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Question

A straight line L with negative slope passes through the point (8,4) and cuts the positive co-ordinate axes at points A and B. The minimum value of OA+OB, as L varies, is (Here O is the origin)

A
12+82
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B
1282
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C
12+42
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D
1242
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Solution

The correct option is A 12+82
m is the slope of line L, m<0
Equation of line L is (y4)=m(x8)
Therefore, intercepts are OA=84m and OB=48m

Let f=OA+OB
f(m)=124m8m
f(m)=4m28
f(m)=0m=12 [m<0]
f′′(12)=8m3>0 as m<0
So minima will lie at m=12
f(12)=12+42+82
Minimum value =12+82

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