Condition for continuity at x = 0 x=0 x = 0
For f f f x = 0 x=0 x = 0
lim x → 0 − f ( x ) = f ( 0 ) = lim x → 0 + f ( x ) . \lim_{x\to 0^-} f(x)=f(0)=\lim_{x\to 0^+} f(x). x → 0 − lim f ( x ) = f ( 0 ) = x → 0 + lim f ( x ) . Here,
f ( x ) = { ( 1 + ∣ sin x ∣ ) 3 a ∣ sin x ∣ , − π 4 < x < 0 b , x = 0 e cot 4 x / cot 2 x , 0 < x < π 4 f(x)=
\begin{cases}
(1+|\sin x|)^{\frac{3a}{|\sin x|}}, & -\frac\pi4<x<0 \\
b, & x=0 \\
e^{\cot 4x/\cot 2x}, & 0<x<\frac\pi4
\end{cases} f ( x ) = ⎩ ⎨ ⎧ ( 1 + ∣ sin x ∣ ) ∣ s i n x ∣ 3 a , b , e c o t 4 x / c o t 2 x , − 4 π < x < 0 x = 0 0 < x < 4 π So we compute the left-hand and right-hand limits.
Left-hand limit x → 0 − x\to 0^- x → 0 −
For x < 0 x<0 x < 0
lim x → 0 − ( 1 + ∣ sin x ∣ ) 3 a ∣ sin x ∣ . \lim_{x\to 0^-}(1+|\sin x|)^{\frac{3a}{|\sin x|}}. x → 0 − lim ( 1 + ∣ sin x ∣ ) ∣ s i n x ∣ 3 a . Let
t = ∣ sin x ∣ . t=|\sin x|. t = ∣ sin x ∣. As x → 0 − x\to 0^- x → 0 − t → 0 + t\to 0^+ t → 0 +
lim t → 0 + ( 1 + t ) 3 a t . \lim_{t\to 0^+}(1+t)^{\frac{3a}{t}}. t → 0 + lim ( 1 + t ) t 3 a . Using the standard limit
lim t → 0 ( 1 + t ) 1 / t = e , \lim_{t\to 0}(1+t)^{1/t}=e, t → 0 lim ( 1 + t ) 1/ t = e , we get
lim t → 0 + ( 1 + t ) 3 a t = e 3 a . \lim_{t\to 0^+}(1+t)^{\frac{3a}{t}}=e^{3a}. t → 0 + lim ( 1 + t ) t 3 a = e 3 a . Hence,
lim x → 0 − f ( x ) = e 3 a . \lim_{x\to 0^-} f(x)=e^{3a}. x → 0 − lim f ( x ) = e 3 a .
Right-hand limit x → 0 + x\to 0^+ x → 0 +
For x > 0 x>0 x > 0
f ( x ) = e cot 4 x / cot 2 x . f(x)=e^{\cot 4x/\cot 2x}. f ( x ) = e c o t 4 x / c o t 2 x . So,
lim x → 0 + f ( x ) = e lim x → 0 + cot 4 x cot 2 x . \lim_{x\to 0^+} f(x)=e^{\lim_{x\to 0^+}\frac{\cot 4x}{\cot 2x}}. x → 0 + lim f ( x ) = e l i m x → 0 + c o t 2 x c o t 4 x . Now,
cot 4 x cot 2 x = cos 4 x sin 4 x cos 2 x sin 2 x = cos 4 x sin 2 x sin 4 x cos 2 x . \frac{\cot 4x}{\cot 2x}
=\frac{\frac{\cos 4x}{\sin 4x}}{\frac{\cos 2x}{\sin 2x}}
=\frac{\cos 4x\,\sin 2x}{\sin 4x\,\cos 2x}. cot 2 x cot 4 x = s i n 2 x c o s 2 x s i n 4 x c o s 4 x = sin 4 x cos 2 x cos 4 x sin 2 x . Using
sin 4 x = 2 sin 2 x cos 2 x , \sin 4x=2\sin 2x\cos 2x, sin 4 x = 2 sin 2 x cos 2 x , we get
cot 4 x cot 2 x = cos 4 x sin 2 x 2 sin 2 x cos 2 2 x = cos 4 x 2 cos 2 2 x . \frac{\cot 4x}{\cot 2x}
=\frac{\cos 4x\,\sin 2x}{2\sin 2x\cos^2 2x}
=\frac{\cos 4x}{2\cos^2 2x}. cot 2 x cot 4 x = 2 sin 2 x cos 2 2 x cos 4 x sin 2 x = 2 cos 2 2 x cos 4 x . As x → 0 x\to 0 x → 0
cos 4 x → 1 , cos 2 x → 1. \cos 4x\to 1, \qquad \cos 2x\to 1. cos 4 x → 1 , cos 2 x → 1. Therefore,
lim x → 0 + cot 4 x cot 2 x = 1 2 . \lim_{x\to 0^+}\frac{\cot 4x}{\cot 2x}=\frac12. x → 0 + lim cot 2 x cot 4 x = 2 1 . So,
lim x → 0 + f ( x ) = e 1 / 2 = e . \lim_{x\to 0^+} f(x)=e^{1/2}=\sqrt e. x → 0 + lim f ( x ) = e 1/2 = e .
Use continuity
Since f f f x = 0 x=0 x = 0
e 3 a = b = e . e^{3a}=b=\sqrt e. e 3 a = b = e . Thus,
e 3 a = e 1 / 2 ⟹ 3 a = 1 2 ⟹ a = 1 6 . e^{3a}=e^{1/2}
\implies 3a=\frac12
\implies a=\frac16. e 3 a = e 1/2 ⟹ 3 a = 2 1 ⟹ a = 6 1 . Also,
b = e . b=\sqrt e. b = e .
Find 6 a + b 2 6a+b^2 6 a + b 2
Compute:
6 a = 6 ⋅ 1 6 = 1 , 6a=6\cdot \frac16=1, 6 a = 6 ⋅ 6 1 = 1 , and
b 2 = ( e ) 2 = e . b^2=(\sqrt e)^2=e. b 2 = ( e ) 2 = e . Therefore,
6 a + b 2 = 1 + e . 6a+b^2=1+e. 6 a + b 2 = 1 + e .
Match with options
1 + e 1+e 1 + e corresponds to Option C .