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Question
If f(x)=4x−x2,x∈R, then f(a+1)−f(a−1) is equal to
If f(x)=4x−x2,x∈R, then f(a+1)−f(a−1) is equal to
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Solution
The correct option is B 4(2−a)
f(x)=4x−x2∴f(a+1)−f(a−1)=4(a+1)−(a+1)2−[4(a−1)−(a−1)2]=8−[(a+1)2−(a−1)2]=8−4a=4(2−a)
f(x)=4x−x2∴f(a+1)−f(a−1)=4(a+1)−(a+1)2−[4(a−1)−(a−1)2]=8−[(a+1)2−(a−1)2]=8−4a=4(2−a)
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