Given:

f(x) is continuous at x = 3 & x = 5

If f(x) to be continuous at x = 3,then,f(3)^– ₌ f(3) ⁺ ₌ f(3)

LHL = f(3)^– =

= 1 ...(1)

RHL = f(3) ⁺ =

⇒ a(3 + 0) + b

⇒ 3a + b ...(2)

Since ,f(x) is continuous at x = 3 and From (1) & (2),we get

3a + b = 1 ...(3)

Similarly ,f(x) is continuous at x = 5

If f(x) to be continuous at x = 5,then, f(5)^– ₌ f(5) ⁺ ₌ f(5)

LHL = f(5)^– =

⇒ a(5–0) + b

⇒ 5a + b ...(4)

RHL = f(5) ⁺ =

7 ...(5)

Since , f(x) is continuous at x = 5 and From (4) & (5),we get,

5a + b = 7 ...(6)

Now equate (3) & (6)

⇒ a = 3

Now Substitute a = 3 in any one of above equation(3) & (6) ,

3a + b = 1

⇒ 3(3) + b = 1

⇒ 9 + b = 1

⇒ b = –8