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# Find the relationship between a and b so that the function f defined by f(x)={(ax + 1, if x ≤ 3) (bx + 3, if x > 3) is continous at x = 3

**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) = {(ax + 1, if x ≤ 3) (bx + 3, if x > 3)

For f to be continuous at x = 3, then

lim_{x→3−} f(x) = lim_{x→3+} f(x) = f(3) …(1)

Also,

lim_{x→3−} f(x) = lim_{x→3− }(ax + 1) = 3a + 1

lim_{x→3+} f(x) = lim_{x→3+ }(bx + 3) = 3b + 3

f(3) = 3a + 1

Therefore,

from (1), we obtain

3a + 1 = 3b + 3 = 3a + 1

⇒ 3a + 1 = 3b + 3

⇒ 3a = 3b + 2

⇒ a = b + 2/3

Therefore, the required relationship is given by, a = b + 2/3

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 17

## Find the relationship between a and b so that the function f defined by f(x)={(ax + 1, if x ≤ 3) (bx + 3, if x > 3) is continous at x = 3

**Summary:**

The relationship between a and b so that the function f defined by f(x)={(ax + 1, if x ≤ 3) (bx + 3, if x > 3) is continous at x = 3 is a = b + 2/3

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