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MHTCET 5 May 2025 Shift 2 Paper

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Question : 106 of 150

Marks: +1, -0

If

\quad f(x)=\left\{\begin{array}{ccc}\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

a+b+c=

$\frac{31}{6}$
$\frac{1}{6}$
$\frac{5}{6}$
$\frac{3}{20}$

Solution:

Concept:

For continuity at

x=0

, the value

f(0)

must equal

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

.

Explanation:

Given that

f(x)

is continuous at

x=0

, we have:

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

Rewrite the numerator as

\left(3^x-1\right)^2

, so:

\lim_{x\to 0} \frac{\left(3^x-1\right)^2}{\log(1+3x)\tan 2x}

Divide numerator and denominator appropriately and use standard limits:

\lim_{x\to 0} \frac{\left(\frac{3^x-1}{x}\right)^2}{\frac{\log(1+3x)}{3x}\cdot 3 \cdot \frac{\tan 2x}{2x}\cdot 2}

Using

\lim_{x\to 0}\frac{3^x-1}{x}=\log 3

,

\lim_{x\to 0}\frac{\log(1+3x)}{3x}=1

, and

\lim_{x\to 0}\frac{\tan 2x}{2x}=1

, we get:

\frac{(\log 3)^2}{3\cdot 2}=\frac{(\log 3)^2}{6}

Hence,

a(\log b)^c=\frac{(\log 3)^2}{6}

.

Comparing both sides gives

a=\frac{1}{6}

,

b=3

,

c=2

.

Therefore,

a+b+c=\frac{1}{6}+3+2=\frac{31}{6}

.

Answer:

\frac{31}{6}

i.e. option A.

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