Question

If the function \(f(x)=\begin{cases}(1+|\cos x|) \frac{\lambda}{|\cos x|} & , 0 < x < \frac{\pi}{2} \\\mu & , \quad x=\frac{\pi}{2} \\\frac{\cot 6 x}{e^{\cot 4 x}} & \frac{\pi}{2}< x< \pi\end{cases}\)is continuous at \(x=\frac{\pi}{2}, then 9 \lambda+6 \log _{ e } \mu+\mu^6- e ^{6 \lambda}\) is equal to

• A. $2 e^4+8$
• B. 11
• C. 8
• D. 10

Answer 1

The Correct Option is D

To ensure continuity of \(f(x)\) at \(x=\frac{\pi}{2}\), the left-hand limit and the right-hand limit as \(x\) approaches \(\frac{\pi}{2}\) must both equal \(f\left(\frac{\pi}{2}\right)=\mu\).

1. Left-Hand Limit (\(x \to \frac{\pi}{2}^{-}\)):

\( \lim_{x \to \frac{\pi}{2}^{-}} (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}} = \lim_{x \to \frac{\pi}{2}^{-}} (1 + \cos x)^{\frac{\lambda}{\cos x}} \)

As \(x \to \frac{\pi}{2}^{-}\), \(\cos x \to 0^{+}\), so:

\( (1 + \cos x)^{\frac{\lambda}{\cos x}} \approx e^{\lambda} \)

Thus,

\( \lim_{x \to \frac{\pi}{2}^{-}} f(x) = e^{\lambda} \)

2. Right-Hand Limit (\(x \to \frac{\pi}{2}^{+}\)):

\( \lim_{x \to \frac{\pi}{2}^{+}} e^{\frac{\cot 6x}{\cot 4x}} = e^{ \lim_{x \to \frac{\pi}{2}^{+}} \frac{\cot 6x}{\cot 4x} } \)

Simplify the exponent:

\( \frac{\cot 6x}{\cot 4x} = \frac{\frac{\cos 6x}{\sin 6x}}{\frac{\cos 4x}{\sin 4x}} = \frac{\cos 6x \cdot \sin 4x}{\cos 4x \cdot \sin 6x} \)

As \(x \to \frac{\pi}{2}^{+}\) :

\( 6x \to 3\pi \Rightarrow \cos 6x = \cos 3\pi = -1, \quad \sin 6x = \sin 3\pi = 0 \)
\( 4x \to 2\pi \Rightarrow \cos 4x = \cos 2\pi = 1, \quad \sin 4x = \sin 2\pi = 0 \)

Applying L'Hôpital's Rule to the indeterminate form:

\( \lim_{x \to \frac{\pi}{2}^{+}} \frac{\cot 6x}{\cot 4x} = \lim_{x \to \frac{\pi}{2}^{+}} \frac{- \csc^2 6x \cdot 6}{-\csc^2 4x \cdot 4} = \lim_{x \to \frac{\pi}{2}^{+}} \frac{6 \sin^2 4x}{4 \sin^2 6x} = \frac{6}{4} \cdot \left(\frac{\sin 4x}{\sin 6x} \right)^2 = \frac{3}{2} \cdot \left( \frac{2}{3} \right)^2 = \frac{3}{2} \cdot \frac{4}{9} = \frac{2}{3} \)

Therefore,

\( \lim_{x \to \frac{\pi}{2}^{+}} f(x) = e^{\frac{2}{3}} \)

3. Continuity Condition:

Thus,

\( e^{\lambda} = \mu = e^{\frac{2}{3}} \)

Thus, \(\lambda = \frac{2}{3}\), \(\mu = e^{\frac{2}{3}}\)

4. Evaluating the Expression:

\( 9\lambda + 6 \ln \mu + \mu^6 - e^{6\lambda} = 9 \left(\frac{2}{3}\right) + 6 \ln \left( e^{\frac{2}{3}} \right) + \left( e^{\frac{2}{3}} \right)^6 - e^{6 \cdot \frac{2}{3}} = 6 + 6 \cdot \frac{2}{3} + e^4 - e^4 = 6 + 4 + 0 = 10 \)

Thus, the correct answer is option (4).

Answer 2

The correct answer is (D) : 10
$\Rightarrow \underset{x\to \frac{{\pi }^{+}}{2}}{\lim }{e}^{\frac{\cot 6x}{\cot 4x}}=\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }{e}^{\frac{\sin 4x\times \cos 6x}{\sin 6x\cdot \cos 4x}}=e^{2/3}$
$\Rightarrow \underset{x\to \frac{{\pi }^{-}}{2}}{\lim }(1+|\cos x|{)}^{\frac{\lambda }{\cos x}\mid }=e^{\lambda }$
$\Rightarrow f(\pi /2)=\mu$
For continuous function $\Rightarrow e^{2/3}=e^{\lambda }=\mu$
$\lambda =\frac{2}{3},\mu =e^{2/3}$
Now, $9\lambda +6{\log }_e\mu +{\mu }^6-e^{6\lambda }=10$