If a function f(a) = max (a, 0) then the smallest integer value of ‘x’ for which the equation f(x - 3) + 2f(x + 1) = 8 holds true is:
Explanation:
Case 1: x > 3 ⇒ f(x - 3) = max(x - 3, 0) = x - 3 ⇒ f(x + 1) = max(x + 1, 0) = x + 1 Given, f(x - 3) + 2f(x + 1) = 8 ⇒ x - 3 + 2(x + 1) = 8 ⇒ 3x = 9 ⇒ x = 3
Case 2: -1 < x ≤ 3 ⇒ f(x - 3) = max(x - 3, 0) = 0 ⇒ f(x + 1) = max(x + 1, 0) = x + 1 Given, f(x - 3) + 2f(x + 1) = 8 ⇒ 0 + 2(x + 1) = 8 ⇒ 2x = 6 ⇒ x = 3
Case 3: x < -1 ⇒ f(x - 3) = max(x - 3, 0) = 0 ⇒ f(x + 1) = max(x + 1, 0) = 0 Given, f(x - 3) + 2f(x + 1) = 8 ⇒ 0 + 0 = 8 Not possible
∴ x = 3 is the only value that satisfies the given function.
Hence, 3.
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