Is the function defined by f(x) = x² − sin x + 5 continuous at x = π?

Solution:

A function is said to be continuous when the graph of the function is a single unbroken curve.

The given function is

f(x) = x² − sin x + 5

It is evident that f is defined at x = π.

At x = π, f(x) = f(π)

= π² − sinπ + 5

= π² − 0 + 5 = π² + 5 

Consider lim_x→π f(x) = lim_x→π (x² − sin x + 5)

Put x = π + h,

it is evident that if x→π, then h→0

⇒ lim_x→π f(x) = lim_x→π (x² − sin x) + 5

= lim _h→0 [(π + h)² − sin(π + h) + 5]

= lim _h→0 (π+h)² − lim _h→0 sin (π+h) + lim _h→0 5

= (π + 0)² − lim _h→0 [sin π cos h + cos π sin h] + 5

= π² − lim _h→0 sin π cos h − lim _h→0 cos π sin h + 5

= π² − sin π cos 0 − cos π sin 0 + 5

= π² − 0(1) − (−1)0 + 5

= π² + 5 = f(π)

Therefore, the given function f is continuous at x = π

---

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 20

Is the function defined by f(x) = x² − sin x + 5 continuous at x = π?

Summary:

Hence we have concluded that the function defined by f(x) = x² − sin x + 5 continuous at x = π