If the function $f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1+\cos x}}, & x \neq 0 \\ a \log_e 2 \log_e 3, & x = 0 \end{cases}$ is continuous at $x = 0$, then the value of $a^2$ is equal to Options: (A) 968 (B) 1152 (C) 746 (D) 1250

### Analyzing the Setup For the function $f(x)$ to be continuous at $x = 0$, the limit of the function as $x$ approaches $0$ must equal the defined value $f(0) = a \ln 2 \ln 3$. This ensures a seamless transition, effectively bridging the gap at the point of discontinuity. Our goal is to evaluate the limit: $$\lim_{x \to 0} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$$ ### The Numerator's Hidden Identity The numerator $72^x - 9^x - 8^x + 1$ can be factored by recognizing that $72 = 9 \cdot 8$. We rewrite the expression as: $$9^x \cdot 8^x - 9^x - 8^x + 1$$ By grouping the terms, we obtain: $$9^x(8^x - 1) - 1(8^x - 1) = (9^x - 1)(8^x - 1)$$ ### The Denominator's Transformation The denominator $\sqrt{2} - \sqrt{1 + \cos x}$ results in an indeterminate form $0/0$ as $x \to 0$. We multiply by the conjugate $\sqrt{2} + \sqrt{1 + \cos x}$ to rationalize it: $$\frac{(\sqrt{2} - \sqrt{1 + \cos x})(\sqrt{2} + \sqrt{1 + \cos x})}{\sqrt{2} + \sqrt{1 + \cos x}} = \frac{2 - (1 + \cos x)}{\sqrt{2} + \sqrt{1 + \cos x}} = \frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}}$$ Using the trigonometric identity $1 - \cos x = 2 \sin^2(x/2)$, the denominator becomes: $$\frac{2 \sin^2(x/2)}{\sqrt{2} + \sqrt{1 + \cos x}}$$ ### The Limit's Grand Finale Combining these results, the limit expression is: $$\lim_{x \to 0} \frac{(9^x - 1)(8^x - 1) \cdot (\sqrt{2} + \sqrt{1 + \cos x})}{2 \sin^2(x/2)}$$ We divide the numerator and denominator by $x^2$ to utilize the standard limits $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x \to 0} \frac{\sin(x/2)}{x/2} = 1$: $$\lim_{x \to 0} \left( \frac{9^x - 1}{x} \right) \left( \frac{8^x - 1}{x} \right) \cdot \frac{\sqrt{2} + \sqrt{1 + \cos x}}{2 \left( \frac{\sin(x/2)}{x/2} \cdot \frac{1}{2} \right)^2}$$ Evaluating the components, we get $(\ln 9)(\ln 8) \cdot \frac{2\sqrt{2}}{2(1/4)} = (\ln 9)(\ln 8) \cdot 4\sqrt{2}$. ### Final Calculation Using logarithmic properties, $\ln 9 = 2 \ln 3$ and $\ln 8 = 3 \ln 2$. Substituting these into our result: $$4\sqrt{2} (2 \ln 3)(3 \ln 2) = 24\sqrt{2} \ln 2 \ln 3$$ Equating this to $f(0) = a \ln 2 \ln 3$, we find $a = 24\sqrt{2}$. The final value requested is $a^2$: $$a^2 = (24\sqrt{2})^2 = 576 \cdot 2 = \mathbf{1152}$$

JEE Main 2024 (04 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f (x) = {\frac{7 2 ^{x} - 9 ^{x} - 8 ^{x} + 1}{2 - 1 + c o s x}, a lo g_{e} 2 lo g_{e} 3, x \neq = 0 x = 0$ is continuous at $x = 0$ , then the value of $a^{2}$ is equal to

Q: How do I know when to use the half-angle identity?

Whenever you see $1 - \cos x$ or $1 + \cos x$ in a limit problem, your brain should immediately trigger the half-angle identities: $1 - \cos x = 2 \sin^2(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$. These are essential for converting cosine terms into sine terms, which are much easier to manage with the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Select Answer:

Visualized Solution

Condition for Continuity at $x = 0$

For $f (x)$ to be continuous at $x = 0$ :
$f (0) = lim_{x \to 0} f (x)$
Given: $f (0) = a lo g_{e} 2 lo g_{e} 3$

Analyzing the Numerator: Factorization

Numerator: $7 2^{x} - 9^{x} - 8^{x} + 1$
Rewrite $7 2^{x}$ as $(9 \cdot 8)^{x} = 9^{x} \cdot 8^{x}$
Numerator $= 9^{x} \cdot 8^{x} - 9^{x} - 8^{x} + 1$

