Let $ f(x)=\left\{\begin{array}{cc}\frac{1+\cos x}{(\pi-x)^{2}} \cdot \frac{\sin ^{2} x}{\log \left(1+\pi^{2}-2 \pi x+x^{2}\right)} & , \quad x \neq \pi \ k & , \quad x=\pi\end{array}\right. $. If $ f(x) $ is continuous functions at $ x=\pi $, then $ k $ is equal to

Question

Let $ f(x)=\left\{\begin{array}{cc}\frac{1+\cos x}{(\pi-x)^{2}} \cdot \frac{\sin ^{2} x}{\log \left(1+\pi^{2}-2 \pi x+x^{2}\right)} & , \quad x \neq \pi \ k & , \quad x=\pi\end{array}\right. $. If $ f(x) $ is continuous functions at $ x=\pi $, then $ k $ is equal to, $ \frac{1}{4} $, $ \frac{1}{2} $, $ \frac{-1}{2} $, $ -\frac{1}{4} $

MARKS App · Accepted Answer

The given function is f x = { 1 + cos x π - x 2 · sin 2 x log 1 + π 2 - 2 π x + x 2 , x ≠ π k , x = π Since, f(x) continuous at x = π . ∴ lim x → π f x = f π ⇒ lim h → 0 f π + h = k ⇒ lim h → 0 1 + cos π + h π - π - h 2 · sin 2 π + h log 1 + π 2 - 2 π π + h + π + h 2 = k ⇒ lim h → 0 1 - cos h π 2 × sin 2 h log 1 + h 2 = k ⇒ lim h → 0 1 2 sin h/2 h/2 2 · h 2 log 1 + h 2 · sin h h 2 = k ⇒ 1 2 = k

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