Test Index

GUJCET 2026 Math Paper

© examsnet.com

Question : 20 of 40

Marks: +1, -0

If function

f

is continuous at point

x = \pi

f(x) = \begin{cases} kx+1; & x \leq \pi \\ \cos x; & x > \pi \end{cases}

then the value of

k

\dots \dots \dots

$\frac{1}{\pi}$
$\frac{1}{2}$
$-\frac{2}{\pi}$
$0$

Solution:

Concept:

A function is continuous at

x=a

if its left-hand limit, right-hand limit, and value at

x=a

are equal.

Explanation:

At

x=\pi

, the function is defined by

kx+1

, so

f(\pi)=k\pi+1

.

Left-hand limit:

\lim_{x \to \pi^-}(kx+1)=k\pi+1

.

Right-hand limit:

\lim_{x \to \pi^+}\cos x=\cos\pi=-1

.

For continuity, LHL

=

RHL.

So,

k\pi+1=-1

.

k\pi=-2

.

k=-\frac{2}{\pi}

.

Answer:

\boxed{-\frac{2}{\pi}}

© examsnet.com

Go to Question:

Prev Question

Next Question