}{243}\right)^{1/5}\approx 3\left(1+\frac{x}{243}\right)

Therefore,

3-(243+5x)^{1/5}\approx 3-3\left(1+\frac{x}{243}\right)=-\frac{x}{81}

Hence,

k=\lim_{x\to 0}\frac{-\frac{x}{6}}{-\frac{x}{81}}=\frac{81}{6}=\frac{27}{2}

Answer:

k=\frac{27}{2}

So the correct option is C.

© examsnet.com

Go to Question:

Prev Question

Next Question