A) 5
B) 6
C) 1
D) 2
Correct Answer: B
Solution :
Let \[f(x)=5+4x-4{{x}^{2}}=y\,\,\,\,\Rightarrow \,\,4{{x}^{2}}-4x-5+y=0\] Since x is real, so \[{{B}^{2}}-4AC\ge 0\] Þ \[16-4.4(-5+y)\ge 0\] Þ \[16-16(-5+y)\ge 0\,\,\Rightarrow \,\,-5+y\le 1\,\,\Rightarrow y\le 6\] Hence maximum value of y is 6.
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