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# Find the values of k so that the function f is continuous at the indicated point f(x) = {(kx + 1, if x ≤ 5) (3x − 5, if x > 5) at x = 5

**Solution:**

The given function is

f(x) = {(kx + 1, if x ≤ 5) (3x − 5, if x > 5)

The given function f is continuous at x = 5,

if f is defined at x = 5 and if the value of the f at x = 5 equals the limit of f at x = 5.

It is evident that f is defined at x = 5 and

f(5) = kx + 1

= 5k + 1

limx→5^{-} f(x) = limx→5^{+ }f(x)

= f(5)

⇒ limx→5^{-} (kx + 1) = limx→5^{+}(3x − 5)

= 5k + 1

= 3(5) − 5 = 5k + 1

⇒ 5k + 1

= 15 − 5 = 5k + 1

⇒ 5k + 1 = 10

⇒ 5k = 9

⇒ k = 9/5

Therefore, the value of k = 9/5

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 29

## Find the values of k so that the function f is continuous at the indicated point f(x) = {(kx + 1, if x ≤ 5) (3x − 5, if x > 5) at x = 5

**Summary:**

The value of k so that the function f is continuous at the indicated point f(x) = {(kx + 1, if x ≤ 5) (3x − 5, if x > 5) at x = 5 is 9/5

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