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प्रश्न
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,
पर्याय
a = `1/2`, b = 3
a = `- 1/2`, b = – 3
a = `1/2`, b = – 3
a = `-1/2`, b = 3
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उत्तर
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
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