QJEEM19Ma5d0d: JEE Main 2019 Mathematics - Class 12 JEEM19Ma5d0d Solution: Let f : R R be a function defined as f ( x ) = { 5 ; x ≤ 1 a +…

Checking

if f(x) is continuous at x = 1 :

f(1 ^$-$ ) = 5

f(1) = 5

f(1⁺) = a + b

if f(x) is continuous at x = 1,

then

f(1 ^$-$ ) = f(1) = f(1⁺)

$\Rightarrow$   5 = 5 = a + b

$\therefore$   a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3 ^$-$ ) = a + 3b

f(3) = b + 15

f(3⁺) = b + 15

if   f(x) = is continuous at x = 3

then,

f(3 ^$-$ ) = f(3) = f(3⁺)

$\Rightarrow$   a + 3b = b + 15 = b + 15

$\Rightarrow$   a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5 ^$-$ ) = b + 25

f(5) = 30

f(5⁺) = 30

if f(x) is continuous at x = 5 then,

f(5 ^$-$ ) = f(5) = f(5⁺)

$\Rightarrow$    b + 25 = 30 = 30

$\Rightarrow$    b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$\Rightarrow$   a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$\therefore$   f is not continuous for any value of a and b.