The value of k which the function f x = 4 5 tan ⁡ 4 x tan ⁡ 5 x , 0 < x < π 2 k + 2 5 , x = π 2 is continuous at x = π 2 , is

Given function is continuous at x = π 2 ⇒ f π 2 = f π + 2 = f π - 2 ⇒ k + 2 5 = 4 5 tan ⁡ 4 x tan ⁡ 5 x ⇒ k + 2 5 = 4 5 tan 2 π cot 5 π 2 ⇒ k + 2 5 = 4 5 0 ⇒ k + 2 5 = 1 ⇒ k = 3 5

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Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityContinuity- MiscellaneousEasy2 minPYQ_2017

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The value of $k$ which the function $f (x) = \{\begin{matrix} {(\frac{4}{5})}^{\frac{\tan 4 x}{\tan 5 x}}, & 0 < x < \frac{π}{2} \\ k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ is

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Answer:

Solution:

Given function is continuous at $x = \frac{π}{2}$

$\Rightarrow f (\frac{π}{2}) = f (\frac{π^{+}}{2}) = f (\frac{π^{-}}{2})$

$\Rightarrow k + \frac{2}{5} = {(\frac{4}{5})}^{\frac{\tan 4 x}{\tan 5 x}}$

$\Rightarrow k + \frac{2}{5} = {(\frac{4}{5})}^{\tan (2 π) \cot (\frac{5 π}{2})}$

$\Rightarrow k + \frac{2}{5} = {(\frac{4}{5})}^{0}$

$\Rightarrow k + \frac{2}{5} = 1$

$\Rightarrow k = \frac{3}{5}$

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Continuity- Miscellaneous

⏱ 2mℹ️ Source: PYQ_2017

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