Let $f:\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \rightarrow R$ be defined as $$ f(x)=\left\{\begin{array}{ccc} (1+|\sin x|)^{\frac{3 a}{|\sin x|}} & , & -\frac{\pi}{4}<x<0 \\ b & , & x=0 \\ e^{\cot 4 x / \cot 2 x} & , & 0<x<\frac{\pi}{4} \end{array}\right. $$ If $f$ is continuous at $x=0$, then the value of $6 a+b^{2}$ is equal to :

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Continuity and Differentiability

Let

f : (- \frac{π}{4}, \frac{π}{4}) \to R

be defined as

f (x) = ⎩ ⎨ ⎧ (1 + ∣ sin x ∣)^{\frac{3 a}{∣ s i n x ∣}} b e^{c o t 4 x / c o t 2 x},,, - \frac{π}{4} < x < 0 x = 0 0 < x < \frac{π}{4}

f

is continuous at

x = 0

, then the value of

6 a + b^{2}

is equal to :

Answer: (C)

Continuity at

x = 0

needs both one-sided limits to equal

b

.
For

x \to 0^{-}

the expression is

(1 + t)^{3 a / t}

with

t = ∣ sin x ∣ \to 0^{+}

, so the left-hand limit is

e^{3 a}

.
For

x \to 0^{+}

\frac{cot 4 x}{cot 2 x} = \frac{tan 2 x}{tan 4 x} = \frac{tan 2 x}{2 x} \cdot \frac{4 x}{tan 4 x} \cdot \frac{1}{2} \to \frac{1}{2}

so the right-hand limit is

e^{1/2}

.
Equating,

e^{3 a} = b = e^{1/2}

, which gives

a = \frac{1}{6}

and

b^{2} = e

, so

6 a + b^{2} = 1 + e

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option