Question

Let $f:(-\frac{\pi}{4}, \frac{\pi}{4}) \rightarrow R$ be defined as 

If f is continuous at $x=0$, then the value of $6a+b^2$ is equal to : 
 

• A. $1+e$
• B. $1-e$
• C. $e$
• D. $e-1$

Hint:

Whenever expressions involve limits of the form $(1+u)^{k/u}$ as $u\to 0$, immediately rewrite them using \[ \lim_{u\to 0} (1+u)^{1/u} = e. \] This avoids unnecessary expansions and saves time in exams.

Answer 1

The Correct Option is A

For the function $f(x)$ to be continuous at $x=0$, the following condition must hold: \[ \lim_{x\to 0^-} f(x) = f(0) = \lim_{x\to 0^+} f(x). \] Given, \[ f(0)=b. \] Step 1: Right-Hand Limit (RHL) \[ \lim_{x\to 0^+} e^{\frac{\cot 4x}{\cot 2x}} = e^{\lim_{x\to 0^+} \frac{\cot 4x}{\cot 2x}} \] Rewrite using $\cot x = \frac{1}{\tan x}$: \[ \frac{\cot 4x}{\cot 2x} = \frac{\tan 2x}{\tan 4x}. \] Using the standard limit $\displaystyle \lim_{t\to 0} \frac{\tan t}{t} = 1$, \[ \lim_{x\to 0} \frac{\tan 2x}{\tan 4x} = \frac{2x}{4x} = \frac{1}{2}. \] Hence, \[ \text{RHL} = e^{1/2} = \sqrt{e}. \] Step 2: Left-Hand Limit (LHL) \[ \lim_{x\to 0^-} (1+|\sin x|)^{\frac{3a}{|\sin x|}}. \] As $x \to 0^-$, $|\sin x| \to 0^+$. Let $y = |\sin x|$. \[ \text{LHL} = \lim_{y\to 0^+} (1+y)^{\frac{3a}{y}} = \left(\lim_{y\to 0^+} (1+y)^{1/y}\right)^{3a}. \] Using the standard limit $\displaystyle \lim_{y\to 0} (1+y)^{1/y} = e$, \[ \text{LHL} = e^{3a}. \] Step 3: Apply Continuity Condition \[ \text{LHL} = \text{RHL} = f(0) \] \[ e^{3a} = \sqrt{e}. \] Comparing powers of $e$: \[ 3a = \frac{1}{2} \quad \Rightarrow \quad a = \frac{1}{6}. \] Also, \[ b = \sqrt{e}. \] Step 4: Compute Required Expression \[ 6a + b^2 = 6\left(\frac{1}{6}\right) + (\sqrt{e})^2 = 1 + e. \] \[ \boxed{6a + b^2 = 1 + e} \]