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NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

lf $f (x) = lim_{x \to 0} \frac{6 ^{x} - 3 ^{x} - 2 ^{x} + 1}{l o g _{e} 9 ( 1 - cos x )}$ is a real number then $lim_{x \to 0} f (x)$

Choose the correct answer:

Explanation

Step 1: Simplify the Numerator

Let's consider the expression inside the limit:

$L = x \to 0 lim \frac{6 ^{x} - 3 ^{x} - 2 ^{x} + 1}{lo g _{e} 9 ( 1 - cos x )}$

First, factorize the numerator by grouping the terms:

$6^{x} - 3^{x} - 2^{x} + 1 = (2 \cdot 3)^{x} - 3^{x} - 2^{x} + 1$

$= 2^{x} \cdot 3^{x} - 3^{x} - 2^{x} + 1$

$= 3^{x} (2^{x} - 1) - 1 (2^{x} - 1)$

$= (3^{x} - 1) (2^{x} - 1)$

Now substitute this back into the limit expression:

$L = x \to 0 lim \frac{( 3 ^{x} - 1 ) ( 2 ^{x} - 1 )}{lo g _{e} 9 ( 1 - cos x )}$

Step 2: Apply Standard Limit Formulas

We will utilize the following standard limit properties as $x \to 0$ :

$lim_{x \to 0} \frac{a ^{x} - 1}{x} = lo g_{e} a$
$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}} = \frac{1}{2}$

To use these standard forms, let's divide both the numerator and the denominator by $x^{2}$ :

$L = \frac{1}{lo g _{e} 9} \cdot x \to 0 lim \frac{\frac{( 3 ^{x} - 1 ) ( 2 ^{x} - 1 )}{x ^{2}}}{\frac{1 - c o s x}{x ^{2}}}$

Separate the terms in the numerator:

$L = \frac{1}{lo g _{e} 9} \cdot \frac{( lim _{x \to 0} \frac{3 ^{x} - 1}{x} ) \cdot ( lim _{x \to 0} \frac{2 ^{x} - 1}{x} )}{lim _{x \to 0} \frac{1 - c o s x}{x ^{2}}}$

Step 3: Calculate the Final Value

Substitute the values from the standard limit properties:

$lim_{x \to 0} \frac{3 ^{x} - 1}{x} = lo g_{e} 3$
$lim_{x \to 0} \frac{2 ^{x} - 1}{x} = lo g_{e} 2$
$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}} = \frac{1}{2}$

Putting it all together:

$L = \frac{1}{lo g _{e} 9} \cdot \frac{( lo g _{e} 3 ) \cdot ( lo g _{e} 2 )}{\frac{1}{2}}$

We know that $lo g_{e} 9 = lo g_{e} (3^{2}) = 2 lo g_{e} 3$ . Substituting this into the denominator:

$L = \frac{1}{2 lo g _{e} 3} \cdot \frac{( lo g _{e} 3 ) \cdot ( lo g _{e} 2 )}{\frac{1}{2}}$

Cancel out $lo g_{e} 3$ from the numerator and denominator:

$L = \frac{lo g _{e} 2}{2 \cdot \frac{1}{2}}$

$L = lo g_{e} 2$

Correct Answer:

Correct Option: C

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