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MHTCET 15 Apr 2026 Shift 2 Paper

Question : 112 of 150

Marks: +1, -0

\lim _{x \rightarrow \infty} \frac{(2 x-1)^{19} \cdot(3 x+2)^{11}}{(6 x-5)^{30}}=2^{a} \cdot 3^{b}

a+b=

Solution:

Concept:

For limits at infinity, only the highest-degree terms matter, and they determine the leading behaviour.

Explanation:

Factor the largest power of

x

from each bracket:

(2x-1)^{19} = \left[x\left(2-\frac{1}{x}\right)\right]^{19} = x^{19}\left(2-\frac{1}{x}\right)^{19}

(3x+2)^{11} = x^{11}\left(3+\frac{2}{x}\right)^{11}

(6x-5)^{30} = x^{30}\left(6-\frac{5}{x}\right)^{30}

Since the powers of

x

cancel, we get:

\lim_{x \to \infty} \frac{\left(2-\frac{1}{x}\right)^{19}\left(3+\frac{2}{x}\right)^{11}}{\left(6-\frac{5}{x}\right)^{30}}

x \to \infty

, the small fractions vanish, giving:

\frac{2^{19} \cdot 3^{11}}{6^{30}} = \frac{2^{19} \cdot 3^{11}}{(2 \cdot 3)^{30}} = \frac{2^{19} \cdot 3^{11}}{2^{30} \cdot 3^{30}}

Using laws of exponents:

= 2^{19-30} \cdot 3^{11-30} = 2^{-11} \cdot 3^{-19}

Thus,

a = -11

and

b = -19

a + b = -11 - 19 = -30

Answer:

0pt

Option A:

-30

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