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If

f(x)={1−tan⁡x4x−π,x≠π4k,x=π4f(x) = \begin{cases}\frac{1- \tan x}{4x-\pi}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4}\end{cases}

is continuous at x=π4x = \frac{\pi}{4}, then the value of k is

Solution

✅ Correct Option: 4

For continuity at x=π4x = \frac{\pi}{4}:

lim⁡x→π4f(x)=f(π4)\lim_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)

lim⁡x→π41−tan⁡x4x−π=k\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{4x-\pi} = k


Substituting x=π4x = \frac{\pi}{4} directly:

Numerator: 1−tan⁡π4=1−1=01 - \tan\frac{\pi}{4} = 1 - 1 = 0

Denominator: 4×π4−π=π−π=04 \times \frac{\pi}{4} - \pi = \pi - \pi = 0

This gives the indeterminate form 00\frac{0}{0}.


Applying L'Hôpital's Rule:

lim⁡x→π41−tan⁡x4x−π=lim⁡x→π4ddx(1−tan⁡x)ddx(4x−π)\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{4x-\pi} = \lim_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\tan x)}{\frac{d}{dx}(4x-\pi)}

=lim⁡x→π4−sec⁡2x4= \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4}


Substituting x=π4x = \frac{\pi}{4}:

k=−sec⁡2π44k = \frac{-\sec^2\frac{\pi}{4}}{4}

Since cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}:

sec⁡π4=2\sec\frac{\pi}{4} = \sqrt{2}

sec⁡2π4=2\sec^2\frac{\pi}{4} = 2


k=−24k = \frac{-2}{4}

k=−12k = -\frac{1}{2}

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