MHT CET 2026 15th April Evening Shift | Limits, Continuity and Differentiability Question 21 | 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

$-\dfrac{1}{8}$

$\dfrac{1}{8}$

$\dfrac{1}{64}$

$8$

MHT CET 2026 15th April Morning Shift

MCQ (Single Correct Answer)

-0

If $\lim\limits_{x \to 0} \dfrac{(4^x - 1)^3}{\tan\left(\dfrac{x}{4}\right)\log\left(1 + \dfrac{x^2}{3}\right)} = 96(\log a)^b$, then $(a + b) = $

$5$

$7$

$3$

$4$

MHT CET 2026 15th April Morning Shift

MCQ (Single Correct Answer)

-0

If the function f is continuous at $x = 1$, where $f(x) = \dfrac{1 + \cos(\pi x)}{\pi(1-x)^2}$, for $x \neq 1$, then the value of $f(1)$ is....

$\dfrac{\pi}{2}$

$\dfrac{\pi}{4}$

$\dfrac{\pi}{6}$

$\dfrac{\pi}{9}$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

If $\quad f(x)=\left\{\begin{array}{cc}\frac{9^x-2 \cdot 3^x+1}{\log (1+3 x) \cdot \tan 2 x} & , \text { if } x \neq 0 \\ a(\log b)^c & , \text { if } x=0\end{array}\right.$ is continuous at $x=0$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}=$

$\frac{31}{6}$

$\frac{1}{6}$

$\frac{5}{6}$

$\frac{3}{20}$

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