Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 154 Question 154 of 401 Medium MCQ
Let f : ( − π 4 , π 4 ) → R f:\left(-\dfrac\pi4,\dfrac\pi4\right)\to\mathbb{R} f : ( − 4 π , 4 π ) → R be defined as
\[f(x)= { ( 1 + ∣ sin x ∣ ) 3 a ∣ sin x ∣ , − π 4 < x < 0 b , x = 0 e cot 4 x / cot 2 x , 0 < x < π 4 \begin{cases}(1+|\sin x|)^{\frac{3a}{|\sin x|}}, & -\frac\pi4<x<0\\b, & x=0\\e^{\cot4x/\cot2x}, & 0<x<\frac\pi4\end{cases} ⎩ ⎨ ⎧ ( 1 + ∣ sin x ∣ ) ∣ s i n x ∣ 3 a , b , e c o t 4 x / c o t 2 x , − 4 π < x < 0 x = 0 0 < x < 4 π \]
If f f f is continuous at x = 0 x=0 x = 0 , then the value of 6 a + b 2 6a+b^2 6 a + b 2 is equal to
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LHL: lim x → 0 − ( 1 + ∣ sin x ∣ ) 3 a / ∣ sin x ∣ = e 3 a \displaystyle\lim_{x\to0^-}(1+|\sin x|)^{3a/|\sin x|}=e^{3a} x → 0 − lim ( 1 + ∣ sin x ∣ ) 3 a /∣ s i n x ∣ = e 3 a .
RHL: lim x → 0 + e cot 4 x / cot 2 x = e 1 / 2 \displaystyle\lim_{x\to0^+}e^{\cot4x/\cot2x}=e^{1/2} x → 0 + lim e c o t 4 x / c o t 2 x = e 1/2 (since cot 4 x cot 2 x → 1 2 \frac{\cot4x}{\cot2x}\to\frac12 c o t 2 x c o t 4 x → 2 1 ).
Continuity: e 3 a = b = e 1 / 2 ⇒ a = 1 6 , b = e e^{3a}=b=e^{1/2}\Rightarrow a=\dfrac16,\ b=\sqrt e e 3 a = b = e 1/2 ⇒ a = 6 1 , b = e .
6 a + b 2 = 1 + e 6a+b^2=1+e 6 a + b 2 = 1 + e . Answer the question to unlock