Completing the Factorization

Group the terms: $9^{x} (8^{x} - 1) - 1 (8^{x} - 1)$
Final Factorized Form: $(9^{x} - 1) (8^{x} - 1)$

Rationalizing the Denominator

Denominator: $2 - 1 + cos x$
Multiply numerator and denominator by conjugate: $2 + 1 + cos x$

Simplifying the Denominator

Denominator becomes: $(2)^{2} - (1 + cos x)^{2}$
$= 2 - (1 + cos x)$
$= 1 - cos x$

Applying Trigonometric Identity

Using half-angle identity: $1 - cos x = 2 sin^{2} (\frac{x}{2})$

Setting up Standard Limits

Limit Expression: $lim_{x \to 0} \frac{( 9 ^{x} - 1 ) ( 8 ^{x} - 1 ) ( 2 + 1 + c o s x )}{2 s i n ^{2} ( \frac{x}{2} )}$
Divide numerator and denominator by $x^{2}$ :
$= lim_{x \to 0} \frac{\frac{9 ^{x} - 1}{x} \cdot \frac{8 ^{x} - 1}{x} \cdot ( 2 + 1 + c o s x )}{\frac{2 s i n ^{2} ( \frac{x}{2} )}{x ^{2}}}$

Evaluating Exponential Limits

Using standard limit: $lim_{x \to 0} \frac{a ^{x} - 1}{x} = lo g_{e} a$
$lim_{x \to 0} \frac{9 ^{x} - 1}{x} = lo g_{e} 9$
$lim_{x \to 0} \frac{8 ^{x} - 1}{x} = lo g_{e} 8$

Evaluating the Trigonometric Limit

Denominator limit: $lim_{x \to 0} \frac{2 s i n ^{2} ( \frac{x}{2} )}{x ^{2}}$
Rewrite as: $lim_{x \to 0} 2 \cdot (\frac{s i n ( \frac{x}{2} )}{x})^{2}$
Multiply and divide inside by $2$ : $2 \cdot (\frac{s i n ( \frac{x}{2} )}{2 \cdot ( \frac{x}{2} )})^{2}$
$= 2 \cdot (1)^{2} \cdot \frac{1}{4} = \frac{1}{2}$

Evaluating the Conjugate Term

Conjugate limit: $lim_{x \to 0} (2 + 1 + cos x)$
Substitute $x = 0$ : $2 + 1 + cos 0$
$= 2 + 1 + 1 = 2 + 2 = 22$

Combining All Limit Results

Total Limit $= \frac{l o g _{e} 9 \cdot l o g _{e} 8 \cdot 2 2}{\frac{1}{2}}$
Multiply by $2$ : $= 42 lo g_{e} 9 lo g_{e} 8$

Simplifying Logarithms

Simplify logs: $lo g_{e} 9 = lo g_{e} (3^{2}) = 2 lo g_{e} 3$
$lo g_{e} 8 = lo g_{e} (2^{3}) = 3 lo g_{e} 2$
Limit $= 42 (2 lo g_{e} 3) (3 lo g_{e} 2)$
$= 242 lo g_{e} 2 lo g_{e} 3$

Equating to $f (0)$ and Solving for $a$

Equate limit to $f (0)$ : $242 lo g_{e} 2 lo g_{e} 3 = a lo g_{e} 2 lo g_{e} 3$
Cancel common log terms from both sides.
Therefore: $a = 242$

Final Answer: Calculating $a^{2}$

The question asks for the value of $a^{2}$ .
$a^{2} = (242)^{2}$
$a^{2} = 2 4^{2} \cdot (2)^{2} = 576 \cdot 2$
$a^{2} = 1152$

00:00 / 00:00

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

For the function

f (x)

to be continuous at

x = 0

, the limit of the function as

x

approaches

0

must equal the defined value

f (0) = a ln 2 ln 3

This ensures a seamless transition, effectively bridging the gap at the point of discontinuity. Our goal is to evaluate the limit:

x \to 0 lim \frac{7 2 ^{x} - 9 ^{x} - 8 ^{x} + 1}{2 - 1 + cos x}

The Numerator's Hidden Identity

The numerator

7 2^{x} - 9^{x} - 8^{x} + 1

can be factored by recognizing that

72 = 9 \cdot 8

. We rewrite the expression as:

9^{x} \cdot 8^{x} - 9^{x} - 8^{x} + 1

By grouping the terms, we obtain:

9^{x} (8^{x} - 1) - 1 (8^{x} - 1) = (9^{x} - 1) (8^{x} - 1)

The Denominator's Transformation

The denominator

2 - 1 + cos x

results in an indeterminate form

0/0

x \to 0

. We multiply by the conjugate

2 + 1 + cos x

to rationalize it:

