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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 $lim_{x \to 0} f (x)$

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Explanation

Evaluating the Limit of $f (x)$
The limit of $f (x)$ as $x \to 0$ is to be evaluated.
Step 1: Simplify the numerator
The numerator, $6^{x} - 3^{x} - 2^{x} + 1$ , can be factored. It is observed that $6^{x} - 3^{x} - 2^{x} + 1 = 3^{x} 2^{x} - 3^{x} - 2^{x} + 1 = 3^{x} (2^{x} - 1) - 1 (2^{x} - 1) = (3^{x} - 1) (2^{x} - 1)$ .
Step 2: Apply standard limits
The limit of the expression can be rewritten using standard limit forms. The expression becomes $lim_{x \to 0} \frac{( 3 ^{x} - 1 ) ( 2 ^{x} - 1 )}{l o g _{e} 9 ( 1 - c o s x )}$ . This can be rearranged as $lim_{x \to 0} \frac{( 3 ^{x} - 1 )}{x} \cdot \frac{( 2 ^{x} - 1 )}{x} \cdot \frac{x ^{2}}{( 1 - c o s x )} \cdot \frac{1}{l o g _{e} 9}$ . It is known that $lim_{x \to 0} \frac{a ^{x} - 1}{x} = lo g_{e} a$ and $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}} = \frac{1}{2}$ .
Step 3: Substitute the standard limits
Substituting the known limits into the expression: The limit is equal to $(lo g_{e} 3) \cdot (lo g_{e} 2) \cdot \frac{1}{\frac{1}{2}} \cdot \frac{1}{l o g _{e} 9}$ . This simplifies to $2 \cdot \frac{( l o g _{e} 3 ) ( l o g _{e} 2 )}{l o g _{e} 9}$ .
Step 4: Simplify the logarithmic terms
Using the property of logarithms $lo g_{e} a^{b} = b lo g_{e} a$ , it is known that $lo g_{e} 9 = lo g_{e} 3^{2} = 2 lo g_{e} 3$ . Substituting this into the expression: The limit becomes $2 \cdot \frac{( l o g _{e} 3 ) ( l o g _{e} 2 )}{2 l o g _{e} 3}$ .
Step 5: Calculate the final value
The term $2 lo g_{e} 3$ in the denominator cancels with $2$ and $lo g_{e} 3$ in the numerator. The final value of the limit is $lo g_{e} 2$ .
Final Answer
The value of $lim_{x \to 0} f (x)$ is $lo g_{e} 2$ .

NIMCET
Let $x$

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NIMCET
If $(x_0, y_0)$

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