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MHTCET 16 Apr 2026 Shift 2 Paper

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Question : 109 of 150

Marks: +1, -0

If

\lim _{x \rightarrow 0} \frac{45^{x}-9^{x}-5^{x}+1}{\left(k^{x}-1\right)\left(3^{x}-1\right)}=2

, then the value of

k

is ...

45
9
5
3

Solution:

Concept:

Use

a^x - 1

factorisation and the standard limit

\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a

.

Explanation:

Since

45 = 9 \times 5

, write

45^x = 9^x \cdot 5^x

.

So the numerator becomes

9^x 5^x - 9^x - 5^x + 1

, which factorises as

(9^x - 1)(5^x - 1)

.

Therefore, the given limit becomes

\lim_{x \to 0} \frac{(9^x - 1)(5^x - 1)}{(k^x - 1)(3^x - 1)} = 2.

Now apply the standard result

\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a

.

We get

\lim_{x \to 0} \frac{9^x - 1}{3^x - 1} = \frac{\ln 9}{\ln 3}

and

\lim_{x \to 0} \frac{5^x - 1}{k^x - 1} = \frac{\ln 5}{\ln k}

.

Thus, the limit evaluates to

\frac{\ln 9}{\ln 3} \cdot \frac{\ln 5}{\ln k} = 2

.

Since

\ln 9 = \ln 3^2 = 2 \ln 3

, we get

\frac{2 \ln 3}{\ln 3} \cdot \frac{\ln 5}{\ln k} = 2

.

This simplifies to

2 \cdot \frac{\ln 5}{\ln k} = 2

, so

\frac{\ln 5}{\ln k} = 1

.

Hence,

\ln k = \ln 5

, which gives

k = 5

.

Answer:

k = 5

Correct option: C. 5

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