NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Mathematics
- Chapter
- Continuity and Differentiability
NTA Abhyas 2020 · Mathematics · Continuity and Differentiability
If f(x) = \ array ll ( 2 ^ x - 1) ^ 2 3x xsi n ^ 2 x λ array array...
If f\left(x\right) = \left\{\right$\begin{array}{ll}0 < x < \dfrac{\pi}{6} \\ x = 0\end{array} is continuous at , then the value of $\dfrac{10\sqrt{3\lambda}}{\ln 2} is equal to
Correct answer
30
Step-by-step solution
\left(x\right)x = 0∴ \lim_{x \to 0}f\left(x\right) = f\left(0\right)∵ \lim_{x \to {0}^{+}}\dfrac{{\left({2}^{x} - 1\right)}^{2}\tan 3x}{x . \sin^{2}x} = \lambda∵ \lim_{x \to {0}^{+}}{\left(\dfrac{{2}^{x} - 1}{x}\right)}^{2}\dfrac{\tan 3x}{\left(3x\right)} . \dfrac{\left(3x\right)}{x}\dfrac{{x}^{2}}{si{n}^{2}x} = \lambda{\left(\ln 2\right)}^{2} . 1.3 . 1 = \lambda\dfrac{10\sqrt{3\lambda}}{\ln 2} = 30