Question

Select the correct answer from the given alternatives:

If f(x) `{:(= "a"x^2 + "b"x + 1",", "for"  |x −1| ≥ 3), (= 4x + 5",", "for"  -2 < x < 4):}` is continuous everywhere then,

विकल्प

• a = `1/2`, b = 3

• a = `- 1/2`, b = – 3

• a = `1/2`, b = – 3

• a = `-1/2`, b = 3

Answer 1

a = `1/2`, b = 3

Explanation;

f(x) `{:(= "a"x^2 + "b"x + 1",",  |x −1| ≥ 3), (= 4x + 5",", -2 < x < 4):}`

The first interval is

|x − 1| ≥ 3

∴ x − 1 ≥ 3  OR   x − 1 ≤ − 3 

∴ x ≥ 4       OR    x ≤ − 2

∴  f(x) is same for x ≤ – 2 as well as x ≥ 4

∴ f(x) is defined as:

f(x) `{:(= "a"x^2 + "b"x + 1";", x ≤  – 2),(= 4x + 5 ";", -2 < x < 4),(= "a"x^2 + "b"x + 1";", x  ≥ 4):}`

f(x) is continuous everywhere

∴ f(x) is continuous at x = – 2 and x = 4

As f(x) is continuous at x = − 2

`lim_(x -> -2^-) "f"(x) = lim_(x -> -2^+) "f"(x)`

∴ `lim_(x -> -2) ("a"x^2 + "b"x + 1) = lim_(x -> -2) (4x + 5)`

∴ a(– 2)² + b(– 2) + 1 = 4(– 2) + 5

∴ 4a – 2b + 1 = – 3

∴ 4a – 2b = – 4

∴ 2a – b = – 2   ...(i)

Also f(x) is continuous at x = 4

∴ `lim_(x -> 4^-) "f"(x) = lim_(x -> 4^+) "f"(x)`

∴ `lim_(x -> 4)(4x + 5) = lim_(x ->4^+) "f"("a"x^2 + "b"x + 1)`

∴ 4(4) + 5 = (4)² + b(4) + 1

∴ 16a + 4b + 1 = 21

∴ 16a + 4b = 20

∴ 4a + b = 5   ...(ii)

Adding (i) and (ii)

6a = 3

∴ a = `1/2`

Substitute a = `1/2` in (ii)

`4(1/2) + "b"` = 5

∴ 2 + b = 5

∴ b = 3

∴ a = `1/2`, b = 3