CUET UG 2025 — Mathematics Calculus

For continuity at $x = \frac{\pi}{4}$x=4π​:

$$\lim_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)$$x→4π​lim​f(x)=f(4π​)

$$\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{4x-\pi} = k$$x→4π​lim​4x−π1−tanx​=k

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Substituting $x = \frac{\pi}{4}$x=4π​ directly:

Numerator: $1 - \tan\frac{\pi}{4} = 1 - 1 = 0$1−tan4π​=1−1=0

Denominator: $4 \times \frac{\pi}{4} - \pi = \pi - \pi = 0$4×4π​−π=π−π=0

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

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Applying L'Hôpital's Rule:

$$\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{4x-\pi} = \lim_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\tan x)}{\frac{d}{dx}(4x-\pi)}$$x→4π​lim​4x−π1−tanx​=x→4π​lim​dxd​(4x−π)dxd​(1−tanx)​

$$= \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4}$$=x→4π​lim​4−sec2x​

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Substituting $x = \frac{\pi}{4}$x=4π​:

$$k = \frac{-\sec^2\frac{\pi}{4}}{4}$$k=4−sec24π​​

Since $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$cos4π​=2​1​:

$$\sec\frac{\pi}{4} = \sqrt{2}$$sec4π​=2​

$$\sec^2\frac{\pi}{4} = 2$$sec24π​=2

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$$k = \frac{-2}{4}$$k=4−2​

$$k = -\frac{1}{2}$$k=−21​