Test Index

MHTCET 15 Apr 2026 Shift 1 Paper

Question : 111 of 150

Marks: +1, -0

\lim _{x \rightarrow 0} \frac{\left(4^{x}-1\right)^{3}}{\tan \left(\frac{x}{4}\right) \log \left(1+\frac{x^{2}}{3}\right)}=96(\log a)^{b}

(a+b)=

Solution:

Concept:

Use standard limit formulas for exponential, trigonometric, and logarithmic functions as

x \to 0

Explanation:

For small

x

, we approximate each factor using standard limits.

4^x - 1 \sim x \log 4

, so

(4^x - 1)^3 \sim x^3 (\log 4)^3

Also,

\tan\left(\frac{x}{4}\right) \sim \frac{x}{4}

And

\log\left(1 + \frac{x^2}{3}\right) \sim \frac{x^2}{3}

Thus, the denominator becomes

\frac{x}{4} \cdot \frac{x^2}{3} = \frac{x^3}{12}

Therefore, the limit is

\frac{x^3(\log 4)^3}{x^3/12} = 12(\log 4)^3

Now,

\log 4 = \log(2^2) = 2\log 2

So,

12(\log 4)^3 = 12 \cdot 8(\log 2)^3 = 96(\log 2)^3

Comparing with

96(\log a)^b

, we get

a = 2

and

b = 3

Hence,

a + b = 2 + 3 = 5

Answer:

Option A,

5

Go to Question:

Prev Question

Next Question