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Question

Consider the function f(x) such that f(x)=x2+10(x+t)f(t)dt, then the minimum value of f(x) is,

A
1093
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B
1096
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C
10912
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D
152
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Solution

The correct option is D 10912
y(x)=x2+ax+b, where a=10y(t)dt,b=10t.y(t)dt

a=10(t2+at+b)dt,b=10t.(t2+at+b)dt=10(t3+at2+bt)dt

a=[t33+at22+bt]10,b=[t44+at33+bt22]10

a=13+a2+b and b=14+a3+b2

a2b=13 and b2a3=14

By solving we get

substituting b=a213 in b2a3=14

we get

b2a3=14

(a213)2a3=14

a416a3=14

a12=1024

a=5

b=a213=5213=176

f(x)=x25x176

=(x52)2254176

=(x52)210912

fmin=10912

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