If the function f defined on π 6 , π 3 by f x = 2 c o s x - 1 c o t x - 1 , x ≠ π 4 k , x = π 4 is continuous, then k is equal to

∵ f x is continuous at x = π 4 ∴ f π 4 = k = lim x → π 4 2 cos x - 1 cot x - 1 → 0 0 form Applying L' Hospital Rule k = l i m x → π 4 2 - sin x - cosec 2 x ⇒ k = 2 sin π 4 cosec 2 π 4 = 2 × 1 2 2 2 = 1 2 .

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Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityContinuity- MiscellaneousEasy2 minPYQ_2019

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If the function $f$ defined on $(\frac{π}{6}, \frac{π}{3})$ by $f (x) = \{\begin{matrix} \frac{\sqrt{2} c o s x - 1}{c o t x - 1}, x \neq \frac{π}{4} \\ k, x = \frac{π}{4} \end{matrix}$ is continuous, then $k$ is equal to

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Solution:

$∵ f (x)$ is continuous at $x = \frac{π}{4}$

$∴ f (\frac{π}{4}) = k = \lim_{x \to \frac{π}{4}} (\frac{\sqrt{2} \cos x - 1}{\cot x - 1}) \to (\frac{0}{0})$ form

Applying L' Hospital Rule

$k = \underset{x \to \frac{π}{4}}{l i m} (\frac{\sqrt{2} (- \sin x)}{- {cosec}^{2} x})$

$\Rightarrow k = \frac{\sqrt{2} \sin \frac{π}{4}}{{cosec}^{2} \frac{π}{4}} = \frac{\sqrt{2} \times (\frac{1}{\sqrt{2}})}{{(\sqrt{2})}^{2}} = \frac{1}{2} .$

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

⏱ 2mℹ️ Source: PYQ_2019

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