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Question

The maximum value of the function f(x)=2x318x2+48x11 over the set S={xR:x2+4213x} is


Your Answer
A
115
Your Answer
B
21
Your Answer
C
29
Correct Answer
D
129

Solution

The correct option is D 129
x2+4213xx213x+420(x6)(x7)0x[6,7]

Now, f(x)=2x318x2+48x11
f(x)=6x236x+48f(x)=6(x26x+8)f(x)=6(x2)(x4)f(x)>0     (x[6,7])

So, f is strictly increasing in [6,7].
So, f(7) is the maximum value.

Hence, the maximum is
f(7)=2×7318×72+48×711       =7(98126+48)11       =7×2011       =129

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