ₓ arrow 2} {3^{x}+3³-x}-12}{3³-x}-3^{x/2}}=

Correct Answer is : −43-4/3−34

Correct Answer is : −43-4/3−34

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MHTCET 15 May 2024 Shift 2 Paper

Question : 146 of 150

Marks: +1, -0

\lim _{x \rightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}=

Solution:

Concept:

Multiply by

3^x

to rewrite the expression in factorisable forms, then cancel the common factor and evaluate the limit.

Explanation:

Let the given limit be

L

L = \lim_{x \rightarrow 2} \frac{3^x + 3^{3-x} - 12}{3^{3-x} - 3^{\frac{x}{2}}}

Multiplying numerator and denominator by

3^x

, we get:

L = \lim_{x \rightarrow 2} \frac{(3^x)^2 - 12(3^x) + 27}{27 - (3^{\frac{x}{2}})^3}

Factorise the numerator:

(3^x)^2 - 12(3^x) + 27 = (3^x - 3)(3^x - 9)

Factorise the denominator using

a^3 - b^3

27 - (3^{\frac{x}{2}})^3 = (3 - 3^{\frac{x}{2}})(9 + 3 \cdot 3^{\frac{x}{2}} + 3^x)

Also,

3^x - 9 = (3^{\frac{x}{2}} - 3)(3^{\frac{x}{2}} + 3)

Since

3 - 3^{\frac{x}{2}} = -(3^{\frac{x}{2}} - 3)

, cancel

(3^{\frac{x}{2}} - 3)

L = -\lim_{x \rightarrow 2} \frac{(3^{\frac{x}{2}} + 3)(3^x - 3)}{3^x + 3 \cdot 3^{\frac{x}{2}} + 9}

x = 2

, we have

3^{\frac{x}{2}} = 3

and

3^x = 9

L = -\frac{(3+3)(9-3)}{9 + 9 + 9} = -\frac{36}{27} = -\frac{4}{3}

Answer:

Option A:

-\frac{4}{3}

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