JEE Main 2022 Mathematics question
Let + bx + c be such that f(1) = 3, f(−2) = and f(3) = 4. If f(0) + f(1) + f(−2) + f(3) = 14, then is equal to
Answer: (4)
Solution
f(1) = a + b + c = 3 and f(3) = 9a + 3b + c = 4, so 8a + 2b = 1. The condition f(0) + f(1) + f(−2) + f(3) = 14 gives c + 3 + (4a − 2b + c) + 4 = 14, i.e. 4a − 2b +
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