\frac{( 2 - 1 + cos x ) ( 2 + 1 + cos x )}{2 + 1 + cos x} = \frac{2 - ( 1 + cos x )}{2 + 1 + cos x} = \frac{1 - cos x}{2 + 1 + cos x}

Using the trigonometric identity

1 - cos x = 2 sin^{2} (x /2)

, the denominator becomes:

\frac{2 sin ^{2} ( x /2 )}{2 + 1 + cos x}

The Limit's Grand Finale

Combining these results, the limit expression is:

x \to 0 lim \frac{( 9 ^{x} - 1 ) ( 8 ^{x} - 1 ) \cdot ( 2 + 1 + cos x )}{2 sin ^{2} ( x /2 )}

We divide the numerator and denominator by

x^{2}

to utilize the standard limits

lim_{x \to 0} \frac{a ^{x} - 1}{x} = ln a

and

lim_{x \to 0} \frac{s i n ( x /2 )}{x /2} = 1

x \to 0 lim (\frac{9 ^{x} - 1}{x}) (\frac{8 ^{x} - 1}{x}) \cdot \frac{2 + 1 + cos x}{2 ( \frac{s i n ( x /2 )}{x /2} \cdot \frac{1}{2} ) ^{2}}

Evaluating the components, we get

(ln 9) (ln 8) \cdot \frac{2 2}{2 ( 1/4 )} = (ln 9) (ln 8) \cdot 42

Final Calculation

Using logarithmic properties,

ln 9 = 2 ln 3

and

ln 8 = 3 ln 2

. Substituting these into our result:

42 (2 ln 3) (3 ln 2) = 242 ln 2 ln 3

Equating this to

f (0) = a ln 2 ln 3

, we find

a = 242

. The final value requested is

a^{2}

a^{2} = (242)^{2} = 576 \cdot 2 = 1152

Common Traps & FAQs

Wait, why can't we just use L'Hopital's rule here? It seems like it would be faster.

While L'Hopital's rule is a powerful tool, applying it to expressions involving square roots of trigonometric functions and exponential terms often leads to a nightmare of chain rule derivatives. In JEE Advanced, the goal is often to test your ability to recognize standard forms and algebraic structures. Factorization and standard limits are much cleaner and less prone to calculation errors than repeated differentiation.

Why do we reject the negative root or worry about the sign when rationalizing?

We aren't rejecting a root; we are simply using the conjugate to eliminate the indeterminate form. The expression

2 - 1 + cos x

approaches

0

x \to 0

. By multiplying by the conjugate, we transform the denominator into

2 - (1 + cos x) = 1 - cos x

, which is a standard form we can easily handle. Always be careful with the signs when expanding

(a - b) (a + b) = a^{2} - b^{2}

How do I know when to use the half-angle identity?

Whenever you see

1 - cos x

1 + cos x

in a limit problem, your brain should immediately trigger the half-angle identities:

1 - cos x = 2 sin^{2} (x /2)

and

1 + cos x = 2 cos^{2} (x /2)

. These are essential for converting cosine terms into sine terms, which are much easier to manage with the standard limit

lim_{θ \to 0} \frac{s i n θ}{θ} = 1

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f (x) = {\frac{7 2 ^{x} - 9 ^{x} - 8 ^{x} + 1}{2 - 1 + c o s x}, a lo g_{e} 2 lo g_{e} 3, x \neq = 0 x = 0$ is continuous at $x = 0$ , then the value of $a^{2}$ is equal to

Select Answer:

Visualized Solution

Condition for Continuity at $x = 0$

Analyzing the Numerator: Factorization

Completing the Factorization

Rationalizing the Denominator

Simplifying the Denominator

Applying Trigonometric Identity

Setting up Standard Limits

Evaluating Exponential Limits

Evaluating the Trigonometric Limit

Evaluating the Conjugate Term

Combining All Limit Results

Simplifying Logarithms

Equating to $f (0)$ and Solving for $a$

Final Answer: Calculating $a^{2}$

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Numerator's Hidden Identity

The Denominator's Transformation

The Limit's Grand Finale

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Condition for Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Analyzing the Numerator: Factorization

Completing the Factorization

Rationalizing the Denominator

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Applying Trigonometric Identity

Setting up Standard Limits

Evaluating Exponential Limits

Evaluating the Trigonometric Limit

Evaluating the Conjugate Term

Combining All Limit Results

Simplifying Logarithms

Equating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0) and Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Final Answer: Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Numerator's Hidden Identity

The Denominator's Transformation

The Limit's Grand Finale

Final Calculation

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Equating to $f (0)$ and Solving for $a$

Final Answer: Calculating $a^{2}$

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