If the function f(x) = cases (1 + | x|)^ / | x|, & 0 < x < 2, & x = 2 e^ 6x 4x, & 2 < x < cases is continuous at x = 2, then 9 + 6 e + ^6 - e^6 is equal to:

Given: The function is piecewise defined and is continuous at x = 2. Find: The value of 9 + 6 e + ^6 - e^6. For continuity at x = 2, the left-hand limit and right-hand limit must both be equal to f ( 2) =. From the left, x 2^- (1 + | x|)^ | x| Since x 2^- gives x 0^+, this becomes x 2^- (1 + x)^ x = e^ So, x 2^- f(x) = e^ From the right, x 2^+ e^ 6x 4x = e^ x 2^+ 6x 4x Apply L'Hôpital's Rule to the exponent: x 2^+ 6x 4x = x 2^+ -6 ^2 6x-4 ^2 4x = x 2^+ 6 ^2 4x4 ^2 6x = 64 ( 4x 6x)^2 = 32 ( 23)^2 = 23 Hence, x 2^+ f(x) = e^ 23 By continuity, e^ = = e^ 23 Therefore, = 23, = e^ 23 Now evaluate: 9 + 6 e + ^6 - e^6 = 9 ( 23) + 6 e (e^ 23) + (e^ 23)^6 - e^6 23 = 6 + 6 ( 23) + e^4 - e^4 = 6 + 4 = 10 Therefore, the correct option is D.

If the function f(x) = cases (1 + | x|)^ / | x|, & 0 < x < 2, & x… — Solution | JEE Mathematics

MCQMediumJEE 2023Continuity

If the function $f(x) = \begin{cases} \left(1 + |\cos x|\right)^{\lambda / |\cos x|}, & 0 < x < \dfrac{\pi}{2} \\ \mu, & x = \dfrac{\pi}{2} \\ e^{\dfrac{\cot 6x}{\cot 4x}}, & \dfrac{\pi}{2} < x < \pi \end{cases}$ is continuous at $x = \dfrac{\pi}{2}$ , then $9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}$ is equal to:

A
$11$
B
$8$
C
$2e^4 + 8$
D
$10$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The function is piecewise defined and is continuous at $x = \dfrac{\pi}{2}$ .

Find: The value of $9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}$ .

For continuity at $x = \dfrac{\pi}{2}$ , the left-hand limit and right-hand limit must both be equal to $f\left(\dfrac{\pi}{2}\right) = \mu$ .

From the left,

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

Since $x \to \dfrac{\pi}{2}^{-}$ gives $\cos x \to 0^{+}$ , this becomes

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

So,

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

From the right,

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

Apply L'Hôpital's Rule to the exponent:

\lim_{x \to \frac{\pi}{2}^{+}} \frac{\cot 6x}{\cot 4x} = \lim_{x \to \frac{\pi}{2}^{+}} \frac{-6\csc^2 6x}{-4\csc^2 4x}

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

Hence,

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

By continuity,

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

Therefore,

\lambda = \frac{2}{3}, \qquad \mu = e^{\frac{2}{3}}

Now evaluate:

9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}

= 9\left(\frac{2}{3}\right) + 6\log_e\left(e^{\frac{2}{3}}\right) + \left(e^{\frac{2}{3}}\right)^6 - e^{6\cdot \frac{2}{3}}

= 6 + 6\left(\frac{2}{3}\right) + e^4 - e^4 = 6 + 4 = 10

Therefore, the correct option is D.

Common mistakes

Using $|\cos x| = -\cos x$ as $x \to \dfrac{\pi}{2}^{-}$ is incorrect, because in this interval $\cos x > 0$ . Here $|\cos x| = \cos x$ , which is needed to evaluate the left-hand limit correctly.
Treating $\lim (1+t)^{\lambda/t}$ as $1$ instead of $e^{\lambda}$ is a standard limit error. Rewrite with $t = \cos x \to 0^{+}$ and use the exponential limit form.
Equating only one-sided limit to $\mu$ is not enough for continuity. Both the left-hand limit and right-hand limit must exist and be equal to the function value at $x = \dfrac{\pi}{2}$ .

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