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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

જો વિધેય \( f:  \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \rightarrow \mathrm{R}\) એ આપેલ મુજબ વ્યાખ્યાયિત છે . \(f(x)=(1+|\sin x|)^{\frac{3 a}{\sin x \mid}} ,\quad -\frac{\pi}{4}\,<\,x\,<\,0\) \(\quad\quad\quad\quad\quad\quad b ,\quad\quad\quad\quad\quad x=0\) \(\quad\quad\quad\quad e^{\cot 4 x / \cot 2 x} ,\quad\quad\quad 0\,<\,x\,<\,\frac{\pi}{4}\) જો \(\mathrm{f}\) એ \(\mathrm{x}=0\) આગળ સતત હોય તો \(6 \mathrm{a}+\mathrm{b}^{2}\) ની કિમંત મેળવો.

  1. A \(e\)
  2. B \(1+\mathrm{e}\)
  3. C \(1-\mathrm{e}\)
  4. D \(\mathrm{e}-1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1+\mathrm{e}\)

Step-by-step Solution

Detailed explanation

\(\lim _{x \rightarrow 0} f(x)=b\) \(\lim _{x \rightarrow 0^{+}} e^{\frac{\cot 4 x}{\cot 2 x}}=e^{\frac{1}{2}}=b\) \(\lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{\frac{3 a}{\sin x \mid}}=e^{3 a}=e^{\frac{1}{2}}\) \(a=\frac{1}{6} \Rightarrow 6 a=1\) \(\left(6 a+b^{2}\right)=(1+e)\)
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