CUET UG 2025 — Mathematics Calculus

For a function to be continuous at a point $x = a$x=a, the limit of the function as $x$x approaches $a$a must equal the value of the function at that point:

\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)$ From the problem statement, we know that $f\left(\frac{\pi}{2}\right) = 3$. Therefore, we need to find $k$ such that:\lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = 3$

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Evaluating the Limit using L'Hôpital's Rule:
Since substituting $x = \frac{\pi}{2}$x=2π​ into the expression results in an indeterminate form of $\frac{0}{0}$00​, we can apply L'Hôpital's Rule (differentiating the numerator and denominator with respect to $x$x):

• The derivative of the numerator $k\cos x$kcosx is $-k\sin x$−ksinx.
• The derivative of the denominator $\pi - 2x$π−2x is $-2$−2.

Applying the rule:

$$\lim_{x \to \frac{\pi}{2}} \frac{-k\sin x}{-2} = 3$$x→2π​lim​−2−ksinx​=3

\frac{-k\sin(\frac{\pi}{2})}{-2} = 3$ Since $\sin(\frac{\pi}{2}) = 1$:\frac{-k(1)}{-2} = 3$$$$\frac{k}{2} = 3$$$$k = 6$$